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I've been reading Brian Conrad's notes on compact Lie groups and came across the assertion $$\chi_{V^*} = \overline{\chi_V} $$ where $V$ is a finite dimensional complex representation of a compact Lie group and $V^*$ is its dual representation (defined in the usual way via $(g\cdot \phi)(v) = \phi(g^{-1}\cdot v)$. He says that the proof is the same as in the finite group case, however when I try to work through the computation in the finite case I get an inverse popping up rather than a conjugate.

This led me to do some exploring, and I found in Dummit and Foote that $\chi(g^{-1})= \overline{\chi(g)}$ for finite groups (and finite dim reps), however the proof relies on the fact that the action of each group element has finite order (hence each minimal polynomial divides $X^k-1$) and then goes on to argue that the group elements just act by diagonal matrices with roots of unity on the diagonal... so I'm not sure how to generalize this to compact Lie groups (or whether its even true outside the finite group setting!).

Is there something basic that I'm missing here? Any help would be greatly appreciated!

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The analogy comes from the following fact, if the group is compact, you have an invariant scalar product $b$. Let $\rho$ be the representation, and $\lambda$ an eigenvalue of $\rho(g)$ whose eigenvector is $v$, with $b$, you define the isomorphism $V\rightarrow V^*$ by $v\rightarrow b(.,v)=v^*$. $\rho(g)(v^*)(u)=v^*(g^{-1}(u))=b(g^{-1}(u),v)=b(gg^{-1}(u),gv)=b(u,g(v))=b(u,\lambda v)=\bar\lambda b(u,v)=\bar\lambda v^*(u)$. This implies that $Trace(g_{V^*})=\bar{Trace(g_{V}})$, and $\chi_{V^*}=\bar{\chi_V}$.

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  • $\begingroup$ Oh, okay, the argument with the invariant (hermitian?) scalar product $b$ makes sense to me, but to arrive at the conclusion $Tr(g_{V^*}) = \overline{Tr(g_V)}$ don't you need a basis of eigenvectors of the transformation $v\mapsto g\cdot v$? I know that finite group representations always act via diagonalizable matrices, but is the same true for general Lie groups? $\endgroup$ – user345730091 Sep 8 '18 at 15:12
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    $\begingroup$ to compute the trace, you just need to know the eigenvalues. $\endgroup$ – Tsemo Aristide Sep 8 '18 at 15:13
  • $\begingroup$ oh, absolutely right! Thanks! $\endgroup$ – user345730091 Sep 8 '18 at 15:16

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