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A finite subgroup $G$ of $GL(n,\mathbb{R})$ is conjugate to a subgroup of $O(n)$.

Proof. Let's define $\beta \colon \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ such that $$ \beta(v,w) = \sum_{g\in G} gv \cdot gw $$ where $\cdot$ is the usual dot product. Then $\beta$ is a bilinear symmetric form. Moreover $$\beta(v,v) = \sum_{g\in G} \|gv\|^2 > 0, \quad \forall v\neq \mathbf 0 $$ and $$ \beta(hv,hw) = \sum_{g\in G} ghv \cdot ghw = \sum_{k\in Gh=G} kv \cdot kw = \beta(v,w)$$ Hence $\beta$ is a $G$-invariant inner product and then $G$ is a subgroup of a conjugate of $O(n)$.

I can't see how the sentences '$\beta$ is a $G$-invariant inner product' and '$G$ is a subgroup of a conjugate of $O(n)$' are related.

user364189
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1 Answers1

2

Big Hint

Better notation can make the problem more clear. Let $\langle\cdot,\cdot\rangle_{I}$ denote the standard inner product on $\mathbb{R}^n$, and let $\langle\cdot,\cdot\rangle_{\beta}$ denote the inner product defined by $\beta$. Let $\mathcal{M}_{\beta}$ denote the matrix of the form. Then we know that $\mathcal{M}_{\beta}=P^tP$ for some invertible matrix $P$.

$O(n)$ is usually defined by $O(n)=\{A\in GL(n,\mathbb{R})\,|\, A^tA=I\}$. An equivalent definition is

$$O(n)=\{A\in GL(n,\mathbb{R})\,|\, \langle Av,Aw\rangle_{I}=\langle v,w\rangle_{I}\}.$$

Consider $G'$ defined by

$$G'=\{A\in GL(n,\mathbb{R})\,|\, \langle Av,Aw\rangle_{\mathbb{\beta}}=\langle v,w\rangle_{\beta}\}.$$

Let $A\in O(n)$, and consider $P^{-1}AP\in GL(n,\mathbb{R})$. Then we have

$$\displaystyle\begin{split}\langle(P^{-1}AP)v,P^{-1}AP)w\rangle_{\beta} &=(P^{-1}APv)^{t}P^tP(P^{-1}APw)\\ &=v^tP^tA^t(P^{-1})^tP^tPP^{-1}APw\\ &=v^tP^tA^tAPw\\ &=v^tP^tPw\\ &=\langle v,w\rangle_{\beta}.\end{split}$$

This shows that $P^{-1}AP\in G'$.

Supplementary

  1. Let $\langle\cdot,\cdot\rangle_{\mathcal{F}}$ be a bilinear form on $\mathbb{R}^n$, and let $\mathcal{M}_{\mathcal{F}}$ be the matrix of the form with respect to a basis $(v_1,\ldots,v_n)$. Then $\langle v,w\rangle_{\mathcal{F}}=v^t\mathcal{M}_{\mathcal{F}}w$ for $v,w\in\mathbb{R}^n$.
  2. If the basis is changed to $(v'_1,\ldots,v'_n)$, then the matrix of the form is changed to $P^t\mathcal{M}_{\mathcal{F}}P$, where $P$ is the base change matrix with respect to the two basis $(v_1,\ldots,v_n)$ and $(v'_1,\ldots,v'_n)$. Thus, if $\mathcal{M}_{\mathcal{F}}$ represents inner product with respect to some basis, then we know that $\mathcal{M}_{\mathcal{F}}=P^tIP=P^tP$ for some $P$, where $I$ is the identity matrix.
Fei Li
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  • How do you know that $\mathcal{M}_{\beta}=P^tP$ for some invertible matrix $P$? – user364189 Aug 28 '16 at 06:57
  • You have proved that the bilinear form $\beta$ is both symmetric and positive definite. Then there exists an orthonormal basis of $\beta$ (namely, a basis $(v_1,\ldots,v_n)$ such that $\langle v_i,v_j\rangle_{\beta}=\delta_{ij}$), in a similar fashion to the spectral theorem for matrices. This means exactly that in terms of the orthonormal basis $(v_1,\ldots,v_n)$, $\langle\cdot,\cdot\rangle_{\beta}$ is an inner product on $\mathbb{R}^n$. Thus the matrix of $\beta$ in this basis $(v_1,\ldots,v_n)$ would be the identity matrix $I$. – Fei Li Aug 28 '16 at 08:04
  • Now use supplementary 2 to change back to our original default basis we are using. – Fei Li Aug 28 '16 at 08:07
  • Thanks, this makes it clear to me. – user364189 Aug 28 '16 at 10:59
  • Is $\phi:A\in O(n)\mapsto P^{-1}AP\in G'$ an isomorphism ? If not, it seems to me that this shows that $G$ is included in $G'$ which contains the subgroup $\text{Im}(\phi)$ which is conjugate to $O(n)$... but then what about $G$ itself ? – P.Fazioli Jul 02 '23 at 21:04