A CANONICAL DESCRIPTION OF THE PLANCHEREL MEASURE FOR A GENERAL TWO-STEP FREE NILPOTENT LIE GROUP

A Dissertation

Submitted to the Graduate Faculty of the Louisiana State University and Agricultural and Mechanical College in partial fulfillment of the requirements for the degree of Doctor of Philosophy

in

The Department of Mathematics

by Chin-Te Chu B.S. in Math., National Chiao Tung University, 1988 M.S., Louisiana State University, 1997 December 2000

Acknowledgments

This dissertation would not be possible without several contributions. The most important of them, without any doubt, was given by my advisor, Professor Leonard F. Richardson. I am grateful for his advice, guidance, dedication and patience. In the ancient East, teachers are not only the men of knowledge, but also the characters of wisdom. From the first semester that I took his class of advanced calculus, I could smell the scent of ancient intellectuals. From now on I was inspired to pursue my studies in this field. This work was motivated by Example 4.3.14 in [1] brought to Dr. Richardson’s attention. During these two years of working on my dissertation, I usually felt frustrated and lost in the ocean. Then doubt rose: can I work out this problem? Can this example be generalized? Dr. Richardson always responded: let faith lead the way. Now I retrospect this doctoral training. His patience brews my endurance, perseverance, courage, and faith. A monk pursuing the truth from his teacher embraces the precious jewelry to go on his dream of education. He may or may not succeed, but thanksgiving towards persons who shape his personality is always fragrant in his heart.

I thank all the people in the Department of Mathematics for providing me with a pleasant working environment. A special thanks to Dr. Ólafsson for his resourcefulness in the construction of norms in Lemma 6.1. and in the finding of an appropriate coordinate patch in Lemma 5.3., and for his help in the correction of this dissertation. What has impressed me the most was how a great mathematician would approach students himself to find out if students had any questions or problems, and if they did, how he was always willing to spend time with students and hold detailed discussions. Since I work on this field, there have been times when I needed support and encouragement. Dr. Ólafsson is always there to listen and to give valuable advice. I felt that I could approach him without any hesitation, and speak to him freely and frankly about anything. I have benefited a lot from him. The proof of Lemma 6.4. was motivated by Dr. Fabec’s intuition. The dust of chalk on the top of nose harmonically danced with the knowledge of a smiling mathematician in class. The unfading impression proves how a devoted teacher is. Thanks also to Dr. Cygan who gave me suggestions after my presentations while working on my dissertation. My gratitude to this department is unsurpassed for giving me an opportunity of teaching. Especially the contribution of Dr. Oxley is invaluable. Lovely Gretchen’s friendship relieved my homesick. Sweet Robin usually brings laughter to cheer me up. Knowledgeable Sam, office mate, helps me to use LaTex. How can I forget the sunrises on Rocky Mountains encouraging me to pursue the ultimate truth?

Sincere appreciation from my heart can not be conveyed in words. This dissertation, the first achievement in my life, as a gift is dedicated to my parents, Yun-Lien and Wen-Hu. They left school at an early age to support their families. After struggling through a tough life, they comprehend that education is the only wealth for their four kids. Letting children get an education as high as possible turns out to be their first goal. As a kid, I like to read poems by mom’s side when she was weaving, and do math homework with daddy after he came home from work. The unforgettable memory is still vivid after I have grown up. Even the wrinkle on my forehead can not erase my memory for mom who has been playing the same role for many lives. Being the eldest son, I owe my family a lot. Finally, heartfelt thanksgiving to beloved Osho, a buddha who gives unconditional and endless love to his disciple, Swami Bodhi Jagran. The seed will do its best to fulfill the responsibility of blossoming.

Table of Contents

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2. Parametrizing Orbits by Automorphisms . . . . . . . . . . . . . . . . . . . . . . 11 3. Relative Invariance of Plancherel Measure . . . . . . . . . . . . . . . . . . . . . 16 4. Nonexistence of Invariant Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5. Explicit Calculations for F_{3,2} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6. General Properties of Free 2-step Nilpotent Lie Algebras on n

Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Abstract

In this work on 𝔤 = F_{n,2}, the free 2-step nilpotent Lie algebra on n generators, we use the group of automorphisms to give a basis-free description of the Fourier Inversion Formula, thereby generalizing and strengthening an example discussed by Corwin and Greenleaf.

In the first charter, Introduction, we begin with a brief survey of traditional viewpoint related to this dissertation, then discuss Example 4.3.14 in Corwin & Greenleaf’s book. It demonstrates how two different bases for F_{3,2} lead to different inversion formulas. But the third ”more” invariant formula describes Plancherel measure on a support expressed in terms of rotations, dilations, and translations. Actually it is not canonical since it still depends on choices of bases for F_{3,2}. Our goal is to re-describe Plancherel measure on a support expressed in terms of Aut^{∗}(𝔤). We accomplish this in the following two chapters.

The second chapter provides a procedure for re-parametrizing the family of generic orbits by establishing a one-to-one correspondence between the maximumdimensional orbits and the quotient space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}).

The third chapter provides background material about relatively invariant measures. Then we prove that Plancherel measure, modeled on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}), is the essentially unique relatively invariant measure corresponding to a specific homomorphism.

The fourth chapter demonstrates that there does not exist an \overline{Aut^{∗}(𝔤)}-invariant measure on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) for the example 𝔤 = F_{3,2}.

Our explicit calculations for F_{3,2} are done in Chapter 5 and are in agreement with the results of the first four chapters. We start by finding an almost global coordinate patch for Ad_{G}\Ad_{G} ·Stab^{∗}(Z_{3}^{∗}), then use this patch to construct a right and left Haar measure on this quotient space. Thus we get its modular function δ. A similar process applies to Ad_{G}\Aut(𝔤), and we obtain its modular function Δ. Hence the ratio of Δ to δ is the restriction of any modular function for any relatively invariant measure on the double coset space Ad_{G}\Aut(𝔤)/Stab^{∗}(Z_{3}^{∗}). Furthermore, we find a cross-section X for Ad_{G}\Aut(𝔤)/Stab^{∗}(l_{0}) with general l_{0} ∈ 𝔤_{Max}^{∗}. Then we use X to verify the relative invariance for the measure corresponding to that given in the example of Corwin and Greenleaf for 𝔤 = F_{3,2}.

In Chapter 6, we illustrate three general properties for F_{n,2} and one additional result for linear algebra. These properties are used in Chapter 5.

1. Introduction

We begin with some general remarks about the subject matter of this dissertation. Harmonic analysis is a very rich subject which can be viewed from many perspectives. One traditional viewpoint is that harmonic analysis permits the recovery of a function from its Fourier transform: For each Schwartz function f ∈ S(R),

f(e) = ∫_{R} f̂(α)dα

One can readily recover also f(x) for general x simply by replacing f in the integral by its translation by x. If one does this, then

f(x) = ∫_{R} f̂(α)e^{2πixα}dα

The Real numbers R are an abelian group under addition. When one investigates this process of Fourier Inversion for general locally compact topological groups, or for Lie groups, it becomes necessary to replace the 1-dimensional characters χ_{α}(x) = e^{2πiαx} by the equivalence classes of irreducible unitary representations in Hilbert space. The collection of these classes is denoted Ĝ. If the group G is neither compact nor abelian, then Ĝ typically includes the classes of many infinite dimensional irreducible representations. There is an abstract Plancherel theorem (due to Segal) which establishes that

f(e) = ∫Ĝ Tr(π_{f})dμ(π)

for all suitable functions f, such as Schwartz functions on a Lie group. Here π_{f} = ∫_{G} f(g)π_{g}dg, and the trace of this operator plays the role of f̂(α) in classical Fourier inversion.

In order to carry out this Fourier inversion in a meaningful way on particular classes of Lie groups, it is necessary first to know the representations in Ĝ and to know explicitly the Plancherel measure and its support within Ĝ. The representations of Lie groups are often constructed using the classical process of Induced Representations, developed by Schur for finite groups and greatly extended by George Mackey to all locally compact topological groups in the 1940’s and 1950’s. (See [7]) Much effort was expended upon the identification of Ĝ for general nilpotent Lie groups until this was finally achieved by A. A. Kirillov in his celebrated dissertation [6] under the direction of I. M. Gelfand. Kirillov showed that if G is nilpotent, then every element of Ĝ is induced by a one-dimensional character of a suitable subgroup M ⊂ G. If l ∈ 𝔤^{∗} one defines the character χ_{l}(m) = e^{2πil(log m)} and χ will induce an irreducible unitary representation of G if and only if M has maximal dimension so as to make χ_{l} a genuine character. Moreover, the unitary equivalence classes in Ĝ are in one-to-one correspondence with the coadjoint orbits of G, also called Kirillov orbits. Thus Ĝ can be identified with 𝔤^{∗}/Ad^{∗}(G). The support of the Plancherel measure μ was shown to be a very large collection of orbits, whose union constitutes a Zariski open subset of 𝔤^{∗}. All orbits in the support of μ are of the greatest possible dimension, but not every such orbit need be in that support. One of the features of classical harmonic analysis which is lost in the study of Lie groups is that Ĝ lacks a group structure. (For abelian groups, Ĝ is a group.) In this dissertation we address that lack. We show that at least for certain types of nilpotent Lie groups one can create a model of the Plancherel measure on a quotient space of the automorphism group, Aut(G). When we do this, we show that the Plancherel measure μ is the essentially unique relatively invariant measure corresponding to the homomorphism which maps each automorphism A → | det(A)|. The idea arose out of our efforts to understand better an example of Corwin and Greenleaf, presented in their book [1]. It is an awkward feature of nilpotent harmonic analysis that the description of the Plancherel measure, and even of the generic orbits on which it is supported, is dependent on the choice of an arbitrary strong Malcev basis. Corwin & Greenleaf illustrate this with the following Example 4.3.14 in [1]: Let 𝔤 = F_{3,2}, 𝔤_{2} = [𝔤, 𝔤], and let 𝔤^{∗} be the linear dual space of 𝔤. Pick a basis \{\overline{Y}_{1}, \overline{Y}_{2}, \overline{Y}_{3}\} for 𝔤/𝔤_{2}, then choose Y_{i} ∈ \overline{Y}_{i}, ∀i = 1, 2, 3. Let

Z_{1} = [Y_{2}, Y_{3}], Z_{2} = [Y_{3}, Y_{1}], Z_{3} = [Y_{1}, Y_{2}],

which are independent of the choice of Y_{i} ∈ \overline{Y}_{i}. Then B := \{Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}\} is a strong Malcev basis of 𝔤. Let B^{∗} := {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}, Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} be the dual basis of B in 𝔤^{∗}, so that B^{∗} is a Jordan-Hölder basis of 𝔤^{∗}. Let 𝔤_{1} = Span_{R}{Y_{1}, Y_{2}, Y_{3}}. We know 𝔤_{2} = Span_{R}{Z_{1}, Z_{2}, Z_{3}}. Let 𝔤_{1}^{∗} = Span_{R}{Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} and 𝔤_{2}^{∗} = Span_{R}{Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}}. Corwin & Greenleaf describe the Plancherel measure with respect to this strong Malcev basis B of 𝔤 as follows. l = ∑_{i=1}^{3} α_{i}Z_{i}^{∗} + ∑_{i=1}^{3} β_{i}Y_{i}^{∗} is generic with respect to this basis if and only if α_{3} ≠ 0, and the Pfaffian for this basis B is given by Pf^{2}(l) = det \overline{B}_{l} = α_{3}^{2}. The Fourier Inversion Theorem says that for f ∈ S(G), the Schwartz functions on G,

f(e) = ∫_{R^{4}} |α_{3}|Trπ_{α_{1}Z_{1}^{∗}+α_{2}Z_{2}^{∗}+α_{3}Z_{3}^{∗}+β_{3}Y_{3}^{∗}}(f)dα_{1}dα_{2}dα_{3}dβ_{3}  (1.1)

On the other hand, if we take the strong Malcev basis to be {Z_{3}, Z_{2}, Z_{1}, Y_{3}, Y_{2}, Y_{1}}, then l is generic if and only if α_{1} ≠ 0. And the corresponding Pfaffian is given by Pf^{2}(l) = α_{1}^{2}. The Fourier Inversion Theorem now says that

f(e) = ∫_{R^{4}} |α_{1}|Trπ_{α_{1}Z_{1}^{∗}+α_{2}Z_{2}^{∗}+α_{3}Z_{3}^{∗}+β_{1}Y_{1}^{∗}}(f)dα_{1}dα_{2}dα_{3}dβ_{1}  (1.2)

Why do these two different formulas, (1.1) and (1.2), give the same result f(e)? To understand this situation, Corwin & Greenleaf describe how the group SO(3, R) acts on 𝔤 (and G) as a group of automorphisms. We discuss on the following pages how their explanation works. Then we will give a preview of how we improve and generalize what they have done. We decompose 𝔤 = 𝔤_{1} ⊕ 𝔤_{2} as a sum of vector spaces. Notice that 𝔤_{1} is not a Lie subalgebra of 𝔤. Since B_{1} := {Y_{1}, Y_{2}, Y_{3}} is a basis of 𝔤_{1}, we regard 𝔤_{1} as isomorphic to R^{3} by the map Y ⟼ [Y]_{B_{1}}. Similarly, 𝔤_{2} is regarded as isomorphic to R^{3} relative to the basis B_{2} = {Z_{1}, Z_{2}, Z_{3}}. For each σ ∈ SO(3, R), we define the map

(σ ⊗ σ) : 𝔤_{1} ⊕ 𝔤_{2} ⟶ 𝔤_{1} ⊕ 𝔤_{2} by (Y, Z) ⟼ (σY, σZ)

Then we claim that σ ⊗ σ is an automorphism of 𝔤. It suffices to show that [σY_{i}, σY_{j}] = σ([Y_{i}, Y_{j}]), for all pairs (i, j), 1 ≤ i, j ≤ 3. We proceed to show this as follows. For each σ ∈ SO(3), we still denote the matrix of σ relative to the standard basis

of R^{3} by σ. Let σ = [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ]. By the definition of SO(3), we have

(σ^{−1})^{t} = σ. Then (\frac{1}{det σ}(adjσ))^{t} = σ, and hence (adjσ)^{t} = σ since det σ = 1. Thus we get the identity

[ \begin{matrix}det ( \begin{matrix}p_{5} & p_{8} \\ p_{6} & p_{9} \end{matrix} ) & − det ( \begin{matrix}p_{2} & p_{8} \\ p_{3} & p_{9} \end{matrix} ) & det ( \begin{matrix}p_{2} & p_{5} \\ p_{3} & p_{6} \end{matrix} ) \\ − det ( \begin{matrix}p_{4} & p_{7} \\ p_{6} & p_{9} \end{matrix} ) & det ( \begin{matrix}p_{1} & p_{7} \\ p_{3} & p_{9} \end{matrix} ) & − det ( \begin{matrix}p_{1} & p_{4} \\ p_{3} & p_{6} \end{matrix} ) \\ det ( \begin{matrix}p_{4} & p_{7} \\ p_{5} & p_{8} \end{matrix} ) & − det ( \begin{matrix}p_{1} & p_{7} \\ p_{2} & p_{8} \end{matrix} ) & det ( \begin{matrix}p_{1} & p_{4} \\ p_{2} & p_{5} \end{matrix} ) \end{matrix} ] = [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

It follows that

[σY_{1}, σY_{2}] = [p_{1}Y_{1} + p_{2}Y_{2} + p_{3}Y_{3}, p_{4}Y_{1} + p_{5}Y_{2} + p_{6}Y_{3}]
= | \begin{matrix}p_{2} & p_{5} \\ p_{3} & p_{6} \end{matrix} | Z_{1} − | \begin{matrix}p_{1} & p_{4} \\ p_{3} & p_{6} \end{matrix} | Z_{2} + | \begin{matrix}p_{1} & p_{4} \\ p_{2} & p_{5} \end{matrix} | Z_{3}
= p_{7}Z_{1} + p_{8}Z_{2} + p_{9}Z_{3}
= σZ_{3}
= σ([Y_{1}, Y_{2}])
[σY_{2}, σY_{3}] = [p_{4}Y_{1} + p_{5}Y_{2} + p_{6}Y_{3}, p_{7}Y_{1} + p_{8}Y_{2} + p_{9}Y_{3}]
= | \begin{matrix}p_{5} & p_{8} \\ p_{6} & p_{9} \end{matrix} | Z_{1} − | \begin{matrix}p_{4} & p_{7} \\ p_{6} & p_{9} \end{matrix} | Z_{2} + | \begin{matrix}p_{4} & p_{7} \\ p_{5} & p_{8} \end{matrix} | Z_{3}
= p_{1}Z_{1} + p_{2}Z_{2} + p_{3}Z_{3}
= σZ_{1}
= σ([Y_{2}, Y_{3}])

[σY_{3}, σY_{1}] = [p_{7}Y_{1} + p_{8}Y_{2} + p_{9}Y_{3}, p_{1}Y_{1} + p_{2}Y_{2} + p_{3}Y_{3}]
= − | \begin{matrix}p_{2} & p_{8} \\ p_{3} & p_{9} \end{matrix} | Z_{1} + | \begin{matrix}p_{1} & p_{7} \\ p_{3} & p_{9} \end{matrix} | Z_{2} − | \begin{matrix}p_{1} & p_{7} \\ p_{2} & p_{8} \end{matrix} | Z_{3}
= p_{4}Z_{1} + p_{5}Z_{2} + p_{6}Z_{3}
= σZ_{2}
= σ([Y_{3}, Y_{1}])

Thus we have proved that σ ⊗ σ is in Aut(𝔤). Similarly, we regard 𝔤_{1}^{∗} and 𝔤_{2}^{∗} as each isomorphic to R^{3} relative to coordinates in the bases {Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} and {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}} respectively. For each σ ∈ SO(3), we define the dual σ^{∗} of σ on 𝔤_{1}^{∗} and 𝔤_{2}^{∗} by its inverse transpose. Since σ ⊗ σ ∈ Aut(𝔤), so is its inverse (σ ⊗ σ)^{−1} = σ^{−1} ⊗ σ^{−1}. Define the dual of σ ⊗ σ by its inverse transpose. So we have (σ ⊗ σ)^{∗} = σ^{∗} ⊗ σ^{∗}. Now we give 𝔤_{1}^{∗} an inner product and a corresponding Euclidean metric by making Y_{1}^{∗}, Y_{2}^{∗}, and Y_{3}^{∗} orthonormal, and similarly for 𝔤_{2}^{∗} by making Z_{1}^{∗}, Z_{2}^{∗}, and Z_{3}^{∗} orthonormal. Define a linear map A : 𝔤_{2}^{∗} ⟶ 𝔤_{1}^{∗} by A(Z_{i}^{∗}) = Y_{i}^{∗}. For every unit vector ω ∈ 𝔤_{2}^{∗}, let ω = α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗}. Then there exists σ ∈ SO(3) such that ω = σ^{∗}Z_{3}^{∗}. It follows that Aω = A(α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗}) = α_{1}Y_{1}^{∗} + α_{2}Y_{2}^{∗} + α_{3}Y_{3}^{∗} = σ^{∗}Y_{3}^{∗}. So rω + tAω = rσ^{∗}Z_{3}^{∗} + tσ^{∗}Y_{3}^{∗} = (σ ⊗ σ)^{∗}(rZ_{3}^{∗} + tY_{3}^{∗}), ∀r > 0, ∀t ∈ R. For l_{0} = rZ_{3}^{∗} + tY_{3}^{∗}, its radical 𝔯_{l_{0}} = 𝔤_{2} ⊕ RY_{3} is an ideal in 𝔤. By Theorem 3.2.3. in [1] we know that O_{l_{0}} = l_{0} + 𝔯_{l_{0}}^{⊥} = l_{0} + Span_{R}{Y_{1}^{∗}, Y_{2}^{∗}}. For l = rω + tAω = (σ ⊗ σ)^{∗}(rZ_{3}^{∗} + tY_{3}^{∗}), its radical 𝔯_{l} = 𝔯_{(σ⊗σ)^{∗}l_{0}} = (σ ⊗ σ)𝔯_{l_{0}} = 𝔤_{2} ⊕ R(σY_{3}) is still an ideal of 𝔤. Then O_{l} = l + 𝔯_{l}^{⊥} = l + Span_{R}{σ^{∗}Y_{1}^{∗}, σ^{∗}Y_{2}^{∗}} = rω + tAω + (Aω)^{⊥} where (Aω)^{⊥} is the plane Span_{R}{σ^{∗}Y_{1}^{∗}, σ^{∗}Y_{2}^{∗}} orthogonal to Aω in 𝔤_{1}^{∗}.

Let O_{Max} = {O_{l} : dim O_{l} is maximal}, and 𝔤_{Max}^{∗} = {l : dim O_{l} is maximal}. We claim that

{rω + tAω|r > 0, t ∈ R, ω ∈ 𝔤_{2}^{∗}, |ω| = 1}

is a cross-section of O_{Max}. Since l ∈ 𝔤_{Max}^{∗} if and only if l|_{𝔤_{2}} ≠ 0, there always exist unique r > 0 and unit vector ω ∈ 𝔤_{2}^{∗} such that l|𝔤_{2} = rω. For each t ∈ R, tAω + (Aω)^{⊥} is a two-dimensional plane parallel to (Aω)^{⊥} in 𝔤_{1}^{∗}. So 𝔤_{1}^{∗} =⋃^{.} {tAω + (Aω)^{⊥} : ∀t ∈ R}, and hence there is a unique t ∈ R such that l|𝔤_{1} ∈ tAω + (Aω)^{⊥}. It follows that l ∈ rω + tAω + (Aω)^{⊥} = O_{rω+tAω}. Therefore {rω + tAω|r > 0, t ∈ R, ω ∈ 𝔤_{2}^{∗}, |ω| = 1} is a cross-section for O_{Max}, as claimed. For ω_{0} = Z_{3}^{∗}, l_{0} = rω_{0} + tAω_{0} lies on the orbit O_{l_{0}} = rω_{0} + tAω_{0} + RY_{1}^{∗} + RY_{2}^{∗}. And relative to the basis {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}}, the Pfaffian of l_{0} is given by P f^{2}(l_{0}) = r^{2}. Then for every f ∈ S(G),

Trπ_{rω_{0}+tAω_{0}}(f) = ∫_{R^{2}} \frac{1}{r}(f ∘ exp)(rω_{0} + tAω_{0} + β_{1}Y_{1}^{∗} + β_{2}Y_{2}^{∗})dβ_{1}dβ_{2}.

Hence

∫_{R} Trπ_{rω_{0}+tAω_{0}}(f)dt = ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ exp)(rω_{0} + λ)dλ

where dλ is normalized Lebesgue measure defined relative to the basis {Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} on 𝔤_{1}^{∗}. Now for each σ ∈ SO(3), since σ ⊗ σ ∈ Aut(𝔤), it follows that

π_{(σ⊗σ)^{∗}(rω_{0}+tAω_{0})} ≅ π_{rω_{0}+tAω_{0}} ∘ (σ ⊗ σ)^{−1}

, and hence

π_{(σ⊗σ)^{∗}(rω_{0}+tAω_{0})}(f) ≅ π_{rω_{0}+tAω_{0}}(f ∘ (σ ⊗ σ)).

So we know that Trπ_{(σ⊗σ)^{∗}(rω_{0}+tAω_{0})}(f) = Trπ_{rω_{0}+tAω_{0}}(f ∘ (σ ⊗ σ)).

Thus, relative to the same basis {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}}, for every general functional l = rω + tAω

∫_{R} Trπ_{rω+tAω}(f)dt = ∫_{R} Trπ_{(σ⊗σ)^{∗}(rω_{0}+tAω_{0})}(f)dt
= ∫_{R} Trπ_{rω_{0}+tAω_{0}}(f ∘ (σ ⊗ σ))dt
= ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ (σ ⊗ σ) ∘ exp)(rω_{0} + λ)dλ
= ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ exp ∘(σ ⊗ σ))(rω_{0} + λ)dλ
= ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ exp)(rσ^{∗}ω_{0} + σ^{∗}λ)dλ
= ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ exp)(rω + λ)dλ

Note that the last equality holds because Lebesgue measure is rotation-invariant. So far we get

∫_{R} Trπ_{rω+tAω}(f)dt = ∫_{𝔤_{1}^{∗}} \frac{1}{r}(f ∘ exp)(rω + λ)dλ.

Since 𝔤_{2}^{∗} is regarded as isomorphic to R^{3},

dv(r, ω) = r^{2}sinφdrdφdθ

is Lebesgue measure on 𝔤_{2}^{∗}. Multiplying both sides of the above equation by r and integrating over 𝔤_{2}^{∗}, we get

∫_{𝔤_{2}^{∗}} ∫_{R} rTrπ_{rω+tAω}(f)dtdv(r, ω)

= ∫_{𝔤_{2}^{∗}} ∫_{𝔤_{1}^{∗}}(f ∘ exp)(rω + λ)dλdv(r, ω)
= ∫_{R^{3}} ∫_{𝔤_{1}^{∗}}(f ∘ exp)(α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗} + λ)dλdα_{1}dα_{2}dα_{3}
= ∫_{𝔤_{2}^{∗}} ∫_{𝔤_{1}^{∗}}(f ∘ exp)(l + λ)dλdl

Hence for every f ∈ S(G), we have f(e) = ∫_{𝔤_{2}^{∗}} ∫_{𝔤_{1}^{∗}}(f ∘ exp)(l + λ)dλdl = ∫_{𝔤_{2}^{∗}} ∫_{R} rTrπ_{rω+tAω}(f)dtdv(r, ω).

We now would like to answer the question posed at the beginning: why do formulas (1.1) and (1.2) give the same result: f(e) ? For the equation (1.1) we start with the Malcev basis {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}} to get V_{T} ∩ U = {α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗} + β_{3}Y_{3}^{∗} : α_{3} ≠ 0}. Since

{rω + tAω|r > 0, t ∈ R, ω ∈ 𝔤_{2}^{∗}, |ω| = 1}

is a cross-section for all orbits of 𝔤^{∗} that correspond to infinite-dimensional representations, for each l ∈ V_{T} ∩ U, we need to know which element rω + tAω from this cross-section such that l ∈ O_{rω+tAω}. Actually for each l = α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗} + β_{3}Y_{3}^{∗} ∈ V_{T} ∩ U, let

r := \sqrt{α_{1}^{2} + α_{2}^{2} + α_{3}^{2}}

and let σ ∈ SO(3) such that

ω := σ^{∗}Z_{3}^{∗} = \frac{α_{1}}{r} Z_{1}^{∗} + \frac{α_{2}}{r} Z_{2}^{∗} + \frac{α_{3}}{r} Z_{3}^{∗}.

Then

Aω = \frac{α_{1}}{r} Y_{1}^{∗} + \frac{α_{2}}{r} Y_{2}^{∗} + \frac{α_{3}}{r} Y_{3}^{∗}.

We want to find t ∈ R such that (α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗}) + β_{3}Y_{3}^{∗} = l ∈ O_{rω+tAω} = (rω) + tAω + (Aω)^{⊥}, i.e., β_{3}Y_{3}^{∗} ∈ tAω + (Aω)^{⊥}. Since the two-dimensional plane tAω + (Aω)^{⊥} in 𝔤_{1}^{∗} is obtained by pushing the plane (Aω)^{⊥} out along the direction Aω, t is the component of the vector β_{3}Y_{3}^{∗} on the direction Aω. In other words, t is the value of inner product of vectors β_{3}Y_{3}^{∗} and Aω in 𝔤_{1}^{∗}, i.e., t = ⟨β_{3}Y_{3}^{∗}, \frac{α_{1}}{r} Y_{1}^{∗} + \frac{α_{2}}{r} Y_{2}^{∗} + \frac{α_{3}}{r} Y_{3}^{∗}⟩ = \frac{α_{3}}{r} β_{3}. It follows that β_{3} = \frac{r}{α_{3}} t, and hence dβ_{3} = \frac{r}{α_{3}} dt. Therefore the equation (1.1) turns out to be

f(e) = ∫_{R^{3}} ∫_{R} |α_{3}|Trπ_{α_{1}Z_{1}^{∗}+α_{2}Z_{2}^{∗}+α_{3}Z_{3}^{∗}+β_{3}Y_{3}^{∗}}(f)dβ_{3}(dα_{1}dα_{2}dα_{3})
= ∫_{𝔤_{2}^{∗}} ∫_{R} |α_{3}|Trπ_{rω+tAω}(f)|\frac{r}{α_{3}} |dtdv(r, ω)
= ∫_{𝔤_{2}^{∗}} ∫_{R} rTrπ_{rω+tAω}(f)dtdv(r, ω)

If we start with the equation (1.2), a similar argument leads to the same formula. This answers the question at the start of this section. Corwin & Greenleaf describe the formula

f(e) = ∫_{𝔤_{2}^{∗}} ∫_{R} rTrπ_{rω+tAω}(f)dtdv(r, ω)

as ‘more’ invariant than the other two forms. But it is surely not canonical. It depends on choices of bases for 𝔤, and on the map A as well. Our goal is to redescribe Plancherel measure on a support expressed in terms of Aut^{∗}(𝔤), and to do this in greater generality.

2. Parametrizing Orbits by Automorphisms

Theorem 2.1. Let 𝔤 = F_{n,2}, the free 2-step nilpotent Lie algebra on n generators, and let 𝔤^{∗} be its dual space. Let G be the corresponding simply connected nilpotent Lie group. Let 𝔤_{Max}^{∗} be the set of all functionals whose orbits have maximal dimension. Let O_{Max} be the set of all orbits of maximal dimension. Fix l_{0} ∈ 𝔤_{Max}^{∗}. Then there is a one-to-one correspondence between the quotient space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) and the set O_{Max}.

Proof. There are two cases depending upon the positive integer n.

Case 1. Suppose n is odd. Let 𝔤_{2} = [𝔤, 𝔤]: then dim(𝔤_{2}) = \frac{n(n−1)}{2}. Pick a Strong Malcev basis B = \{X_{1}, X_{2}, ⋯, X_{m}\} for 𝔤 through 𝔤_{2} where m = \frac{n(n−1)}{2} + n. Let B^{∗} = {X_{1}^{∗}, X_{2}^{∗}, ⋯, X_{m}^{∗}} be its dual basis in 𝔤^{∗}. Then B^{∗} is a Jordan-Hölder basis for 𝔤^{∗}. Let V_{j} = Span_{R}{X_{j+1}^{∗}, ⋯, X_{m}^{∗}} and V_{m} = {0}. Since 𝔤_{2} is the center and 𝔤 is two-step, coadjoint orbits in 𝔤^{∗} can not have dimension greater than n. And by Lemma 1.3.2. in [1], coadjoint orbits are always of even dimension. Since n is odd, the maximal dimension of orbits can not be greater than n-1. By definition generic orbits have maximal dimension in each quotient space 𝔤^{∗}/V_{j}. In particular, each generic orbit has the maximal possible dimension in 𝔤^{∗}, though the converse is false. It follows that each generic orbit has dimension n − 1 since we can achieve the maximum possible dimension n-1 by assigning the basis element X_{\frac{n(n−1)}{2} +1} to 𝔯_{l} and giving l values on the center such that the

non-degenerate bilinear form has a block-diagonal matrix with block [ \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ].

Now for each l and l′ ∈ 𝔤_{Max}^{∗}, dim O_{l} = n − 1 = dim O_{l′}. It follows that dim 𝔯_{l} = \frac{n(n−1)}{2} + 1 = dim 𝔯_{l′} since dim O_{l} = dim(G/R_{l}) = dim 𝔤 − dim 𝔯_{l}. Let A_{n}^{l} andA_{n}^{l′} be picked such that 𝔯_{l} = 𝔤_{2} ⊕ RA_{n}^{l} and 𝔯_{l′} = 𝔤_{2} ⊕ RA_{n}^{l′}. Let B_{l} and B_{l′} be the non-degenerate skew-symmetric bilinear forms corresponding to l and l′ living on 𝔤/𝔯_{l} and 𝔤/𝔯_{l′} respectively. By Theorem 6 of Chapter 10 in [5], there exists a basis \{\overline{A_{1}^{l}}, \overline{A_{2}^{l}}, ⋯, \overline{A_{n−1}^{l}}\} for 𝔤/𝔯_{l} such that with respect to this basis the non-degenerate form B_{l} has the matrix diag(B_{1}, B_{2,} ⋯, B_{\frac{n−1}{2}} ) where each B_{i} is a

2 × 2 matrix [ \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ], and similarly for B_{l′} using the basis \{\overline{A_{1}^{l′}}, \overline{A_{2}^{l′}}, ⋯, \overline{A_{n−1}^{l′}}\}.

Since l and l′ ∈ 𝔤_{Max}^{∗}, we know l |_{𝔤_{2}}≠ 0 and l |_{𝔤_{2}}≠ 0. Then let’s subtract multiples of a central vector from each A_{i}^{l} and A_{i}^{l′} to make l(A_{i}^{l} − c_{i}Z^{l}) = l′(A_{i}^{l′} − c′_{i}Z^{l′}), and still denote A_{i}^{l} − c_{i}Z^{l} and A_{i}^{l′} − c′_{i}Z^{l′} by A_{i}^{l} and A_{i}^{l′}. Thus we have

l(A_{i}^{l}) = l′(A_{i}^{l′})

∀i = 1, 2, ⋯, n. Now let’s define two bases for 𝔤 through 𝔤_{2}

B^{l} = {[A_{i}^{l}, A_{j}^{l}] : 1 ≤ i < j ≤ n} ∪ {A_{1}^{l}, A_{2}^{l}, ⋯, A_{n}^{l}}

and

B^{l′} = {[A_{i}^{l′}, A_{j}^{l′}] : 1 ≤ i < j ≤ n} ∪ {A_{1}^{l′}, A_{2}^{l′}, ⋯, A_{n}^{l′}}

Then define the mapping A : 𝔤 ⟶ 𝔤 by

A_{i}^{l} ⟼ A_{i}^{l′}

and

[A_{i}^{l}, A_{j}^{l}] ⟼ [A_{i}^{l′}, A_{j}^{l′}]

1 ≤ i < j ≤ n. It follows that A ∈ Aut(𝔤) and A𝔯_{l} = 𝔯_{l′}. Hence we have the following identities

(A^{∗}l)([A_{i}^{l′}, A_{j}^{l′}]) = l([A^{−1}A_{i}^{l′}, A^{−1}A_{j}^{l′}]) = l([A_{i}^{l}, A_{j}^{l}]) = B_{l}(\overline{A_{i}^{l}}, \overline{A_{j}^{l}}) = B_{l′}(\overline{A_{i}^{l′}}, \overline{A_{j}^{l′}})
= l′([A_{i}^{l′}, A_{j}^{l′}])

1 ≤ i < j ≤ n − 1. If j=n, since A_{n}^{l} ∈ 𝔯_{l} and A_{n}^{l′} ∈ 𝔯_{l′}, we still have (A^{∗}l)([A_{i}^{l′}, A_{j}^{l′}]) = l([A_{i}^{l}, A_{j}^{l}]) = 0 = l′([A_{i}^{l′}, A_{j}^{l′}]) So (A^{∗}l)|_{𝔤_{2}} = l′|_{𝔤_{2}}. And we know also that

(A^{∗}l)(A_{i}^{l′}) = l(A^{−1}A_{i}^{l′}) = l(A_{i}^{l}) = l′(A_{i}^{l′})

∀i = 1, ⋯, n. Thus we get the relation A^{∗}l = l′. Next we claim that A^{∗}O_{l} = O_{l′}. Here we abuse notation to denote by A the two corresponding automorphisms of 𝔤 and G respectively. For every x ∈ G, the map i_{x} : g ⟼ xgx^{−1} is an inner automorphism of G. Then we have

(A ∘ i_{x})(g) = A(xgx^{−1}) = (Ax)(Ag)(Ax)^{−1} = (i_{Ax} ∘ A)(g)

∀g ∈ G. So A ∘ i_{x} = i_{Ax} ∘ A. Then taking its differential at the unit element we get the identity A ∘ Ad_{x} = Ad_{Ax} ∘ A. By using the duality on both sides it follows that

A^{∗}Ad_{x}^{∗} = Ad_{Ax}^{∗}A^{∗},

for all x ∈ G. Applying both maps to the functional l we get the relation A^{∗}O_{l} = O_{A^{∗}l} = O_{l′} as desired. In fact, the relation A^{∗}O_{l} = O_{A^{∗}l} holds for all automorphisms A. Therefore for every l, l′ ∈ 𝔤_{Max}^{∗}, there exists an automorphism A of 𝔤 such that A^{∗}l = l′ and A^{∗}O_{l} = O_{l′}. Hence Aut^{∗}(𝔤) maps 𝔤_{Max}^{∗} onto itself.

