orthogonality relations

First orthogonality relations:Let ${\rho }_{\alpha }:G\to V_{\alpha }$ and ${\rho }_{\beta }:G\to V_{\beta }$ be irreducible representations of a finite group $G$ over the field $ℂ$. Then

$$\frac{1}{|G|}\sum _{g\in G}\overline{{\rho }_{ij}^{(\alpha )}(g)}{\rho }_{kl}^{(\beta )}(g)=\frac{{\delta }_{\alpha \beta }{\delta }_{ik}{\delta }_{jl}}{dim{V}_{\alpha }}.$$

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 actionof the group. Then ${\mathrm{Hom}}_G(k,V_2)\cong V_2^G$, the fixed points.On the other hand, consider the map

$$\varphi =\frac{1}{|G|}\sum _{g\in G}g:V_2\to V_2$$

(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 restrictedto $V_2^G$. Thus, it is a projection with image $V_2^G$. Now, the rank ofa projection (over a field of characteristic 0) is its trace. Thus,

$${dim}_k{\mathrm{Hom}}_G(k,V_2)=dim{V}_2^G=\mathrm{tr}(\varphi )=\frac{1}{|G|}\sum {\chi }_2(g)$$

which is exactly the orthogonality formulafor $V_1=k$.

Now, in general, $\mathrm{Hom}(V_1,V_2)\cong V_1^{*}\otimes V_2$ isa 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

where $D$ is a division algebra. In particular, non-isomorphic irreduciblerepresentations 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

$$V\cong V_1^{\oplus n_1}\oplus \mathrm{\cdots }\oplus V_m^{\oplus n_m}$$

where $V_i$ ranges over irreducible representations of $G$ over $k$, can bedetermined 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 positivecharacteristic.

If the field $k$ is algebraically closed,the only finite division algebra over $k$ is $k$ itself, sothe characters of irreducible representations form an orthonormal basis forthe vector space of class functions with respect to this inner product.Since $({\chi }_i,{\chi }_i)=1$ for all irreducibles, the multiplicity formulaabove 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

where the sum is over the characters of irreducible representations, and $C_G(g)$ is the centralizer of $g$.

Proof.

Let ${\chi }_1,\mathrm{\dots },{\chi }_n$ be the characters of the irreducible representations,and let $g_1,\mathrm{\dots },g_n$ 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, $A{A}^{*}=|G|I$ (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 expressionwe 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.