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Problem of the Week #264 - Jan 23, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
Here is this week's POTW:

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Let $G$ be a compact Lie group, and let $V$ be a finite-dimensional representation of $G$. Prove that if $\chi$ is the character associated with $V$, then $\int_G \chi(g)\, dg = \operatorname{dim}(V^G)$ where $V^G\subset V$ is the subspace of $G$-invariants of $V$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
No one answered this week’s problem. You can read my solution below.


Consider the linear operator $T$ on $V$ given by $$Tv= \int_G gv \, dg$$
By invariance of the Haar measure on $G$, $Th = hT$ for all $h\in G$, so $T$ is $G$-equivariant. Furthermore, $T$ projects $V$ onto $V^G$. To see this, note that if $w = Tv$, then for every $h\in G$ $$hw = hTv = T(hv) = \int_G (hg)v\, dg = \int_G gv\, dg = Tv = w$$ showing that $w\in V^G$. On the other hand if $w\in V^G$, then $Tw = w$ since the Haar measure of $G$ is $1$. So $T$ projects onto $V^G$; consequently the trace of $T$ is the rank of $T$, which is $\operatorname{dim}(V^G)$. Thus, if $\rho : G \to \operatorname{Aut}(V)$ is the representation mapping, $$\int_G \chi(g)\, dg = \int_G \operatorname{trace}[\rho(g)]\, dg = \operatorname{trace}(T) = \operatorname{dim}(V^G)\quad\blacksquare$$
 
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