Case 2. Suppose n is even. By arguments similar to those in the beginning of Case 1., since n is even, each generic orbit has dimension n: that is n is the maximal dimension for all orbits in 𝔤^{∗}. Now for each l, l′ ∈ 𝔤_{Max}^{∗}, we have dim 𝔯_{l} = dim 𝔯_{l′} = \frac{n(n−1)}{2} = dim 𝔤_{2}. Then 𝔯_{l} = 𝔤_{2} = 𝔯_{l′} since the center 𝔤_{2} ⊆ 𝔯_{l}. Now let B_{l} and B_{l′} be the non-degenerate skew-symmetric bilinear forms corresponding to l and l′ living on 𝔤/𝔤_{2} respectively. Then there exists a basis \{\overline{A_{1}^{l}}, \overline{A_{2}^{l}}, ⋯, \overline{A_{n}^{l}}\} for 𝔤/𝔤_{2} giving a canonical matrix for B_{l}, and similarly for B_{l′}. Let’s subtract multiples of a central vector from each A_{i}^{l} and A_{i}^{l′} to make l(A_{i}^{l} − c_{i}Z^{l}) = l′(A_{i}^{l′} − c′_{i}Z^{l′}), and still denote A_{i}^{l} − c_{i}Z^{l} and A_{i}^{l′} − c′_{i}Z^{l′} by A_{i}^{l} and A_{i}^{l′}. Thus we have l(A_{i}^{l}) = l′(A_{i}^{l′}), ∀i = 1, 2, ⋯, n. Then define the map A : 𝔤 ⟶ 𝔤 by

A_{i}^{l} ⟼ A_{i}^{l′}

and

[A_{i}^{l}, A_{j}^{l}] ⟼ [A_{i}^{l′}, A_{j}^{l′}],

for 1 ≤ i < j ≤ n. Then A ∈ Aut(𝔤) such that A^{∗}l = l′ and A^{∗}O_{l} = O_{l′} as before. Thus, by cases 1 and 2 we know that Aut^{∗}(𝔤) maps 𝔤_{Max}^{∗} onto itself. By hypothesis l_{0} ∈ 𝔤_{Max}^{∗}, for every A, B ∈ Aut(𝔤) we know A^{∗}l_{0} = B^{∗}l_{0} ⇔ B^{∗^{−1}}A^{∗}l_{0} = l_{0} ⇔ B^{∗^{−1}}A^{∗} ∈ Stab(l_{0}) ⇔ A^{∗}Stab(l_{0}) = B^{∗}Stab(l_{0}). So we obtain that Aut^{∗}(𝔤)/Stab(l_{0}) maps l_{0} one-to-one and onto 𝔤_{Max}^{∗}. Now let’s prove the mapping

Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) ⟶ O_{Max}

Ad_{G}^{∗}A^{∗}Stab(l_{0}) ⟼ Ad_{G}^{∗}A^{∗}l_{0}

is one-to-one.

For each double coset Ad_{G}^{∗}A^{∗}Stab(l_{0}) we get an orbit of maximal dimension, Ad_{G}^{∗}A^{∗}l_{0} ∈ O_{Max}. Suppose Ad_{G}^{∗}A^{∗}l_{0} = Ad_{G}^{∗}B^{∗}l_{0}, then there exists a group element g ∈ G such that

Ad_{g}^{∗}A^{∗}l_{0} = B^{∗}l_{0} ⇔ l_{0} = A^{∗^{−1}}Ad_{g^{−1}}^{∗}B^{∗}l_{0} ⇔ A^{∗^{−1}}Ad_{g^{−1}}^{∗}B^{∗} ∈ Stab(l_{0})

⇔ B^{∗} ∈ Ad_{g}^{∗}A^{∗}Stab(l_{0}).

So B^{∗} ∈ Ad_{G}^{∗}A^{∗}Stab(l_{0}), and hence A^{∗} and B^{∗} are in the same double coset. Therefore the mapping

Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) ⟶ O_{Max}

is one-to-one. This proves the theorem.

3. Relative Invariance of Plancherel Measure

Let S be a topological space and let G be a topological group acting on S. A measure μ on S is called a G-invariant measure if for every g ∈ G and for every measurable subset E of S, the set gE is measurable and μ(gE) = μ(E). It is very helpful if a non-trivial G-invariant Borel measure on a locally compact space S exists. But it is possible that no non-trivial invariant measure exists on S. This can happen already in the special case when G is a locally compact group, H is a closed subgroup of G and S is the homogeneous space G/H. A measure μ on a homogeneous space G/H is called relatively invariant provided D(g)μ(gE) ≡ μ(E) ∀g ∈ G and ∀E ⊂ G/H with E measurable. Then it is known that D is a homomorphism mapping G ⟶ R^{+} and that μ corresponds uniquely up to a constant factor to the homomorphism D. We call D the modular function for μ. Theorems (3.1) and (3.2) are proven in Chapter 5 of [4]. Theorem (3.1) Let Δ be the modular function of the locally compact group G, and let δ be the modular function of the closed subgroup H. If there exists a relatively invariant measure μ on G/H, then its modular function D must be a continuous real character such that D(h) = (Δ/δ)(h) for every h ∈ H. Theorem (3.2) A necessary and sufficient condition for the existence of a non-trivial relatively invariant measure μ on G/H with modular function D is that D is a continuous real character on G such that D(h) = (Δ/δ)(h) for every h ∈ H. If D satisfies the requirements, the relatively invariant measure associated with it is essentially uniquely determined by D. Now we are ready to state and prove the main result of this chapter.

Theorem (3.3). Let 𝔤 = F_{n,2}. Fix l_{0} ∈ 𝔤_{Max}^{∗}, let μ be a copy of Plancherel measure on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}). Then μ is the essentially unique relatively invariant measure corresponding to the homomorphism Ad_{G}^{∗}\Aut^{∗}(𝔤) ⟶ R^{+} given by \overline{A^{∗}} ⟼| det \overline{A^{∗}} |. Note. det(Ad_{g}^{∗}) = 1 for all g ∈ G, so det \overline{A^{∗}} ≡ det A^{∗}, ∀A ∈ Aut(𝔤).

Proof: For every automorphism A ∈ Aut(𝔤), we want to show that

μ(\overline{A^{∗}}E) =| det A^{∗} | μ(E)

for all measurable subsets E of Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}). For each A ∈ Aut(𝔤) we also denote its corresponding automorphism of G by A. For every l ∈ 𝔤^{∗} we claim that

π_{A^{−1^{∗}}l} ≅ π_{l} ∘ A

For every l ∈ 𝔤^{∗}, choose a polarizer M for l, and let M = exp(M). Then π_{l} = Ind_{M}^{G}χ_{l,M} where χ_{l,M} is a character defined on M. It follows that

π_{l} ∘ A = (Ind_{M}^{G}χ_{l,M}) ∘ A
≅ Ind_{A^{−1}(M)}^{G}(χ_{l,M} ∘ A)

by Lemma 2.1.3. in [1]. For every X ∈ M, since (χ_{l,M} ∘ A)(exp X) = χ_{l,M}(A(exp X)) = χ_{l,M}(exp(AX)) = e^{2πil(AX)} = e^{2πi(A^{−1^{∗}}l)(X)} = χ_{A^{−1^{∗}}l,M}(exp X), it follows further

π_{l} ∘ A ≅ Ind_{A^{−1}(M)}^{G}χ_{A^{−1^{∗}}l,M}
= π_{A^{−1^{∗}}l}

This proves the claim. Then for every ϕ ∈ S(G) we have

π_{A^{−1^{∗}}l}(ϕ) = ∫_{G} ϕ(g)π_{A^{−1^{∗}}l}(g)dg
≅ ∫_{G} ϕ(g)(π_{l} ∘ A)(g)dg
= ∫_{G} ϕ(g)π_{l}(Ag)dg
= ∫_{G} ϕ(A^{−1}x)π_{l}(x)d(A^{−1}x)
= ∫_{G}(ϕ ∘ A^{−1})(x)π_{l}(x)| det A^{−1}|dx
= | det A^{∗}| ∫_{G}(ϕ ∘ A^{−1})(x)π_{l}(x)dx
= | det A^{∗}|π_{l}(ϕ ∘ A^{−1})

and hence

Trπ_{A^{−1^{∗}}l}(ϕ) = | det A^{∗}|Trπ_{l}(ϕ ∘ A^{−1})

∀A ∈ Aut(𝔤), l ∈ 𝔤^{∗}, ϕ ∈ S(G). Now let D = Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}), note that the group Ad_{G}^{∗}\Aut^{∗}(𝔤) acts on the left on D. By hypothesis l_{0} ∈ 𝔤_{Max}^{∗}, we fix any strong Malcev basis B of 𝔤 such that l_{0} ∈ 𝔤_{gen}^{∗} with respect to B. Let O_{gen} = {O_{l}| l ∈ 𝔤_{gen}^{∗}}. We know ∪_{O∈O_{gen}}O is a Zariski-open set in 𝔤^{∗}, and Plancherel measure μ, as in [1], is supported on O_{gen} ⫋ O_{Max}. Then define a copy of μ on O_{Max} by letting μ(O_{Max}\O_{gen}) = 0. By Theorem 2.1. we know there is a one-to-one correspondence between O_{Max} and D. So copy μ again onto D, in other words, μ is a copy of Plancherel measure on D. Let E_{0} = \{\overline{A^{∗}} ∈ D| A^{∗}l_{0} ∈ 𝔤_{Max}^{∗}\𝔤_{gen}^{∗}\}, then E_{0} is a μ-nullset. For every measurable subset E ⊆ D, and every A ∈ Aut(𝔤) , we claim that

μ(\overline{A^{∗}}E) = | det A^{∗}|μ(E)

∫_{D} Trπ_{C^{∗}l_{0}}(ϕ)dμ(\overline{A^{∗}} C^{∗}^{.}) = ∫_{D} Trπ_{C^{∗}l_{0}}(ϕ)dμ((A^{∗}C^{∗})^{.})
= ∫_{D} Trπ_{A^{−1^{∗}}C^{∗}l_{0}}(ϕ)dμ(C^{∗}^{.})
= ∫_{D} | det A^{∗}|Trπ_{C^{∗}l_{0}}(ϕ ∘ A^{−1})dμ(C^{∗}^{.})
= | det A^{∗}| ∫_{D} Trπ_{C^{∗}l_{0}}(ϕ ∘ A^{−1})dμ(C^{∗}^{.})
= | det A^{∗}|(ϕ ∘ A^{−1})(e)
= | det A^{∗}|ϕ(e)
= ∫_{D} Trπ_{C^{∗}l_{o}}(ϕ)| det A^{∗}|dμ(C^{∗}^{.})

∀ ϕ ∈ S(G), where \overline{A^{∗}} means coset Ad_{G}^{∗}A^{∗} ∈ Ad_{G}^{∗}\Aut^{∗}(𝔤) and C^{∗}^{.} means coset Ad_{G}^{∗}C^{∗}Stab(l_{0}) ∈ D, and hence

\overline{A^{∗}} C^{∗}^{.}= Ad_{G}^{∗}A^{∗}Ad_{G}^{∗}C^{∗}Stab(l_{0}) = Ad_{G}^{∗}(A^{∗}C^{∗})Stab(l_{0}) =(A^{∗}C^{∗})^{.}.

Thus we get the identity

∫_{D} Trπ_{C^{∗}l_{0}}(ϕ)dμ(\overline{A^{∗}} C^{∗}^{.}) = ∫_{D} Trπ_{C^{∗}l_{0}}(ϕ) | det A^{∗} | dμ(C^{∗}^{.}).

Then by the Abstract Plancherel Theorem in Chapter 7 of [8], or in [2], one concludes, since μ is essentially unique, that

μ(\overline{A^{∗}}E) =| det A^{∗} | μ(E)

for all A ∈ Aut(𝔤) and all measurable subsets E ⊆ D. Therefore μ is the essentially unique relatively invariant measure on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) corresponding to the homomorphism given by \overline{A^{∗}} ⟼| det A^{∗} |=| det \overline{A^{∗}} |.

Remark 1. This theorem can be generalized readily as follows. Let G be any group to which ’the Abstract Plancherel Theorem’ applies. For every A ∈ Aut(𝔤), define A^{∗}π = π ∘ A^{−1}, ∀π ∈ Ĝ. Then Tr((A^{∗}π)(ϕ)) = | det A|Tr(π(ϕ ∘ A)). Now suppose further that for every π in the support of Plancherel measure μ, the set {A^{∗}π| A ∈ Aut(G)} contains the support of Plancherel measure. Let Stab(π) = {A ∈ Aut(G)| A^{∗}π ≅ π}. Then μ can be transported to a new domain Aut(G)/Stab(π), where μ is 0 on the complement of its support. Then the same argument as at the top of page 15 shows that μ(\overline{A^{∗}}E) ≡ | det \overline{A^{∗}}|μ(E). Thus μ displays the same relative invariance whenever it can be transported to the domain Aut(G)/Stab(π). In the case of F_{n,2}, we have exploited the symmetry of the group to prove that Aut^{∗}(G)π covers the support of the Plancherel measure whenever π is in that support. Remark 2. Consider the example of the 2n + 1 dimensional Heisenberg group H_{2n+1}, which is not free if n > 1. Let 𝔥_{2n+1} be its Heisenberg algebra which is spanned by {Z, Y_{1}, ⋯, Y_{n}, X_{1}, ⋯, X_{n}} with non-trivial brackets [X_{i}, Y_{i}] = Z, ∀i = 1, ⋯ n. Consider the generic representations π_{l} requiring l(Z) ≠ 0. Let δ_{r} be the usual dilation automorphism, and δ_{r}^{∗} be its dual, i.e.,

δ_{r} : \{ \begin{matrix}X_{i} & ⟼ \frac{1}{\sqrt{r}}X_{i} \\ Y_{i} & ⟼ \frac{1}{\sqrt{r}}Y_{i} \\ Z & ⟼ \frac{1}{r}Z \end{matrix} \} and δ_{r}^{∗} : \{ \begin{matrix}X_{i}^{∗} & ⟼ \sqrt{r}X_{i}^{∗} \\ Y_{i}^{∗} & ⟼ \sqrt{r}Y_{i}^{∗} \\ Z^{∗} & ⟼ rZ^{∗} \end{matrix} \}

Then δ_{r}^{∗}(π_{l}) ≅ π_{δ_{r}^{∗}l} and (δ_{r}^{∗}l)(Z) = rl(Z) ≠ 0. Therefore there exist sufficient automorphisms for the first remark to apply to all these (non-free!) groups as well. Remark 3. Consider the example of 3-step Chain group G generated by X, Y_{1}, Y_{2}, Z where [X, Y_{1}] = Y_{2} and [X, Y_{2}] = Z generate all non-trivial brackets. This is neither 2-step nor free. In Example (3.1.12) of [1] it is shown that all the generic representations π_{l} are given by l = zZ^{∗} + y_{1}Y_{1}^{∗} with z ≠ 0 and y_{1} ∈ R. Its radical 𝔯_{l} = Span_{R}{Z, Y_{1}}, hence its polarizer is three-dimensional since the dimension of polarizer is the average of dimensions of whole algebra and radical. Since 𝔪 = Span_{R}{Z, Y_{2}, Y_{1}} is an abelian ideal in 𝔤 of the correct dimension, it is a polarizer for each l = zZ^{∗} + y_{1}Y_{1}^{∗}. Now consider the following automorphism T_{A,e} indexed by

A = [ \begin{matrix}a_{1} & a_{2} \\ 0 & a_{3} \end{matrix} ] ∈ GL(2, R) and e ∈ R given by

T_{A,e} : \{ \begin{matrix}X & ⟼ a_{1}X \\ Y_{1} & ⟼ a_{2}X + a_{3}Y_{1} + eZ \\ Y_{2} & ⟼ (det A)Y_{2} \\ Z & ⟼ (a_{1} det A)Z \end{matrix} \}

By Kirillov theory we know each generic π_{z,y_{1}} is induced by a character χ_{z,y_{1}} defined on polarizer 𝔪. Then π_{z,y_{1}} ∘ T_{A,e} ≅ π_{T_{A,e}^{−1^{∗}}l_{z,y_{1}}} = π_{(a_{1} det A)z,a_{3}y_{1}+ez}. So any generic π can be carried into any other such representation by a suitable automorphism. Thus, Remark 1. applies to this case too.

4. Nonexistence of Invariant Measure

Proposition 4.1. Let 𝔤 = F_{3,2}. For l_{0} ∈ 𝔤_{Max}^{∗}, there is no \overline{Aut^{∗}(𝔤)}-invariant measure on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}). Proof. Pick a basis B = {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}} for 𝔤 just as we did before in the Introduction, and let B^{∗} = {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}, Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} be its dual basis in 𝔤^{∗}. We start with the simple case in which l_{0} = Z_{3}^{∗}, then we will show that the result in the general case follows immediately. We begin by identifying Stab^{∗}(l_{0}) explicitly. By the definition

Stab^{∗}(l_{0}) = {A ∈ Aut(𝔤)| A^{∗}l_{0} = l_{0}}
= {A ∈ Aut(𝔤)| l_{0} = A^{−1^{∗}}l_{0}}
= {A ∈ Aut(𝔤)| l_{0}(X) = (A^{−1^{∗}}l_{0})(X), ∀X ∈ 𝔤}
= {A ∈ Aut(𝔤)| l_{0}(X) = l_{0}(AX), ∀X ∈ 𝔤}
= {A ∈ Aut(𝔤)| l_{0}((A − I)(X)) = 0, ∀X ∈ 𝔤}
= {A ∈ Aut(𝔤)| (A − I) : 𝔤 → ker(l_{0})}

This says Stab^{∗}(l_{0}) is the subgroup of Aut(𝔤) such that the operator after subtracting the identity from each element maps 𝔤 into the kernel of l_{0}. For l_{0} = Z_{3}^{∗}, its kernel ker(l_{0}) = Span_{R}{Y_{1}, Y_{2}, Y_{3}, Z_{1}, Z_{2}}. Then for each A ∈ Stab^{∗}(l_{0}), we verify two properties:

AZ_{3} ∈ Z_{3} + Span_{R}{Z_{1}, Z_{2}}  (4.1)
A(ker(l_{0})) = ker(l_{0})  (4.2)

Since A ∈ Stab^{∗}(l_{0}), we have (A − I)(Z_{3}) ∈ ker(l_{0}), so AZ_{3} ∈ Z_{3} + ker(l_{0}). Since each automorphism of 𝔤 maps 𝔤_{2} onto 𝔤_{2}, it follows AZ_{3} ∈ (Z_{3} + ker(l_{0})) ∩ 𝔤_{2} = Z_{3} + Span_{R}{Z_{1}, Z_{2}}. This proves property (4.1).

For each K ∈ ker(l_{0}) we have l_{0}(AK) = (A^{−1^{∗}}l_{0})(K) = l_{0}(K) = 0, so A(ker(l_{0})) ⊆ ker(l_{0}). By hypothesis A ∈ Stab^{∗}(l_{0}), so is A^{−1}. Using the above argument we get A^{−1}(ker(l_{0})) ⊆ ker(l_{0}), and hence ker(l_{0}) ⊆ A(ker(l_{0})). Thus we obtain A(ker(l_{0})) = ker(l_{0}) as desired. For each A ∈ Stab^{∗}(l_{0}), since A is also an automorphism of 𝔤, we know that A maps 𝔤_{2} onto 𝔤_{2}. This reason and the property A(ker(l_{0})) = ker(l_{0}) together imply that

A|_{Span_{R}{Z_{1},Z_{2}}} : Span_{R}{Z_{1}, Z_{2}} ⟶ Span_{R}{Z_{1}, Z_{2}}  (4.3)

By properties (4.1), (4.2), and (4.3) we get a 6 × 6 matrix

[A]_{B} = [ \begin{matrix}∗ & ∗ & ∗ & ∗ & ∗ & ∗ \\ ∗ & ∗ & ∗ & ∗ & ∗ & ∗ \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & p_{7} \\ 0 & 0 & 0 & p_{2} & p_{5} & p_{8} \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ] =: [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ]

Then det A = (det Q)(det P) = (det P)^{2}(det P) = (det P)^{3} by Lemma 6.3. Since A ∈ Aut(𝔤), A is invertible, and hence so is the matrix P. In other words, {AY_{1}, AY_{2}, AY_{3}} is linearly independent modulo 𝔤_{2}. Since A ∈ Aut(𝔤), this means A([Y_{i}, Y_{j}]) = [AY_{i}, AY_{j}]. Explicitly,

A(Z_{1}) = A([Y_{2}, Y_{3}]) = [AY_{2}, AY_{3}] = [p_{4}Y_{1} + p_{5}Y_{2} + p_{6}Y_{3}, p_{7}Y_{1} + p_{8}Y_{2} + p_{9}Y_{3}]
= (p_{5}p_{9} − p_{6}p_{8})Z_{1} − (p_{4}p_{9} − p_{6}p_{7})Z_{2} + (p_{4}p_{8} − p_{5}p_{7})Z_{3}
A(Z_{2}) = A([Y_{3}, Y_{1}]) = [AY_{3}, AY_{1}] = [p_{7}Y_{1} + p_{8}Y_{2} + p_{9}Y_{3}, p_{1}Y_{1} + p_{2}Y_{2} + p_{3}Y_{3}]
= −(p_{2}p_{9} − p_{3}p_{8})Z_{1} + (p_{1}p_{9} − p_{3}p_{7})Z_{2} − (p_{1}p_{8} − p_{2}p_{7})Z_{3}

A(Z_{3}) = A([Y_{1}, Y_{2}]) = [AY_{1}, AY_{2}] = [p_{1}Y_{1} + p_{2}Y_{2} + p_{3}Y_{3}, p_{4}Y_{1} + p_{5}Y_{2} + p_{6}Y_{3}]
= (p_{2}p_{6} − p_{3}p_{5})Z_{1} − (p_{1}p_{6} − p_{3}p_{4})Z_{2} + (p_{1}p_{5} − p_{2}p_{4})Z_{3}

Then we get the corresponding matrix

Q = [ \begin{matrix}det ( \begin{matrix}p_{5} & p_{8} \\ p_{6} & p_{9} \end{matrix} ) & − det ( \begin{matrix}p_{2} & p_{8} \\ p_{3} & p_{9} \end{matrix} ) & det ( \begin{matrix}p_{2} & p_{5} \\ p_{3} & p_{6} \end{matrix} ) \\ − det ( \begin{matrix}p_{4} & p_{7} \\ p_{6} & p_{9} \end{matrix} ) & det ( \begin{matrix}p_{1} & p_{7} \\ p_{3} & p_{9} \end{matrix} ) & − det ( \begin{matrix}p_{1} & p_{4} \\ p_{3} & p_{6} \end{matrix} ) \\ det ( \begin{matrix}p_{4} & p_{7} \\ p_{5} & p_{8} \end{matrix} ) & − det ( \begin{matrix}p_{1} & p_{7} \\ p_{2} & p_{8} \end{matrix} ) & det ( \begin{matrix}p_{1} & p_{4} \\ p_{2} & p_{5} \end{matrix} ) \end{matrix} ] =: [ \begin{matrix}q_{1} & q_{4} & q_{7} \\ q_{2} & q_{5} & q_{8} \\ q_{3} & q_{6} & q_{9} \end{matrix} ]

Since the third row of Q is [0, 0, 1], we obtain three conditions:

det ( \begin{matrix}p_{4} & p_{7} \\ p_{5} & p_{8} \end{matrix} ) = 0, − det ( \begin{matrix}p_{1} & p_{7} \\ p_{2} & p_{8} \end{matrix} ) = 0, det ( \begin{matrix}p_{1} & p_{4} \\ p_{2} & p_{5} \end{matrix} ) = 1.

We analyze them by two cases.

Case 1. Suppose p_{1} ≠ 0. Since we have the conditions

p_{4}p_{8} − p_{5}p_{7} = 0, p_{1}p_{8} − p_{2}p_{7} = 0, p_{1}p_{5} − p_{2}p_{4} = 1,

it follows that p_{8} = \frac{p_{2}p_{7}}{p_{1}} and p_{5} = \frac{1+p_{2}p_{4}}{p_{1}} , and hence the first condition 0 = p_{4}p_{8} − p_{5}p_{7} = p_{4} \frac{p_{2}p_{7}}{p_{1}} − \frac{1+p_{2}p_{4}}{p_{1}} p_{7} = −\frac{p_{7}}{p_{1}} . Then p_{7} = 0 and p_{8} = 0. So we get the matrix

[A]_{B} = [ \begin{matrix}q_{1} & q_{4} & q_{7} & r_{1} & r_{4} & r_{7} \\ q_{2} & q_{5} & q_{8} & r_{2} & r_{5} & r_{8} \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & 0 \\ 0 & 0 & 0 & p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

Since the matrix P is invertible, it follows det P = p_{9} ≠ 0. Hence for each A ∈ Stab^{∗}(Z_{3}^{∗}) with p_{1} ≠ 0, A has the above form of 6 × 6 matrix relative to the ordered basis B with the property det A = (det P)^{3} = p_{9}^{3} ≠ 0. Thus, the matrix Q is completely determined by matrix P, and matrices P and R have a total of 12 variables. Hence dim(Stab^{∗}(Z_{3}^{∗})) = 12.

Case 2. Suppose p_{1} = 0. The second and third condition follow that p_{2}p_{7} = 0 and p_{2}p_{4} = −1. Then p_{2} ≠ 0, and hence p_{7} = 0. So the first condition follows that p_{4}p_{8} = 0, then p_{8} = 0 since p_{4} = −\frac{1}{p_{2}} ≠ 0. Thus we get the matrix

[A]_{B} = [ \begin{matrix}q_{1} & q_{4} & q_{7} & r_{1} & r_{4} & r_{7} \\ q_{2} & q_{5} & q_{8} & r_{2} & r_{5} & r_{8} \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −\frac{1}{p_{2}} & 0 \\ 0 & 0 & 0 & p_{2} & p_{5} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

with det A = (det P)^{3} = p_{9}^{3} ≠ 0. So the case p_{1} = 0 corresponds to an 11-dimensional submanifold of the 12-dimensional manifold Stab^{∗}(Z_{3}^{∗}). From the above two cases we see that we can always use p_{1} as one of the coordinates in every local chart for Stab^{∗}(Z_{3}^{∗}). If p_{1} ≠ 0, we get the local coordinates given in the first case. If p_{1} = 0, we get an 11-dimensional submanifold of Stab^{∗}(Z_{3}^{∗}) corresponding to p_{1} = 0.

From Theorem 3.3. we know that the modular function D of the double coset space D := Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}) is D(\overline{A^{∗}}) = (\frac{Δ}{δ})(\overline{A^{∗}}) = | det \overline{A^{∗}}|^{−1} = | det A| for every \overline{A^{∗}} ∈ Ad_{G}^{∗}\Ad_{G}^{∗} · Stab(l_{0}), i.e., A ∈ Stab^{∗}(l_{0}). And from the previous two cases, we already got | det A| = |p_{9}^{3}|, ∀A ∈ Stab^{∗}(l_{0}), which is not identically 1. Therefore there does not exist an invariant measure on D since the modular functions Δ and δ of Ad_{G}^{∗}\Aut^{∗}(𝔤) and Ad_{G}^{∗}\Ad_{G}^{∗} · Stab(l_{0}), respectively, do not agree on the subgroup Ad_{G}^{∗}\Ad_{G}^{∗} · Stab(l_{0}) for l_{0} = Z_{3}^{∗}. We next show the same result for general functionals l_{0} ∈ 𝔤_{Max}^{∗}. For each l_{0} ∈ 𝔤_{Max}^{∗}, we know already that there exists an automorphism A_{0} of 𝔤 such that A_{0}^{∗}Z_{3}^{∗} = l_{0}. Then it follows that

Stab^{∗}(l_{0}) = {A ∈ Aut(𝔤)| A^{∗}l_{0} = l_{0}}
= {A ∈ Aut(𝔤)| A^{∗}A_{0}^{∗}Z_{3}^{∗} = A_{0}^{∗}Z_{3}^{∗}}
= {A_{0}(A_{0}^{−1}AA_{0})A_{0}^{−1}| (A_{0}^{−1}AA_{0})^{∗}Z_{3}^{∗} = Z_{3}^{∗}}
= A_{0}{A_{0}^{−1}AA_{0}| (A_{0}^{−1}AA_{0})^{∗}Z_{3}^{∗} = Z_{3}^{∗}}A_{0}^{−1}
= A_{0}{B ∈ Aut(𝔤)| B^{∗}Z_{3}^{∗} = Z_{3}^{∗}}A_{0}^{−1}
= A_{0}Stab^{∗}(Z_{3}^{∗})A_{0}^{−1}

So for every A ∈ Stab^{∗}(l_{0}), A = A_{0}CA_{0}^{−1} for some C ∈ Stab^{∗}(Z_{3}^{∗}). And hence | det A| = | det(A_{0}CA_{0}^{−1})| = | det C| = |p_{9}^{3}|, which is not identically 1 either. Therefore we obtain the same conclusion that there is no invariant measure on the double coset space Ad^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}).

5. Explicit Calculations for F_{3,2}

Brief Outline. There are seven lemmas and one verification in this long chapter for 𝔤 = F_{3,2}. We make a brief outline here. By using the matrix format in the Proposition 4.1. , we find an almost global coordinate patch for Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}) in Lemma 5.1. In Lemma 5.2. we find a right and left Haar measure on Ad_{G}\Ad_{G} · Stab^{∗}(l_{0}) with respect to the coordinate patch in Lemma 5.1. So its modular function is δ = |\frac{1}{p_{9}^{5}} | = |\frac{1}{(det P)^{5}} | In Lemma 5.3. we find an almost global coordinate patch for Ad_{G}\Aut(𝔤). Then with respect to this patch, we figure out a right and left Haar measure on Ad_{G}\Aut(𝔤) in Lemma 5.4. Hence its modular function is Δ = |\frac{1}{(det P)^{2}} | In Lemma 5.5. we construct automorphisms ρ_{t}^{σ} of 𝔤 related to automorphisms σ ⊗ σ. Then we find a cross-section X for Ad_{G}\Aut(𝔤)/Stab^{∗}(l_{0}) in Lemma 5.6. Furthermore, in Lemma 5.7. we show to which element of X the composition of two elements of X corresponds since X is not a group. Then we use this concrete cross-section to verify the relative invariance for 𝔤 = F_{3,2}. This is the purpose of Chapter 5. Throughout this chapter we have two ways to construct bases for F_{3,2}. One involves norms, another does not. We build them here, and later just cite them for the hypothesis of each lemma. Construction 1. Let 𝔤 = F_{3,2}, 𝔤_{2} = [𝔤, 𝔤], and let 𝔤^{∗} be the dual space of 𝔤. Choose a basis {Ȳ_{1}, Ȳ_{2}, Ȳ_{3}} for 𝔤/𝔤_{2}. Pick Y_{i} ∈ Ȳ_{i}, ∀i = 1, 2, 3. Let Z_{1} = [Y_{2}, Y_{3}], Z_{2} = [Y_{3}, Y_{1}], Z_{3} = [Y_{1}, Y_{2}], and these are independent of the choice of Y_{i} ∈ Ȳ_{i}. Then B := {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}} is a strong Malcev basis of 𝔤 through 𝔤_{2}.

Let B^{∗} := {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}, Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} be the dual basis of B in 𝔤^{∗}, so B^{∗} is a Jordan-Hölder basis of 𝔤^{∗}. We know 𝔤_{2} = Span_{R}{Z_{1}, Z_{2}, Z_{3}}, let 𝔤_{1} := Span_{R}{Y_{1}, Y_{2}, Y_{3}}, 𝔤_{2}^{∗} := Span_{R}{Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}}, and 𝔤_{1}^{∗} := Span_{R}{Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}}. Construction 2. Let 𝔤 = F_{3,2}, 𝔤_{2} = [𝔤, 𝔤], and let 𝔤^{∗} be the dual space of 𝔤. Give 𝔤 a norm respecting Lie bracket, and use the operator norm on 𝔤^{∗}. Let 𝔤_{1} be the orthogonal complement of 𝔤_{2} in 𝔤. Choose an orthonormal basis {Y_{1}, Y_{2}, Y_{3}} for 𝔤_{1}, then let Z_{1} = [Y_{2}, Y_{3}], Z_{2} = [Y_{3}, Y_{1}], and Z_{3} = [Y_{1}, Y_{2}]. Since 𝔤 is equipped with a norm respecting bracket (which is defined at the first page of Chapter 6) , by Lemma 6.1. it follows that B := {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}} is an orthonormal basis for 𝔤 through 𝔤_{2}. Let B^{∗} := {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}, Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} be the dual basis of B in 𝔤^{∗}, then B^{∗} is also an orthonormal basis for 𝔤^{∗} by Lemma 6.2. We know 𝔤_{2} = Span_{R}{Z_{1}, Z_{2}, Z_{3}} and 𝔤_{1} = Span_{R}{Y_{1}, Y_{2}, Y_{3}}, let 𝔤_{2}^{∗} := Span_{R}{Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}} and 𝔤_{1}^{∗} := Span_{R}{Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}}. Note. If the hypothesis does not require a norm of 𝔤 to respect the Lie bracket, then we can just build a basis B^{∗} for 𝔤^{∗} by Construction 1., then give 𝔤^{∗} a Euclidean metric and corresponding norm by making B^{∗} orthonormal. Lemma 5.1. Let 𝔤 = F_{3,2}, and pick a basis B for 𝔤 just as we did in Construction 1. Let C_{S} be the set of all A ∈ Aut(𝔤) such that A relative to the basis B has 6 × 6 matrix

[ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) & r_{1} & r_{4} & 0 \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} & 0 & r_{5} & 0 \\ 0 & 0 & 1 & r_{3} & r_{6} & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & 0 \\ 0 & 0 & 0 & p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

with p_{i} ∈ R, p_{1}p_{9} ≠ 0, ∀i = 1, 2, 3, 4, 6, 9, and r_{j} ∈ R, ∀j = 1, 3, 4, 5, 6. We claim that C_{S} is an almost global coordinate patch for Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}).

