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I am reading a textbook "Representation theory" by Fulton and Harris and I have a question.

They proved the following theorem on page 16.

With an Hermitian inner product on a set of class function, the characters of the irreducible representation of a finite group $G$ are orthonormal.

For a corollary of this theorem, they mentions that

Corollary: The number of irreducible representation of $G$ is less than or equal to the number of conjugacy classes.

I don't know how to prove this corollary. Could you give me some advice, please?

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  • $\begingroup$ Do you already know that, if two representations have the same character, then these two representations are in fact equal? If so, then I think this corollary is clear, since there is a basis for the space of class-functions, which has cardinality $=$ the number of conjugacy classes of $G$. This basis consists of class-sums in the group-algebra. $\endgroup$ – awllower Jul 16 '13 at 14:38
  • $\begingroup$ @awllower For this corollary it suffices that different irreducible representations have different characters, and this follows from the orthonormality. In Fulton & Harris, the statement that arbitrary representations are determined by their characters is deduced as another corollary (and stated after this one). $\endgroup$ – mdp Jul 16 '13 at 14:51
  • $\begingroup$ @MattPressland Sorry for the irrespect of the book. Then ignore my comment. :P $\endgroup$ – awllower Jul 16 '13 at 14:56
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    $\begingroup$ @allower That's fine - if you don't have the book then there's no way you could possibly know! $\endgroup$ – mdp Jul 16 '13 at 14:56
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You should note that the dimension of the space of class functions is equal to the number of conjugacy classes, and that orthonormal vectors in a Hermitian inner product space are linearly independent.

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  • $\begingroup$ But how do we know they span the space? $\endgroup$ – Jack M Jan 6 '17 at 18:09
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To prove that the characters of the irreducible representations span the centrum of the associated group algebra it is enough to prove that every central function can be spanned by the characters of the irreducible representations (by their orthonormality they are linearly independent). We already know from the Peter Weyl theorem that the functions $\pi_{ij}$, i.e. the coefficients of the matrix representations $\pi$ of the irreducible representations of the group generate the group algebra. Then, let $f$ be a central function of $\mathbb{C}(G)$. Let $\widetilde{G}$ be the set of irreducible non equivalent representations of $G$. Since the $\pi_{ij}$ for all $\pi\in G$ are a basis $\mathbb{C}(G)$ we can write $$ f(g)=\sum_{\pi\in \widetilde{G}}\sum_{i,j=1}^{d_\pi}f_{\pi,ij}\pi_{ij}(g),$$ where $d_\pi$ is the dimension of the irreducible representation $\pi$. Since $f$ is central: $$ f(g)=f(hgh^{-1})=\sum_{\pi\in \widetilde{G}}\sum_{i,j=1}^{d_\pi}f_{\pi,ij}\pi_{ij}(hgh^{-1}),\qquad \forall h\in G. $$ Hence, we can obtain by assuming $e_{\pi,i}$ for $i=1,\ldots,d_\pi$ being a basis of the irreducible representation $\pi$: $$ f(g)=\sum_{\pi\in \widetilde{G}}\frac{1}{|G|}\sum_{h\in G}\sum_{i,j=1}^{d_\pi}f_{\pi,ij}\pi_{ij}(hgh^{-1})=\sum_{\pi\in \widetilde{G}}\sum_{i,j=1}^{d_\pi}f_{\pi,ij}(e_{\pi,i}\widetilde{\pi(g)}|e_{\pi,j}). $$ The mapping $\widetilde {\pi(g)}$ is a morphism of the representation $\pi$ and by the Schur theorem is proportional to the identity. It is simple to prove that $$ f(g)=\sum_{\pi\in \widetilde{G}}\sum_{i=1}^{d_\pi}f_{\pi,ii}\frac{\chi_{\pi}(g)}{d_\pi}. $$ Hence, the central function is spanned by the characters of irreducible representations.

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  • $\begingroup$ This gives the opposite inequality compared to what was asked about. $\endgroup$ – Tobias Kildetoft Dec 16 '17 at 22:06
  • $\begingroup$ I answered to Jack M. The initial uestion was already answered. $\endgroup$ – Javier Dec 17 '17 at 8:54
  • $\begingroup$ That is not what the answer box is for. $\endgroup$ – Tobias Kildetoft Dec 17 '17 at 8:58
  • $\begingroup$ So, how I should have answered to Jack M? With the 'add comment'? Is there any way to rearrange the position of my answer now? $\endgroup$ – Javier Dec 17 '17 at 9:28

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