Abstract Nonsense

Crushing one theorem at a time

Second Orthogonality Relation For Irreducible Characters

Point of post: In this post we discuss the second orthogonality relation for the irreducible characters. In particular, we prove that \displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}=c(g,h)\#\left(\bold{C}_G(g)\right) where c(g,h) takes the value one if g and h are conjugate and zero otherwise and \bold{C}_G(g) is the centralizer of g in G.


In the past we’ve seen that if we ‘fix \alpha,\beta\in\widehat{G} and let g ‘vary’ over G (in the form of the sum) that there is an interesting orthogonality relation. Namely, this is just the orthogonality relation \displaystyle \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\overline{\chi^{(\beta)}(g)}=\delta_{\alpha,\beta}. In this post we explore what happens if the ‘fixing’ and ‘varying’ are reversed. In other words, we compute \displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}. This shall serve as an interesting tool inall that comes. We call this the second orthogonality relation for the irreducible characters.

Second Orthogonality Relation for the Irreducible Characters

Theorem (Second Orthogonality Relation for the Irreducible Characters): For a finite group G and g\in G denote the centralizer of k by \bold{C}_G(g). Then, for g,h\in G let c(g,h) take the value one if g and h are conjugate  and zero otherwise. Then,

\displaystyle \sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(g)\overline{\chi^{(\alpha)}(h)}=c(g,h)\#\left(\bold{C}_G(g)\right)

Proof: Let \mathcal{C}_1,\cdots,\mathcal{C}_k be the number of conjugacy classes in G and let g_1,\cdots,g_k be a set of representatives such that g_j\in \mathcal{C}_j for j=1,\cdots,k. Recall then that \#\left(\widehat{G}\right)=k and so the elements of G can be labeled \alpha_1,\cdots,\alpha_k. We then consider the k\times k matrix M=[M_{i,j}] where M_{i,j}=\chi^{(\alpha_i)}(g_j). We then let D denote the matrix \text{diag }\left(\#\left(\mathcal{C}_1\right),\cdots,\#\left(\mathcal{C}_k\right)\right). Recall though from the first orthogonality relation that

\displaystyle |G|\delta_{i,j}=\sum_{g\in G}\chi^{(\alpha_i)}(g)\overline{\chi^{(\alpha_j)}(g)}


But, recalling that the irreducible characters are class functions and thus constant on conjugacy classes it easily follows that this may be rewritten

\displaystyle |G|\delta_{i,j}=\sum_{r=1}^{k}\#\left(\mathcal{C}_k\right)\chi^{(\alpha_i)}(g_r)\overline{\chi^{(\alpha_j)}(g_r)}


This then implies that

\displaystyle |G|I_k=MD M^{\ast}


But, recall the fact that for square matrices a left inverse is necessarily a right inverse, and thus the above implies that

|G|I=DM^\ast M


But, the general (i,j)^{\text{th}} entry of this right hand matrix is

\displaystyle \sum_{r=1}^{k}\#\left(\mathcal{C}_i\right)\overline{\chi^{(\alpha_k)}(g_i)}\chi^{(\alpha_k)}(g_j)


Dividing both sides and recalling by the orbit stabilizer theorem that \displaystyle \#\left(\bold{C}_G(g_i)\right)=\frac{|G|}{\#\left(\mathcal{C}_i\right)} we see that

\displaystyle \#\left(\bold{C}_G(g_i)\right)=\sum_{r=1}^{k}\chi^{(\alpha_k)}(g_j)\overline{\chi^{(\alpha_k)}(g_i)}


And recalling all of our notation and taking the conjugate of both sides gives the desired result. \blacksquare




1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.


March 7, 2011 - Posted by | Algebra, Representation Theory | , , , ,


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