Proof. Since Ad_{G} is normal in Aut(𝔤), and Stab^{∗}(Z_{3}^{∗}) is a subgroup of Aut(𝔤), it follows their product Ad_{G} · Stab^{∗}(Z_{3}^{∗}) is a group. we first need to re-parametrize cosets Ad_{G}s ∈ Ad_{G} · Stab^{∗}(Z_{3}^{∗}). For every X ∈ 𝔤, let X = ∑_{i=1}^{3} z_{i}Z_{i} + ∑_{i=1}^{3} y_{i}Y_{i}. Since 𝔤 is two-step, we know Ad_{exp(X)} = exp(ad_{X}) = I + ad_{X}. Since ad_{X}Y_{1} = [X, Y_{1}] = y_{3}Z_{2} − y_{2}Z_{3}, ad_{X}Y_{2} = [X, Y_{2}] = −y_{3}Z_{1} + Y_{1}Z_{3}, ad_{X}Y_{3} = [X, Y_{3}] = y_{2}Z_{1} − y_{1}Z_{2}, and (ad_{X})|_{𝔤_{2}} = 0, it follows

[Ad_{exp(X)}]_{B} = [ \begin{matrix}1 & 0 & 0 & 0 & −y_{3} & y_{2} \\ 0 & 1 & 0 & y_{3} & 0 & −y_{1} \\ 0 & 0 & 1 & −y_{2} & y_{1} & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ] =: [ \begin{matrix}I_{3×3} & S_{3×3} \\ 0_{3×3} & I_{3×3} \end{matrix} ]

In Proposition 4.1. we have shown already that for almost all s ∈ Stab^{∗}(Z_{3}^{∗})

[s]_{B} = [ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) & r_{1} & r_{4} & r_{7} \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} & r_{2} & r_{5} & r_{8} \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & 0 \\ 0 & 0 & 0 & p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

=: [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ]

Then

Ad_{G}s
= {Ad_{exp(X)}s| X = ∑_{i=1}^{3} z_{i}Z_{i} + ∑_{i=1}^{3} y_{i}Y_{i}, ∀z_{i}, y_{i} ∈ R}
= \{Ad_{exp(X)}s|[Ad_{exp(X)}s]_{B} = [ \begin{matrix}I & S \\ 0 & I \end{matrix} ] [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] = [ \begin{matrix}Q & R + SP \\ 0 & P \end{matrix} ] , ∀y_{i} ∈ R\}

where the 3 × 3 matrix

R + SP
= [ \begin{matrix}r_{1} & r_{4} & r_{7} \\ r_{2} & r_{5} & r_{8} \\ 0 & 0 & 0 \end{matrix} ] + [ \begin{matrix}0 & −y_{3} & y_{2} \\ y_{3} & 0 & −y_{1} \\ −y_{2} & y_{1} & 0 \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= [ \begin{matrix}r_{1} + p_{3}y_{2} − p_{2}y_{3} & r_{4} + p_{6}y_{2} − \frac{1+p_{2}p_{4}}{p_{1}} y_{3} & r_{7} + p_{9}y_{2} \\ r_{2} − p_{3}y_{1} + p_{1}y_{3} & r_{5} − p_{6}y_{1} + p_{4}y_{3} & r_{8} − p_{9}y_{1} \\ p_{2}y_{1} − p_{1}y_{2} & \frac{1+p_{2}p_{4}}{p_{1}} y_{1} − p_{4}y_{2} & 0 \end{matrix} ]

Since the condition of s ∈ Stab^{∗}(Z_{3}^{∗}) is p_{1}p_{9} ≠ 0, we re-coordinate the matrix R + SP by letting

s_{1} = r_{2} − p_{3}y_{1} + p_{1}y_{3}
s_{2} = r_{7} + p_{9}y_{2}
s_{3} = r_{8} − p_{9}y_{1}

Then we have unique solution

y_{1} = ( \frac{1}{p_{9}} r_{8}) − \frac{1}{p_{9}} s_{3}
y_{2} = ( −\frac{1}{p_{9}} r_{7}) + \frac{1}{p_{9}} s_{2}
y_{3} = (−\frac{1}{p_{9}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8}) + \frac{1}{p_{1}} s_{1} − \frac{p_{3}}{p_{1}p_{9}} s_{3}

We get y_{3} by means of y_{1} and y_{2} since s_{1} = r_{2} − p_{3}y_{1} + p_{1}y_{3} = r_{2} − \frac{p_{3}}{p_{9}} r_{8} + \frac{p_{3}}{p_{9}} s_{3} + p_{1}y_{3}, it follows p_{1}y_{3} = (−r_{2} + \frac{p_{3}}{p_{9}} r_{8}) + s_{1} − \frac{p_{3}}{p_{9}} s_{3}. Let

P⃗ = P⃗ (p_{1}, p_{2}, p_{3}, p_{4}, p_{6}, p_{9})
R⃗ = R⃗(r_{1}, r_{2}, r_{4}, r_{5}, r_{7}, r_{8})
S⃗ = S⃗(s_{1}, s_{2}, s_{3})

for each P = [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ], R = [ \begin{matrix}r_{1} & r_{4} & r_{7} \\ r_{2} & r_{5} & r_{8} \\ 0 & 0 & 0 \end{matrix} ], and triple (s_{1}, s_{2}, s_{3}).

Then we replace y_{1}, y_{2}, and y_{3} by their corresponding identities in each entry of 3 × 3 matrix R + SP to obtain

R + SP := [ \begin{matrix}R_{1}(P⃗ , R⃗, S⃗) & R_{4}(P⃗ , R⃗, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗ , R⃗, S⃗) & s_{3} \\ R_{3}(P⃗ , R⃗, S⃗) & R_{6}(P⃗ , R⃗, S⃗) & 0 \end{matrix} ]

Let’s figure out the explicit formula for each R_{i}(P⃗ , R⃗, S⃗), i = 1, 3, 4, 5, 6.

R_{1}(P⃗ , R⃗, S⃗) = r_{1} + p_{3}y_{2} − p_{2}y_{3} = r_{1} + p_{3}(−\frac{1}{p_{9}} r_{7} + \frac{1}{p_{9}} s_{2}) − p_{2}(−\frac{1}{p_{1}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8} + \frac{1}{p_{1}} s_{1} − \frac{p_{3}}{p_{1}p_{9}} s_{3}) = (r_{1} + \frac{p_{2}}{p_{1}} r_{2} − \frac{p_{3}}{p_{9}} r_{7} − \frac{p_{2}p_{3}}{p_{1}p_{9}} r_{8}) − \frac{p_{2}}{p_{1}} s_{1} + \frac{p_{3}}{p_{9}} s_{2} + \frac{p_{2}p_{3}}{p_{1}p_{9}} s_{3}, R_{3}(P⃗ , R⃗, S⃗)

= p_{2}Y_{1} − p_{1}y_{2}

= p_{2}(\frac{1}{p_{9}} r_{8} − \frac{1}{p_{9}} s_{3}) − p_{1}(−\frac{1}{p_{9}} r_{7} + \frac{1}{p_{9}} s_{2})

= (\frac{p_{1}}{p_{9}} r_{7} + \frac{p_{2}}{p_{9}} r_{8}) − \frac{p_{1}}{p_{9}} s_{2} − \frac{p_{2}}{p_{9}} s_{3},

R_{4}(P⃗ , R⃗, S⃗)

= r_{4} + p_{6}y_{2} − \frac{1+p_{2}p_{4}}{p_{1}} y_{3}

= r_{4} + p_{6}(−\frac{1}{p_{9}} r_{7} + \frac{1}{p_{9}} s_{2}) − \frac{1+p_{2}p_{4}}{p_{1}} (−\frac{1}{p_{1}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8} + \frac{1}{p_{1}} s_{1} − \frac{p_{3}}{p_{1}p_{9}} s_{3})

= (\frac{1+p_{2}p_{4}}{p_{1}^{2}} r_{2} + r_{4} − \frac{p_{6}}{p_{9}} r_{7} − \frac{p_{3}(1+p_{2}p_{4})}{p_{1}^{2}p_{9}}r_{8}) − \frac{1+p_{2}p_{4}}{p_{1}^{2}} s_{1} + \frac{p_{6}}{p_{9}} s_{2} + \frac{p_{3}(1+p_{2}p_{4})}{p_{1}^{2}p_{9}}s_{3},

R_{5}(P⃗ , R⃗, S⃗)

= r_{5} − p_{6}y_{1} + p_{4}y_{3}

= r_{5} − p_{6}(\frac{1}{p_{9}} r_{8} − \frac{1}{p_{9}} s_{3}) + p_{4}(−\frac{1}{p_{1}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8} + \frac{1}{p_{1}} s_{1} − \frac{p_{3}}{p_{1}p_{9}} s_{3})

= (−\frac{p_{4}}{p_{1}} r_{2} + r_{5} − \frac{p_{1}p_{6}−p_{3}p_{4}}{p_{1}p_{9}} r_{8}) + \frac{p_{4}}{p_{1}} s_{1} + \frac{p_{1}p_{6}−p_{3}p_{4}}{p_{1}p_{9}} s_{3},

R_{6}(P⃗ , R⃗, S⃗)

= \frac{1+p_{2}p_{4}}{p_{1}} y_{1} − p_{4}y_{2}

= \frac{1+p_{2}p_{4}}{p_{1}} (\frac{1}{p_{9}} r_{8} − \frac{1}{p_{9}} s_{3}) − p_{4}(−\frac{1}{p_{9}} r_{7} + \frac{1}{p_{9}} s_{2})

= (\frac{p_{4}}{p_{9}} r_{7} + \frac{1+p_{2}p_{4}}{p_{1}p_{9}} r_{8}) − \frac{p_{4}}{p_{9}} s_{2} − \frac{1+p_{2}p_{4}}{p_{1}p_{9}} s_{3}.

Let S⃗ = 0, i.e., ( \begin{matrix}s_{1} \\ s_{2} \\ s_{3} \end{matrix} ) = ( \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ). Then we have unique solution

y_{1} = \frac{1}{p_{9}} r_{8}
y_{2} = −\frac{1}{p_{9}} r_{7}
y_{3} = −\frac{1}{p_{1}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8}

to make a unique automorphism Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s in the coset Ad_{G}s with

[Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s]_{B}

= [ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) & R_{1}(P⃗ , R⃗) & R_{4}(P⃗ , R⃗) & 0 \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} & 0 & R_{5}(P⃗ , R⃗) & 0 \\ 0 & 0 & 1 & R_{3}(P⃗ , R⃗) & R_{6}(P⃗ , R⃗) & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & 0 \\ 0 & 0 & 0 & p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

where each R_{j}(P⃗ , R⃗) := R_{j}(P⃗ , R⃗,0), ∀j = 1, 3, 4, 5, 6, i.e.,

R_{1}(P⃗ , R⃗) = r_{1} + \frac{p_{2}}{p_{1}} r_{2} − \frac{p_{3}}{p_{9}} r_{7} − \frac{p_{2}p_{3}}{p_{1}p_{9}} r_{8}
R_{3}(P⃗ , R⃗) = \frac{p_{1}}{p_{9}} r_{7} + \frac{p_{2}}{p_{9}} r_{8}
R_{4}(P⃗ , R⃗) = \frac{1 + p_{2}p_{4}}{p_{1}^{2}} r_{2} + r_{4} − \frac{p_{6}}{p_{9}} r_{7} − \frac{p_{3}(1 + p_{2}p_{4})}{p_{1}^{2}p_{9}}r_{8}
R_{5}(P⃗ , R⃗) = −\frac{p_{4}}{p_{1}} r_{2} + r_{5} − \frac{p_{1}p_{6} − p_{3}p_{4}}{p_{1}p_{9}} r_{8}
R_{6}(P⃗ , R⃗) = \frac{p_{4}}{p_{9}} r_{7} + \frac{1 + p_{2}p_{4}}{p_{1}p_{9}} r_{8}

Notice that R_{k}(P⃗ , R⃗, S⃗) = R_{k}(P⃗ , R⃗) + { a linear function in s_{i} with rational

coefficients in p_{j} }

In summary, for each s ∈ Stab^{∗}(Z_{3}^{∗}) with [s]_{B} = [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] such that

p_{1}p_{9} ≠ 0 where P = [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ],

Q = [ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} \\ 0 & 0 & 1 \end{matrix} ], and R = [ \begin{matrix}r_{1} & r_{4} & r_{7} \\ r_{2} & r_{5} & r_{8} \\ 0 & 0 & 0 \end{matrix} ],

we have the coset

Ad_{G}s
= {Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s| y_{i} ∈ R, i = 1, 2, 3}
= \{Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s| \{ \begin{matrix}y_{1} & = & \frac{1}{p_{9}} r_{8} − \frac{1}{p_{9}} s_{3} \\ y_{2} & = & −\frac{1}{p_{9}} r_{7} + \frac{1}{p_{9}} s_{2} \\ y_{3} & = & −\frac{1}{p_{1}} r_{2} + \frac{p_{3}}{p_{1}p_{9}} r_{8} + \frac{1}{p_{1}} s_{1} − \frac{p_{3}}{p_{1}p_{9}} s_{3} \end{matrix} \} s_{i} ∈ R, ∀i = 1, 2, 3\}

with [Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s]_{B} = [ \begin{matrix}Q_{3×3} & R̃_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] where

R̃ = [ \begin{matrix}R_{1}(P⃗ , R⃗, S⃗) & R_{4}(P⃗ , R⃗, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗ , R⃗, S⃗) & s_{3} \\ R_{3}(P⃗ , R⃗, S⃗) & R_{6}(P⃗ , R⃗, S⃗) & 0 \end{matrix} ]. Let S⃗ = 0, then we have unique

(y_{1}, y_{2}, y_{3}) to make unique Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s ∈ Ad_{G}s with

[Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s]_{B} = [ \begin{matrix}Q_{3×3} & R̃_{3×3}^{0} \\ 0_{3×3} & P_{3×3} \end{matrix} ] where

R̃^{0} := [ \begin{matrix}R_{1}(P⃗ , R⃗) & R_{4}(P⃗ , R⃗) & 0 \\ 0 & R_{5}(P⃗ , R⃗) & 0 \\ R_{3}(P⃗ , R⃗) & R_{6}(P⃗ , R⃗) & 0 \end{matrix} ]. Thus for each s ∈ Stab^{∗}(Z_{3}^{∗}), we get unique

automorphism Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}s ∈ Ad_{G}s. Since each entry p_{i} of P_{3×3} and r_{j} of R_{3×3} varies in R with p_{1}p_{9} ≠ 0, from the formula we know each R_{k}(P⃗ , R⃗) also varies in R, ∀i = 1, 2, 3, 4, 6, 9, j = 1, 2, 4, 5, 7, 8, k = 1, 3, 4, 5, 6. Since each entry in Q_{3×3} depends on p_{i}, let all p_{i} and r_{j} vary in R (p_{1}p_{9} ≠ 0), we get almost the whole group Stab^{∗}(Z_{3}^{∗}) except a subset with measure zero. Since we already got unique automorphism from each coset Ad_{G}s, which automorphism depends on each p_{i} and R_{k}, let C_{S} be the collection of all those unique automorphisms. Explicitly,

C_{S} := {A ∈ Aut(𝔤)|p_{i}, R_{k} ∈ R (p_{1}p_{9} ≠ 0), ∀i = 1, 2, 3, 4, 6, 9, ∀k = 1, 3, 4, 5, 6}

with

[A]_{B} = [ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) & R_{1} & R_{4} & 0 \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} & 0 & R_{5} & 0 \\ 0 & 0 & 1 & R_{3} & R_{6} & 0 \\ 0 & 0 & 0 & p_{1} & p_{4} & 0 \\ 0 & 0 & 0 & p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

We next verify that different elements of C_{S} lie on different cosets. If this is done, then we obtain an almost global coordinate patch for Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}). For every distinct A, A′ ∈ C_{S}, we know A ∈ Ad_{G}s and A′ ∈ Ad_{G}s′ for some

s, s′ ∈ Stab^{∗}(Z_{3}^{∗}). Let [A]_{B} = [ \begin{matrix}Q_{3×3} & R̃_{3×3}^{0} \\ 0_{3×3} & P_{3×3} \end{matrix} ] and [A′]_{B} = [ \begin{matrix}Q′_{3×3} & R̃_{3×3}^{′0} \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

as above matrix. Then Ad_{G}s = \{Ad_{g}s|[Ad_{g}s]_{B} = [ \begin{matrix}Q_{3×3} & R̃_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] , ∀S⃗ ∈ R^{3}\} and

Ad_{G}s′ = \{Ad_{g}s′|[Ad_{g}s′]_{B} = [ \begin{matrix}Q′_{3×3} & R̃′_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ] , ∀S⃗ ∈ R^{3}\}. Since A and A′ are

distinct, it requires that at least one p_{i} ≠ p′_{i} or R_{j} ≠ R′_{j}. If p_{i} ≠ p′_{i} for some i ∈ {1, 2, 3, 4, 6, 9}, then P_{3×3} ≠ P′_{3×3}, so Ad_{G}s ≠ Ad_{G}s′. Suppose p_{i} = p′_{i}, ∀i = 1, 2, 3, 4, 6, 9, i.e., P⃗ = P⃗′. If R_{j} ≠ R′_{j} for some j ∈ {1, 3, 4, 5, 6}. For example, let R_{1} ≠ R′_{1}, i.e., R_{1}(P⃗ , R⃗,0) ≠ R′_{1}(P⃗′, R⃗′,0) = R′_{1}(P⃗ , R⃗′,0). By formula of R_{1}(P⃗ , R⃗, S⃗) we get

R_{1}(P⃗ , R⃗, S⃗) = R_{1}(P⃗ , R⃗,0) − \frac{p_{2}}{p_{1}} s_{1} + \frac{p_{3}}{p_{9}} s_{2} + \frac{p_{2}p_{3}}{p_{1}p_{9}} s_{3}
≠ R′_{1}(P⃗ , R⃗′,0) − \frac{p_{2}}{p_{1}} s_{1} + \frac{p_{3}}{p_{9}} s_{2} + \frac{p_{2}p_{3}}{p_{1}p_{9}} s_{3}
= R′_{1}(P⃗ , R⃗′, S⃗)
= R′_{1}(P⃗′, R⃗′, S⃗)

So R_{1}(P⃗ , R⃗, S⃗) ≠ R′_{1}(P⃗′, R⃗′, S⃗), then R̃_{3×3} ≠ R̃′_{3×3}, and hence Ad_{G}s ≠ Ad_{G}s′. Similarly for other cases R_{i} ≠ R′_{i} with i = 3, 4, 5, 6, we obtain the same result. Therefore different elements of C_{S} always lie on different cosets. One concludes that C_{S} is an almost global coordinate patch for Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}).

Lemma 5.2. The previous lemma says that C_{S} is an almost global coordinate patch for Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}). With respect to C_{S} we claim that

\frac{1}{p_{9}^{6}} dp_{1}dp_{2}dp_{3}dp_{4}dp_{6}dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}

is a left Haar measure and

\frac{1}{p_{9}} dp_{1}dp_{2}dp_{3}dp_{4}dp_{6}dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}

is a right Haar measure on Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}).

Proof. Given A_{0} ∈ C_{S}, for each A ∈ C_{S}, let Ã = A_{0}A. Since C_{S} is not a group, Ã may not be in C_{S}. We want to get the corresponding automorphism A′ associated with Ã such that A′ ∈ C_{S}. Then we can calculate the determinant of the Jacobian of the map A ⟼ A′ at the point A in order to figure out a left Haar measure on the quotient group Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}).

Given A_{0} ∈ C_{S}, for each A ∈ C_{S}, we know A_{0} ∈ Ad_{G}s_{0} and A ∈ Ad_{G}s for some s_{0}, s ∈ Stab^{∗}(Z_{3}^{∗}). Then Ã ∈ (Ad_{G}s_{0})(Ad_{G}s) = Ad_{G}(s_{0}s). Let

[A_{0}]_{B} = [ \begin{matrix}B_{3×3} & C_{3×3} \\ 0_{3×3} & A_{3×3} \end{matrix} ] and [A]_{B} = [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ]

where A = [ \begin{matrix}a_{1} & a_{4} & 0 \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{3} & a_{6} & a_{9} \end{matrix} ], C = [ \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & 0 \end{matrix} ], P = [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ],

R = [ \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & 0 \end{matrix} ], and B and Q are determined by A and P respectively,

just like matrix formula in the collection C_{S}. Then

[Ã]_{B} = [ \begin{matrix}B & C \\ 0 & A \end{matrix} ] [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] = [ \begin{matrix}BQ & BR + CP \\ 0 & AP \end{matrix} ] =: [ \begin{matrix}Q′_{3×3} & R̃_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

Since Ã ∈ Ad_{G}(s_{0}s) with [Ã]_{B} = [ \begin{matrix}Q′ & R̃ \\ 0 & P′ \end{matrix} ], let [s_{0}s]_{B} = [ \begin{matrix}Q′ & R′ \\ 0 & P′ \end{matrix} ]

where R′ is an unknown 3 × 3 matrix [ \begin{matrix}r′_{1} & r′_{4} & r′_{7} \\ r′_{2} & r′_{5} & r′_{8} \\ 0 & 0 & 0 \end{matrix} ]. Then

Ad_{G}(s_{0}s)
= \{Ad_{g}(s_{0}s)|[Ad_{g}(s_{0}s)]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′, S⃗) & R_{4}(P⃗′, R⃗′, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗′, R⃗′, S⃗) & s_{3} \\ R_{3}(P⃗′, R⃗′, S⃗) & R_{6}(P⃗′, R⃗′, S⃗) & 0 \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ] , ∀S⃗ ∈ R^{3}\}

Thus, the corresponding automorphism A′ ∈ Ad_{G}(s_{0}s) associated with Ã in the coordinate patch C_{S} has the matrix

[A′]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′,0) & R_{4}(P⃗′, R⃗′,0) & 0 \\ 0 & R_{5}(P⃗′, R⃗′,0) & 0 \\ R_{3}(P⃗′, R⃗′,0) & R_{6}(P⃗′, R⃗′,0) & 0 \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

Note that we still have not known each R_{i}(P⃗′, R⃗′,0) yet, since 3 × 3 matrix R′ is unknown. But we will figure out later. First, let’s calculate matrices P′ and R̃.

P′
= A_{3×3}P_{3×3}
= [ \begin{matrix}a_{1} & a_{4} & 0 \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{3} & a_{6} & a_{9} \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= [ \begin{matrix}a_{1}p_{1} + a_{4}p_{2} & a_{1}p_{4} + a_{4} \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ a_{2}p_{1} + \frac{1+a_{2}a_{4}}{a_{1}} p_{2} & a_{2}p_{4} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ a_{3}p_{1} + a_{6}p_{2} + a_{9}p_{3} & a_{3}p_{4} + a_{6} \frac{1+p_{2}p_{4}}{p_{1}} + a_{9}p_{6} & a_{9}p_{9} \end{matrix} ]

= : [ \begin{matrix}p′_{1} & p′_{4} & 0 \\ p′_{2} & \frac{1+p′_{2}p′_{4}}{p′_{1}} & 0 \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ]

For the entry \frac{1+p′_{2}p′_{4}}{p′_{1}} of the last matrix, we need to check it equals a_{2}p_{4} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} .

\frac{1 + p′_{2}p′_{4}}{p′_{1}}

= \frac{1 + (a_{2}p_{1} + \frac{1+a_{2}a_{4}}{a_{1}} p_{2})(a_{1}p_{4} + a_{4} \frac{1+p_{2}p_{4}}{p_{1}} )}{a_{1}p_{1} + a_{4}p_{2}}
= \frac{a_{1}a_{2}p_{1}p_{4} + a_{2}a_{4}p_{2}p_{4}}{a_{1}p_{1} + a_{4}p_{2}} + \frac{1 + a_{2}a_{4} + (1 + a_{2}a_{4})p_{2}p_{4} + a_{4}p_{2} \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1} + a_{4}p_{2}}
= a_{2}p_{4} + \frac{1 + a_{2}a_{4}}{a_{1}} \frac{1 + p_{2}p_{4}}{p_{1}}

For the matrix R̃

R̃

= BR + CP
= [ \begin{matrix}\frac{a_{9}}{a_{1}} (1 + a_{2}a_{4}) & −a_{2}a_{9} & a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}) \\ −a_{4}a_{9} & a_{1}a_{9} & a_{3}a_{4} − a_{1}a_{6} \\ 0 & 0 & 1 \end{matrix} ] [ \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & 0 \end{matrix} ]

+ [ \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & 0 \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= : [ \begin{matrix}r̃′_{1} & r̃′_{4} & 0 \\ r̃′_{2} & r̃′_{5} & 0 \\ r̃′_{3} & r̃′_{6} & 0 \end{matrix} ]

where r̃′_{1} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{1} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{3} + c_{1}p_{1} + c_{4}p_{2} r̃′_{2} = −a_{4}a_{9}r_{1} + (a_{3}a_{4} − a_{1}a_{6})r_{3} + c_{5}p_{2} r̃′_{3} = r_{3} + c_{3}p_{1} + c_{6}p_{2}

r̃′_{4} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{4} − a_{2}a_{9}r_{5} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{6} + c_{1}p_{4} + c_{4} \frac{1+p_{2}p_{4}}{p_{1}}

r̃′_{5} = −a_{4}a_{9}r_{4} + a_{1}a_{9}r_{5} + (a_{3}a_{4} − a_{1}a_{6})r_{6} + c_{5} \frac{1+p_{2}p_{4}}{p_{1}}

r̃′_{6} = r_{6} + c_{3}p_{4} + c_{6} \frac{1+p_{2}p_{4}}{p_{1}}

Since Ã ∈ Ad_{G}(s_{0}s) with [Ã]_{B} = [ \begin{matrix}Q′ & R̃ \\ 0 & P′ \end{matrix} ], it follows

R̃ ∈ \{ [ \begin{matrix}R_{1}(P⃗′, R⃗′, S⃗) & R_{4}(P⃗′, R⃗′, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗′, R⃗′, S⃗) & s_{3} \\ R_{3}(P⃗′, R⃗′, S⃗) & R_{6}(P⃗′, R⃗′, S⃗) & 0 \end{matrix} ] : S⃗ = ( \begin{matrix}s_{1} \\ s_{2} \\ s_{3} \end{matrix} ) ∈ R^{3}\}

Thus if we let S⃗ = ( \begin{matrix} & r̃′_{2} \\ 0 \\ 0 \end{matrix} ), we should get five equations: r̃′_{k} = R_{k}(P⃗′, R⃗′, S⃗),

k = 1, 3, 4, 5, 6. Next, let’s try to get R_{k}(P⃗′, R⃗′,0), and recall that R_{k}(P⃗′, R⃗′, S⃗) = R_{k}(P⃗′, R⃗′,0) + {linear function in s_{i}}, ∀i = 1, 2, 3.

Since r̃′_{1} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{1} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{3} + c_{1}p_{1} + c_{4}p_{2} = R_{1}(P⃗′, R⃗′, S⃗) = R_{1}(P⃗′, R⃗′,0) − \frac{p′_{2}}{p′_{1}} s_{1} + \frac{p′_{3}}{p′_{9}} · 0 + \frac{p′_{2}p′_{3}}{p′_{1}p′_{9}} · 0, it follows R_{1}(P⃗′, R⃗′,0) = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{1} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{3} + c_{1}p_{1} + c_{4}p_{2} + \frac{p′_{2}}{p′_{1}} s_{1} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{1} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{3} + c_{1}p_{1} + c_{4}p_{2} +\frac{a_{2}p_{1}+ \frac{1+a_{2}a_{4}}{a_{1}} p_{2}}{a_{1}p_{1}+a_{4}p_{2}} (−a_{4}a_{9}r_{1} + (a_{3}a_{4} − a_{1}a_{6})r_{3} + c_{5}p_{2}) = (\frac{a_{9}}{a_{1}} (1 + a_{2}a_{4}) − a_{4}a_{9} \frac{a_{2}p_{1}+ \frac{1+a_{2}a_{4}}{a_{1}} p_{2}}{a_{1}p_{1}+a_{4}p_{2}} )r_{1} +(a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}) + (a_{3}a_{4} − a_{1}a_{6})\frac{a_{2}p_{1}+ \frac{1+a_{2}a_{4}}{a_{1}} p_{2}}{a_{1}p_{1}+a_{4}p_{2}} )r_{3} + rational(P⃗) = \frac{a_{9}p_{1}}{a_{1}p_{1}+a_{4}p_{2}} r_{1} − \frac{a_{3}p_{1}+a_{6}p_{2}}{a_{1}p_{1}+a_{4}p_{2}} r_{3} + rational(P⃗).

Since r̃′_{3} = r_{3} + c_{3}p_{1} + c_{6}p_{2} = R_{3}(P⃗′, R⃗′, S⃗) = R_{3}(P⃗′, R⃗′,0) − \frac{p′_{1}}{p′_{9}} · 0 − \frac{p′_{2}}{p′_{9}} · 0, it follows R_{3}(P⃗′, R⃗′,0) = r_{3} + c_{3}p_{1} + c_{6}p_{2} = r_{3} + rational(P⃗).

Since r̃′_{4} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{4} − a_{2}a_{9}r_{5} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{6} + c_{1}p_{4} + c_{4} \frac{1+p_{2}p_{4}}{p_{1}} = R_{4}(P⃗′, R⃗′, S⃗) = R_{4}(P⃗′, R⃗′,0) − \frac{1+p′_{2}p′_{4}}{p_{1}^{′2}} s_{1} + \frac{p′_{6}}{p′_{9}} · 0 + \frac{p′_{3}(1+p′_{2}p′_{4})}{p_{1}^{′2}p′_{9}} · 0 and we know \frac{1+p′_{2}p′_{4}}{p′_{1}} = a_{2}p_{4} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} , it follows R_{4}(P⃗′, R⃗′,0) = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{4} − a_{2}a_{9}r_{5} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{6} + c_{1}p_{4} + c_{4} \frac{1+p_{2}p_{4}}{p_{1}} + \frac{1+p′_{2}p′_{4}}{p_{1}^{′2}} s_{1} = \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{4} − a_{2}a_{9}r_{5} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{6} + c_{1}p_{4} + c_{4} \frac{1+p_{2}p_{4}}{p_{1}} +\frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} (−a_{4}a_{9}r_{1} + (a_{3}a_{4} − a_{1}a_{6})r_{3} + c_{5}p_{2}) = −a_{4}a_{9} \frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} r_{1} + (a_{3}a_{4} − a_{1}a_{6})\frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} r_{3} +\frac{a_{9}}{a_{1}} (1 + a_{2}a_{4})r_{4} − a_{2}a_{9}r_{5} + (a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}))r_{6} + rational(P⃗).

Since r̃′_{5} = −a_{4}a_{9}r_{4} + a_{1}a_{9}r_{5} + (a_{3}a_{4} − a_{1}a_{6})r_{6} + c_{5} \frac{1+p_{2}p_{4}}{p_{1}} = R_{5}(P⃗′, R⃗, S⃗) = R_{5}(P⃗′, R⃗′,0) + \frac{p′_{4}}{p′_{1}} s_{1} + \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{p′_{1}p′_{9}} · 0, it follows R_{5}(P⃗′, R⃗′,0) = −a_{4}a_{9}r_{4} + a_{1}a_{9}r_{5} + (a_{3}a_{4} − a_{1}a_{6})r_{6} + c_{5} \frac{1+p_{2}p_{4}}{p_{1}} − \frac{p′_{4}}{p′_{1}} s_{1} = −a_{4}a_{9}r_{4} + a_{1}a_{9}r_{5} + (a_{3}a_{4} − a_{1}a_{6})r_{6} + c_{5} \frac{1+p_{2}p_{4}}{p_{1}} −\frac{a_{1}p_{4}+a_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} (−a_{4}a_{9}r_{1} + (a_{3}a_{4} − a_{1}a_{6})r_{3} + c_{5}p_{2}) = a_{4}a_{9} \frac{a_{1}p_{4}+a_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} r_{1} − (a_{3}a_{4} − a_{1}a_{6})\frac{a_{1}p_{4}+a_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} r_{3} −a_{4}a_{9}r_{4} + a_{1}a_{9}r_{5} + (a_{3}a_{4} − a_{1}a_{6})r_{6} + rational(P⃗).

Since r̃′_{6} = r_{6} + c_{3}p_{4} + c_{6} \frac{1+p_{2}p_{4}}{p_{1}} = R_{6}(P⃗′, R⃗′, S⃗) = R_{6}(P⃗′, R⃗′,0) − \frac{p′_{4}}{p′_{9}} · 0 − \frac{1+p′_{2}p′_{4}}{p′_{1}p′_{9}} · 0, it follows R_{6}(P⃗′, R⃗′,0) = r_{6} + c_{3}p_{4} + c_{6} \frac{1+p_{2}p_{4}}{p_{1}} = r_{6} + rational(P⃗).

Thus we obtain all R_{k}(P⃗′, R⃗′,0), and hence we get the corresponding A′ ∈ Ad_{G}(s_{0}s) associated with Ã= A_{0}A such that A′ ∈ C_{S}.

