6
$\begingroup$

I am trying to get my head around representation theory and was hoping for some help. I will write out some text and then ask questions and make comments after.

Let $G$ be a compact connected Lie group with Haar measure $\nu$, and suppose that $\mathbb{R}^n$ is a representation of $G$. We may always assume without loss of generality that the image of $\pi:G\to \mathbb{R}^{n,n}$ lies in $O(n)$, the $n\times n$ orthogonal matrices. Indeed, take any inner product $\langle\cdot,\cdot\rangle$ on $\mathbb{R}^n$. Define the inner product $$[x,y]=\int_G \langle \pi(g)x,\pi(g)y\rangle\, d\nu(g)$$ on $\mathbb{R}^n$. This is $\pi(G)$-invariant since $\nu$ is the Haar measure.

Questions/Comments:

1) My understanding is as follows. Given a representation as above, one can find an inner product for which this representation is invariant. Then the image of this representation lies in the orthogonal group with respect to this inner product. Why does this let us assume $\pi(G)\subset O(n)$? According to Wikipedia $O(n)$ consists of $n\times n$ matrices for which the inverse=transpose. Does $[\pi(g)x,\pi(g)y]=[x,y]$ imply $\pi(g)^{-1}=\pi(g)^T$? If $[\cdot,\cdot]$ is the usual Euclidean inner product then it is clear, but why is it true for some arbitrary inner product?

2) This is sort of a chicken and egg question. Specifically, if $\pi(G)$ is assumed to lie in $O(n)$, then the Euclidean norm on $\mathbb{R}^n$ satisfies $|\pi(g)v|=|v|$ for all $g\in G$ and $v\in\mathbb{R}^n$. So then there is no need to construct such an inner product as above. However, in some notes I am reading, the author assumes $\pi(G)\subset O(n)$, but still chooses a $G$-invariant inner product. What is the point in this? Is the whole point of the above construction not to say "WLOG, $\pi(G)\subset O(n)$ and so we can work with the Euclidean norms/inner products"?

Apologies if these are very basic!

$\endgroup$
8
  • $\begingroup$ If $G$ is compact then $\pi(G)$ is compact in $GL_n$ and $O(n)$ is a maximal compact and the other maximal compacts are conjugate to it, ie. $ M \pi(G) M^{-1}\subset O(n)$. The proof you gave it : if $H$ is a maximal compact then $(x,y) = \int_H <hx,hy>d\mu(h)$ is $H$-invariant inner product and taking a $(.,.)$ orthonormal basis we get an isometry $<.,.> \to (.,.)$ which is our linear map $M : O(n)\to H$. $\endgroup$
    – reuns
    Oct 2, 2019 at 2:09
  • $\begingroup$ If $\pi: G \to \operatorname{GL}(V)$ is a representation of $G$ on an $n$-dimensional vector space, we can fix an isomorphism $f: \mathbb{R}^n \to V$ and define $\rho(g) = f^{-1} \circ \pi(g) \circ f$. Then $\rho: G \to \operatorname{GL}_n$ is a representation isomorphic to $\pi$, but we have simplified our lives since $\rho$ now takes values in matrices. This is what a statement like “we may assume $\pi(G)$ is a matrix subgroup” means. Note that $f$ is precisely a choice of basis in $V$. Can you extrapolate from there to the corresponding statements for the orthogonal matrices? $\endgroup$
    – Joppy
    Oct 2, 2019 at 2:38
  • $\begingroup$ @Joppy, if $V=\mathbb{R}^n$ then there is no need to do that change of basis since things are matrices anyways. Once we have this matrix subgroup, I am failing to see the connection between the orthogonal group $O(n)=\{A\in GL_n: AA^T=A^TA=I_n\}$ and the orthogonal group corresponding to the inner product $[\cdot,\cdot]$ constructed in my question. I.e. the group $\{A\in GL_n: [Ax,Ay]=[x,y]\text{ for all }x,y\in \mathbb{R}^n\}$. $\endgroup$
    – Artur
    Oct 2, 2019 at 13:22
  • 1
    $\begingroup$ Given a finite-dimensional inner product space $(V, [-, -])$, there is an orthogonal subgroup $O(V, [-, -]) \subseteq \operatorname{GL}(V)$, defined as the subgroup of invertible linear transformations $g$ satisfying $[gv, gw] = [v, w]$ for all $v, w \in V$. After picking an orthonormal basis for $V$, the matrix of any $g \in O(V, [-, -])$ will be an orthogonal matrix. But often in problems a choice of basis is not canonical, and so we might prefer to work with $V$ and $[-, -]$ rather than $\mathbb{R}^n$ and the dot product. $\endgroup$
    – Joppy
    Oct 2, 2019 at 13:30
  • $\begingroup$ That makes more sense @Joppy. I have some remarks: "After picking an orthonormal basis for $V$, the matrix of any $g\in O(V,[--,--])$ will be an orthogonal matrix". So by this, you mean $g^T=g^{-1}$. But then this means $\langle gx,gy\rangle=\langle x,y\rangle$ where $\langle \cdot,\cdot\rangle$ is the usual dot product. That is, the usual dot product is invariant for $G$ (under some basis change). Is this precisely what is meant? In the case of $V=\mathbb{R}^n$, does Gram-Schmidt not always make it possible to find such bases? $\endgroup$
    – Artur
    Oct 2, 2019 at 16:04

1 Answer 1

Reset to default
1
$\begingroup$

  1. "Without loss of generality" here means more specifically "up to change of basis." When we find a $G$-invariant inner product we can choose to further work in terms of an orthonormal basis wrt that inner product, rather than in terms of the standard basis of $\mathbb{R}^n$.

  2. I don't now, that seems like a strange thing to do. Can you give more context about the text?

$\endgroup$
2
  • $\begingroup$ For 2. it is just that taking an orthonormal basis of the $G$-invariant inner product we get that $\Pi = B\pi(.)B^{-1}$ is contained in $O(n)$ and its $G$-invariant inner product $[x,y] = \int_G <\Pi(g)x,\Pi(g)y>d\nu(g)$ is the euclidean inner product. $\endgroup$
    – reuns
    Oct 2, 2019 at 21:45
  • $\begingroup$ @reuns are $x$ and $y$ written with respect to the orthonormal basis for the $G$-invariant inner product which was constructed by Weyl's trick? (recall the original representation is $\mathbb{R}^n$) $\endgroup$
    – Artur
    Oct 3, 2019 at 0:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .