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Let $V = \mathbb R^n$, and $Q : V\times V\to \mathbb R$ be a symmetric and positive definite bilinear form. I define $SO(n)$ to be $$ SO(n) := \left\{ A \in SL_n(\mathbb R) ~|~ A^T A = I_n \right\}. $$ Fulton and Harris claim, in page 96 of their book on Representation Theory, that the subgroup of $SL_n(\mathbb R)$ preserving $Q$ is $SO(n)$. It isn't clear to me that this is true, and I would like to prove it.

Fixing a basis for $V$, we get that $Q$ is given by a symmetric, positive definite matrix $M$ and we wish to prove that

$$ SO(n) \cong \left\{ A \in SL_n(\mathbb R) ~|~ A^T M A = M \right\}. $$

We may assume $M$ is diagonal since every real symmetric matrix is diagonalizable.

Here's what I have so far:

Write $A = (a_{ij})$ and $M = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$ with $\lambda_i > 0$ for all $1 \leq i \leq n.$ Then $A^TMA = M$ gives us the following system of equations:

$$\sum_{i} \left(A^T\right)_{si} \left(MA\right)_{it} = \left(M\right)_{st}$$ $$ \iff \sum_i a_{is} \lambda_i a_{it} = \lambda_s \delta_{st}, $$ for all $1 \leq s,t \leq n$. So the entries $a_{ij}$ of $A$ satisfy the equations we obtain above.

I would like to construct a map $\varphi$ which takes any such $A$ to $\varphi(A) \in SO(n)$, i.e whose entries $\tilde a_{ij}$ satisfy $$ \sum_i \tilde a_{is} \tilde a_{it} = \delta_{st}, $$ for all $1\leq s,t \leq n$, and then eventually argue that $\varphi$ is an isomorphism. I don't see how I can construct such a $\varphi$, any advice?

ABanerjee
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    Let $\sqrt{M}:=\mathrm{diag}(\sqrt{\lambda_1},\dots,\sqrt{\lambda_n})$. What can you say about the map $\varphi(A):=\sqrt{M}A\sqrt{M}^{-1}$ where $A$ is a matrix with $A^{T}MA=M$? – Levent Mar 31 '23 at 10:52
  • I think that $\varphi$ works perfectly @Levent ! It preserves determinants, so $\varphi(A) \in SL_n\mathbb R$, and $\varphi(A)$ satisfies $\varphi(A)^T \varphi(A) = I_n$ for all $A$ satisfying $A^TMA = M$. It is clearly injective, and the inverse is easy to write down as well - so we have the desired isomorphism! Thank you very much! – ABanerjee Mar 31 '23 at 11:07
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    I'd suggest you to learn about the square roots of positive definite matrices, see https://math.stackexchange.com/questions/349721/square-root-of-positive-definite-matrix . It is a very useful construction when one studies the problems such as yours. – Levent Mar 31 '23 at 11:15
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    That's good to know, thank you for the advice :) – ABanerjee Mar 31 '23 at 11:21
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    “...the subgroup of $SL_n(\mathbb R)$ preserving $Q$ is $SO(n)$”. This does not sound right. I guess the authors meant the subgroup is isomorphic to $SO(n)$ rather than identical to $SO(n)$. – user1551 Mar 31 '23 at 12:18
  • I wholeheartedly agree @user1551, although the authors do not make this clarification in their book as far as I know. – ABanerjee Mar 31 '23 at 12:52

1 Answers1

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Thanks to Levent for the map $\varphi$.

Let $G = \left\{ A \in SL_n\mathbb R ~|~ A^TMA = M \right\},$ and let $\sqrt M = \textrm{diag}\left(\sqrt \lambda_1, \ldots, \sqrt \lambda_n\right)$.

We take $\varphi : G \to SO(n)$ to be the map given by $\varphi(A) = \sqrt M A {\sqrt M}^{-1}$. Indeed, we have $$ \varphi(A)^T \varphi(A) = \sqrt{M}^{-1} A^T \sqrt M \sqrt{M}A\sqrt{M}^{-1} = \sqrt M^{-1} M \sqrt M^{-1} = I_n, $$ so $\varphi(A) \in SO(n)$ for all $A \in G$.

Note that $\varphi$ is an injective homomorphism with inverse $\psi(B) = \sqrt M^{-1} B \sqrt M$, and for $B \in SO(n)$ $$ \psi(B)^T M \psi(B) = \sqrt M B^T \sqrt M^{-1} M \sqrt M^{-1} B \sqrt M = \sqrt M B^T B \sqrt M = M. $$

Hence $\varphi : G \to SO(n)$ is an isomorphism.

ABanerjee
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