Then the determinant of the Jacobian of the map A ⟼ A′ at point A is

det [\frac{∂(p′_{1}, p′_{2}, p′_{3}, p′_{4}, p′_{6}, p′_{9}, R_{1}, R_{3}, R_{4}, R_{5}, R_{6})}{∂(p_{1}, p_{2}, p_{3}, p_{4}, p_{6}, p_{9}, r_{1}, r_{3}, r_{4}, r_{5}, r_{6})}] = det [ \begin{matrix}(\frac{∂p′_{i}}{∂p_{j}} )6×6 & (\frac{∂p′_{i}}{∂r_{j}} )6×5 \\ (\frac{∂R_{i}}{∂p_{j}} )5×6 & (\frac{∂R_{i}}{∂r_{j}} )5×5 \end{matrix} ]

= det [ (\frac{∂p′_{i}}{∂p_{j}} )_{6×6} 0_{6×5}_{(\frac{∂R_{i}}{∂p_{j}} )5×6 (\frac{∂R_{i}}{∂r_{j}} )5×5} ] = det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6} det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}

Let’s calculate both determinants.

det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6}
= det [\frac{∂(p′_{1}, p′_{2}, p′_{3}, p′_{4}, p′_{6}, p′_{9})}{∂(p_{1}, p_{2}, p_{3}, p_{4}, p_{6}, p_{9})}]

= det [ \begin{matrix}a_{1} & a_{4} & 0 & 0 & 0 & 0 \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} & 0 & 0 & 0 & 0 \\ a_{3} & a_{6} & a_{9} & 0 & 0 & 0 \\ −a_{4} \frac{1+p_{2}p_{4}}{p_{1}^{2}} & a_{4} \frac{p_{4}}{p_{1}} & 0 & a_{1} + a_{4} \frac{p_{2}}{p_{1}} & 0 & 0 \\ −a_{6} \frac{1+p_{2}p_{4}}{p_{1}^{2}} & a_{6} \frac{p_{4}}{p_{1}} & 0 & a_{3} + a_{6} \frac{p_{2}}{p_{1}} & a_{9} & 0 \\ 0 & 0 & 0 & 0 & 0 & a_{9} \end{matrix} ]

= a_{9} · a_{9} · (a_{1} + a_{4} \frac{p_{2}}{p_{1}} ) · a_{9} · det [ \begin{matrix}a_{1} & a_{4} \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ]

= a_{9}^{3} \frac{a_{1}p_{1} + a_{4}p_{2}}{p_{1}}
= (det A)^{3} \frac{p′_{1}}{p_{1}}
= \frac{(det P′)^{3}p′_{1}}{(det P)^{3}p_{1}}

det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}
= det [\frac{∂(R_{1}, R_{3}, R_{4}, R_{5}, R_{6})}{∂(r_{1}, r_{3}, r_{4}, r_{5}, r_{6})}]

= det [ \begin{matrix}\frac{a_{9}p_{1}}{a_{1}p_{1}+a_{4}p_{2}} & ∗ & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ −a_{4}a_{9} \frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} & ∗ & \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4}) & −a_{2}a_{9} & a_{2}a_{6} − \frac{a_{3}}{a_{1}} (1 + a_{2}a_{4}) \\ a_{4}a_{9} \frac{a_{1}p_{4}+a_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} & ∗ & −a_{4}a_{9} & a_{1}a_{9} & a_{3}a_{4} − a_{1}a_{6} \\ 0 & 0 & 0 & 0 & 1 \end{matrix} ]

= det [ \begin{matrix}\frac{a_{9}p_{1}}{a_{1}p_{1}+a_{4}p_{2}} & 0 & 0 \\ −a_{4}a_{9} \frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} & \frac{a_{9}}{a_{1}} (1 + a_{2}a_{4}) & −a_{2}a_{9} \\ a_{4}a_{9} \frac{a_{1}p_{4}+a_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} & −a_{4}a_{9} & a_{1}a_{9} \end{matrix} ]
= \frac{a_{9}p_{1}}{a_{1}p_{1} + a_{4}p_{2}} (a_{9}^{2}(1 + a_{2}a_{4}) − a_{2}a_{4}a_{9}^{2})
= a_{9}^{3} \frac{p_{1}}{a_{1}p_{1} + a_{4}p_{2}}
= (det A)^{3} \frac{p_{1}}{p′_{1}}
= \frac{(det P′)^{3}p_{1}}{(det P)^{3}p′_{1}}

Hence

det [\frac{∂(p′_{1}, p′_{2}, p′_{3}, p′_{4}, p′_{6}, p′_{9}, R_{1}, R_{3}, R_{4}, R_{5}, R_{6})}{∂(p_{1}, p_{2}, p_{3}, p_{4}, p_{6}, p_{9}, r_{1}, r_{3}, r_{4}, r_{5}, r_{6})}] = det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6} det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}

= \frac{(det P′)^{3}p′_{1}}{(det P)^{3}p_{1}} \frac{(det P′)^{3}p_{1}}{(det P)^{3}p′_{1}} = \frac{(det P′)^{6}}{(det P)^{6}} = \frac{p_{9}^{′6}}{p_{9}^{6}}

By the explanation at page 251. in [4], one concludes that

\frac{1}{p_{9}^{6}} dp_{1}dp_{2}dp_{3}dp_{4}dp_{6}dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}

is a left Haar measure on Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}). Next step we would like to find a right Haar measure with respect to the same patch C_{S}.

Given A_{0} ∈ C_{S}, for each A ∈ C_{S}, let Ã = AA_{0}. But Ã may not be in C_{S}, we want to get the corresponding A′ associated with Ã such that A′ ∈ C_{S}. Since A ∈ Ad_{G}s and A_{0} ∈ Ad_{G}s_{0} for some s, s_{0} ∈ Stab^{∗}(Z_{3}^{∗}), we have Ã ∈ Ad_{G}(ss_{0}).

Let [A]_{B} = [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] and [A_{0}]_{B} = [ \begin{matrix}B_{3×3} & C_{3×3} \\ 0_{3×3} & A_{3×3} \end{matrix} ] as before. Then

[Ã]_{B} = [AA_{0}]_{B} = [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] [ \begin{matrix}B & C \\ 0 & A \end{matrix} ] = [ \begin{matrix}QB & QC + RA \\ 0 & P A \end{matrix} ] =:

[ \begin{matrix}Q′_{3×3} & R̃_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ],

and hence we can let [ss_{0}]_{B} = [ \begin{matrix}Q′_{3×3} & R′_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ] where R′_{3×3} is an unknown matrix

with the form [ \begin{matrix}r′_{1} & r′_{4} & r′_{7} \\ r′_{2} & r′_{5} & r′_{8} \\ 0 & 0 & 0 \end{matrix} ]. Then

Ad_{G}(ss_{0}) = {Ad_{g}(ss_{0})|[Ad_{g}(ss_{0})]_{B} =

[ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′, S⃗) & R_{4}(P⃗′, R⃗′, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗′, R⃗′, S⃗) & s_{3} \\ R_{3}(P⃗′, R⃗′, S⃗) & R_{6}(P⃗′, R⃗′, S⃗) & 0 \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ] , S⃗ ∈ R^{3}\}

And the corresponding A′ ∈ C_{S} associated with Ã has the matrix

[A′]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′,0) & R_{4}(P⃗′, R⃗′,0) & 0 \\ 0 & R_{5}(P⃗′, R⃗′,0) & 0 \\ R_{3}(P⃗′, R⃗′,0) & R_{6}(P⃗′, R⃗′,0) & 0 \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

Notice that we still have not known each R_{k}(P⃗′, R⃗′,0) yet, but we will figure out later. First, let’s calculate matrices P′ and R̃.

P′
= P A
= [ \begin{matrix}p_{1} & p_{4} & 0 \\ p_{2} & \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & 0 \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{3} & a_{6} & a_{9} \end{matrix} ]

= [ \begin{matrix}a_{1}p_{1} + a_{2}p_{4} & a_{4}p_{1} + \frac{1+a_{2}a_{4}}{a_{1}} p_{4} & 0 \\ a_{1}p_{2} + a_{2} \frac{1+p_{2}p_{4}}{p_{1}} & a_{4}p_{2} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} & 0 \\ a_{1}p_{3} + a_{2}p_{6} + a_{3}p_{9} & a_{4}p_{3} + \frac{1+a_{2}a_{4}}{a_{1}} p_{6} + a_{6}p_{9} & a_{9}p_{9} \end{matrix} ]

= : [ \begin{matrix}p′_{1} & p′_{4} & 0 \\ p′_{2} & \frac{1+p′_{2}p′_{4}}{p′_{1}} & 0 \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ]

We need to check the entry \frac{1+p′_{2}p′_{4}}{p′_{1}} of the last matrix equal to a_{4}p_{2} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} .

\frac{1 + p′_{2}p′_{4}}{p′_{1}}

= \frac{1 + (a_{1}p_{2} + a_{2} \frac{1+p_{2}p_{4}}{p_{1}} )(a_{4}p_{1} + \frac{1+a_{2}a_{4}}{a_{1}} p_{4})}{a_{1}p_{1} + a_{2}p_{4}}
= \frac{1 + a_{1}a_{4}p_{1}p_{2} + a_{2}a_{4}(1 + p_{2}p_{4}) + (1 + a_{2}a_{4})p_{2}p_{4} + a_{2} \frac{1+a_{2}a_{4}}{a_{1}} p_{4} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1} + a_{2}p_{4}}
= \frac{a_{4}p_{2}(a_{1}p_{1} + a_{2}p_{4}) + (1 + a_{2}a_{4})(1 + p_{2}p_{4}) + a_{2}p_{4} \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1} + a_{2}p_{4}}
= a_{4}p_{2} + \frac{1 + a_{2}a_{4}}{a_{1}} \frac{1 + p_{2}p_{4}}{p_{1}}

For the 3 × 3 matrix R̃

R̃

= QC + RA
= [ \begin{matrix}\frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) & −p_{2}p_{9} & p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4}) \\ −p_{4}p_{9} & p_{1}p_{9} & p_{3}p_{4} − p_{1}p_{6} \\ 0 & 0 & 1 \end{matrix} ] [ \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & 0 \end{matrix} ]

+ [ \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & 0 \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & 0 \\ a_{2} & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{3} & a_{6} & a_{9} \end{matrix} ]

= : [ \begin{matrix}r̃′_{1} & r̃′_{4} & 0 \\ r̃′_{2} & r̃′_{5} & 0 \\ r̃′_{3} & r̃′_{6} & 0 \end{matrix} ]

where r̃′_{1} = a_{1}r_{1} + a_{2}r_{4} + c_{1} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) + c_{3}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) r̃′_{2} = a_{2}r_{5} − c_{1}p_{4}p_{9} + c_{3}(p_{3}p_{4} − p_{1}p_{6}) r̃′_{3} = a_{1}r_{3} + a_{2}r_{6} + c_{3} r̃′_{4} = a_{4}r_{1} + \frac{1+a_{2}a_{4}}{a_{1}} r_{4} + c_{4} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) − c_{5}p_{2}p_{9} + c_{6}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) r̃′_{5} = \frac{1+a_{2}a_{4}}{a_{1}} r_{5} − c_{4}p_{4}p_{9} + c_{5}p_{1}p_{9} + c_{6}(p_{3}p_{4} − p_{1}p_{6}) r̃′_{6} = a_{4}r_{3} + \frac{1+a_{2}a_{4}}{a_{1}} r_{6} + c_{6}

Since Ã ∈ Ad_{G}(ss_{0}) with [Ã]_{B} = [ \begin{matrix}Q′ & R̃ \\ 0 & P′ \end{matrix} ], it follows

R̃ ∈ \{ [ \begin{matrix}R_{1}(P⃗′, R⃗′, S⃗) & R_{4}(P⃗′, R⃗′, S⃗) & s_{2} \\ s_{1} & R_{5}(P⃗′, R⃗′, S⃗) & s_{3} \\ R_{3}(P⃗′, R⃗′, S⃗) & R_{6}(P⃗′, R⃗′, S⃗) & 0 \end{matrix} ] : S⃗ = ( \begin{matrix}s_{1} \\ s_{2} \\ s_{3} \end{matrix} ) ∈ R^{3}\}

Thus if we let S⃗ = ( \begin{matrix} & r̃′_{2} \\ 0 \\ 0 \end{matrix} ), we should get five equations: r̃′_{k} = R_{k}(P⃗′, R⃗′, S⃗),

k = 1, 3, 4, 5, 6. From these equations we can get R_{k}(P⃗′, R⃗′,0).

Since r̃′_{1} = a_{1}r_{1} + a_{2}r_{4} + c_{1} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) + c_{3}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) = R_{1}(P⃗′, R⃗′, S⃗) = R_{1}(P⃗′, R⃗′,0) − \frac{p′_{2}}{p′_{1}} s_{1} + \frac{p′_{3}}{p′_{9}} · 0 + \frac{p′_{2}p′_{3}}{p′_{1}p′_{9}} · 0, it follows R_{1}(P⃗′, R⃗′,0) = a_{1}r_{1} + a_{2}r_{4} + c_{1} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) + c_{3}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) +\frac{a_{1}p_{2}+a_{2} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{2}p_{4}} (a_{2}r_{5} − c_{1}p_{4}p_{9} + c_{3}(p_{3}p_{4} − p_{1}p_{6})) = a_{1}r_{1} + a_{2}r_{4} + \frac{a_{1}a_{2}p_{1}p_{2}+a_{2}^{2}(1+p_{2}p_{4})}{p_{1}(a_{1}p_{1}+a_{2}p_{4})}r_{5} + rational(P⃗).

Since r̃′_{3} = a_{1}r_{3} + a_{2}r_{6} + c_{3} = R_{3}(P⃗′, R⃗′, S⃗) = R_{3}(P⃗′, R⃗′,0) − \frac{p′_{1}}{p′_{9}} · 0 − \frac{p′_{2}}{p′_{9}} · 0, it follows R_{3}(P⃗′, R⃗′,0) = a_{1}r_{3} + a_{2}r_{6} + c_{3}.

Since r̃′_{4} = a_{4}r_{1} + \frac{1+a_{2}a_{4}}{a_{1}} r_{4} + c_{4} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) − c_{5}p_{2}p_{9} + c_{6}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) = R_{4}(P⃗′, R⃗′, S⃗) =R_{4}(P⃗′, R⃗′,0) − \frac{1+p′_{2}p′_{4}}{p_{1}^{′2}} s_{1} + \frac{p′_{6}}{p′_{9}} · 0 + \frac{p′_{3}(1+p′_{2}p′_{4})}{p_{1}^{′2}p′_{9}} · 0 and we know \frac{1+p′_{2}p′_{4}}{p′_{1}} = a_{4}p_{2} + \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}} , it follows R_{4}(P⃗′, R⃗′,0) = a_{4}r_{1} + \frac{1+a_{2}a_{4}}{a_{1}} r_{4} + c_{4} \frac{p_{9}}{p_{1}} (1 + p_{2}p_{4}) − c_{5}p_{2}p_{9} + c_{6}(p_{2}p_{6} − \frac{p_{3}}{p_{1}} (1 + p_{2}p_{4})) +\frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} (a_{2}r_{5} − c_{1}p_{4}p_{9} + c_{3}(p_{3}p_{4} − p_{1}p_{6})) = a_{4}r_{1} + \frac{1+a_{2}a_{4}}{a_{1}} r_{4} + a_{2} \frac{a_{2}p_{4}+ \frac{1+a_{2}a_{4}}{a_{1}} \frac{1+p_{2}p_{4}}{p_{1}}}{a_{1}p_{1}+a_{4}p_{2}} r_{5} + rational(P⃗)

Since r̃′_{5}= \frac{1+a_{2}a_{4}}{a_{1}} r_{5} − c_{4}p_{4}p_{9} + c_{5}p_{1}p_{9} + c_{6}(p_{3}p_{4} − p_{1}p_{6}) = R_{5}(P⃗′, R⃗′, S⃗) =R_{5}(P⃗′, R⃗′,0) + \frac{p′_{4}}{p′_{1}} s_{1} + \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{p′_{1}p′_{9}} · 0, it follows R_{5}(P⃗′, R⃗′,0) = \frac{1+a_{2}a_{4}}{a_{1}} r_{5} − c_{4}p_{4}p_{9} + c_{5}p_{1}p_{9} + c_{6}(p_{3}p_{4} − p_{1}p_{6}) −\frac{a_{4}p_{1}+ \frac{1+a_{2}a_{4}}{a_{1}} p_{4}}{a_{1}p_{1}+a_{2}p_{4}} (a_{2}r_{5} − c_{1}p_{4}p_{9} + c_{3}(p_{3}p_{4} − p_{1}p_{6})) = (\frac{1+a_{2}a_{4}}{a_{1}} − a_{2} \frac{a_{4}p_{1}+ \frac{1+a_{2}a_{4}}{a_{1}} p_{4}}{a_{1}p_{1}+a_{2}p_{4}} )r_{5} + rational(P⃗) = \frac{p_{1}}{a_{1}p_{1}+a_{2}p_{4}} r_{5} + rational(P⃗).

Since r̃′_{6} = a_{4}r_{3} + \frac{1+a_{2}a_{4}}{a_{1}} r_{6} + c_{6} = R_{6}(P⃗′, R⃗′, S⃗) = R_{6}(P⃗′, R⃗′,0) − \frac{p′_{4}}{p′_{9}} · 0 − \frac{1+p′_{2}p′_{4}}{p′_{1}p′_{9}} · 0, it follows R_{6}(P⃗′, R⃗′,0) = a_{4}r_{3} + \frac{1+a_{2}a_{4}}{a_{1}} r_{6} + c_{6}.

Thus we obtain all R_{k}(P⃗′, R⃗′,0), k = 1, 3, 4, 5, 6, and hence we get the

corresponding A′ ∈ Ad_{G}(ss_{0}) associated with Ã = AA_{0} such that A′ ∈ C_{S}.

Let’s calculate both det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6} and det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}.

det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6}

= det [ \begin{matrix}a_{1} & 0 & 0 & a_{2} & 0 & 0 \\ −a_{2} \frac{1+p_{2}p_{4}}{p_{1}^{2}} & a_{1} + a_{2} \frac{p_{4}}{p_{1}} & 0 & a_{2} \frac{p_{2}}{p_{1}} & 0 & 0 \\ 0 & 0 & a_{1} & 0 & a_{2} & a_{3} \\ a_{4} & 0 & 0 & \frac{1+a_{2}a_{4}}{a_{1}} & 0 & 0 \\ 0 & 0 & a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} & a_{6} \\ 0 & 0 & 0 & 0 & 0 & a_{9} \end{matrix} ]

= a_{9}(a_{1} + a_{2} \frac{p_{4}}{p_{1}} ) det [ \begin{matrix}a_{1} & 0 & a_{2} & 0 \\ 0 & a_{1} & 0 & a_{2} \\ a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ 0 & a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ]

= a_{9}(a_{1} + a_{2} \frac{p_{4}}{p_{1}} )\{a_{1} det [ \begin{matrix}a_{1} & 0 & a_{2} \\ 0 & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ] + a_{2} det [ \begin{matrix}0 & a_{1} & a_{2} \\ a_{4} & 0 & 0 \\ 0 & a_{4} & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ] \}

= a_{9}(a_{1} + a_{2} \frac{p_{4}}{p_{1}} )\{a_{1}(\frac{(1 + a_{2}a_{4})^{2}}{a_{1}} − \frac{a_{2}a_{4}(1 + a_{2}a_{4})}{a_{1}}) + a_{2}(a_{2}a_{4}^{2} − a_{4}(1 + a_{2}a_{4}))\}
= a_{9} \frac{a_{1}p_{1} + a_{2}p_{4}}{p_{1}} \{(1 + a_{2}a_{4}) − a_{2}a_{4}\}
= (det A)\frac{p′_{1}}{p_{1}}
= \frac{det P′}{det P} \frac{p′_{1}}{p_{1}}

det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}

= det [ \begin{matrix}a_{1} & 0 & a_{2} & ∗ & 0 \\ 0 & a_{1} & 0 & 0 & a_{2} \\ a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} & ∗ & 0 \\ 0 & 0 & 0 & \frac{p_{1}}{a_{1}p_{1}+a_{2}p_{4}} & 0 \\ 0 & a_{4} & 0 & 0 & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ]

= \frac{p_{1}}{a_{1}p_{1} + a_{2}p_{4}} det [ \begin{matrix}a_{1} & 0 & a_{2} & 0 \\ 0 & a_{1} & 0 & a_{2} \\ a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ 0 & a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ]

= \frac{p_{1}}{a_{1}p_{1} + a_{2}p_{4}} \{a_{1} det [ \begin{matrix}a_{1} & 0 & a_{2} \\ 0 & \frac{1+a_{2}a_{4}}{a_{1}} & 0 \\ a_{4} & 0 & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ] + a_{2} det [ \begin{matrix}0 & a_{1} & a_{2} \\ a_{4} & 0 & 0 \\ 0 & a_{4} & \frac{1+a_{2}a_{4}}{a_{1}} \end{matrix} ] \}
= \frac{p_{1}}{a_{1}p_{1} + a_{2}p_{4}}
= \frac{p_{1}}{p′_{1}}

Hence the determinant of the Jacobian of the map A ⟼ A′ at each point A is

det [\frac{∂(p′_{1}, p′_{2}, p′_{3}, p′_{4}, p′_{6}, p′_{9}, R_{1}, R_{3}, R_{4}, R_{5}, R_{6})}{∂(p_{1}, p_{2}, p_{3}, p_{4}, p_{6}, p_{9}, r_{1}, r_{3}, r_{4}, r_{5}, r_{6})}] = det [\frac{∂p′_{i}}{∂p_{j}} ]_{6×6} det [\frac{∂R_{i}}{∂r_{j}} ]_{5×5}

= \frac{(det P′)p′_{1}}{(det P)p_{1}} \frac{p_{1}}{p′_{1}} = \frac{det P′}{det P} = \frac{p′_{9}}{p_{9}}

Therefore \frac{1}{p_{9}} dp_{1}dp_{2}dp_{3}dp_{4}dp_{6}dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6} is a right Haar measure on Ad_{G}\Ad_{G} · Stab^{∗}(Z_{3}^{∗}).

Lemma 5.3. Let 𝔤 = F_{3,2}, choose a basis B for 𝔤 like we did in Construction 1. Let C_{A} be the collection of all automorphisms A such that

[A]_{B} = [ \begin{matrix}p_{5}p_{9} − p_{6}p_{8} & −(p_{2}p_{9} − p_{3}p_{8}) & p_{2}p_{6} − p_{3}p_{5} & r_{1} & r_{4} & 0 \\ −(p_{4}p_{9} − p_{6}p_{7}) & p_{1}p_{9} − p_{3}p_{7} & −(p_{1}p_{6} − p_{3}p_{4}) & 0 & r_{5} & 0 \\ p_{4}p_{8} − p_{5}p_{7} & −(p_{1}p_{8} − p_{2}p_{7}) & p_{1}p_{5} − p_{2}p_{4} & r_{3} & r_{6} & r_{9} \\ 0 & 0 & 0 & p_{1} & p_{4} & p_{7} \\ 0 & 0 & 0 & p_{2} & p_{5} & p_{8} \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

∀p_{i}, r_{j} ∈ R with p_{9}(p_{1}p_{9} − p_{3}p_{7})(det P) ≠ 0, i = 1, 2, ⋯, 9, j = 1, 3, 4, 5, 6, 9. Claim that C_{A} is an almost global coordinate patch for Ad_{G}\Aut(𝔤).

Proof. For every X ∈ 𝔤, let X = ∑_{i=1}^{3} z_{i}Z_{i} + ∑_{i=1}^{3} y_{i}Y_{i}. Since 𝔤 is two-step, we showed already in Lemma 4.1. relative to the basis B, Ad_{exp(X)} has matrix

[Ad_{exp(X)}]_{B} = [ \begin{matrix}I_{3×3} & S_{3×3} \\ 0_{3×3} & I_{3×3} \end{matrix} ] where S_{3×3} = [ \begin{matrix}0 & −y_{3} & y_{2} \\ y_{3} & 0 & −y_{1} \\ −y_{2} & y_{1} & 0 \end{matrix} ].

For each A ∈ Aut(𝔤), let [A]_{B} = [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] where P_{3×3} = [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ],

Q_{3×3} = [ \begin{matrix}p_{5}p_{9} − p_{6}p_{8} & −(p_{2}p_{9} − p_{3}p_{8}) & p_{2}p_{6} − p_{3}p_{5} \\ −(p_{4}p_{9} − p_{6}p_{7}) & p_{1}p_{9} − p_{3}p_{7} & −(p_{1}p_{6} − p_{3}p_{4}) \\ p_{4}p_{8} − p_{5}p_{7} & −(p_{1}p_{8} − p_{2}p_{7}) & p_{1}p_{5} − p_{2}p_{4} \end{matrix} ], and

R_{3×3} = [ \begin{matrix}r_{1} & r_{4} & r_{7} \\ r_{2} & r_{5} & r_{8} \\ r_{3} & r_{6} & r_{9} \end{matrix} ]. Then

[Ad_{exp(X)}A]_{B} = [ \begin{matrix}I & S \\ 0 & I \end{matrix} ] [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] = [ \begin{matrix}Q & R + SP \\ 0 & P \end{matrix} ],

and hence Ad_{G}A = {Ad_{exp(X)}A| X = ∑_{i=1}^{3} z_{i}Z_{i} + ∑_{i=1}^{3} y_{i}Y_{i}, ∀y_{i}, z_{i} ∈ R}

= \{Ad_{exp(X)}A| [Ad_{exp(X)}A]_{B} = [ \begin{matrix}Q & R + SP \\ 0 & P \end{matrix} ] , ∀y_{i} ∈ R, i = 1, 2, 3\}

For the 3 × 3 matrix R + SP

= [ \begin{matrix}r_{1} & r_{4} & r_{7} \\ r_{2} & r_{5} & r_{8} \\ r_{3} & r_{6} & r_{9} \end{matrix} ] + [ \begin{matrix}0 & −y_{3} & y_{2} \\ y_{3} & 0 & −y_{1} \\ −y_{2} & y_{1} & 0 \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= [ \begin{matrix}r_{1} + p_{3}y_{2} − p_{2}y_{3} & r_{4} + p_{6}y_{2} − p_{5}y_{3} & r_{7} + p_{9}y_{2} − p_{8}y_{3} \\ r_{2} − p_{3}y_{1} + p_{1}y_{3} & r_{5} − p_{6}y_{1} + p_{4}y_{2} & r_{8} − p_{9}y_{1} + p_{7}y_{3} \\ r_{3} + p_{2}y_{1} − p_{1}y_{2} & r_{6} + p_{5}y_{1} − p_{4}y_{2} & r_{9} + p_{8}y_{1} − p_{7}y_{2} \end{matrix} ]

In order to re-coordinate the matrix R + SP, we need to put restrictions on some entries of P. We choose the condition for P is

p_{9} ≠ 0 and δ := det [ \begin{matrix}p_{1} & p_{7} \\ p_{3} & p_{9} \end{matrix} ] ≠ 0

Then let

s_{1} = r_{2} − p_{3}y_{1} + p_{1}y_{3}
s_{2} = r_{8} − p_{9}y_{1} + p_{7}y_{3}
s_{3} = r_{7} + p_{9}y_{2} − p_{8}y_{3}

Under the restrictions p_{9} ≠ 0 and δ = p_{1}p_{9} − p_{3}p_{7} ≠ 0, we get

y_{1} = (\frac{−p_{7}}{δ} r_{2} + \frac{p_{1}}{δ} r_{8}) + \frac{p_{7}}{δ} s_{1} − \frac{p_{1}}{δ} s_{2}
y_{2} = (\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3}
y_{3} = (\frac{−p_{9}}{δ} r_{2} + \frac{p_{3}}{δ} r_{8}) + \frac{p_{9}}{δ} s_{1} − \frac{p_{3}}{δ} s_{2}

Thus, we replace y_{1}, y_{2}, and y_{3} by the above corresponding identity in each entry of R + SP, and denote

P⃗ := P⃗ (p_{1}, p_{2}, ⋯, p_{9}) corresponding to matrix P_{3×3},

R⃗ := R⃗(r_{1}, r_{2}, ⋯, r_{9}) corresponding to matrix R_{3×3},
S⃗ := S⃗(s_{1}, s_{2}, s_{3}) corresponding to triple (s_{1}, s_{2}, s_{3}).

Then we can denote matrix

R + SP = [ \begin{matrix}R_{1}(P⃗ , R⃗, S⃗) & R_{4}(P⃗ , R⃗, S⃗) & s_{3} \\ s_{1} & R_{5}(P⃗ , R⃗, S⃗) & s_{2} \\ R_{3}(P⃗ , R⃗, S⃗) & R_{6}(P⃗ , R⃗, S⃗) & R_{9}(P⃗ , R⃗, S⃗) \end{matrix} ]

, and hence Ad_{G}A

=^{∗} \{Ad_{g}A| [Ad_{g}A]_{B} = [ \begin{matrix}Q & ( \begin{matrix}R_{1}(P⃗ , R⃗, S⃗) & R_{4}(P⃗ , R⃗, S⃗) & s_{3} \\ s_{1} & R_{5}(P⃗ , R⃗, S⃗) & s_{2} \\ R_{3}(P⃗ , R⃗, S⃗) & R_{6}(P⃗ , R⃗, S⃗) & R_{9}(P⃗ , R⃗, S⃗) \end{matrix} ) \\ 0 & P \end{matrix} ] , S⃗ ∈

R^{3}}

Let’s figure out the explicit formula for each R_{k}(P⃗ , R⃗, S⃗), k = 1, 3, 4, 5, 6, 9.

R_{1}(P⃗ , R⃗, S⃗)

= r_{1} + p_{3}y_{2} − p_{2}y_{3}

= r_{1} + p_{3}((\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3})

−p_{2}((\frac{−p_{9}}{δ} r_{2} + \frac{p_{3}}{δ} r_{8}) + \frac{p_{9}}{δ} s_{1} − \frac{p_{3}}{δ} s_{2})

= (r_{1} + \frac{p_{2}p_{9}−p_{3}p_{8}}{δ} r_{2} − \frac{p_{3}}{p_{9}} r_{7} − \frac{p_{3}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}δ}r_{8})

−\frac{p_{2}p_{9}−p_{3}p_{8}}{δ} s_{1} + \frac{p_{3}}{p_{9}} s_{3} + \frac{p_{3}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}δ}s_{2},

R_{3}(P⃗ , R⃗, S⃗)

= r_{3} + p_{2}y_{1} − p_{1}y_{2}

= r_{3} + p_{2}((\frac{−p_{7}}{δ} r_{2} + \frac{p_{1}}{δ} r_{8}) + \frac{p_{7}}{δ} s_{1} − \frac{p_{1}}{δ} s_{2})

−p_{1}((\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3})

= (\frac{p_{1}p_{8}−p_{2}p_{7}}{δ} r_{2} + r_{3} + \frac{p_{1}}{p_{9}} r_{7} + \frac{p_{1}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}δ}r_{8})

−\frac{p_{1}p_{8}−p_{2}p_{7}}{δ} s_{1} − \frac{p_{1}}{p_{9}} s_{3} − \frac{p_{1}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}δ}s_{2},

R_{4}(P⃗ , R⃗, S⃗)

= r_{4} + p_{6}y_{2} − p_{5}y_{3}

= r_{4} + p_{6}((\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3})

−p_{5}((\frac{−p_{9}}{δ} r_{2} + \frac{p_{3}}{δ} r_{8}) + \frac{p_{9}}{δ} s_{1} − \frac{p_{3}}{δ} s_{2})

= (\frac{p_{5}p_{9}−p_{6}p_{8}}{δ} r_{2} + r_{4} − \frac{p_{6}}{p_{9}} r_{7} − \frac{p_{3}(p_{5}p_{9}−p_{6}p_{8})}{p_{9}δ}r_{8})

−\frac{p_{5}p_{9}−p_{6}p_{8}}{δ} s_{1} + \frac{p_{6}}{p_{9}} s_{3} + \frac{p_{3}(p_{5}p_{9}−p_{6}p_{8})}{p_{9}δ}s_{2},

R_{5}(P⃗ , R⃗, S⃗)

= r_{5} − p_{6}y_{1} + p_{4}y_{3}

= r_{5} − p_{6}((\frac{−p_{7}}{δ} r_{2} + \frac{p_{1}}{δ} r_{8}) + \frac{p_{7}}{δ} s_{1} − \frac{p_{1}}{δ} s_{2})

+p_{4}((\frac{−p_{9}}{δ} r_{2} + \frac{p_{3}}{δ} r_{8}) + \frac{p_{9}}{δ} s_{1} − \frac{p_{3}}{δ} s_{2})

= (−\frac{p_{4}p_{9}−p_{6}p_{7}}{δ} r_{2} + r_{5} − \frac{p_{1}p_{6}−p_{3}p_{4}}{δ} r_{8})

+\frac{p_{4}p_{9}−p_{6}p_{7}}{δ} s_{1} + \frac{p_{1}p_{6}−p_{3}p_{4}}{δ} s_{2}, R_{6}(P⃗ , R⃗, S⃗) = r_{6} + p_{5}y_{1} − p_{4}y_{2} = r_{6} + p_{5}((\frac{−p_{7}}{δ} r_{2} + \frac{p_{1}}{δ} r_{8}) + \frac{p_{7}}{δ} s_{1} − \frac{p_{1}}{δ} s_{2}) −p_{4}((\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3}) = (\frac{p_{4}p_{8}−p_{5}p_{7}}{δ} r_{2} + r_{6} + \frac{p_{4}}{p_{9}} r_{7} + \frac{p_{1}p_{5}p_{9}−p_{3}p_{4}p_{8}}{p_{9}δ} r_{8}) −\frac{p_{4}p_{8}−p_{5}p_{7}}{δ} s_{1} − \frac{p_{4}}{p_{9}} s_{3} − \frac{p_{1}p_{5}p_{9}−p_{3}p_{4}p_{8}}{p_{9}δ} s_{2},

R_{9}(P⃗ , R⃗, S⃗) = r_{9} + p_{8}y_{1} − p_{7}y_{2} = r_{9} + p_{8}((\frac{−p_{7}}{δ} r_{2} + \frac{p_{1}}{δ} r_{8}) + \frac{p_{7}}{δ} s_{1} − \frac{p_{1}}{δ} s_{2}) −p_{7}((\frac{−p_{8}}{δ} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}δ} r_{8}) + \frac{p_{8}}{δ} s_{1} + \frac{−p_{3}p_{8}}{p_{9}δ} s_{2} + \frac{1}{p_{9}} s_{3}) = (\frac{p_{7}}{p_{9}} r_{7} + \frac{p_{8}}{p_{9}} r_{8} + r_{9}) − \frac{p_{8}}{p_{9}} s_{2} − \frac{p_{7}}{p_{9}} s_{3}.

