# orthogonality relations

: Let $\rho_{\alpha}\colon G\to V_{\alpha}$ and $\rho_{\beta}\colon G\to V_{\beta}$ be irreducible representations of a finite group $G$ over the field $\mathbb{C}$. Then

 $\frac{1}{|G|}\sum_{g\in G}\overline{\rho^{(\alpha)}_{ij}(g)}\rho^{(\beta)}_{kl% }(g)=\frac{\delta_{\alpha\beta}\delta_{ik}\delta_{jl}}{\dim V_{\alpha}}.$

We have the following useful corollary. Let $\chi_{1}$, $\chi_{2}$ be characters of representations $V_{1}$, $V_{2}$ of a finite group $G$ over a field $k$ of characteristic $0$. Then

 $(\chi_{1},\chi_{2})=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_{1}(g)}\chi_{2}(g% )=\dim(\mathrm{Hom}(V_{1},V_{2})).$
###### Proof.

First of all, consider the special case where $V=k$ with the trivial action of the group. Then $\mathrm{Hom}_{G}(k,V_{2})\cong V_{2}^{G}$, the fixed points. On the other hand, consider the map

 $\phi=\frac{1}{|G|}\sum_{g\in G}g\colon V_{2}\to V_{2}$

(with the sum in $\mathrm{End}(V_{2})$). Clearly, the image of this map is contained in $V_{2}^{G}$, and it is the identity restricted to $V_{2}^{G}$. Thus, it is a projection with image $V_{2}^{G}$. Now, the rank of a projection (over a field of characteristic 0) is its trace. Thus,

 $\dim_{k}\mathrm{Hom}_{G}(k,V_{2})=\dim V_{2}^{G}=\mathrm{tr}(\phi)=\frac{1}{|G% |}\sum\chi_{2}(g)$

which is exactly the orthogonality formula for $V_{1}=k$.

Now, in general, $\mathrm{Hom}(V_{1},V_{2})\cong V_{1}^{*}\otimes V_{2}$ is a representation, and $\mathrm{Hom}_{G}(V_{1},v_{2})=(\mathrm{Hom}(V_{1},V_{2}))^{G}$. Since $\chi_{V^{*}_{1}\otimes V_{2}}=\overline{\chi_{1}}\chi_{2}$,

 $\dim_{k}\mathrm{Hom}_{G}(V_{1},V_{2})=\dim_{k}(\mathrm{Hom}(V_{1},V_{2}))^{G}=% \sum_{g\in G}\overline{\chi_{1}}\chi_{2}$

which is exactly the relation we desired. ∎

In particular, if $V_{1},V_{2}$ irreducible, by Schur’s Lemma

 $\mathrm{Hom}(V_{1},V_{2})=\begin{cases}D&V_{1}\cong V_{2}\\ 0&V_{1}\ncong V_{2}\end{cases}$

where $D$ is a division algebra. In particular, non-isomorphic irreducible representations have orthogonal characters. Thus, for any representation $V$, the multiplicities $n_{i}$ in the unique decomposition of $V$ into the direct sum (http://planetmath.org/DirectSum) of irreducibles

 $V\cong V_{1}^{\oplus n_{1}}\oplus\cdots\oplus V_{m}^{\oplus n_{m}}$

where $V_{i}$ ranges over irreducible representations of $G$ over $k$, can be determined in terms of the character inner product:

 $n_{i}=\frac{(\psi,\chi_{i})}{(\chi_{i},\chi_{i})}$

where $\psi$ is the character of $V$ and $\chi_{i}$ the character of $V_{i}$. In particular, representations over a field of characteristic zero are determined by their character. Note: This is not true over fields of positive characteristic.

If the field $k$ is algebraically closed, the only finite division algebra over $k$ is $k$ itself, so the characters of irreducible representations form an orthonormal basis for the vector space of class functions with respect to this inner product. Since $(\chi_{i},\chi_{i})=1$ for all irreducibles, the multiplicity formula above reduces to $n_{i}=(\psi,\chi_{i})$.

Second orthogonality relations: We assume now that $k$ is algebraically closed. Let $g,g^{\prime}$ be elements of a finite group $G$. Then

 $\sum_{\chi}\chi(g)\overline{\chi(g^{\prime})}=\begin{cases}|C_{G}(g_{1})|&g% \sim g^{\prime}\\ 0&g\nsim g^{\prime}\end{cases}$

where the sum is over the characters of irreducible representations, and $C_{G}(g)$ is the centralizer of $g$.

###### Proof.

Let $\chi_{1},\ldots,\chi_{n}$ be the characters of the irreducible representations, and let $g_{1},\ldots,g_{n}$ be representatives of the conjugacy classes.

Let $A$ be the matrix whose $ij$th entry is $\sqrt{|G:C_{G}(g_{j})|}(\overline{\chi_{i}(g_{j})})$. By first orthogonality, $AA^{*}=|G|I$ (here $*$ denotes conjugate transpose), where $I$ is the identity matrix. Since left inverses (http://planetmath.org/MatrixInverse) are right , $A^{*}A=|G|I$. Thus,

 $\sqrt{|G:C_{G}(g_{i})||G:C_{G}(g_{k})|}\sum_{j=1}^{n}\chi_{j}(g_{i})\overline{% \chi_{j}(g_{k})}=|G|\delta_{ik}.$

Replacing $g_{i}$ or $g_{k}$ with any conjuagate will not change the expression above. thus, if our two elements are not conjugate, we obtain that $\sum_{\chi}\chi(g)\overline{\chi(g^{\prime})}=0$. On the other hand, if $g\sim g^{\prime}$, then $i=k$ in the sum above, which reduced to the expression we desired. ∎

A special case of this result, applied to $1$ is that $|G|=\sum_{\chi}\chi(1)^{2}$, that is, the sum of the squares of the dimensions (http://planetmath.org/Dimension) of the irreducible representations of any finite group is the order of the group.

Title orthogonality relations OrthogonalityRelations 2013-03-22 13:21:27 2013-03-22 13:21:27 mhale (572) mhale (572) 16 mhale (572) Theorem msc 20C15