Since S⃗ varying all R^{3} makes a coset Ad_{G}A, pick S⃗ = 0, then we have unique solution (y_{1}, y_{2}, y_{3}) where

y_{1} = \frac{−p_{7}}{p_{1}p_{9} − p_{3}p_{7}} r_{2} + \frac{p_{1}}{p_{1}p_{9} − p_{3}p_{7}} r_{8}
y_{2} = \frac{−p_{8}}{p_{1}p_{9} − p_{3}p_{7}} r_{2} − \frac{1}{p_{9}} r_{7} + \frac{p_{3}p_{8}}{p_{9}(p_{1}p_{9} − p_{3}p_{7})}r_{8}
y_{3} = \frac{−p_{9}}{p_{1}p_{9} − p_{3}p_{7}} r_{2} + \frac{p_{3}}{p_{1}p_{9} − p_{3}p_{7}} r_{8}

to make a unique automorphism Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}A ∈ Ad_{G}A for each A ∈ Aut(𝔤) satisfying the restriction p_{9}(p_{1}p_{9} − p_{3}p_{7}) ≠ 0 with

[Ad_{exp(∑_{i=1}^{3} y_{i}Y_{i})}A]_{B} = [ \begin{matrix}Q_{3×3} & ( \begin{matrix}R_{1}(P⃗ , R⃗) & R_{4}(P⃗ , R⃗) & 0 \\ 0 & R_{5}(P⃗ , R⃗) & 0 \\ R_{3}(P⃗ , R⃗) & R_{6}(P⃗ , R⃗) & R_{9}(P⃗ , R⃗) \end{matrix} ) \\ 0_{3×3} & P_{3×3} \end{matrix} ]

where each R_{k}(P⃗ , R⃗) = R_{k}(P⃗ , R⃗,0), k = 1, 3, 4, 5, 6, 9.

R_{1}(P⃗ , R⃗) = r_{1} + \frac{p_{2}p_{9}−p_{3}p_{8}}{p_{1}p_{9}−p_{3}p_{7}} r_{2} − \frac{p_{3}}{p_{9}} r_{7} − \frac{p_{3}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}r_{8} R_{3}(P⃗ , R⃗) = \frac{p_{1}p_{8}−p_{2}p_{7}}{p_{1}p_{9}−p_{3}p_{7}} r_{2} + r_{3} + \frac{p_{1}}{p_{9}} r_{7} + \frac{p_{1}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}r_{8} R_{4}(P⃗ , R⃗) = \frac{p_{5}p_{9}−p_{6}p_{8}}{p_{1}p_{9}−p_{3}p_{7}} r_{2} + r_{4} − \frac{p_{6}}{p_{9}} r_{7} − \frac{p_{3}(p_{5}p_{9}−p_{6}p_{8})}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}r_{8} R_{5}(P⃗ , R⃗) = −\frac{p_{4}p_{9}−p_{6}p_{7}}{p_{1}p_{9}−p_{3}p_{7}} r_{2} + r_{5} − \frac{p_{1}p_{6}−p_{3}p_{4}}{p_{1}p_{9}−p_{3}p_{7}} r_{8} R_{6}(P⃗ , R⃗) = \frac{p_{4}p_{8}−p_{5}p_{7}}{p_{1}p_{9}−p_{3}p_{7}} r_{2} + r_{6} + \frac{p_{4}}{p_{9}} r_{7} + \frac{p_{1}p_{5}p_{9}−p_{3}p_{4}p_{8}}{p_{9}(p_{1}p_{9}−p_{3}p_{7})} r_{8} R_{9}(P⃗ , R⃗) = \frac{p_{7}}{p_{9}} r_{7} + \frac{p_{8}}{p_{9}} r_{8} + r_{9} Thus, as P⃗ (p_{1}, p_{2}, ⋯, p_{9}) and R⃗(r_{1}, r_{2}, ⋯, r_{9}) vary over all R^{9} satisfying p_{9}(p_{1}p_{9} − p_{3}p_{7})(det P) ≠ 0, each R_{k}(P⃗ , R⃗) varies over all R, k = 1, 3, 4, 5, 6, 9. Now let C_{A} be the collection of all automorphisms A of 𝔤 such that relative to the basis B, A has matrix

[A]_{B} = [ \begin{matrix}p_{5}p_{9} − p_{6}p_{8} & −(p_{2}p_{9} − p_{3}p_{8}) & p_{2}p_{6} − p_{3}p_{5} & R_{1} & R_{4} & 0 \\ −(p_{4}p_{9} − p_{6}p_{7}) & p_{1}p_{9} − p_{3}p_{7} & −(p_{1}p_{6} − p_{3}p_{4}) & 0 & R_{5} & 0 \\ p_{4}p_{8} − p_{5}p_{7} & −(p_{1}p_{8} − p_{2}p_{7}) & p_{1}p_{5} − p_{2}p_{4} & R_{3} & R_{6} & R_{9} \\ 0 & 0 & 0 & p_{1} & p_{4} & p_{7} \\ 0 & 0 & 0 & p_{2} & p_{5} & p_{8} \\ 0 & 0 & 0 & p_{3} & p_{6} & p_{9} \end{matrix} ]

∀p_{i}, R_{k} ∈ R satisfying p_{9}(p_{1}p_{9} − p_{3}p_{7}) ≠ 0, i = 1, 2, ⋯, 9, k = 1, 3, 4, 5, 6, 9. Then we verify that different elements of C_{S} lie on different cosets. If this is done, that means C_{S} is an almost global coordinate patch for Ad_{G}\Aut(𝔤). For every distinct c, c′ ∈ C_{A}, we know c ∈ Ad_{G}A and c′ ∈ Ad_{G}A′ for some A, A′ ∈ Aut(𝔤). Let

[c]_{B} = [ \begin{matrix}Q_{3×3} & ( \begin{matrix}R_{1} & R_{4} & 0 \\ 0 & R_{5} & 0 \\ R_{3} & R_{6} & R_{9} \end{matrix} ) \\ 0_{3×3} & P_{3×3} \end{matrix} ] and [c′]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R′_{1} & R′_{4} & 0 \\ 0 & R′_{5} & 0 \\ R′_{3} & R′_{6} & R′_{9} \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

like the above matrix in C_{A}. Then Ad_{G}A is the set of all Ad_{g}A such that

[Ad_{g}A]_{B} = [ \begin{matrix}Q & ( \begin{matrix}R_{1}(P⃗ , R⃗, S⃗) & R_{4}(P⃗ , R⃗, S⃗) & s_{3} \\ s_{1} & R_{5}(P⃗ , R⃗, S⃗) & s_{2} \\ R_{3}(P⃗ , R⃗, S⃗) & R_{6}(P⃗ , R⃗, S⃗) & R_{9}(P⃗ , R⃗, S⃗) \end{matrix} ) \\ 0 & P \end{matrix} ] , ∀S⃗ ∈ R^{3}.

And it’s similar for Ad_{G}A′. Since c ≠ c′, it requires at least one p_{i} ≠ p′_{i} or R_{j} ≠ R′_{j}. If p_{i} ≠ p′_{i} for some i ∈ {1, 2, ⋯, 9}, then Ad_{G}A ≠ Ad_{G}A′. Suppose p_{i} = p′_{i} for all i, i.e., P⃗ = P⃗′. If R_{j} ≠ R′_{j} for some j ∈ {1, 3, 4, 5, 6, 9}. For example, let R_{1} ≠ R′_{1}. This means R_{1}(P⃗ , R⃗,0) ≠ R′_{1}(P⃗′, R⃗′,0), then R_{1}(P⃗ , R⃗,0) ≠ R′_{1}(P⃗ , R⃗′,0). By the formula of R_{1}(P⃗ , R⃗, S⃗), we get

R_{1}(P⃗ , R⃗, S⃗) = R_{1}(P⃗ , R⃗,0)−\frac{p_{2}p_{9}−p_{3}p_{8}}{p_{1}p_{9}−p_{3}p_{7}} s_{1} + \frac{p_{3}}{p_{9}} s_{3} + \frac{p_{3}(p_{2}p_{9}−p_{3}p_{8})}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}s_{2}
≠ R′_{1}(P⃗ , R⃗′,0) − \frac{p_{2}p_{9} − p_{3}p_{8}}{p_{1}p_{9} − p_{3}p_{7}} s_{1} + \frac{p_{3}}{p_{9}} s_{3} + \frac{p_{3}(p_{2}p_{9} − p_{3}p_{8})}{p_{9}(p_{1}p_{9} − p_{3}p_{7})}s_{2}
= R′_{1}(P⃗ , R⃗′, S⃗)
= R′_{1}(P⃗′, R⃗′, S⃗)

So R_{1}(P⃗ , R⃗, S⃗) ≠ R′_{1}(P⃗′, R⃗′, S⃗), and hence Ad_{G}A ≠ Ad_{G}A′. Similarly for other cases R_{j} ≠ R′_{j}, j = 3, 4, 5, 6, 9, we obtain the same result. Therefore different elements of C_{A} lie on different cosets. One concludes C_{A} is an almost global coordinate patch for Ad_{G}\Aut(𝔤).

Lemma 5.4. The assumption is the same as Lemma 5.3. With respect to the coordinate patch C_{A}, we claim that

\frac{|p_{9}(p_{1}p_{9}−p_{3}p_{7})|}{(det P)^{8}}dp_{1}dp_{2} ⋯ dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}dr_{9}

is a left Haar measure and

\frac{|p_{9}(p_{1}p_{9}−p_{3}p_{7})|}{(det P)^{6}}dp_{1}dp_{2} ⋯ dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}dr_{9}

is a right Haar measure on Ad_{G}\Aut(𝔤).

Proof. Given A_{0} ∈ C_{A}, for each A ∈ C_{A}, let A′ = A_{0}A. But A′ may not be in the patch C_{A}, we would like to find the corresponding Ã ∈ Ad_{G}A′ such that Ã is in C_{A}. Then we can calculate the determinant of the Jacobian of the map A ⟼ Ã at point A in order to figure out a left Haar measure on Ad_{G}\Aut(𝔤). Since A_{0}, A ∈ C_{A}, let

[A_{0}]_{B} = [ \begin{matrix}B_{3×3} & ( \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & c_{9} \end{matrix} ) \\ 0_{3×3} & (a_{ij})_{3×3} \end{matrix} ] =: [ \begin{matrix}B_{3×3} & C_{3×3} \\ 0_{3×3} & A_{3×3} \end{matrix} ] and

[A]_{B} = [ \begin{matrix}Q_{3×3} & ( \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & r_{9} \end{matrix} ) \\ 0_{3×3} & (p_{ij})_{3×3} \end{matrix} ] =: [ \begin{matrix}Q_{3×3} & R_{3×3} \\ 0_{3×3} & P_{3×3} \end{matrix} ] where matrices B and Q

are determined by A and P respectively. Then

[A′]_{B} = [ \begin{matrix}B & C \\ 0 & A \end{matrix} ] [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] = [ \begin{matrix}BQ & BR + CP \\ 0 & AP \end{matrix} ] =: [ \begin{matrix}Q′_{3×3} & R′_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

and hence from previous lemma we know the unique Ã ∈ Ad_{G}A′ such that Ã is in C_{A} which its matrix is

\left[{ \atop Ã}\right]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′) & R_{4}(P⃗′, R⃗′) & 0 \\ 0 & R_{5}(P⃗′, R⃗′) & 0 \\ R_{3}(P⃗′, R⃗′) & R_{6}(P⃗′, R⃗′) & R_{9}(P⃗′, R⃗′) \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

So we need to figure out matrices P′ and R′, then we will know each R_{i}(P⃗′, R⃗′), i = 1, 3, 4, 5, 6, 9. For the matrices

P′

= AP
= [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= [ \begin{matrix}a_{1}p_{1} + a_{4}p_{2} + a_{7}p_{3} & a_{1}p_{4} + a_{4}p_{5} + a_{7}p_{6} & a_{1}p_{7} + a_{4}p_{8} + a_{7}p_{9} \\ a_{2}p_{1} + a_{5}p_{2} + a_{8}p_{3} & a_{2}p_{4} + a_{5}p_{5} + a_{8}p_{6} & a_{2}p_{7} + a_{5}p_{8} + a_{8}p_{9} \\ a_{3}p_{1} + a_{6}p_{2} + a_{9}p_{3} & a_{3}p_{4} + a_{6}p_{5} + a_{9}p_{6} & a_{3}p_{7} + a_{6}p_{8} + a_{9}p_{9} \end{matrix} ]

= : [ \begin{matrix}p′_{1} & p′_{4} & p′_{7} \\ p′_{2} & p′_{5} & p′_{8} \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ]
R′
= BR + CP
= [ \begin{matrix}b_{1} & b_{4} & b_{7} \\ b_{2} & b_{5} & b_{8} \\ b_{3} & b_{6} & b_{9} \end{matrix} ] [ \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & r_{9} \end{matrix} ] + [ \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & c_{9} \end{matrix} ] [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ]

= : [ \begin{matrix}r′_{1} & r′_{4} & r′_{7} \\ r′_{2} & r′_{5} & r′_{8} \\ r′_{3} & r′_{6} & r′_{9} \end{matrix} ]

where r′_{1} = b_{1}r_{1} + b_{7}r_{3} + c_{1}p_{1} + c_{4}p_{2}, r′_{2} = b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}, r′_{3} = b_{3}r_{1} + b_{9}r_{3} + c_{3}p_{1} + c_{6}p_{2} + c_{9}p_{3}, r′_{4} = b_{1}r_{4} + b_{4}r_{5} + b_{7}r_{6} + c_{1}p_{4} + c_{4}p_{5}, r′_{5} = b_{2}r_{4} + b_{5}r_{5} + b_{8}r_{6} + c_{5}p_{5}, r′_{6} = b_{3}r_{4} + b_{6}r_{5} + b_{9}r_{6} + c_{3}p_{4} + c_{6}p_{5} + c_{9}p_{6}, r′_{7} = b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}, r′_{8} = b_{8}r_{9} + c_{5}p_{8}, r′_{9} = b_{9}r_{9} + c_{3}p_{7} + c_{6}p_{8} + c_{9}p_{9}.

Let δ′ := p′_{1}p′_{9} − p′_{3}p′_{7}. By formulas in previous lemma we get each R_{i}(P⃗′, R⃗′):

R_{1}(P⃗′, R⃗′) = r′_{1} + \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} r′_{2} − \frac{p′_{3}}{p′_{9}} r′_{7} − \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}r′_{8} = (b_{1}r_{1} + b_{7}r_{3} + c_{1}p_{1} + c_{4}p_{2}) + \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} (b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}) −\frac{p′_{3}}{p′_{9}} (b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}) − \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}(b_{8}r_{9} + c_{5}p_{8}) = (b_{1} + b_{2} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′})r_{1} + (b_{7} + b_{8} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′})r_{3} −(b_{7} \frac{p′_{3}}{p′_{9}} + b_{8} \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′})r_{9} + rational(P⃗),

R_{3}(P⃗′, R⃗′) = \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} r′_{2} + r′_{3} + \frac{p′_{1}}{p′_{9}} r′_{7} + \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}r′_{8} = \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} (b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}) + (b_{3}r_{1} + b_{9}r_{3} + c_{3}p_{1} + c_{6}p_{2} + c_{9}p_{3}) +\frac{p′_{1}}{p′_{9}} (b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}) + \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}(b_{8}r_{9} + c_{5}p_{8}) = (b_{2} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + b_{3})r_{1} + (b_{8} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + b_{9})r_{3} +(b_{7} \frac{p′_{1}}{p′_{9}} + b_{8} \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′})r_{9} + rational(P⃗), R_{4}(P⃗′, R⃗′)

= \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} r′_{2} + r′_{4} − \frac{p′_{6}}{p′_{9}} r′_{7} − \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′}r′_{8}

= \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} (b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}) + (b_{1}r_{4} + b_{4}r_{5} + b_{7}r_{6} + c_{1}p_{4} + c_{4}p_{5})

−\frac{p′_{6}}{p′_{9}} (b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}) − \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′}(b_{8}r_{9} + c_{5}p_{8})

= (b_{2} \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′})r_{1} + (b_{8} \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′})r_{3} + b_{1}r_{4} + b_{4}r_{5} + b_{7}r_{6}

−(b_{7} \frac{p′_{6}}{p′_{9}} + b_{8} \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′})r_{9} + rational(P⃗),

R_{5}(P⃗′, R⃗′)

= −\frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} r′_{2} + r′_{5} − \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′} r′_{8}

= −\frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} (b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}) + (b_{2}r_{4} + b_{5}r_{5} + b_{8}r_{6} + c_{5}p_{5})

−\frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′} (b_{8}r_{9} + c_{5}p_{8})

= (−b_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′})r_{1} + (−b_{8} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′})r_{3} + b_{2}r_{4} + b_{5}r_{5} + b_{8}r_{6}

+(−b_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′})r_{9} + rational(P⃗),

R_{6}(P⃗′, R⃗′)

= \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} r′_{2} + r′_{6} + \frac{p′_{4}}{p′_{9}} r′_{7} + \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′} r′_{8}

= \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} (b_{2}r_{1} + b_{8}r_{3} + c_{5}p_{2}) + (b_{3}r_{4} + b_{6}r_{5} + b_{9}r_{6} + c_{3}p_{4} + c_{6}p_{5} + c_{9}p_{6})

+\frac{p′_{4}}{p′_{9}} (b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}) + \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′} (b_{8}r_{9} + c_{5}p_{8})

= (b_{2} \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′})r_{1} + (b_{8} \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′})r_{3} + b_{3}r_{4} + b_{6}r_{5} + b_{9}r_{6}

+(b_{7} \frac{p′_{4}}{p′_{9}} + b_{8} \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′})r_{9} + rational(P⃗),

R_{9}(P⃗′, R⃗′)

= \frac{p′_{7}}{p′_{9}} r′_{7} + \frac{p′_{8}}{p′_{9}} r′_{8} + r′_{9}

= \frac{p′_{7}}{p′_{9}} (b_{7}r_{9} + c_{1}p_{7} + c_{4}p_{8}) + \frac{p′_{8}}{p′_{9}} (b_{8}r_{9} + c_{5}p_{8}) + (b_{9}r_{9} + c_{3}p_{7} + c_{6}p_{8} + c_{9}p_{9})

= (b_{7} \frac{p′_{7}}{p′_{9}} + b_{8} \frac{p′_{8}}{p′_{9}} + b_{9})r_{9} + rational(P⃗).

Thus we obtain all R_{i}(P⃗′, R⃗′), and hence we get unique Ã ∈ Ad_{G}A′ such that Ã is in the patch C_{A}. Now let’s calculate both det [\frac{∂p′_{i}}{∂p_{j}} ]_{9×9} and det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6}.

det [\frac{∂p′_{i}}{∂p_{j}} ]_{9×9}

= det [ \begin{matrix}a_{1} & a_{4} & a_{7} & 0 & 0 & 0 & 0 & 0 & 0 \\ a_{2} & a_{5} & a_{8} & 0 & 0 & 0 & 0 & 0 & 0 \\ a_{3} & a_{6} & a_{9} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{1} & a_{4} & a_{7} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{2} & a_{5} & a_{8} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{3} & a_{6} & a_{9} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{1} & a_{4} & a_{7} \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{2} & a_{5} & a_{8} \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{3} & a_{6} & a_{9} \end{matrix} ]

= (det A_{3×3})^{3} = \frac{(det P′)^{3}}{(det P)^{3}}

det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6}

= det [ \begin{matrix}b_{1} + b_{2} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} & b_{7} + b_{8} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} & 0 & 0 & 0 & ∗ \\ b_{2} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + b_{3} & b_{8} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + b_{9} & 0 & 0 & 0 & ∗ \\ b_{2} \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} & b_{8} \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} & b_{1} & b_{4} & b_{7} & ∗ \\ −b_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} & −b_{8} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} & b_{2} & b_{5} & b_{8} & ∗ \\ b_{2} \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} & b_{8} \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} & b_{3} & b_{6} & b_{9} & ∗ \\ 0 & 0 & 0 & 0 & 0 & b_{7} \frac{p′_{7}}{p′_{9}} + b_{8} \frac{p′_{8}}{p′_{9}} + b_{9} \end{matrix} ]

= (b_{7} \frac{p′_{7}}{p′_{9}} + b_{8} \frac{p′_{8}}{p′_{9}} + b_{9})(det B_{3×3}) det [ \begin{matrix}b_{1} + b_{2} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} & b_{7} + b_{8} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} \\ b_{3} + b_{2} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} & b_{9} + b_{8} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} \end{matrix} ]

= \frac{1}{p′_{9}} (b_{7}p′_{7} + b_{8}p′_{8} + b_{9}p′_{9})(det B_{3×3}) ·\{(b_{1} + b_{2} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′})(b_{9} + b_{8} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′}) − (b_{3} + b_{2} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′})(b_{9} + b_{8} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′})\} Since b_{7}p′_{7} + b_{8}p′_{8} + b_{9}p′_{9}

= | \begin{matrix}a_{2} & a_{5} \\ a_{3} & a_{6} \end{matrix} | (a_{1}p_{7} + a_{4}p_{8} + a_{7}p_{9}) − | \begin{matrix}a_{1} & a_{4} \\ a_{3} & a_{6} \end{matrix} | (a_{2}p_{7} + a_{5}p_{8} + a_{8}p_{9})

+ | \begin{matrix}a_{1} & a_{4} \\ a_{2} & a_{5} \end{matrix} | (a_{3}p_{7} + a_{6}p_{8} + a_{9}p_{9})

= (a_{1} | \begin{matrix}a_{2} & a_{5} \\ a_{3} & a_{6} \end{matrix} | − a_{2} | \begin{matrix}a_{1} & a_{4} \\ a_{3} & a_{6} \end{matrix} | + a_{3} | \begin{matrix}a_{1} & a_{4} \\ a_{2} & a_{5} \end{matrix} |)p_{7}

+(a_{4} | \begin{matrix}a_{2} & a_{5} \\ a_{3} & a_{6} \end{matrix} | − a_{5} | \begin{matrix}a_{1} & a_{4} \\ a_{3} & a_{6} \end{matrix} | + a_{6} | \begin{matrix}a_{1} & a_{4} \\ a_{2} & a_{5} \end{matrix} |)p_{8}

+(a_{7} | \begin{matrix}a_{2} & a_{5} \\ a_{3} & a_{6} \end{matrix} | − a_{8} | \begin{matrix}a_{1} & a_{4} \\ a_{3} & a_{6} \end{matrix} | + a_{9} | \begin{matrix}a_{1} & a_{4} \\ a_{2} & a_{5} \end{matrix} |)p_{9}

= | \begin{matrix}a_{1} & a_{4} & a_{1} \\ a_{2} & a_{5} & a_{2} \\ a_{3} & a_{6} & a_{3} \end{matrix} | p_{7} + | \begin{matrix}a_{1} & a_{4} & a_{4} \\ a_{2} & a_{5} & a_{5} \\ a_{3} & a_{6} & a_{6} \end{matrix} | p_{8} + | \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} | p_{9}

= (det A_{3×3})p_{9}, it follows further that

det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6}

= \frac{1}{p′_{9}} ((det A)p_{9})(det B)\{ | \begin{matrix}b_{1} & b_{7} \\ b_{3} & b_{9} \end{matrix} | + | \begin{matrix}b_{1} & b_{7} \\ b_{2} & b_{8} \end{matrix} | \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + | \begin{matrix}b_{2} & b_{8} \\ b_{3} & b_{9} \end{matrix} | \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} \}

= (det A)(det A)^{2} \frac{p_{9}}{p′_{9}} \{\frac{1}{δ′} (a_{5}(det A) | \begin{matrix}p′_{1} & p′_{7} \\ p′_{3} & p′_{9} \end{matrix} | − a_{6}(det A) | \begin{matrix}p′_{1} & p′_{7} \\ p′_{2} & p′_{8} \end{matrix} | −

a_{4}(det A) | \begin{matrix}p′_{2} & p′_{8} \\ p′_{3} & p′_{9} \end{matrix} | )\}

= (det A)^{4} \frac{p_{9}}{p′_{9}} \frac{1}{δ′} ( − a_{4} | \begin{matrix}p′_{2} & p′_{8} \\ p′_{3} & p′_{9} \end{matrix} | + a_{5} | \begin{matrix}p′_{1} & p′_{7} \\ p′_{3} & p′_{9} \end{matrix} | − a_{6} | \begin{matrix}p′_{1} & p′_{7} \\ p′_{2} & p′_{8} \end{matrix} | )

= (det A)^{4} \frac{p_{9}}{p′_{9}} \frac{1}{δ′} det [ \begin{matrix}p′_{1} & a_{4} & p′_{7} \\ p′_{2} & a_{5} & p′_{8} \\ p′_{3} & a_{6} & p′_{9} \end{matrix} ]

= (det A)^{4} \frac{p_{9}}{p′_{9}} \frac{1}{δ′} det ([ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ] [ \begin{matrix}p_{1} & 0 & p_{7} \\ p_{2} & 1 & p_{8} \\ p_{3} & 0 & p_{9} \end{matrix} ])

= (det A)^{5} \frac{p_{9}}{p′_{9}} \frac{p_{1}p_{9}−p_{3}p_{7}}{p′_{1}p′_{9}−p′_{3}p′_{7}} = \frac{(det P′)^{5}}{(det P)^{5}} \frac{p_{9}}{p′_{9}} \frac{p_{1}p_{9}−p_{3}p_{7}}{p′_{1}p′_{9}−p′_{3}p′_{7}}

Hence

det [\frac{∂(p′_{1},p′_{2},··· ,p′_{9},R_{1},R_{3},R_{4},R_{5},R_{6},R_{9})}{∂(p_{1},p_{2},··· ,p_{9},r_{1},r_{3},r_{4},r_{5},r_{6},r_{9})}]

= det [\frac{∂p′_{i}}{∂p_{j}} ]_{9×9} det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6} = \frac{(det P′)^{3}}{(det P)^{3}} \frac{(det P′)^{5}}{(det P)^{5}} \frac{p_{9}}{p′_{9}} \frac{p_{1}p_{9}−p_{3}p_{7}}{p′_{1}p′_{9}−p′_{3}p′_{7}} = \frac{\frac{(det P′)^{8}}{p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})}}{\frac{(det P)^{8}}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}}

Therefore

\frac{|p_{9}(p_{1}p_{9} − p_{3}p_{7})|}{(det P)^{8}}dp_{1}dp_{2} ⋯ dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}dr_{9}

is a left Haar measure on Ad_{G}\Aut(𝔤). Next step we want to find a right Haar measure. Given A_{0} ∈ C_{A}, for each A ∈ C_{A}, let A′ = AA_{0}. But A′ may not be in the patch C_{A}. We want to find the corresponding Ã ∈ Ad_{G}A′ such that Ã is in C_{A}. Let the matrices of A and A_{0} relative to the basis B are the same as before. Then

[A′]_{B} = [AA_{0}]_{B} = [ \begin{matrix}Q & R \\ 0 & P \end{matrix} ] [ \begin{matrix}B & C \\ 0 & A \end{matrix} ] = [ \begin{matrix}QB & QC + RA \\ 0 & P A \end{matrix} ] =:

[ \begin{matrix}Q′_{3×3} & R′_{3×3} \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

From previous lemma we know Ã has matrix

\left[{ \atop Ã}\right]_{B} = [ \begin{matrix}Q′_{3×3} & ( \begin{matrix}R_{1}(P⃗′, R⃗′) & R_{4}(P⃗′, R⃗′) & 0 \\ 0 & R_{5}(P⃗′, R⃗′) & 0 \\ R_{3}(P⃗′, R⃗′) & R_{6}(P⃗′, R⃗′) & R_{9}(P⃗′, R⃗′) \end{matrix} ) \\ 0_{3×3} & P′_{3×3} \end{matrix} ]

So we need to know matrices P′ and R′, then each R_{i}(P⃗′, R⃗′) follows by formula. P′ = P A

= [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ]

= [ \begin{matrix}a_{1}p_{1} + a_{2}p_{4} + a_{3}p_{7} & a_{4}p_{1} + a_{5}p_{4} + a_{6}p_{7} & a_{7}p_{1} + a_{8}p_{4} + a_{9}p_{7} \\ a_{1}p_{2} + a_{2}p_{5} + a_{3}p_{8} & a_{4}p_{2} + a_{5}p_{5} + a_{6}p_{8} & a_{7}p_{2} + a_{8}p_{5} + a_{9}p_{8} \\ a_{1}p_{3} + a_{2}p_{6} + a_{3}p_{9} & a_{4}p_{3} + a_{5}p_{6} + a_{6}p_{9} & a_{7}p_{3} + a_{8}p_{6} + a_{9}p_{9} \end{matrix} ]

=: [ \begin{matrix}p′_{1} & p′_{4} & p′_{7} \\ p′_{2} & p′_{5} & p′_{8} \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ]

R′ = QC + RA

= [ \begin{matrix}q_{1} & q_{4} & q_{7} \\ q_{2} & q_{5} & q_{8} \\ q_{3} & q_{6} & q_{9} \end{matrix} ] [ \begin{matrix}c_{1} & c_{4} & 0 \\ 0 & c_{5} & 0 \\ c_{3} & c_{6} & c_{9} \end{matrix} ] + [ \begin{matrix}r_{1} & r_{4} & 0 \\ 0 & r_{5} & 0 \\ r_{3} & r_{6} & r_{9} \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ]

=: [ \begin{matrix}r′_{1} & r′_{4} & r′_{7} \\ r′_{2} & r′_{5} & r′_{8} \\ r′_{3} & r′_{6} & r′_{9} \end{matrix} ]

where r′_{1} = a_{1}r_{1} + a_{2}r_{4} + c_{1}q_{1} + c_{3}q_{7} r′_{2} = a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8} r′_{3} = a_{1}r_{3} + a_{2}r_{6} + a_{3}r_{9} + c_{1}q_{3} + c_{3}q_{9} r′_{4} = a_{4}r_{1} + a_{5}r_{4} + c_{4}q_{1} + c_{5}q_{4} + c_{6}q_{7} r′_{5} = a_{5}r_{5} + c_{4}q_{2} + c_{5}q_{5} + c_{6}q_{8} r′_{6} = a_{4}r_{3} + a_{5}r_{6} + a_{6}r_{9} + c_{4}q_{3} + c_{5}q_{6} + c_{6}q_{9} r′_{7} = a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7} r′_{8} = a_{8}r_{5} + c_{9}q_{8} r′_{9} = a_{7}r_{3} + a_{8}r_{6} + a_{9}r_{9} + c_{9}q_{9}.

Let δ′ := p′_{1}p′_{9} − p′_{3}p′_{7}. By formula in previous lemma, each R_{i}(P⃗′, R⃗′) is the following:

R_{1}(P⃗′, R⃗′)

= r′_{1} + \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} r′_{2} − \frac{p′_{3}}{p′_{9}} r′_{7} − \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}r′_{8}

= (a_{1}r_{1} + a_{2}r_{4} + c_{1}q_{1} + c_{3}q_{7}) + \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} (a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8})

−\frac{p′_{3}}{p′_{9}} (a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7}) − \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}(a_{8}r_{5} + c_{9}q_{8})

= (a_{1} − a_{7} \frac{p′_{3}}{p′_{9}} )r_{1} + (a_{2} − a_{8} \frac{p′_{3}}{p′_{9}} )r_{4} + (a_{2} \frac{p′_{2}p′_{9}−p′_{3}p′_{8}}{δ′} − a_{8} \frac{p′_{3}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′})r_{5} + rational(P⃗),

R_{3}(P⃗′, R⃗′)

= \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} r′_{2} + r′_{3} + \frac{p′_{1}}{p′_{9}} r′_{7} + \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}r′_{8}

= \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} (a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8}) + (a_{1}r_{3} + a_{2}r_{6} + a_{3}r_{9} + c_{1}q_{3} + c_{3}q_{9})

+\frac{p′_{1}}{p′_{9}} (a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7}) + \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′}(a_{8}r_{5} + c_{9}q_{8})

= (a_{7} \frac{p′_{1}}{p′_{9}} )r_{1} + a_{1}r_{3} + (a_{8} \frac{p′_{1}}{p′_{9}} )r_{4} + (a_{2} \frac{p′_{1}p′_{8}−p′_{2}p′_{7}}{δ′} + a_{8} \frac{p′_{1}(p′_{2}p′_{9}−p′_{3}p′_{8})}{p′_{9}δ′})r_{5}

+a_{2}r_{6} + a_{3}r_{9} + rational(P⃗),

R_{4}(P⃗′, R⃗′)

= \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} r′_{2} + r′_{4} − \frac{p′_{6}}{p′_{9}} r′_{7} − \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′}r′_{8}

= \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} (a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8}) + (a_{4}r_{1} + a_{5}r_{4} + c_{4}q_{1} + c_{5}q_{4} + c_{6}q_{7})

−\frac{p′_{6}}{p′_{9}} (a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7}) − \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′}(a_{8}r_{5} + c_{9}q_{8})

= (a_{4} − a_{7} \frac{p′_{6}}{p′_{9}} )r_{1} + (a_{5} − a_{8} \frac{p′_{6}}{p′_{9}} )r_{4} + (a_{2} \frac{p′_{5}p′_{9}−p′_{6}p′_{8}}{δ′} − a_{8} \frac{p′_{3}(p′_{5}p′_{9}−p′_{6}p′_{8})}{p′_{9}δ′})r_{5} + rational(P⃗),

R_{5}(P⃗′, R⃗′)

= −\frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} r′_{2} + r′_{5} − \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′} r′_{8}

= −\frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} (a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8}) + (a_{5}r_{5} + c_{4}q_{2} + c_{5}q_{5} + c_{6}q_{8}) − \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′} (a_{8}r_{5} + c_{9}q_{8})

= (−a_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} + a_{5} − a_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′})r_{5} + rational(P⃗),

R_{6}(P⃗′, R⃗′)

= \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} r′_{2} + r′_{6} + \frac{p′_{4}}{p′_{9}} r′_{7} + \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′} r′_{8}

= \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′} (a_{2}r_{5} + c_{1}q_{2} + c_{3}q_{8}) + (a_{4}r_{3} + a_{5}r_{6} + a_{6}r_{9} + c_{4}q_{3} + c_{5}q_{6} + c_{6}q_{9})

+\frac{p′_{4}}{p′_{9}} (a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7}) + \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′} (a_{8}r_{5} + c_{9}q_{8})

= (a_{7} \frac{p′_{4}}{p′_{9}} )r_{1} + a_{4}r_{3} + (a_{8} \frac{p′_{4}}{p′_{9}} )r_{4} + (a_{2} \frac{p′_{4}p′_{8}−p′_{5}p′_{7}}{δ′}) + a_{8} \frac{p′_{1}p′_{5}p′_{9}−p′_{3}p′_{4}p′_{8}}{p′_{9}δ′})r_{5}

+a_{5}r_{6} + a_{6}r_{9} + rational(P⃗), R_{9}(P⃗′, R⃗′) = \frac{p′_{7}}{p′_{9}} r′_{7} + \frac{p′_{8}}{p′_{9}} r′_{8} + r′_{9} = \frac{p′_{7}}{p′_{9}} (a_{7}r_{1} + a_{8}r_{4} + c_{9}q_{7}) + \frac{p′_{8}}{p′_{9}} (a_{8}r_{5} + c_{9}q_{8}) + (a_{7}r_{3} + a_{8}r_{6} + a_{9}r_{9} + c_{9}q_{9}) = (a_{7} \frac{p′_{7}}{p′_{9}} )r_{1} + a_{7}r_{3} + (a_{8} \frac{p′_{7}}{p′_{9}} )r_{4} + (a_{8} \frac{p′_{8}}{p′_{9}} )r_{5} + a_{8}r_{6} + a_{9}r_{9} + rational(P⃗). Thus we obtain all R_{i}(P⃗′, R⃗′), and hence we get Ã ∈ Ad_{G}A′ such that Ã is in the patch C_{A}. Now let’s calculate two determinants.

det [\frac{∂p′_{i}}{∂p_{j}} ]_{9×9}

= det [ \begin{matrix}a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} & 0 & 0 \\ 0 & a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} & 0 \\ 0 & 0 & a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} \\ a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} & 0 & 0 \\ 0 & a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} & 0 \\ 0 & 0 & a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} \\ a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} & 0 & 0 \\ 0 & a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} & 0 \\ 0 & 0 & a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} \end{matrix} ]

= − det [ \begin{matrix}a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} & 0 & 0 \\ a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} & 0 & 0 \\ a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} & 0 & 0 \\ 0 & a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} & 0 \\ 0 & a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} & 0 \\ 0 & a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} & 0 \\ 0 & 0 & a_{1} & 0 & 0 & a_{2} & 0 & 0 & a_{3} \\ 0 & 0 & a_{4} & 0 & 0 & a_{5} & 0 & 0 & a_{6} \\ 0 & 0 & a_{7} & 0 & 0 & a_{8} & 0 & 0 & a_{9} \end{matrix} ]

= det [ \begin{matrix}a_{1} & a_{2} & a_{3} & 0 & 0 & 0 & 0 & 0 & 0 \\ a_{4} & a_{5} & a_{6} & 0 & 0 & 0 & 0 & 0 & 0 \\ a_{7} & a_{8} & a_{9} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{1} & a_{2} & a_{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{4} & a_{5} & a_{6} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{7} & a_{8} & a_{9} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{1} & a_{2} & a_{3} \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{4} & a_{5} & a_{6} \\ 0 & 0 & 0 & 0 & 0 & 0 & a_{7} & a_{8} & a_{9} \end{matrix} ]

= (det A)^{3}

= \frac{(det P′)^{3}}{(det P)^{3}}

det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6}

= det [ \begin{matrix}a_{1} − a_{7} \frac{p′_{3}}{p′_{9}} & 0 & a_{2} − a_{8} \frac{p′_{3}}{p′_{9}} & ∗ & 0 & 0 \\ a_{7} \frac{p′_{1}}{p′_{9}} & a_{1} & a_{8} \frac{p′_{1}}{p′_{9}} & ∗ & a_{2} & a_{3} \\ a_{4} − a_{7} \frac{p′_{6}}{p′_{9}} & 0 & a_{5} − a_{8} \frac{p′_{6}}{p′_{9}} & ∗ & 0 & 0 \\ 0 & 0 & 0 & −a_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} + a_{5} − a_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′} & 0 & 0 \\ a_{7} \frac{p′_{4}}{p′_{9}} & a_{4} & a_{8} \frac{p′_{4}}{p′_{9}} & ∗ & a_{5} & a_{6} \\ a_{7} \frac{p′_{7}}{p′_{9}} & a_{7} & a_{8} \frac{p′_{7}}{p′_{9}} & ∗ & a_{8} & a_{9} \end{matrix} ]

= (−a_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} + a_{5} − a_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′}) det [ \begin{matrix}a_{1} − a_{7} \frac{p′_{3}}{p′_{9}} & 0 & a_{2} − a_{8} \frac{p′_{3}}{p′_{9}} & 0 & 0 \\ a_{7} \frac{p′_{1}}{p′_{9}} & a_{1} & a_{8} \frac{p′_{1}}{p′_{9}} & a_{2} & a_{3} \\ a_{4} − a_{7} \frac{p′_{6}}{p′_{9}} & 0 & a_{5} − a_{8} \frac{p′_{6}}{p′_{9}} & 0 & 0 \\ a_{7} \frac{p′_{4}}{p′_{9}} & a_{4} & a_{8} \frac{p′_{4}}{p′_{9}} & a_{5} & a_{6} \\ a_{7} \frac{p′_{7}}{p′_{9}} & a_{7} & a_{8} \frac{p′_{7}}{p′_{9}} & a_{8} & a_{9} \end{matrix} ]

= (−a_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} + a_{5} − a_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′}) det [ \begin{matrix}a_{1} − a_{7} \frac{p′_{3}}{p′_{9}} & a_{2} − a_{8} \frac{p′_{3}}{p′_{9}} & 0 & 0 & 0 \\ a_{4} − a_{7} \frac{p′_{6}}{p′_{9}} & a_{5} − a_{8} \frac{p′_{6}}{p′_{9}} & 0 & 0 & 0 \\ a_{7} \frac{p′_{1}}{p′_{9}} & a_{8} \frac{p′_{1}}{p′_{9}} & a_{1} & a_{2} & a_{3} \\ a_{7} \frac{p′_{4}}{p′_{9}} & a_{8} \frac{p′_{4}}{p′_{9}} & a_{4} & a_{5} & a_{6} \\ a_{7} \frac{p′_{7}}{p′_{9}} & a_{8} \frac{p′_{7}}{p′_{9}} & a_{7} & a_{8} & a_{9} \end{matrix} ]

= (−a_{2} \frac{p′_{4}p′_{9}−p′_{6}p′_{7}}{δ′} + a_{5} − a_{8} \frac{p′_{1}p′_{6}−p′_{3}p′_{4}}{δ′})(det A) ·\{(a_{1} − a_{7} \frac{p′_{3}}{p′_{9}} )(a_{5} − a_{8} \frac{p′_{6}}{p′_{9}} ) − (a_{2} − a_{8} \frac{p′_{3}}{p′_{9}} )(a_{4} − a_{7} \frac{p′_{6}}{p′_{9}} )\}

= (det A)\frac{1}{δ′} ( − a_{2} det [ \begin{matrix}p′_{4} & p′_{7} \\ p′_{6} & p′_{9} \end{matrix} ] + a_{5} det [ \begin{matrix}p′_{1} & p′_{7} \\ p′_{3} & p′_{9} \end{matrix} ] − a_{8} det [ \begin{matrix}p′_{1} & p′_{4} \\ p′_{3} & p′_{6} \end{matrix} ] )

·\{(a_{1}a_{5} − a_{2}a_{4}) − (a_{1}a_{8} − a_{2}a_{7})\frac{p′_{6}}{p′_{9}} + (a_{4}a_{8} − a_{5}a_{7})\frac{p′_{3}}{p′_{9}} \}

= (det A)\frac{1}{p′_{9}δ′} det [ \begin{matrix}p′_{1} & p′_{4} & p′_{7} \\ a_{2} & a_{5} & a_{8} \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ] det [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ p′_{3} & p′_{6} & p′_{9} \end{matrix} ]

=

\frac{det A}{p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})} det ([ \begin{matrix}p_{1} & p_{4} & p_{7} \\ 0 & 1 & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ]) det ([ \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ p_{3} & p_{6} & p_{9} \end{matrix} ] [ \begin{matrix}a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix} ])

= (det A)^{3} \frac{p_{9}(p_{1}p_{9}−p_{3}p_{7})}{p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})} = \frac{(det P′)^{3}p_{9}(p_{1}p_{9}−p_{3}p_{7})}{(det P)^{3}p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})} Then

det [\frac{∂(p′_{1},p′_{2},··· ,p′_{9},R_{1},R_{3},R_{4},R_{5},R_{6},R_{9})}{∂(p_{1},p_{2},··· ,p_{9},r_{1},r_{3},r_{4},r_{5},r_{6},r_{9})}]

= det [\frac{∂p′_{i}}{∂p_{j}} ]_{9×9} det [\frac{∂R_{i}}{∂r_{j}} ]_{6×6} = \frac{(det P′)^{3}}{(det P)^{3}} \frac{(det P′)^{3}p_{9}(p_{1}p_{9}−p_{3}p_{7})}{(det P)^{3}p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})} = \frac{\frac{(det P′)^{6}}{p′_{9}(p′_{1}p′_{9}−p′_{3}p′_{7})}}{\frac{(det P)^{6}}{p_{9}(p_{1}p_{9}−p_{3}p_{7})}} Therefore

\frac{|p_{9}(p_{1}p_{9} − p_{3}p_{7})|}{(det P)^{6}}dp_{1}dp_{2} ⋯ dp_{9}dr_{1}dr_{3}dr_{4}dr_{5}dr_{6}dr_{9}

is a right Haar measure measure on Ad_{G}\Aut(𝔤).

Motivation. From the introductory example for 𝔤 = F_{3,2}, we know that σ ⊗ σ is an automorphism of 𝔤 for all σ in SO(3, R). We call it the rotation-automorphism. We now define dilation-automorphisms of 𝔤 by

τ_{r} : 𝔤_{2} ⊕ 𝔤_{1} ⟶ 𝔤_{2} ⊕ 𝔤_{1}
(Z, Y) ⟼ (\frac{1}{r}Z, \frac{1}{\sqrt{r}}Y)

∀r > 0. Then τ_{r} indeed is an automorphism of 𝔤 since τ_{r}([Y_{i}, Y_{j}]) = \frac{1}{r}[Y_{i}, Y_{j}] = [\frac{1}{\sqrt{r}}Y_{i}, \frac{1}{\sqrt{r}}Y_{j}] = [τ_{r}Y_{i}, τ_{r}Y_{j}] for all pair (i, j). And define its dual τ_{r}^{∗} by the inverse transpose, then

τ_{r}^{∗}(Z^{∗}, Y^{∗}) = (rZ^{∗}, \sqrt{r}Y^{∗})

since (τ_{r}^{∗}(Z^{∗}, Y^{∗}))(Z′, Y′) = (Z^{∗}, Y^{∗})(τ_{r}^{−1}(Z′, Y′)) = (Z^{∗}, Y^{∗})(rZ′, \sqrt{r}Y′) = (rZ^{∗}, \sqrt{r}Y^{∗})(Z′, Y′) , ∀(Z′, Y′) ∈ 𝔤_{2} ⊕ 𝔤_{1}. Now let l_{0} ∈ 𝔤_{Max}^{∗} be fixed, for each l ∈ 𝔤_{Max}^{∗}, there exist r > 0 and σ ∈ SO(3) such that rσ^{∗}(l_{0}|_{𝔤_{2}}) = l|_{𝔤_{2}} since any two non-zero vectors can be rotated and dilated into one another. Thus we get an automorphism τ_{r} ∘ (σ ⊗ σ) with (τ_{r}^{∗} ∘ (σ ⊗ σ)^{∗})l_{0} = rσ^{∗}(l_{0}|_{𝔤_{2}}) + \sqrt{r}σ^{∗}(l_{0}|_{𝔤_{1}}), and hence we have the following relations of orbits O_{τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} ∥ O_{rσ^{∗}(l_{0}|_{𝔤_{2}})} = O_{l|_{𝔤_{2}}} ∥ O_{l}. The purpose of next lemma is to find an automorphism ρ_{t}^{σ} associated with σ such that its dual ρ_{t}^{σ^{∗}} moves the orbit O_{τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} parallel to O_{l} along the direction in 𝔤_{1}^{∗} related to σ. Then we call ρ_{t}^{σ} the translation-automorphism of 𝔤.

Lemma 5.5. For 𝔤 = F_{3,2}, give g a norm respecting the bracket. Let B = {Z_{1}, Z_{2}, Z_{3}, Y_{1}, Y_{2}, Y_{3}} be an orthonormal basis for 𝔤 built as in Construction 2, and let B^{∗} = {Z_{1}^{∗}, Z_{2}^{∗}, Z_{3}^{∗}, Y_{1}^{∗}, Y_{2}^{∗}, Y_{3}^{∗}} be its dual basis in 𝔤^{∗}. Fix l_{0} = a_{1}Z_{1}^{∗} + a_{2}Z_{2}^{∗} + a_{3}Z_{3}^{∗} + b_{1}Y_{1}^{∗} + b_{2}Y_{2}^{∗} + b_{3}Y_{3}^{∗} ∈ 𝔤_{Max}^{∗}, and denote it by (a⃗, b⃗). Choose a cross-section for SO(3)/Stab_{SO(3)}(a⃗). For each l ∈ 𝔤_{Max}^{∗}, let l = z_{1}Z_{1}^{∗} + z_{2}Z_{2}^{∗} + z_{3}Z_{3}^{∗} + y_{1}Y_{1}^{∗} + y_{2}Y_{2}^{∗} + y_{3}Y_{3}^{∗} =: (z⃗, y⃗). We show there exists an automorphism ρ_{t}^{σ} of 𝔤 with σ ∈ Σ such that ρ_{t}^{σ^{∗}}l = l + t(z_{1}Y_{1}^{∗} + z_{2}Y_{2}^{∗} + z_{3}Y_{3}^{∗}), ∀t ∈ R, where σ^{∗} rotate a⃗ into the direction of z⃗. Moreover, let u⃗ = \frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}). For every (α⃗, β⃗) ∈ 𝔤_{2}^{∗} ⊕ 𝔤_{1}^{∗}, ρ_{t}^{σ^{∗}}(α⃗, β⃗) = (α⃗, β⃗ + \frac{t}{∥a⃗∥^{2}} ⟨σ^{∗}a⃗, α⃗⟩σ^{∗}(∥a⃗∥u⃗)).

Proof. For fixed l_{0} = Z_{3}^{∗} ∈ 𝔤_{Max}^{∗}, let ρ̃_{t}^{e} be the map

ρ̃_{t}^{e} : 𝔤 ⟶ 𝔤
Z_{i} ⟼ Z_{i}, i = 1, 2, 3,
Y_{j} ⟼ Y_{j}, j = 1, 2,
Y_{3} ⟼ Y_{3} − tZ_{3}

then ρ̃_{t}^{e} is an automorphism of 𝔤, and its dual ρ̃_{t}^{e^{∗}} is

ρ̃_{t}^{e^{∗}} : 𝔤^{∗} ⟶ 𝔤^{∗}
Y_{i}^{∗} ⟼ Y_{i}^{∗}, i = 1, 2, 3,
Z_{j}^{∗} ⟼ Z_{j}^{∗}, j = 1, 2,
Z_{3}^{∗} ⟼ Z_{3}^{∗} + tY_{3}^{∗}

We verify the dual ρ̃_{t}^{e^{∗}} as follows:

(ρ̃_{t}^{e^{∗}}(Y_{i}^{∗}))(Y_{j}) = Y_{i}^{∗}(ρ̃_{t}^{e^{−1}}Y_{j}) = Y_{i}^{∗}(Y_{j}), ∀j = 1, 2, (ρ̃_{t}^{e^{∗}}(Y_{i}^{∗}))(Y_{3}) = Y_{i}^{∗}(ρ̃_{t}^{e^{−1}}Y_{3}) = Y_{i}^{∗}(Y_{3} + tZ_{3}) = Y_{i}^{∗}(Y_{3}) (ρ̃_{t}^{e^{∗}}(Y_{i}^{∗}))(Z_{j}) = Y_{i}^{∗}(ρ̃_{t}^{e^{−1}}Z_{j}) = Y_{i}^{∗}(Z_{j}), ∀j = 1, 2, 3. So ρ̃_{t}^{e^{∗}} : Y_{i}^{∗} ⟼ Y_{i}^{∗}, ∀i = 1, 2, 3.

For i = 1, 2, we have (ρ̃_{t}^{e^{∗}}(Z_{i}^{∗}))(Y_{j}) = Z_{i}^{∗}(ρ̃_{t}^{e^{−1}}Y_{j}) = Z_{i}^{∗}(Y_{j}), ∀j = 1, 2, (ρ̃_{t}^{e^{∗}}(Z_{i}^{∗}))(Y_{3}) = Z_{i}^{∗}(ρ̃_{t}^{e^{−1}}Y_{3}) = Z_{i}^{∗}(Y_{3} + tZ_{3}) = Z_{i}^{∗}(Y_{3}), (ρ̃_{t}^{e^{∗}}(Z_{i}^{∗}))(Z_{j}) = Z_{i}^{∗}(ρ̃_{t}^{e^{−1}}Z_{j}) = Z_{i}^{∗}(Z_{j}), ∀j = 1, 2, 3. So ρ̃_{t}^{e^{∗}} : Z_{i}^{∗} ⟼ Z_{i}^{∗}, ∀i = 1, 2.

(ρ̃_{t}^{e^{∗}}(Z_{3}^{∗}))(Y_{j}) = Z_{3}^{∗}(ρ̃_{t}^{e^{−1}}Y_{j}) = Z_{3}^{∗}(Y_{j}) = (Z_{3}^{∗} + tY_{3}^{∗})(Y_{j}), ∀j = 1, 2, (ρ̃_{t}^{e^{∗}}(Z_{3}^{∗}))(Y_{3}) = Z_{3}^{∗}(ρ̃_{t}^{e^{−1}}Y_{3}) = Z_{3}^{∗}(Y_{3} + tZ_{3}) = t = (Z_{3}^{∗} + tY_{3}^{∗})(Y_{3}), (ρ̃_{t}^{e^{∗}}(Z_{3}^{∗}))(Z_{j}) = Z_{3}^{∗}(ρ̃_{t}^{e^{−1}}Z_{j}) = Z_{3}^{∗}(Z_{j}) = (Z_{3}^{∗} + tY_{3}^{∗})(Z_{j}), ∀j = 1, 2, 3. So ρ̃_{t}^{e^{∗}} : Z_{3}^{∗} ⟼ Z_{3}^{∗} + tY_{3}^{∗}. Thus, we get the dual ρ̃_{t}^{e^{∗}} as desired.

For each σ ∈ SO(3), let σ = [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ], define

ρ̃_{t}^{σ} = (σ ⊗ σ) ∘ ρ̃_{t}^{e} ∘ (σ ⊗ σ)^{−1}

Then we get

ρ̃_{t}^{σ} : 𝔤 ⟶ 𝔤
Z_{i} ⟼ Z_{i}, ∀i = 1, 2, 3,
Y_{1} ⟼ Y_{1} − tp_{7}(σZ_{3})
Y_{2} ⟼ Y_{2} − tp_{8}(σZ_{3})
Y_{3} ⟼ Y_{3} − tp_{9}(σZ_{3})

since ρ̃_{t}^{σ}(Y_{1}) = (σ ⊗ σ)ρ̃_{t}^{e}(σ^{−1}Y_{1}) = (σ ⊗ σ)ρ̃_{t}^{e}(σ^{t}Y_{1}) = (σ ⊗ σ)ρ̃_{t}^{e}(p_{1}Y_{1} + p_{4}Y_{2} + p_{7}Y_{3}) = σ(p_{1}Y_{1} + p_{4}Y_{2} + p_{7}Y_{3}) − tp_{7}(σZ_{3}) = Y_{1} − tp_{7}(σZ_{3}), and it’s similar for Y_{2} and Y_{3}. We also get the dual ρ̃_{t}^{σ^{∗}} = (σ ⊗ σ)^{∗} ∘ ρ̃_{t}^{e^{∗}} ∘ (σ^{−1} ⊗ σ^{−1})^{∗},

ρ̃_{t}^{σ^{∗}} : 𝔤^{∗} ⟶ 𝔤^{∗}
Y_{i}^{∗} ⟼ Y_{i}^{∗}, ∀i = 1, 2, 3,
Z_{1}^{∗} ⟼ Z_{1}^{∗} + tp_{7}(σ^{∗}Y_{3}^{∗})
Z_{2}^{∗} ⟼ Z_{2}^{∗} + tp_{8}(σ^{∗}Y_{3}^{∗})
Z_{3}^{∗} ⟼ Z_{3}^{∗} + tp_{9}(σ^{∗}Y_{3}^{∗})

since ρ̃_{t}^{σ^{∗}}(Z_{1}^{∗}) = (σ ⊗ σ)^{∗}ρ̃_{t}^{e^{∗}}(σ^{−1^{∗}}Z_{1}^{∗}) = (σ ⊗ σ)^{∗}ρ̃_{t}^{e^{∗}}(p_{1}Z_{1}^{∗} + p_{4}Z_{2}^{∗} + p_{7}Z_{3}^{∗}) = σ^{∗}(p_{1}Z_{1}^{∗} + p_{4}Z_{2}^{∗} + p_{7}Z_{3}^{∗}) + tp_{7}(σ^{∗}Y_{3}^{∗}) = Z_{1}^{∗} + tp_{7}(σ^{∗}Y_{3}^{∗}) , and it’s similar for Z_{2}^{∗} and Z_{3}^{∗}. Since l_{0} = (a⃗, b⃗) ∈ 𝔤_{Max}^{∗}, a⃗ ≠ 0. Then there exists unique σ_{a} ∈ Σ such that σ_{a}^{∗}(\frac{1}{∥a⃗∥}a⃗) = Z_{3}^{∗}, i.e., ∥a⃗∥σ_{a}^{−1^{∗}}Z_{3}^{∗} = a⃗. So we have ∥a⃗∥σ_{a}^{−1^{∗}}Y_{3}^{∗} = a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}, and ρ̃_{t}^{σ_{a}^{−1}} = (σ_{a}^{−1} ⊗ σ_{a}^{−1})ρ̃_{t}^{e}(σ_{a} ⊗ σ_{a}). Hence

ρ̃_{t}^{σ_{a}^{−1^{∗}}}a⃗ = (σ_{a}^{−1} ⊗ σ_{a}^{−1})^{∗}ρ̃_{t}^{e^{∗}}(σ_{a} ⊗ σ_{a})^{∗}(∥a⃗∥σ_{a}^{−1^{∗}}Z_{3}^{∗})
= ∥a⃗∥(σ_{a}^{−1} ⊗ σ_{a}^{−1})^{∗}ρ̃_{t}^{e^{∗}}Z_{3}^{∗}
= ∥a⃗∥(σ_{a}^{−1^{∗}}Z_{3}^{∗} + tσ_{a}^{−1^{∗}}Y_{3}^{∗})
= a⃗ + t(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗})

For each l = (z⃗, y⃗) ∈ 𝔤_{Max}^{∗}, z⃗ = l|_{𝔤_{2}} ≠ 0. Then there exists unique (r, σ) ∈ R^{+} × Σ such that z⃗ = rσ^{∗}a⃗. So we have z_{1}Z_{1}^{∗} + z_{2}Z_{2}^{∗} + z_{3}Z_{3}^{∗} = rσ^{∗}(a_{1}Z_{1}^{∗} + a_{2}Z_{2}^{∗} + a_{3}Z_{3}^{∗}) and z_{1}Y_{1}^{∗} + z_{2}Y_{2}^{∗} + z_{3}Y_{3}^{∗} = rσ^{∗}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}). Now we define

ρ_{t}^{σ} = (σ ⊗ σ)ρ̃_{t}^{σ_{a}^{−1}}(σ ⊗ σ)^{−1}

Then

ρ_{t}^{σ^{∗}}l = ρ_{t}^{σ^{∗}}(z⃗ + y⃗)
= ρ_{t}^{σ^{∗}}(rσ^{∗}a⃗ + y⃗)
= rρ_{t}^{σ^{∗}}(σ^{∗}a⃗) + ρ_{t}^{σ^{∗}}(y⃗)
= r(σ ⊗ σ)^{∗}ρ̃_{t}^{σ_{a}^{−1^{∗}}}(σ ⊗ σ)^{−1^{∗}}(σ^{∗}a⃗) + y⃗
= r(σ ⊗ σ)^{∗}ρ̃_{t}^{σ_{a}^{−1^{∗}}}(a⃗) + y⃗
= r(σ ⊗ σ)^{∗}(a⃗ + t(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗})) + y⃗
= (rσ^{∗}a⃗ + y⃗) + trσ^{∗}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗})
= l + t(z_{1}Y_{1}^{∗} + z_{2}Y_{2}^{∗} + z_{3}Y_{3}^{∗})

Since

ρ_{t}^{σ} = (σ ⊗ σ)ρ̃_{t}^{σ_{a}^{−1}}(σ ⊗ σ)^{−1}
= (σ ⊗ σ)(σ_{a}^{−1} ⊗ σ_{a}^{−1})ρ̃_{t}^{e}(σ_{a} ⊗ σ_{a})(σ ⊗ σ)^{−1}
= (σσ_{a}^{−1} ⊗ σσ_{a}^{−1})ρ̃_{t}^{e}(σσ_{a}^{−1} ⊗ σσ_{a}^{−1})^{−1}

let σσ_{a}^{−1} = [ \begin{matrix}p_{1} & p_{4} & p_{7} \\ p_{2} & p_{5} & p_{8} \\ p_{3} & p_{6} & p_{9} \end{matrix} ], then we get ρ_{t}^{σ^{∗}}(Z_{1}^{∗}) = Z_{1}^{∗} + tp_{7}(σσ_{a}^{−1})^{∗}Y_{3}^{∗},

ρ_{t}^{σ^{∗}}(Z_{2}^{∗}) = Z_{2}^{∗} + tp_{8}(σσ_{a}^{−1})^{∗}Y_{3}^{∗}, and ρ_{t}^{σ^{∗}}(Z_{3}^{∗}) = Z_{3}^{∗} + tp_{9}(σσ_{a}^{−1})^{∗}Y_{3}^{∗}. Hence for every (α⃗, β⃗) ∈ 𝔤_{2}^{∗} ⊕ 𝔤_{1}^{∗}, let α⃗ = α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗}, and β⃗ = β_{1}Y_{1}^{∗} + β_{2}Y_{2}^{∗} + β_{3}Y_{3}^{∗}. Then

ρ_{t}^{σ^{∗}}(α⃗, β⃗) = α_{1}ρ_{t}^{σ^{∗}}(Z_{1}^{∗}) + α_{2}ρ_{t}^{σ^{∗}}(Z_{2}^{∗}) + α_{3}ρ_{t}^{σ^{∗}}(Z_{3}^{∗}) + ρ_{t}^{σ^{∗}}(β⃗)
= α_{1}Z_{1}^{∗} + α_{2}Z_{2}^{∗} + α_{3}Z_{3}^{∗} + t(p_{7}α_{1} + p_{8}α_{2} + p_{9}α_{3})(σσ_{a}^{−1})^{∗}Y_{3}^{∗} + β⃗
= (α⃗, β⃗ + t(p_{7}α_{1} + p_{8}α_{2} + p_{9}α_{3})(σσ_{a}^{−1})^{∗}Y_{3}^{∗})
= (α⃗, β⃗ + t⟨σ^{∗}σ_{a}^{−1^{∗}}Z_{3}^{∗}, α⃗⟩σ^{∗}(\frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗})))
= (α⃗, β⃗ + t⟨σ^{∗}(\frac{1}{∥a⃗∥}a⃗), α⃗⟩σ^{∗}u⃗)
= (α⃗, β⃗ + \frac{t}{∥a⃗∥^{2}} ⟨σ^{∗}a⃗, α⃗⟩σ^{∗}(∥a⃗∥u⃗))

This proves the lemma.

Lemma 5.6. Fix l_{0} = a_{1}Z_{1}^{∗} + a_{2}Z_{2}^{∗} + a_{3}Z_{3}^{∗} + b_{1}Y_{1}^{∗} + b_{2}Y_{2}^{∗} + b_{3}Y_{3}^{∗} ∈ 𝔤_{Max}^{∗}, and denote it by (a⃗, b⃗). Let Σ be a cross-section of SO(3)/Stab_{SO(3)}(a⃗). Show

X := {ρ_{t}^{σ} ∘ τ_{r} ∘ (σ ⊗ σ)| (t, r, σ) ∈ R × R^{+} × Σ}

is a cross-section of the double coset space Ad_{G}\Aut(𝔤)/Stab^{∗}(l_{0}).

Proof. Since fixed l_{0} = (a⃗, b⃗) ∈ 𝔤_{Max}^{∗}, there is some coefficient a_{i} ≠ 0, 1 ≤ i ≤ 3. Without loss of generality we let a_{3} ≠ 0. Then claim 𝔯_{l_{0}} = 𝔤_{2} ⊕ R(a_{1}Y_{1} + a_{2}Y_{2} + a_{3}Y_{3}). Since 𝔯_{l_{0}} = {Z + Y | l_{0}([Z + Y, 𝔤]) = 0}, let Y = y_{1}Y_{1} + y_{2}Y_{2} + y_{3}Y_{3}, then it follows l_{0}([y_{1}Y_{1} + y_{2}Y_{2} + y_{3}Y_{3}, Y_{i}]) = 0, ∀i = 1, 2, 3. If i = 1, then −a_{3}y_{2} + a_{2}y_{3} = 0; if i = 2, then a_{3}y_{1} − a_{1}y_{3} = 0; if i = 3, then −a_{2}y_{1} + a_{1}y_{2} = 0. Since a_{3} ≠ 0, we get y_{1} = \frac{a_{1}}{a_{3}} y_{3} and y_{2} = \frac{a_{2}}{a_{3}} y_{3}, so y_{1}Y_{1} + y_{2}Y_{2} + y_{3}Y_{3} = \frac{a_{1}}{a_{3}} y_{3}Y_{1} + \frac{a_{2}}{a_{3}} y_{3}Y_{2} + y_{3}Y_{3} = \frac{y_{3}}{a_{3}} (a_{1}Y_{1} + a_{2}Y_{2} + a_{3}Y_{3}). Hence 𝔯_{l_{0}} = \{Z + \frac{y_{3}}{a_{3}} (a_{1}Y_{1} + a_{2}Y_{2} + a_{3}Y_{3})| Z ∈ 𝔤_{2}, y_{3} ∈ R\} = 𝔤_{2} ⊕ R(a_{1}Y_{1} + a_{2}Y_{2} + a_{3}Y_{3}), and we see 𝔯_{l_{0}} is an ideal in 𝔤. Then by Theorem 3.2.3. in [1], we know O_{l_{0}} = l_{0} + 𝔯_{l_{0}}^{⊥} where the annihilator of 𝔯_{l_{0}},

𝔯_{l_{0}}^{⊥} = 𝔯⊥ ; a⃗ = Sp{a_{3}Y_{1}^{∗} − a_{1}Y_{3}^{∗}, a_{3}Y_{2}^{∗} − a_{2}Y_{3}^{∗}}, is a two-dimensional plane in 𝔤_{1}^{∗}.

Hence we can choose a unit vector u⃗ = \frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}) ∈ 𝔤_{1}^{∗} perpendicular to this plane 𝔯⊥ ; a⃗. Before we go further, let’s verify one fact. For each A ∈ Aut(𝔤) and l ∈ 𝔤^{∗}, claim 𝔯_{A^{∗}l}^{⊥} = A^{∗}(𝔯_{l}^{⊥}). Proof of claim:

𝔯_{A^{∗}l}^{⊥} = {f ∈ 𝔤^{∗}| f(𝔯_{A^{∗}l}) = 0}
= {f ∈ 𝔤^{∗}| f(A𝔯_{l}) = 0}
= {f = f̃∘ A^{−1}| f̃(𝔯_{l}) = 0}
= {f = A^{∗}f̃| f̃(𝔯_{l}) = 0}
= A^{∗}{f̃| f̃(𝔯_{l}) = 0}
= A^{∗}𝔯_{l}^{⊥}

For each σ ∈ Σ, since (σ ⊗ σ) ∈ Aut(𝔤), we get σ^{∗}𝔯⊥ ; a⃗ = (σ ⊗ σ)^{∗}𝔯⊥ ; a⃗ = 𝔯_{σ^{∗}a⃗}^{⊥}. Since fixed l_{0} = (a⃗, b⃗) ∈ 𝔤_{Max}^{∗}, a⃗ = l_{0}|_{𝔤_{2}} ≠ 0. For each A ∈ Aut(𝔤), we know A^{∗}l_{0} ∈ 𝔤_{Max}^{∗}, so (A^{∗}l_{0})|_{𝔤_{2}} ≠ 0. Then there exists \underline{unique} (r, σ) ∈ R^{+} × Σ such that rσ^{∗}a⃗ = (A^{∗}l_{0})|_{𝔤_{2}}. Thus we get unique automorphism τ_{r} ∘ (σ ⊗ σ) with τ_{r}^{∗}(σ ⊗ σ)^{∗}l_{0} = τ_{r}^{∗}(σ^{∗}a⃗ + σ^{∗}b⃗) = rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗, and its orbit

O_{τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} = (rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗) + 𝔯_{(rσ^{∗}a⃗+\sqrt{r}σ^{∗}b⃗)}^{⊥} = rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + 𝔯_{σ^{∗}a⃗}^{⊥}

Let z_{1}Z_{1}^{∗} + z_{2}Z_{2}^{∗} + z_{3}Z_{3}^{∗} = z⃗ := σ^{∗}a⃗, and y_{1}Y_{1}^{∗} + y_{2}Y_{2}^{∗} + y_{3}Y_{3}^{∗} = y⃗ := σ^{∗}b⃗. Then we have z_{1}Y_{1}^{∗} + z_{2}Y_{2}^{∗} + z_{3}Y_{3}^{∗} = σ^{∗}(∥a⃗∥u⃗). Since σ^{∗} rotate a⃗ into the direction of z⃗, by previous lemma we know there exists ρ_{t}^{σ} ∈ Aut(𝔤) such that ρ_{t}^{σ^{∗}}(τ_{r}^{∗}(σ ⊗ σ)^{∗}l_{0}) = ρ_{t}^{σ^{∗}}(rz⃗ + \sqrt{r}y⃗) = (rz⃗ + \sqrt{r}y⃗) + t(rz_{1}Y_{1}^{∗} + rz_{2}Y_{2}^{∗} + rz_{3}Y_{3}^{∗}) = rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)), and its orbit is parallel to the orbit of τ_{r}^{∗}(σ ⊗ σ)^{∗}l_{0}, in other words,

O_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} = rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)) + 𝔯_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}}^{⊥}
= rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)) + 𝔯_{σ^{∗}a⃗}^{⊥}
∥ rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + 𝔯_{σ^{∗}a⃗}^{⊥}
= O_{τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}}

Thus, for each t ∈ R we have the following relations of orbits

O_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} ∥ O_{τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} = O_{rσ^{∗}a⃗+\sqrt{r}σ^{∗}b⃗} ∥ O_{rσ^{∗}a⃗} = O_{(A^{∗}l_{0})|_{𝔤_{2}}} ∥ O_{A^{∗}l_{0}}

We want to show that there exists unique t ∈ R such that O_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} = O_{A^{∗}l_{0}}.

Since u⃗ = \frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}) ∉ 𝔯_{l_{0}}^{⊥} = 𝔯⊥ ; a⃗ in 𝔤_{1}^{∗}, for each t ∈ R we have

(tr∥a⃗∥u⃗ + \sqrt{r}b⃗ + 𝔯⊥ ; a⃗) ∥ 𝔯⊥ ; a⃗, and hence σ^{∗}(tr∥a⃗∥u⃗ + \sqrt{r}b⃗ + 𝔯⊥ ; a⃗) ∥ σ^{∗}𝔯⊥ ; a⃗ in 𝔤_{1}^{∗}. So

𝔤_{1}^{∗} = ∪^{.} \{σ^{∗}(tr∥a⃗∥u⃗ + \sqrt{r}b⃗ + 𝔯⊥ ; a⃗)| t ∈ R\}
= ∪^{.} \{t(rσ^{∗}(∥a⃗∥u⃗) + \sqrt{r}σ^{∗}b⃗ + σ^{∗}𝔯⊥ ; a⃗| t ∈ R\}
= ∪^{.} \{\sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)) + 𝔯_{σ^{∗}a⃗}^{⊥}| t ∈ R\}

Hence there exists \underline{unique} t ∈ R such that (A^{∗}l_{0})|_{𝔤_{1}} ∈ (\sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)) + 𝔯_{σ^{∗}a⃗}^{⊥}). Then

A^{∗}l_{0} = (A^{∗}l_{0})|_{𝔤_{2}} + (A^{∗}l_{0})|_{𝔤_{1}}
= rσ^{∗}a⃗ + (A^{∗}l_{0})|_{𝔤_{1}}
∈ rσ^{∗}a⃗ + \sqrt{r}σ^{∗}b⃗ + t(rσ^{∗}(∥a⃗∥u⃗)) + 𝔯_{σ^{∗}a⃗}^{⊥}
= O_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}}

Hence we get O_{A^{∗}l_{0}} = O_{ρ_{t}^{σ^{∗}}τ_{r}^{∗}(σ⊗σ)^{∗}l_{0}} as desired.

Thus we know for each A ∈ Aut(𝔤) there exists unique (t, r, σ) ∈ R × R^{+} × Σ such that O_{A^{∗}l_{0}} = O_{(ρ_{t}^{σ}τ_{r}(σ⊗σ))^{∗}l_{0}}. And by Theorem 2.1. we know for each O_{l} ∈ O_{Max} there exists A ∈ Aut(𝔤) such that O_{l} = O_{A^{∗}l_{0}}. Now define T : X^{∗} ⟶ O_{Max} by T (x^{∗}) = O_{x^{∗}l_{0}}, ∀x ∈ X. Then T is bijective. Since there is one-to-one correspondence between O_{Max} and D := Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(l_{0}), it follows that there exists uniquely one x^{∗} ∈ X^{∗} from each coset d ∈ D. It concludes that X^{∗} is a cross-section of D, in other words, X is a cross-section of Ad_{G}\Aut(𝔤)/Stab^{∗}(l_{0}).

Motivation. Let X = {ρ_{t}^{σ} ∘ τ_{r} ∘ (σ ⊗ σ)| ∀(t, r, σ) ∈ R × R^{+} × Σ}, we have shown that X is a cross-section of Ad_{G}\Aut(𝔤)/Stab^{∗}(l_{0}). But X is not a group, i.e., it is not closed under composition. We show next to which element of X the composition of two elements of X corresponds.

Lemma 5.7. Fix l_{0} ∈ 𝔤_{Max}^{∗}. Let l_{0} = a_{1}Z_{1}^{∗} + a_{2}Z_{2}^{∗} + a_{3}Z_{3}^{∗} + b_{1}Y_{1}^{∗} + b_{2}Y_{2}^{∗} + b_{3}Y_{3}^{∗}, and denote it by (a⃗, b⃗) ∈ 𝔤_{2}^{∗} ⊕ 𝔤_{1}^{∗}. For each ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}), ρ_{t}^{σ}τ_{r}(σ ⊗ σ) ∈ X, we have the equality

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}l_{0} = Ad_{G}^{∗}(ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ}τ_{r_{0}r}(σ_{0}σ ⊗ σ_{0}σ))^{∗}l_{0}

where ψ is the angle between vectors a⃗ and σ^{∗}a⃗, i.e., ψ is the angle of rotation for σ^{∗}.

Proof. For fixed l_{0} = (a⃗, b⃗), we showed already in the beginning of Lemma5.6, its radical is 𝔯_{l_{0}} = 𝔤_{2} ⊕ R(a_{1}Y_{1} + a_{2}Y_{2} + a_{3}Y_{3}). So dim(𝔯_{l_{0}}) = 4, and hence dim(𝔯_{l_{0}}^{⊥}) = 2. Then 𝔯_{l_{0}}^{⊥} = Span_{R}{a_{3}Y_{1}^{∗} − a_{1}Y_{3}^{∗}, a_{3}Y_{2}^{∗} − a_{2}Y_{3}^{∗}} since it contains the two listed independent vectors.

Now we choose a unit vector u⃗ ∈ 𝔤_{1}^{∗} perpendicular to the plane 𝔯_{l_{0}}^{⊥},

u⃗ = \frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗})

For each (α⃗, β⃗) ∈ 𝔤_{2}^{∗} ⊕ 𝔤_{1}^{∗}, let’s recall that ρ_{t}^{σ^{∗}}(α⃗, β⃗) = (α⃗, β⃗ + \frac{t}{∥a⃗∥^{2}} ⟨σ^{∗}a⃗, α⃗⟩σ^{∗}(∥a⃗∥u⃗)

and Ad_{G}^{∗}(α⃗, β⃗) = (α⃗, β⃗) + 𝔯⊥ ; α⃗. Then for each ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}), σ_{t}^{σ}τ_{r}(σ ⊗ σ) ∈ X, the

orbit of the dual of their composition acting on l_{0} is

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}(a⃗, b⃗)
= Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r})^{∗}(σ^{∗}a⃗, σ^{∗}b⃗)
= Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ})^{∗}(rσ^{∗}a⃗, \sqrt{r}σ^{∗}b⃗)
= Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}))^{∗}(rσ^{∗}a⃗, \sqrt{r}σ^{∗}b⃗ + \frac{t}{∥a⃗∥^{2}} ⟨σ^{∗}a⃗, rσ^{∗}a⃗⟩σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}))^{∗}(rσ^{∗}a⃗, \sqrt{r}σ^{∗}b⃗ + trσ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}})^{∗}(rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r}σ_{0}^{∗}σ^{∗}b⃗ + trσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}ρ_{t_{0}}^{σ_{0}^{∗}}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + t\sqrt{r_{0}}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}ρ_{t_{0}}^{σ_{0}^{∗}}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + \frac{t_{0}}{∥a⃗∥^{2}} ⟨σ_{0}^{∗}a⃗, r_{0}rσ_{0}^{∗}σ^{∗}a⃗⟩σ_{0}^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + \frac{t_{0}}{∥a⃗∥^{2}} r_{0}r⟨a⃗, σ^{∗}a⃗⟩σ_{0}^{∗}(∥a⃗∥u⃗))

Let ψ be the angle between vectors a⃗ and σ^{∗}a⃗, then it follows

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}(a⃗, b⃗)
= Ad_{G}^{∗}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + t_{0}(cosψ)r_{0}rσ_{0}^{∗}(∥a⃗∥u⃗))
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + t_{0}(cosψ)r_{0}rσ_{0}^{∗}(∥a⃗∥u⃗)) + 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}

Since unit vector u⃗ = \frac{1}{∥a⃗∥}(a_{1}Y_{1}^{∗} + a_{2}Y_{2}^{∗} + a_{3}Y_{3}^{∗}) perpendicular to two-dimensional plane 𝔯_{l_{0}}^{⊥} = 𝔯⊥ ; a⃗ in 𝔤_{1}^{∗} ≅ R^{3}, it implies that the unit vector σ_{0}^{∗}σ^{∗}u⃗ perpendicular to two-dimensional plane σ_{0}^{∗}σ^{∗}𝔯⊥ ; a⃗ = 𝔯_{σ_{0}^{∗}σ^{∗}a⃗}^{⊥} = 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥} in 𝔤_{1}^{∗}. Then the vector σ_{0}^{∗}(∥a⃗∥u⃗) ∈ 𝔤_{1}^{∗} can be written as the addition ⟨σ_{0}^{∗}(∥a⃗∥u⃗), σ_{0}^{∗}σ^{∗}u⃗⟩σ_{0}^{∗}σ^{∗}u⃗ + v⃗ for some vector v⃗ in the plane 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}. And hence it follows that

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}(a⃗, b⃗)
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗)
+t_{0}(cosψ)r_{0}r(⟨σ_{0}^{∗}(∥a⃗∥u⃗), σ_{0}^{∗}σ^{∗}u⃗⟩σ_{0}^{∗}σ^{∗}u⃗ + v⃗)) + 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗)
+t_{0}(cosψ)r_{0}r∥a⃗∥⟨u⃗, σ^{∗}u⃗⟩σ_{0}^{∗}σ^{∗}u⃗ + t_{0}(cosψ)r_{0}rv⃗) + 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + t_{0}(cosψ)r_{0}r⟨u⃗, σ^{∗}u⃗⟩σ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
+𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{t}{\sqrt{r_{0}}} r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗) + t_{0}(cosψ)^{2}r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗)) + 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}

The last equality holds because ψ is the angle between a⃗ and σ^{∗}a⃗, meanwhile, it is the angle between u⃗ and σ^{∗}u⃗. Furthermore, it follows

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}(a⃗, b⃗)
= (r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + (\frac{t}{\sqrt{r_{0}}} + t_{0}cos^{2}ψ)r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗)) + 𝔯_{r_{0}rσ_{0}^{∗}σ^{∗}a⃗}^{⊥}
= Ad_{G}^{∗}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + (\frac{t}{\sqrt{r_{0}}} + t_{0}cos^{2}ψ)r_{0}rσ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗ + \frac{\frac{t}{\sqrt{r_{0}}} + t_{0}cos^{2}ψ}{∥a⃗∥^{2}} ⟨σ_{0}^{∗}σ^{∗}a⃗, r_{0}rσ_{0}^{∗}σ^{∗}a⃗⟩σ_{0}^{∗}σ^{∗}(∥a⃗∥u⃗))
= Ad_{G}^{∗}ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ^{∗}}(r_{0}rσ_{0}^{∗}σ^{∗}a⃗, \sqrt{r_{0}r}σ_{0}^{∗}σ^{∗}b⃗)
= Ad_{G}^{∗}(ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ}τ_{r_{0}r})^{∗}(σ_{0}^{∗}σ^{∗}a⃗, σ_{0}^{∗}σ^{∗}b⃗)
= Ad_{G}^{∗}(ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ}τ_{r_{0}r}(σ_{0}σ ⊗ σ_{0}σ))^{∗}(a⃗, b⃗)

Therefore we get the desired equality

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}(a⃗, b⃗) = Ad_{G}^{∗}(ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ}τ_{r_{0}r}(σ_{0}σ ⊗ σ_{0}σ))^{∗}(a⃗, b⃗)

for every ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}), ρ_{t}^{σ}τ(σ ⊗ σ) ∈ X.

Verification of Relative Invariance for F_{3,2}

Let 𝔤 = F_{3,2}, choose bases B for 𝔤 and B^{∗} for 𝔤^{∗} just like in Construction 2. Define a linear map A : 𝔤_{2}^{∗} ⟶ 𝔤_{1}^{∗} by AZ_{i}^{∗} = Y_{i}^{∗}, i = 1, 2, 3. From Example 4.3.14. in [1], we know that Ω := {r̃ω + t̃Aω| r̃ > 0, t̃∈ R, ω ∈ 𝔤_{2}^{∗} with ∥ω∥ = 1} gives a cross-section for O_{Max}. For each r̃ω + t̃Aω ∈ 𝔤_{Max}^{∗}, there exists A ∈ Aut(𝔤) such that A^{∗}Z_{3}^{∗} = r̃ω + t̃Aω. Since X^{∗} = {(ρ_{t}^{σ} ∘ τ_{r} ∘ (σ ⊗ σ))^{∗}| (t, r, σ) ∈ R × R^{+} × Σ_{0}} gives a cross-section for Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(Z_{3}^{∗}), there exists unique (t, r, σ) ∈ R × R^{+} × Σ_{0} such that O_{(ρ_{t}^{σ}τ_{r}(σ⊗σ))^{∗}Z_{3}^{∗}} = O_{A^{∗}Z_{3}^{∗}}. So we have O_{(ρ_{t}^{σ}τ_{r}(σ⊗σ))^{∗}Z_{3}^{∗}} = O_{r̃ω+t̃Aω}, and hence there exists X ∈ 𝔤 such that

r̃ω + t̃Aω = Ad_{exp(X)}^{∗}(ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}Z_{3}^{∗}
= Ad_{exp(X)}^{∗}(ρ_{t}^{σ}τ_{r})^{∗}(σ^{∗}Z_{3}^{∗})
= Ad_{exp(X)}^{∗}ρ_{t}^{σ^{∗}}(rσ^{∗}Z_{3}^{∗})
= Ad_{exp(X)}^{∗}(rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗})
= (rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗}) + (ad_{X}^{∗}(rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗})
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + (ad_{X}^{∗} ∘ (σ ⊗ σ)^{∗})(rZ_{3}^{∗} + trY_{3}^{∗})
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + ((σ ⊗ σ)^{∗} ∘ ad_{(σ⊗σ)^{−1}X}^{∗})(rZ_{3}^{∗} + trY_{3}^{∗})
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + (σ ⊗ σ)^{∗}ad_{(σ⊗σ)^{−1}X}^{∗}(rZ_{3}^{∗})
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + r(σ ⊗ σ)^{∗}ad_{(σ⊗σ)^{−1}X}^{∗}Z_{3}^{∗}
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + r(σ ⊗ σ)^{∗}(c_{1}Y_{1}^{∗} + c_{2}Y_{2}^{∗})
= rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + rc_{1}σ^{∗}Y_{1}^{∗} + rc_{2}σ^{∗}Y_{2}^{∗}

for some constants c_{1}, c_{2} ∈ R. The seventh equality holds since we proved in Theorem 2.1 the identity AAd_{x} = Ad_{Ax}A, ∀x ∈ G, A ∈ Aut(𝔤) (and Aut(G)), it follows Aad_{X} = ad_{AX}A. By replacing A by A^{−1}, we get Aad_{A^{−1}X} = ad_{X}A, and hence A^{∗}ad_{A^{−1}X}^{∗} = ad_{X}^{∗}A^{∗}. Thus we have r̃ω + t̃Aω = rσ^{∗}Z_{3}^{∗} + trσ^{∗}Y_{3}^{∗} + rc_{1}σ^{∗}Y_{1}^{∗} + rc_{2}σ^{∗}Y_{2}^{∗}. it contains two conditions r̃ω = rσ^{∗}Z_{3}^{∗} ∈ 𝔤_{2}^{∗} and t̃Aω = trσ^{∗}Y_{3}^{∗} + rc_{1}σ^{∗}Y_{1}^{∗} + rc_{2}σ^{∗}Y_{2}^{∗} ∈ 𝔤_{1}^{∗}. The first condition follows ω = σ^{∗}Z_{3}^{∗} and \underline{r̃ = r}. Since ω = σ^{∗}Z_{3}^{∗} and A : Z_{i}^{∗} ⟼ Y_{i}^{∗} is linear, we get Aω = σ^{∗}Y_{3}^{∗}. So the second condition follows t̃σ^{∗}Y_{3}^{∗} = trσ^{∗}Y_{3}^{∗} + rc_{1}σ^{∗}Y_{1}^{∗} + rc_{2}σ^{∗}Y_{2}^{∗}. Since {σ^{∗}Y_{1}^{∗}, σ^{∗}Y_{2}^{∗}, σ^{∗}Y_{3}^{∗}} are linear independent, we get \underline{t̃= tr} and c_{1} = c_{2} = 0. Hence there is a one-to-one correspondence between cross-section Ω of O_{Max} and X^{∗} of Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(Z_{3}^{∗}) with relations r̃ = r and t̃= tr.

From Example 4.3.14. in [1], we know r̃^{3}sinφdt̃dr̃dφdθ is the Plancherel measure on O_{Max} viewed on cross-section Ω. Since 𝔤_{2}^{∗} is algebraically isomorphic to R^{3}, give the rotation group on 𝔤_{2}^{∗} the measure dσ which is area on the unit sphere. Thus dσ = sinφdφdθ. This is indeed invariant under all rotations centered at the origin. Since t̃= tr, it follows dt̃= rdt, and hence

r̃^{3}sinφdt̃dr̃dφdθ = r^{4}sinφdtdrdφdθ
= r^{4}dtdr(sinφdφdθ)
= r^{4}dtdrdσ

Then

dμ(t, r, σ) := r^{4}dtdrdσ

is the Plancherel measure on Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(Z_{3}^{∗}) viewed on cross-section X^{∗}. Next we verify the relative invariance of μ. For every ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}), ρ_{t}^{σ}τ_{r}(σ ⊗ σ) ∈ X, we know already

Ad_{G}^{∗}(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0})ρ_{t}^{σ}τ_{r}(σ ⊗ σ))^{∗}Z_{3}^{∗} = Ad_{G}^{∗}(ρ_{\frac{t}{\sqrt{r_{0}}} +t_{0}cos^{2}ψ}^{σ_{0}σ}τ_{r_{0}r}(σ_{0}σ ⊗ σ_{0}σ))^{∗}Z_{3}^{∗} It follows that

dμ((t_{0}, r_{0}, σ_{0})(t, r, σ)) = dμ((\frac{t}{\sqrt{r_{0}}} + t_{0}cos^{2}ψ, r_{0}r, σ_{0}σ))
= (r_{0}r)^{4}d(\frac{t}{\sqrt{r_{0}}} + t_{0}cos^{2}ψ)d(r_{0}r)d(σ_{0}σ)
= r_{0}^{4}r^{4}(\frac{1}{\sqrt{r_{0}}} dt)(r_{0}dr)dσ
= r0^{\frac{9}{2}} (r^{4}dtdrdσ)
= | det(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}))^{∗}|dμ(t, r, σ)

Note that the equality d(σ_{0}σ) = dσ holds because dσ = sinφdφdθ is rotation-invariant. Thus we know

μ((ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}))^{∗}E) = | det(ρ_{t_{0}}^{σ_{0}}τ_{r_{0}}(σ_{0} ⊗ σ_{0}))^{∗}|μ(E)

for every measurable subset E. Therefore Plancherel measure μ = r^{4}dtdrdσ on the double coset space Ad_{G}^{∗}\Aut^{∗}(𝔤)/Stab(Z_{3}^{∗}) is relatively invariant for 𝔤 = F_{3,2}.

6. General Properties of Free 2-step Nilpotent Lie Algebras on n Generators

For any arbitrary nilpotent Lie algebra 𝔤, let 𝔤^{(2)} = [𝔤, 𝔤], and 𝔤^{(3)} = [𝔤^{(2)}, 𝔤]. If 𝔤^{(3)} = \{0\}, dim 𝔤^{(2)} = \frac{n(n−1)}{2}, and dim 𝔤 = n + \frac{n(n−1)}{2}, then 𝔤 is algebraically isomorphic to F_{n,2}, free nilpotent Lie algebra of two steps on n generators. The reason is the following: Since dim(𝔤/𝔤^{(2)}) = dim 𝔤 − dim 𝔤^{(2)} = n, pick a basis {Ȳ_{1}, Ȳ_{2}, ⋯, Ȳ_{n}} for 𝔤/𝔤^{(2)}. Choose Y_{i} ∈ Ȳ_{i}, ∀i = 1, ⋯, n. Let B_{2} := \{Z_{k}| k = 1, 2, ⋯, \frac{n(n−1)}{2}\} = \{[Y_{i}, Y_{j}]| 1 ≤ i < j ≤ n\}. Then B_{2} is a basis of 𝔤^{(2)} since they span a space which is given as \frac{n(n−1)}{2} dimension. And hence B_{2} ∪ {Y_{1}, Y_{2}, ⋯, Y_{n}} is a basis for 𝔤. Therefore 𝔤 is isomorphic to F_{n,2}. The above argument suggests all generators play the same role or they look similar with each other in some sense. So we expect their brackets look similar too. Let’s explain this explicitly. Let 𝔤 = F_{n,2}, and 𝔤_{2} = [𝔤, 𝔤]. Give any arbitrary norm on 𝔤, let 𝔤_{1} be the orthogonal complement of 𝔤_{2}. Pick a set of n orthonormal generators {Y_{i}|i = 1, ⋯, n} in 𝔤_{1}. Then a question is raised: does the bracket preserve the norm and the orthogonality? In fact, it is not necessary. But we may adjust the norm in order to get a positive answer. The following lemma gives the answer in the affirmative. First, let’s give a definition. A norm is said to \underline{respect} the Lie bracket if given any orthogonal map σ_{1} of 𝔤_{1} there exists an orthogonal map σ_{2} of 𝔤_{2} such that σ_{1} ⊗ σ_{2} defines an automorphism of 𝔤.

Lemma 6.1. Let 𝔤 = F_{n,2} (n > 1), there exist norms on 𝔤 respecting the bracket. In other words, for any set of n orthonormal generators in the orthogonal complement of [𝔤, 𝔤], all nontrivial brackets of any two generators are orthonormal.

Proof. Give any arbitrary norm ∥·∥_{1} to 𝔤. Let 𝔤_{2} = [𝔤, 𝔤], and let 𝔤_{1} be the orthogonal complement of 𝔤_{2}. Pick a set {Y_{1}, Y_{2}, ⋯, Y_{n}} of orthonormal generators in 𝔤_{1}, and let {Z_{1}, Z_{2}, ⋯, Z_{m}} = {[Y_{i}, Y_{j}]| 1 ≤ i < j ≤ n} where m = \frac{n(n−1)}{2}. Then 𝔤_{1} = Sp_{R}\{Y_{1}, ⋯, Y_{n}\} and 𝔤_{2} = Sp_{R}\{Z_{1}, ⋯, Z_{m}\}. Keep the given norm ∥·∥_{1} in 𝔤_{1}, and use Euclidean norm ∥·∥_{E} in 𝔤_{2} respecting {Z_{1}, ⋯, Z_{m}}. This means the corresponding inner product ⟨ , ⟩_{E} defined by Euclidean norm ∥·∥_{E} makes ⟨Z_{i}, Z_{j}⟩_{E} = δ_{ij}, ∀i, j = 1, 2, ⋯, n. Then define the map ∥·∥ : 𝔤 ⟶ \overline{R^{+}} by

∥Y + Z∥ = \sqrt{∥Y ∥_{1}^{2} + ∥Z∥_{E}^{2}}

∀Y ∈ 𝔤_{1} and Z ∈ 𝔤_{2}. We claim that ∥·∥ is a norm on 𝔤.

(1) For every W ∈ 𝔤 = 𝔤_{1} ⊕ 𝔤_{2}, let W = Y + Z, then ∥cW ∥ = ∥c(Y + Z)∥ = ∥cY + cZ∥ = \sqrt{∥cY ∥_{1}^{2} + ∥cZ∥_{E}^{2}} = |c|∥Y + Z∥ = |c|∥W ∥, ∀c ∈ R.

(2) ∥W + W′∥^{2} = ∥(Y + Z) + (Y′ + Z′)∥^{2}
= ∥(Y + Y′) + (Z + Z′)∥^{2}
= ∥Y + Y′∥_{1}^{2} + ∥Z + Z′∥_{E}^{2}
≤ (∥Y ∥_{1} + ∥Y′∥_{1})^{2} + (∥Z∥_{E} + ∥Z′∥_{E})^{2}
= ∥Y ∥_{1}^{2} + ∥Y′∥_{1}^{2} + ∥Z∥_{E}^{2} + ∥Z′∥_{E}^{2} + 2(∥Y ∥_{1}∥Y′∥_{1} + ∥Z∥_{E}∥Z′∥_{E})
≤ ∥Y ∥_{1}^{2} + ∥Y′∥_{1}^{2} + ∥Z∥_{E}^{2} + ∥Z′∥_{E}^{2} + 2\sqrt{∥Y ∥_{1}^{2} + ∥Z∥_{E}^{2}}\sqrt{∥Y′∥_{E}^{2} + ∥Z′∥_{E}^{2}}
= (\sqrt{∥Y ∥_{1}^{2} + ∥Z∥_{E}^{2}} + \sqrt{∥Y′∥_{E}^{2} + ∥Z′∥_{E}^{2}})^{2}
= (∥W ∥ + ∥W′∥)^{2},

so we get ∥W + W′∥ ≤ ∥W ∥ + ∥W′∥, ∀W, W′ ∈ 𝔤.

(3) ∥W ∥ = 0 ⇔ \sqrt{∥Y ∥_{1}^{2} + ∥Z∥_{E}^{2}} = 0 ⇔ Y = 0 = Z ⇔ W = 0, ∀W = Y + Z ∈ 𝔤 = 𝔤_{1} ⊕ 𝔤_{2}. By (1), (2), and (3) we know ∥ · ∥ : 𝔤 ⟶ \overline{R^{+}} is a norm on 𝔤. Hence {Y_{1}, ⋯, Y_{n}, Z_{1}, ⋯ Z_{m}} is an orthonormal basis in (𝔤, ∥ · ∥) since {Y_{1}, ⋯, Y_{n}} and {Z_{1}, ⋯ Z_{m}} are orthonormal bases in (𝔤_{1}, ∥ · ∥_{1}) and (𝔤_{2}, ∥ · ∥_{E}) respectively. Now choose any other orthonormal generators Y′_{1}, Y′_{2}, ⋯, Y′_{n} in (𝔤_{1}, ∥ · ∥), let {Z′_{k}| k = 1, ⋯, m} = {[Y′_{i}, Y′_{j}]| 1 ≤ i < j ≤ n}. We claim that Z′_{1}, Z′_{2}, ⋯, Z′_{m} are orthonormal in (𝔤, ∥ · ∥). It is sufficient to show that

(1) [Y′_{1}, Y′_{2}] is perpendicular to [Y′_{3}, Y′_{4}],

(2) [Y′_{1}, Y′_{2}] is perpendicular to [Y′_{2}, Y′_{3}],

(3) [Y′_{1}, Y′_{2}] is a unit vector. Since 𝔤_{1} is spanned by {Y_{1}, ⋯, Y_{n}}, let

Y′_{1} = a_{1}Y_{1} + a_{2}Y_{2} + ⋯ + a_{n}Y_{n}
Y′_{2} = b_{1}Y_{1} + b_{2}Y_{2} + ⋯ + b_{n}Y_{n}
Y′_{3} = c_{1}Y_{1} + c_{2}Y_{2} + ⋯ + c_{n}Y_{n}
Y′_{4} = d_{1}Y_{1} + d_{2}Y_{2} + ⋯ + d_{n}Y_{n}

, ∀a_{i}, b_{i}, c_{i}, d_{i} ∈ R, i = 1, 2, ⋯, n. By assumption Y′_{1}, Y′_{2}, Y′_{3}, Y′_{4} are orthonormal. There are some helpful identities: ∑_{j=1}^{n} a_{j}b_{j} = 0, ∑_{j=1}^{n} a_{j}c_{j} = 0, ∑_{j=1}^{n} a_{j}d_{j} = 0, ∑_{j=1}^{n} b_{j}c_{j} = 0, ∑_{j=1}^{n} b_{j}d_{j} = 0, ∑_{j=1}^{n} a_{j}^{2} = 1, and ∑_{j=1}^{n} b_{j}^{2} = 1.

(1) We first verify that the inner product of [Y′_{1}, Y′_{2}] and [Y′_{3}, Y′_{4}] is zero. ⟨[Y′_{1}, Y′_{2}], [Y′_{3}, Y′_{4}]⟩ = ⟨[∑_{i=1}^{n} a_{i}Y_{i}, ∑_{j=1}^{n} b_{j}Y_{j}] , [∑_{i=1}^{n} c_{i}Y_{i}, ∑_{j=1}^{n} d_{j}Y_{j}]⟩ = ⟨∑_{i≠j} a_{i}b_{j}[Y_{i}, Y_{j}], ∑_{i≠j} c_{i}d_{j}[Y_{i}, Y_{j}]⟩

= ⟨∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})[Y_{i}, Y_{j}], ∑_{i<j}(c_{i}d_{j} − d_{i}c_{j})[Y_{i}, Y_{j}]⟩

= ∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})(c_{i}d_{j} − d_{i}c_{j})

= ∑_{i=1}^{n−1} ∑_{j=i+1}^{n}(a_{i}c_{i}b_{j}d_{j} + b_{i}d_{i}a_{j}c_{j} − b_{i}c_{i}a_{j}d_{j} − a_{i}d_{i}b_{j}c_{j})

= ∑_{i=1}^{n−1} {a_{i}c_{i} ∑_{j=i+1}^{n} b_{j}d_{j} + b_{i}d_{i} ∑_{j=i+1}^{n} a_{j}c_{j} − b_{i}c_{i} ∑_{j=i+1}^{n} a_{j}d_{j} − a_{i}d_{i} ∑_{j=i+1}^{n} b_{j}c_{j}}

+a_{n}c_{n} · 0 + b_{n}d_{n} · 0 − b_{n}c_{n} · 0 − a_{n}d_{n} · 0

= ∑_{i=1}^{n−1} {a_{i}c_{i}(− ∑_{j=1}^{i} b_{j}d_{j}) + b_{i}d_{i}(− ∑_{j=1}^{i} a_{j}c_{j}) − b_{i}c_{i}(− ∑_{j=1}^{i} a_{j}d_{j}) −

a_{i}d_{i}(− ∑_{j=1}^{i} b_{j}c_{j})}

+a_{n}c_{n}(−∑_{j=1}^{n} b_{j}d_{j})+b_{n}d_{n}(− ∑_{j=1}^{n} a_{j}c_{j})−b_{n}c_{n}(− ∑_{j=1}^{n} a_{j}d_{j})−a_{n}d_{n}(− ∑_{j=1}^{n} b_{j}c_{j})

= {a_{1}c_{1}(−b_{1}d_{1}) + b_{1}d_{1}(−a_{1}c_{1}) − b_{1}c_{1}(−a_{1}d_{1}) − a_{1}d_{1}(−b_{1}c_{1})}

+ ∑_{i=2}^{n} {a_{i}c_{i}(− ∑_{j=1}^{i} b_{j}d_{j}) + b_{i}d_{i}(− ∑_{j=1}^{i} a_{j}c_{j}) − b_{i}c_{i}(− ∑_{j=1}^{i} a_{j}d_{j}) −

a_{i}d_{i}(− ∑_{j=1}^{i} b_{j}c_{j})}

= {0} + ∑_{i=2}^{n} {a_{i}c_{i}(− ∑_{j=1}^{i−1} b_{j}d_{j}) + b_{i}d_{i}(− ∑_{j=1}^{i−1} a_{j}c_{j}) − b_{i}c_{i}(− ∑_{j=1}^{i−1} a_{j}d_{j}) −

a_{i}d_{i}(− ∑_{j=1}^{i−1} b_{j}c_{j})}

+ ∑_{i=2}^{n} {a_{i}c_{i}(−b_{i}d_{i}) + b_{i}d_{i}(−a_{i}c_{i}) − b_{i}c_{i}(−a_{i}d_{i}) − a_{i}d_{i}(−b_{i}c_{i})}

= − ∑_{i=2}^{n} ∑_{j=1}^{i−1} {a_{i}c_{i}b_{j}d_{j} + b_{i}d_{i}a_{j}c_{j} − b_{i}c_{i}a_{j}d_{j} − a_{i}d_{i}b_{j}c_{j}}

= − ∑_{j=1}^{n−1} ∑_{i=j+1}^{n} {a_{i}c_{i}b_{j}d_{j} + b_{i}d_{i}a_{j}c_{j} − b_{i}c_{i}a_{j}d_{j} − a_{i}d_{i}b_{j}c_{j}}

= − ∑_{i=1}^{n−1} ∑_{j=i+1}^{n} {a_{j}c_{j}b_{i}d_{i} + b_{j}d_{j}a_{i}c_{i} − b_{j}c_{j}a_{i}d_{i} − a_{j}d_{j}b_{i}c_{i}}

= − ∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})(c_{i}d_{j} − d_{i}c_{j})

= −⟨[Y′_{1}, Y′_{2}], [Y′_{3}, Y′_{4}]⟩

Thus, we get ⟨[Y′_{1}, Y′_{2}], [Y′_{3}, Y′_{4}]⟩ = −⟨[Y′_{1}, Y′_{2}], [Y′_{3}, Y′_{4}]⟩, and hence

⟨[Y′_{1}, Y′_{2}], [Y′_{3}, Y′_{4}]⟩ = 0.

(2) We now verify that the inner product of [Y′_{1}, Y′_{2}] and [Y′_{2}, Y′_{3}] is zero. Actually

we replace Y′_{4} by −Y′_{2} in (1), in other words, replace d_{j} by −b_{j} in (1), then we get

⟨[Y′_{1}, Y′_{2}], [Y′_{2}, Y′_{3}]⟩ = ⟨[Y′_{1}, Y′_{2}], [Y′_{3}, −Y′_{2}]⟩ = − ∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})(c_{i}(−b_{j}) − (−b_{i})c_{j})

= − ∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})(b_{i}c_{j} − c_{i}b_{j}) = −⟨[Y′_{1}, Y′_{2}], [Y′_{2}, Y′_{3}]⟩

So we have ⟨[Y′_{1}, Y′_{2}], [Y′_{2}, Y′_{3}]⟩ = −⟨[Y′_{1}, Y′_{2}], [Y′_{2}, Y′_{3}]⟩, and hence ⟨[Y′_{1}, Y′_{2}], [Y′_{2}, Y′_{3}]⟩ = 0.

(3) We now verify that ∥[Y′_{1}, Y′_{2}]∥^{2} = 1. ∥[Y′_{1}, Y′_{2}]∥^{2} = ∥[∑_{i=1}^{n} a_{i}Y_{i}, ∑ b_{j}Y_{j}]∥^{2} = ∥∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})[Y_{i}, Y_{j}]∥^{2} = ∑_{i<j}(a_{i}b_{j} − b_{i}a_{j})^{2} = ∑_{i=1}^{n−1} ∑_{j=i+1}^{n}(a_{i}^{2}b_{j}^{2} + b_{i}^{2}a_{j}^{2} − 2a_{i}b_{i}a_{j}b_{j}) = ∑_{i=1}^{n−1} {a_{i}^{2} ∑_{j=i+1}^{n} b_{j}^{2} + b_{i}^{2} ∑_{j=i+1}^{n} a_{j}^{2} − 2a_{i}b_{i} ∑_{j=i+1}^{n} a_{j}b_{j}} + {a_{n}^{2} · 0 + b_{n}^{2} · 0 − 2a_{n}b_{n} · 0} = ∑_{i=1}^{n−1} {a_{i}^{2}(1 − ∑_{j=1}^{i} b_{j}^{2}) + b_{i}^{2}(1 − ∑_{j=1}^{i} a_{j}^{2}) − 2a_{i}b_{i}(− ∑_{j=1}^{i} a_{j}b_{j})} + {a_{n}^{2} (1 − ∑_{j=1}^{n} b_{j}^{2}) + b_{n}^{2} (1 − ∑_{j=1}^{n} a_{j}^{2}) − 2a_{n}b_{n} (− ∑_{j=1}^{n} a_{j}b_{j})} = {a_{1}^{2} (1 − b_{1}^{2}) + b_{1}^{2} (1 − a_{1}^{2}) − 2a_{1}b_{1} (−a_{1}b_{1})} + ∑_{i=2}^{n} {a_{i}^{2}(1 − ∑_{j=1}^{i} b_{j}^{2}) + b_{i}^{2}(1 − ∑_{j=1}^{i} a_{j}^{2}) − 2a_{i}b_{i}(− ∑_{j=1}^{i} a_{j}b_{j})} = ∑_{i=1}^{n} a_{i}^{2} + ∑_{i=1}^{n} b_{i}^{2} − ∑_{i=2}^{n} {a_{i}^{2} ∑_{j=1}^{i} b_{j}^{2} + b_{i}^{2} ∑_{j=1}^{i} a_{j}^{2} − 2a_{i}b_{i} ∑_{j=1}^{i} a_{j}b_{j}} = 1 + 1 − ∑_{i=2}^{n} ∑_{j=1}^{i} {a_{i}^{2}b_{j}^{2} + b_{i}^{2}a_{j}^{2} − 2a_{i}b_{i}a_{j}b_{j}} = 2 − ∑_{i=2}^{n} ∑_{j=1}^{i−1} {a_{i}^{2}b_{j}^{2} + b_{i}^{2}a_{j}^{2} − 2a_{i}b_{i}a_{j}b_{j}} = 2 − ∑_{j=1}^{n−1} ∑_{i=j+1}^{n} {a_{i}^{2}b_{j}^{2} + b_{i}^{2}a_{j}^{2} − 2a_{i}b_{i}a_{j}b_{j}} = 2 − ∑_{i=1}^{n−1} ∑_{j=i+1}^{n} {a_{j}^{2}b_{i}^{2} + b_{j}^{2}a_{i}^{2} − 2a_{j}b_{j}a_{i}b_{i}} = 2 − ∑_{i<j} (a_{i}b_{j} − b_{i}a_{j})^{2} = 2 − ∥[Y′_{1}, Y′_{2}]∥^{2} Thus we get ∥[Y′_{1}, Y′_{2}]∥^{2} = 2 − ∥[Y′_{1}, Y′_{2}]∥^{2}, and hence ∥[Y′_{1}, Y′_{2}]∥^{2} = 1. We verified (1), (2), and (3), then this proved the claim. Therefore for any orthonormal generators {Y′_{1}, Y′_{2}, ⋯, Y′_{n}}, their brackets {[Y′_{i}, Y′_{j}] : 1 ≤ i < j ≤ n} are also orthonormal with respect to the norm ∥ · ∥.

Motivation. Let 𝔤 = F_{n,2}. The previous lemma guarantees that there always exist norms on 𝔤 respecting the bracket. So now we give a norm on 𝔤 respecting the bracket. Let 𝔤^{∗} be the dual space of 𝔤, and use the operator norm on 𝔤^{∗}. Choose an orthonormal basis B for 𝔤, let B^{∗} be its dual basis in 𝔤^{∗}. Then the question is that whether B^{∗} is orthonormal or not? The following lemma answers this question saying ’yes’. And more than that is the orthonormality of both bases are equivalent for any finite-dimensional vector space 𝔤 and its dual space 𝔤^{∗}.

Lemma 6.2. If 𝔤 is an n-dimensional real vector space, let 𝔤^{∗} be its dual space. Give any arbitrary norm on 𝔤, and use the corresponding operator norm on 𝔤^{∗}. If B = {X_{1}, X_{2}, ⋯, X_{n}} is a basis of 𝔤, let B^{∗} = {X_{1}^{∗}, X_{2}^{∗}, ⋯, X_{n}^{∗}} be its dual basis. Then the orthonormality of B and B^{∗} in 𝔤 and 𝔤^{∗} are equivalent.

Proof. We first claim that if B is an orthonormal basis of 𝔤, then so is B^{∗} for 𝔤^{∗}. Let ∥ · ∥ be any given norm on 𝔤, and let ∥ · ∥_{op} be the corresponding operator norm on 𝔤^{∗}. Since B = {X_{1}, ⋯, X_{n}} is a basis of 𝔤, for each X ∈ 𝔤, let X = ∑_{j=1}^{n} α_{j}X_{j}. Since B is orthonormal, it follows ∥X∥ = \sqrt{∑_{j=1}^{n} α_{j}^{2}}. Since B^{∗} is the dual basis of B, it means X_{i}^{∗}(X_{j}) = δ_{ij}, ∀i, j = 1, 2, ⋯ n. So X_{i}^{∗}(X) = X_{i}^{∗}(∑_{j=1}^{n} α_{j}X_{j}) = ∑_{j=1}^{n} α_{j}X_{i}^{∗}(X_{j}) = α_{i}, ∀i = 1, 2, ⋯, n. Then ∥X_{i}^{∗}∥_{op} = sup_{∥X∥=1}|X_{i}^{∗}(X)| = sup_{\sqrt{∑_{j=1}^{n} α_{j}^{2}}=1}|α_{i}| = 1, ∀i = 1, 2, ⋯, n. Thus, the orthonormality of B implies that the unit length of every basis element of B^{∗}. We now want to verify the orthogonality of B^{∗}. It suffices to show that ∥X_{i}^{∗} + X_{j}^{∗}∥_{op} = ∥X_{i}^{∗} − X_{j}^{∗}∥_{op}, ∀1 ≤ i ≠ j ≤ n. If this is done, it means that the inner product of any distinct basis elements is zero. In other words, all basis elements of B^{∗} are mutually perpendicular.

For every X ∈ 𝔤, let X = ∑_{k=1}^{n} α_{k}X_{k}. Then we have ∥X_{i}^{∗} + X_{j}^{∗}∥_{op} = sup_{∥X∥=1}|(X_{i}^{∗} + X_{j}^{∗})(X)| = sup_{\sqrt{∑_{k=1}^{n} α_{k}^{2}}=1}|α_{i} + α_{j}| = \sqrt{2} when α_{i} = α_{j} =−^{+} \frac{1}{\sqrt{2}} and other α_{k}’s are zero. Meanwhile, we also have ∥X_{i}^{∗} − X_{j}^{∗}∥_{op} = sup_{∥X∥=1}|(X_{i}^{∗} − X_{j}^{∗})(X)| = sup_{\sqrt{∑_{k=1}^{n} α_{k}^{2}}=1}|α_{i} − α_{j}| = \sqrt{2} when α_{i} = −α_{j} =−^{+} \frac{1}{\sqrt{2}} and other α_{k}’s are zero. So we get ∥X_{i}^{∗} + X_{j}^{∗}∥_{op} = ∥X_{i}^{∗} − X_{j}^{∗}∥_{op} as desired, and hence we know the orthonormality of B in (𝔤, ∥ · ∥) implies the orthonormality of B^{∗} in (𝔤^{∗}, ∥ · ∥_{op}). Next we claim that if B^{∗} is an orthonormal basis of 𝔤^{∗}, then so is B in 𝔤. Let 𝔤^{∗^{∗}} be the dual space of 𝔤^{∗}. For every X ∈ 𝔤, define X^{∗^{∗}}(l) = l(X), ∀l ∈ 𝔤^{∗}. Then the map X ⟼ X^{∗^{∗}} is an isomorphism from 𝔤 onto 𝔤^{∗^{∗}} since 𝔤 is finite-dimensional. Hence we can identify X^{∗^{∗}} with X, ∀X^{∗^{∗}} ∈ 𝔤^{∗^{∗}}. Let B^{∗^{∗}} = {X_{1}^{∗^{∗}}, X_{2}^{∗^{∗}}, ⋯, X_{n}^{∗^{∗}}} be the dual basis of B^{∗} = {X_{1}^{∗}, X_{2}^{∗}, ⋯, X_{n}^{∗}}. This means X_{i}^{∗^{∗}}(X_{j}^{∗}) = δ_{ij}, i.e., X_{j}^{∗}(X_{i}) = δ_{ij}, ∀1 ≤ i, j ≤ n. We denote the operator norm on 𝔤^{∗^{∗}} by ∥ · ∥_{sup}, so

∥X^{∗^{∗}} ∥_{sup} = sup_{∥l∥_{op}=1} |X^{∗^{∗}}(l)| = sup_{∥l∥_{op}=1} |l(X)|

By applying the argument of the first claim we know that the orthonormality of B^{∗} in (𝔤^{∗}, ∥ · ∥_{op}) implies the orthonormality of B^{∗^{∗}} in (𝔤^{∗^{∗}}, ∥ · ∥_{sup}). Since every X^{∗^{∗}} ∈ 𝔤^{∗^{∗}} can be identified with X ∈ 𝔤, it follows that the orthonormality of B^{∗} in (𝔤^{∗}, ∥ · ∥_{op}) implies the orthonormality of B in (𝔤, ∥ · ∥_{sup}). So now the question is that whether the operator norm ∥ · ∥_{sup} on 𝔤 obtained by identifying 𝔤 with 𝔤^{∗^{∗}} is the same as the original given norm ∥ · ∥ on 𝔤 or not? The answer is positive. Since B = {X_{1}, ⋯, X_{n}} is an orthonormal basis of (𝔤, ∥ · ∥), for every X ∈ 𝔤, let X = ∑_{j=1}^{n} α_{j}X_{j}, then ∥X∥ = ∥ ∑_{j=1}^{n} α_{j}X_{j}∥ = \sqrt{∑_{j=1}^{n} α_{j}^{2}}. Meanwhile, we also know B is an orthonormal basis of (𝔤, ∥ · ∥_{sup}) by the identification of 𝔤 with 𝔤^{∗^{∗}}.

So ∥X∥_{sup} = ∥∑_{j=1}^{n} α_{j}X_{j}∥_{sup} = \sqrt{∑_{j=1}^{n} α_{j}^{2}}. Thus, we get ∥X∥ = ∥X∥_{sup}, ∀X ∈ 𝔤. Hence the operator norm on 𝔤 ≅ 𝔤^{∗^{∗}} is the same as arbitrary given norm on 𝔤. By both claims we conclude that the orthonormality of B in (𝔤, ∥ · ∥) and B^{∗} in (𝔤^{∗}, ∥ · ∥_{op}) are equivalent for any arbitrary norm ∥ · ∥ on 𝔤 and the corresponding operator norm ∥ · ∥_{op} on 𝔤^{∗}.

Motivation. For 𝔤 = F_{n,2}, the dimension of its center is m = \frac{n(n−1)}{2}. Let Y_{1}, Y_{2}, ⋯, Y_{n} be n generators of 𝔤, and let {Z_{1}, Z_{2}, ⋯, Z_{m}} := {[Y_{i}, Y_{j}]| 1 ≤ i < j ≤ n}, then B := {Z_{1}, ⋯, Z_{m}, Y_{1}, ⋯, Y_{n}} is a basis of 𝔤. For every A ∈ Aut(𝔤), A has the (m + n) × (m + n) matrix relative to the basis B

[A]_{B} := [ \begin{matrix}Q_{m×m} & R_{m×n} \\ 0_{n×m} & P_{n×n} \end{matrix} ]

and each entry of Q_{m×m} is a polynomial in the variables of entries of P_{n×n}. We next show the relation between determinants of P_{n×n} and Q_{m×m}.

Lemma 6.3. For 𝔤 = F_{n,2}, let Y_{1}, Y_{2}, ⋯, Y_{n} be n generators of 𝔤, and let the space S be spanned by all generators Y_{i}. For each invertible linear operator P on S, define the map Q : [𝔤, 𝔤] ⟶ [𝔤, 𝔤] by Q([Y_{i}, Y_{j}]) = [P Y_{i}, P Y_{j}], ∀1 ≤ i < j ≤ n, then we have

det Q = (det P)^{n−1}

Proof. Let B_{2} := {[Y_{i}, Y_{j}]| 1 ≤ i < j ≤ n}, then B_{2} is a basis of [𝔤, 𝔤]. There are two cases depending upon the positive integer n.

Case 1. Suppose n is even.

For each invertible linear operator P on S, let detQ be viewed as a function of n column vectors P Y_{1}, P Y_{2}, ⋯, P Y_{n}. Define a function F : S^{n} ⟶ R by

F (P Y_{1}, P Y_{2}, ⋯, P Y_{n}) = (det Q)^{\frac{1}{n−1}}

then claim that F = det P. If this is done, then the result det Q = (det P)^{n−1} follows immediately. Actually it is equivalent to show that F is an alternating n-linear form on S which maps (Y_{1}, ⋯, Y_{n}) to 1.

(1) If P is the identity operator on S, by definition Q([Y_{i}, Y_{j}]) = [P Y_{i}, P Y_{j}] = [Y_{i}, Y_{j}]. Then Q is the identity operator on [𝔤, 𝔤], and hence F (Y_{1}, ⋯, Y_{n}) = (det I_{m×m})^{\frac{1}{n−1}} = 1.

(2) Let P′ : S ⟶ S be the operator such that P′Y_{k} = cP Y_{k} for some non-zero constant c, and P′Y_{i} = P Y_{i}, ∀i ≠ k. And let Q′ be the corresponding operator of P′ by definition. Since P′Y_{k} appears in brackets with exactly n-1 other vectors P′Y_{i}, it follows that det Q′ = c^{n−1} det Q, and hence

F (P Y_{1}, ⋯, P Y_{k−1}, cP Y_{k}, P Y_{k+1}, ⋯, P Y_{n}) = (c^{n−1} det Q)^{\frac{1}{n−1}} = c(det Q)^{\frac{1}{n−1}} = cF (P Y_{1}, ⋯, P Y_{n})

(3) Let P′ : S ⟶ S be the operator such that P′Y_{i} = P Y_{j} for some i ≠ j, and P′Y_{s} = P Y_{s} for all s ≠ i. Let Q′ be the corresponding operator of P′, then Q′([Y_{i}, Y_{j}]) = [P′Y_{i}, P′Y_{j}] = [P Y_{j}, P Y_{j}] = 0. So one column of matrix [Q′]_{B_{2}} is zero vector, and hence det Q′ = 0. It follows

F (P Y_{1}, ⋯, \underbrace{P Y_{j}}_{the i \underline{th} term}, ⋯ P Y_{j}, ⋯, P Y_{n}) =

F (P′Y_{1}, ⋯, P′Y_{i}, ⋯ P′Y_{j}, ⋯ P′Y_{n}) = (det Q′)^{\frac{1}{n−1}} = 0

(4) In order to show linearity of F in each variable, we verify F (P Y_{1}, ⋯, P Y_{i−1}, Y + cY′, P Y_{i+1}, ⋯, P Y_{n}) = F (P Y_{1}, ⋯, P Y_{i−1}, Y, P Y_{i+1}, ⋯, P Y_{n}) + cF (P Y_{1}, ⋯ P Y_{i−1}, Y′, P Y_{i+1}, ⋯, P Y_{n}) for every Y, Y′ ∈ S and c ∈ R where Y + cY′, Y, and Y′ are in the i \underline{th} component. Since P is bijective on the space S = R − span{Y_{1}, ⋯, Y_{n}}, every vector in S can be written as ∑_{j=1}^{n} c_{j}P Y_{j}. So it is sufficient to show property (2)

and F (P Y_{1}, ⋯, \underbrace{P Y_{i} + P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n}) =

F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{n}) + F (P Y_{1}, ⋯, \underbrace{P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n}). Since we already

finished the proof of property (2), let’s verify the above identity. Let P′ : S ⟶ S be the operator such that P′Y_{i} = P Y_{i} + P Y_{j} for some i ≠ j, and P′Y_{s} = P Y_{s} for all s ≠ i. And let Q′ be the corresponding operator of P′. Then by definition F (P Y_{1}, ⋯, P Y_{i−1}, P Y_{i} + P Y_{j}, P Y_{i+1}, ⋯, P Y_{n}) = F (P′Y_{1}, ⋯, P′Y_{n}) = (det Q′)^{\frac{1}{n−1}} Since the difference between [Q′]_{B_{2}} and [Q]_{B_{2}} are the columns of the brackets of P Y_{i} + P Y_{j} with other P Y_{k}’s for k ≠ i, we next figure out what det Q′ is by considering each bracket of P Y_{i} + P Y_{j} with P Y_{k} for k ≠ i. For k = 1 the bracket [P Y_{1}, P Y_{i} + P Y_{j}] = [P Y_{1}, P Y_{i}] + [P Y_{1}, P Y_{j}], it follows that det Q′ = det Q′_{1} + det Q′_{2} where [Q′_{1}]_{B_{2}} and [Q′_{2}]_{B_{2}} are obtained by replacing the column Q′([P Y_{1}, P Y_{i} + P Y_{j}]) of [Q′]_{B_{2}} by Q′([P Y_{1}, P Y_{i}]) and Q′([P Y_{1}, P Y_{j}]) respectively. Notice that the vector P Y_{i} + P Y_{j} is on the i \underline{th} column of matrix [P′]_{B_{1}} where basis B_{1} = {Y_{1}, ⋯, Y_{n}}. So [Q′_{2}]_{B_{2}} has two identical columns Q′_{2}([P Y_{1}, P Y_{j}]), and hence det Q′_{2} = 0. Thus we get det Q′ = det Q′_{1}, and note that [Q′_{1}]_{B_{2}} has the same columns as [Q]_{B_{2}} except the columns Q′_{1}([P Y_{i} + P Y_{j}, P Y_{s}]) for s = 2, ⋯, i − 1, i + 1, ⋯, n. For k = 2 the bracket [P Y_{2}, P Y_{i} + P Y_{j}] = [P Y_{2}, P Y_{i}] + [P Y_{2}, P Y_{j}], it follows that det Q′_{1} = det Q′′_{1} + det Q′′_{2} where Q′′_{1} and Q′′_{2} are obtained by replacing the column Q′_{1}([P Y_{2}, P Y_{i} + P Y_{j}]) of [Q′_{1}]_{B_{2}} by Q′_{1}([P Y_{2}, P Y_{i}]) and Q′_{1}([P Y_{2}, P Y_{j}]) respectively. Then [Q′′_{2}]_{B_{2}} has two identical columns Q′′_{2}([P Y_{2}, P Y_{j}]), and hence det Q′′_{2} = 0. Thus we get det Q′ = det Q′_{1} = det Q′′_{1}, and note that [Q′′_{1}]_{B_{2}} has the same columns as [Q]_{B_{2}} except the columns Q′′_{1}([P Y_{i} + P Y_{j}, P Y_{s}]) for s = 3, ⋯, i − 1, i + 1, ⋯, n. Similar procedure applies to k = 3, ⋯, i − 1, i + 1, ⋯, n, then we get det Q′ = det Q′_{1} = det Q′′_{1} = ⋯ = det Q_{1}^{(n−1)}, and [Q_{1}^{(n−1)}]_{B_{2}} has the same columns as [Q]_{B_{2}}. Hence det Q′ = det Q. Thus F (P Y_{1}, ⋯, P Y_{i−1}, P Y_{i} + P Y_{j}, P Y_{i+1}, ⋯, P Y_{n}) = (det Q′)^{\frac{1}{n−1}} = (det Q)^{\frac{1}{n−1}} = F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{j}, ⋯, P Y_{n}) = F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{j}, ⋯, P Y_{n}) + F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{j}, ⋯, P Y_{n}) since F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{j}, ⋯, P Y_{n}) = 0 by property (3). Therefore F is an n-linear function.

(5) We now verify F is alternating.

0 = F (P Y_{1}, ⋯, \underbrace{P Y_{i} + P Y_{j}}_{the i \underline{th} term}, ⋯, \underbrace{P Y_{i} + P Y_{j}}_{the j \underline{th} term}, ⋯, P Y_{n})

= F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{i}, ⋯ P Y_{n}) + F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{j}, ⋯ P Y_{n}) +F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{i}, ⋯ P Y_{n}) + F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{j}, ⋯ P Y_{n}) = F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{j}, ⋯ P Y_{n}) + F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{i}, ⋯ P Y_{n}) It follows that F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{j}, ⋯ P Y_{n}) = −F (P Y_{1}, ⋯, P Y_{j}, ⋯, P Y_{i}, ⋯ P Y_{n}) Hence F is alternating. Thus we know F is an alternating n-linear function which maps (Y_{1}, ⋯, Y_{n}) to 1, so F (P Y_{1}, ⋯, P Y_{n}) = det P. Therefore det Q = (det P)^{n−1}.

Case 2. Suppose n is odd. For each invertible linear operator P on S, define a function F : S^{n} ⟶ R by F (P Y_{1}, ⋯, P Y_{n}) = (det P det Q)^{\frac{1}{n}} where Q is the corresponding operator of P defined as before. We claim that F = det P. If this is done, then we get (det P det Q)^{\frac{1}{n}} = det P, and hence det P det Q = (det P)^{n}. So det Q = (det P)^{n−1}.

(1) If P is the identity operator on S, so is Q. Then F (Y_{1}, ⋯, Y_{n}) = (det I_{n×n} det I_{\frac{n(n−1)}{2} × \frac{n(n−1)}{2}} )^{\frac{1}{n}} = 1.

(2) The same assumption as in the property (2) of the first case, if P′Y_{k} = cP Y_{k} for some index k, we obtain F (P Y_{1}, ⋯ P Y_{k−1}, cP Y_{k}, P Y_{k+1}, ⋯ P Y_{n}) = F (P′Y_{1}, ⋯, P′Y_{n}) = (det P′ det Q′)^{\frac{1}{n}} = ((c det P)(c^{n−1} det Q))^{\frac{1}{n}} = c(det P det Q)^{\frac{1}{n}} = cF (P Y_{1}, ⋯, P Y_{k}, ⋯, P Y_{n})

(3) Let P′ : S ⟶ S be the operator such that P′Y_{i} = P Y_{j} for some i ≠ j, and P′Y_{s} = P Y_{s} for all s ≠ i. Then det P′ = 0 since matrix [P′]_{B_{1}} has two identical column P Y_{j} where B_{1} = {Y_{1}, ⋯, Y_{n}}. Let Q′ be the corresponding operator of P′. Then

F (P Y_{1}, ⋯, \underbrace{P Y_{j}}_{the i \underline{th} term}, ⋯, \underbrace{P Y_{j}}_{the j \underline{th} term}, ⋯ P Y_{n}) = F (P′Y_{1}, ⋯, P′Y_{n}) =

(det P′ det Q′)^{\frac{1}{n}} = 0

(4) In order to show the linearity of F, it suffices to verify property (2) and

F (P Y_{1}, ⋯, \underbrace{P Y_{i} + P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n})

= F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{n}) + F (P Y_{1}, ⋯, \underbrace{P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n})

Let P′ : S ⟶ S be the operator such that P′Y_{i} = P Y_{i} + P Y_{j} for some i ≠ j, and P′Y_{s} = P Y_{s} for all s ≠ i. Then det P′ = det P + det P_{0} where [P_{0}]_{B_{1}} is the matrix after replacing the i \underline{th} column P Y_{i} of [P]_{B_{1}} by P Y_{j}. So det P_{0} = 0, and hence det P′ = det P. Let Q′ be the corresponding operator of P′. The same reason as property (4) of the first case leads to the fact det Q′ = det Q. Hence

F (P Y_{1}, ⋯, \underbrace{P Y_{i} + P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n}) = F (P′Y_{1}, ⋯, P′Y_{i}, ⋯, P′Y_{n})

= (det P′ det Q′)^{\frac{1}{n}} = (det P det Q)^{\frac{1}{n}} = F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{n})

= F (P Y_{1}, ⋯, P Y_{i}, ⋯, P Y_{n}) + F (P Y_{1}, ⋯, \underbrace{P Y_{j}}_{the i \underline{th} term}, ⋯, P Y_{n})

Therefore F is an n-linear function.

(5) The same argument as in the property (5) of the first case leads to the fact that F is alternating. Hence F is an alternating n-linear function which maps (Y_{1}, ⋯, Y_{n}) to 1. So F (P Y_{1}, ⋯, P Y_{n}) = det P, i.e., (det P det Q)^{\frac{1}{n}} = det P. Therefore det Q = (det P)^{n−1}.

Lemma 6.4. For each n × n matrix R and almost all n × n matrix P, there exists a unique n × n skew-symmetric matrix S such that R + SP is an upper-triangular matrix. Proof. Given R = (r_{ij})_{n×n}, P = (p_{ij})_{n×n}, let

S = [ \begin{matrix}0 & −s_{2,1} & −s_{3,1} & ⋯ & −s_{n,1} \\ s_{2,1} & 0 & −s_{3,2} & ⋯ & −s_{n,2} \\ s_{3,1} & s_{3,2} & 0 & ... & ⋮ \\ ⋮ & ⋮ & ... & ... & −s_{n,n−1} \\ s_{n,1} & s_{n,2} & ⋯ & s_{n,n−1} & 0 \end{matrix} ]

Consider the last row of the n × n matrix SP + R, let (SP + R)_{n,i} = 0, ∀i = 1, 2, ⋯, n − 1. Then we have n-1 linear equations with n-1 variables s_{n,1}, s_{n,2}, ⋯, s_{n,n−1}:

s_{n,1}p_{1,1} + s_{n,2}p_{2,1} + ⋯ + s_{n,n−1}p_{n−1,1} + 0 · p_{n,1} = −r_{n,1}
s_{n,1}p_{1,2} + s_{n,2}p_{2,2} + ⋯ + s_{n,n−1}p_{n−1,2} + 0 · p_{n,2} = −r_{n,2}
⋮ ⋮ ⋮
s_{n,1}p_{1,n−1} + s_{n,2}p_{2,n−1} + ⋯ + s_{n,n−1}p_{n−1,n−1} + 0 · p_{n,n−1} = −r_{n,n−1}

Let

P_{n−1} := [ \begin{matrix}p_{1,1} & p_{2,1} & ⋯ & p_{n−1,1} \\ p_{1,2} & p_{2,2} & ⋯ & p_{n−1,2} \\ ⋮ & ⋮ & ... & ⋮ \\ p_{1,n−1} & p_{2,n−1} & ⋯ & p_{n−1,n−1} \end{matrix} ]

If det P_{n−1} ≠ 0, then there exist unique solutions (s_{n,1}, s_{n,2}, ⋯, s_{n,n−1}) with

s_{n,i} = \begin{matrix}det & P_{n−1}^{i} \\ det & P_{n−1} \end{matrix} ∀i = 1, 2, ⋯, n − 1, where each P_{n−1}^{i} is the (n − 1) × (n − 1)

matrix of replacing the i \underline{th} column of P_{n−1} by ( −r_{n,1} ; −r_{n,2} ; ⋮ ; −r_{n,n−1} )

Consider the n-1 \underline{th} row of n × n matrix SP + R, let (SP + R)_{n−1,i} = 0, ∀i = 1, 2, ⋯, n − 2. Then we have n-2 linear equations with n-2 variables s_{n−1,1}, s_{n−1,2}, ⋯, s_{n−1,n−2}:

s_{n−1,1}p_{1,1} + ⋯ + s_{n−1,n−2}p_{n−2,1} + 0 · p_{n−1,1} − s_{n,n−1}p_{n,1} + r_{n−1,1} = 0
s_{n−1,1}p_{1,2} + ⋯ + s_{n−1,n−2}p_{n−2,2} + 0 · p_{n−1,2} − s_{n,n−1}p_{n,2} + r_{n−1,2} = 0
⋮ ⋮ ⋮
s_{n−1,1}p_{1,n−2} + ⋯ + s_{n−1,n−2}p_{n−2,n−2} + 0 · p_{n−1,n−2} − s_{n,n−1}p_{n,n−2} + r_{n−1,n−2} = 0

It follows that

s_{n−1,1}p_{1,1} + ⋯ + s_{n−1,n−2}p_{n−2,1} = s_{n,n−1}p_{n,1} − r_{n−1,1}
s_{n−1,1}p_{1,2} + ⋯ + s_{n−1,n−2}p_{n−2,2} = s_{n,n−1}p_{n,2} − r_{n−1,2}
⋮ ⋮ ⋮
s_{n−1,1}p_{1,n−2} + ⋯ + s_{n−1,n−2}p_{n−2,n−2} = s_{n,n−1}p_{n,n−2} − r_{n−1,n−2}

Note that s_{n,n−1} is a known number from the previous procedure of considering the n \underline{th} row of SP + R. Let

P_{n−2} := [ \begin{matrix}p_{1,1} & p_{2,1} & ⋯ & p_{n−2,1} \\ p_{1,2} & p_{2,2} & ⋯ & p_{n−2,2} \\ ⋮ & ⋮ & ... & ⋮ \\ p_{1,n−2} & p_{2,n−2} & ⋯ & p_{n−2,n−2} \end{matrix} ]

If det P_{n−2} ≠ 0, then there exist unique solutions (s_{n−1,1}, s_{n−1,2}, ⋯, s_{n−1,n−2})

with s_{n−1,i} = \begin{matrix}det & P_{n−2}^{i} \\ det & P_{n−2} \end{matrix} ∀i = 1, 2, ⋯, n − 2 where each P_{n−2}^{i} is the (n − 2) × (n − 2)

matrix of replacing the i \underline{th} column of P_{n−2} by ( s_{n,n−1}p_{n,1} − r_{n−1,1} ; s_{n,n−1}p_{n,2} − r_{n−1,2} ; ⋮ ; s_{n,n−1}p_{n,n−2} − r_{n−1,n−2} )

Consider the n-2 \underline{th} row of n × n matrix SP + R, ⋯ ⋮ Consider the third row of n × n matrix SP + R, let (SP + R)_{3,i} = 0, i = 1, 2. Then we have two linear equations with two variables s_{3,1}, s_{3,2}:

s_{3,1}p_{1,1} + s_{3,2}p_{2,1} + 0 · p_{3,1} − s_{4,3}p_{4,1} − s_{5,3}p_{5,1} − ⋯ − s_{n,3}p_{n,1} + r_{3,1} = 0
s_{3,1}p_{1,2} + s_{3,2}p_{2,2} + 0 · p_{3,2} − s_{4,3}p_{4,2} − s_{5,3}p_{5,2} − ⋯ − s_{n,3}p_{n,2} + r_{3,2} = 0

It follows that

s_{3,1}p_{1,1} + s_{3,2}p_{2,1} = s_{4,3}p_{4,1} + s_{5,3}p_{5,1} + ⋯ + s_{n,3}p_{n,1} − r_{3,1}
s_{3,1}p_{1,2} + s_{3,2}p_{2,2} = s_{4,3}p_{4,2} + s_{5,3}p_{5,2} + ⋯ + s_{n,3}p_{n,2} − r_{3,2}

Note that s_{4,3}, s_{5,3}, ⋯, s_{n,3} are known numbers from the procedures of considering the n, n − 1, ⋯, 5, 4 \underline{th} row of matrix SP + R. Let

P_{2} := [ \begin{matrix}p_{1,1} & p_{2,1} \\ p_{1,2} & p_{2,2} \end{matrix} ]

If det P_{2} ≠ 0, then there exist unique solutions (s_{3,1}, s_{3,2}) with s_{3,i} = \begin{matrix}det & P_{2}^{i} \\ det & P_{2} \end{matrix} , i = 1, 2, where each P_{2}^{i} is the 2 × 2 matrix of replacing the i \underline{th} column of P_{2} by

( \begin{matrix}s_{4,3}p_{4,1} + s_{5,3}p_{5,1} + ⋯ + s_{n,3}p_{n,1} − r_{3,1} \\ s_{4,3}p_{4,2} + s_{5,3}p_{5,2} + ⋯ + s_{n,3}p_{n,2} − r_{3,2} \end{matrix} )

Consider the second row of n × n matrix SP + R, let (SP + R)_{2,1} = 0. Then we have one linear equation with one variable s_{2,1}:

s_{2,1}p_{1,1} + 0 · p_{2,1} − s_{3,2}p_{3,1} − s_{4,2}p_{4,1} − ⋯ − s_{n,2}p_{n,1} + r_{2,1} = 0

It follows that

s_{2,1}p_{1,1} = s_{3,2}p_{3,1} + s_{4,2}p_{4,1} + ⋯ + s_{n,2}p_{n,1} − r_{2,1}

Let 1 × 1 matrix P_{1} := (p_{1,1}). If det P_{1} ≠ 0, then there exists a unique solution s_{2,1} = \frac{1}{p_{1,1}} (s_{3,2}p_{3,1} + s_{4,2}p_{4,1} + ⋯ + s_{n,2}p_{n,1} − r_{2,1}) such that (SP + R)_{2,1} = 0. In summary, if det P_{i} ≠ 0, ∀i = 1, 2, ⋯, n − 1, then there exist unique solutions (s_{2,1}), (s_{3,1}, s_{3,2}), ⋯, (s_{n,1}, s_{n,2}, ⋯, s_{n,n−1}) such that SP + R is an n × n upper-triangular matrix. In other words, for almost all P = (p_{ij})_{n×n} and all R = (r_{ij})_{n×n}, there exists a unique skew-symmetric matrix

S = [ \begin{matrix}0 & −s_{2,1} & −s_{3,1} & ⋯ & −s_{n−1,1} & −s_{n,1} \\ s_{2,1} & 0 & −s_{3,2} & ⋯ & −s_{n−1,2} & −s_{n,2} \\ s_{3,1} & s_{3,2} & 0 & ⋯ & −s_{n−1,3} & −s_{n,3} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ s_{n−1,1} & s_{n−1,2} & s_{n−1,3} & ⋯ & 0 & −s_{n,n−1} \\ s_{n,1} & s_{n,2} & s_{n,3} & ⋯ & s_{n,n−1} & 0 \end{matrix} ]

such that SP + R is an n × n upper-triangular matrix. This proves the lemma.

References

[1] L. Corwin and F. P. Greenleaf, Representations of nilpotent Lie groups and their applications, Cambridge studies in advanced mathematics, 18, Cambridge University Press, 1990.

[2] Raymond C.Fabec, Fundamentals of Infinite Dimensional Representation Theory, 114, Chapman and Hall/CRC, 2000.

[3] Gerald B. Folland, A Course in Abstract Harmonic Analysis, CRC Press, 1995.

[4] S. A. Gaal, Linear Analysis and Representation Theory, Die Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen, 198, Springer-Verlag, 1973.

[5] Kenneth Hoffman and Ray Kunze, Linear Algebra, Prentice-Hall, Inc., 1971.

[6] A. A. Kirillov, Unitary Representations of Nilpotent Lie Groups, Russin Math Surveys 17, 53-104, 1962.

[7] George W. Mackey, The Theory of Unitary Group Representations, Univ. of Chicago Press, 1976.

[8] Garth Warner, Harmonic Analysis on Semi-Simple Lie Groups II, Springer-Verlag, 1972.

Vita

Chin-Te Chu was born on October 21 1965, in Kaohsiung City, Taiwan. He finished his undergraduate studies at National Chiao Tung University in May 1988. He earned a master of science degree in mathematics from Louisiana State University in May 1997. In October 1992 he came to the United States, then pursued graduate studies in mathematics at Louisiana State University in January 1993. He is currently a candidate for the degree of Doctor of Philosophy in mathematics, which will be awarded in December 2000.