PlanetMath: orthogonality relations

First 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

over the field $\mathbb{C}$ . Then

We have the following useful corollary. Let $\chi_1$ , $\chi_2$ be characters of representations

of a finite group

over a field

of characteristic 0. Then

Proof. First of all, consider the special case where

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

$\displaystyle \phi=\frac{1}{\vert G\vert}\sum_{g\in G} g\colon V_2\to V_2$

(with the sum in $\mathrm{End}(V_2)$ ). Clearly, the image of this map is contained in

, and it is the identity restricted to

. Thus, it is a projection with image

. Now, the rank of a projection (over a field of characteristic 0) is its trace. Thus,

$\displaystyle \dim_k \mathrm{Hom}_G(k,V_2)=\dim V_2^G=\mathrm{tr}(\phi)= \frac{1}{\vert G\vert}\sum\chi_2(g)$

which is exactly the orthogonality formula for

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 $\chi_{V^*_1\otimes V_2}=\overline{\chi_1}\chi_2$ ,

$\displaystyle \dim_k\mathrm{Hom}_G(V_1,V_2)=\dim_k (\mathrm{Hom}(V_1,V_2))^G= \sum_{g\in G}\overline{\chi_1}\chi_2$

which is exactly the relation we desired. $\qedsymbol$

where

ranges over irreducible representations of

over

, can be determined in terms of the character inner product:

where $\psi$ is the character of

and $\chi_i$ the character of

. 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

is algebraically closed, the only finite division algebra over

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

is algebraically closed. Let

be elements of a finite group

. Then

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 be the matrix whose th entry is $\sqrt{\vert G:C_G(g_j)\vert}(\overline{\chi_i(g_j)})$ . By first orthogonality, $AA^{*}=\vert G\vert I$ (here denotes conjugate transpose), where is the identity matrix. Since left inverses are right inverses, $A^{*}A=\vert G\vert I$ . Thus,

$\displaystyle \sqrt{\vert G:C_G(g_i)\vert\vert G:C_G(g_k)\vert}\sum_{j=1}^n\chi_j(g_i)\overline{\chi_j(g_k)}=\vert G\vert\delta_{ik}.$

Replacing

with any conjuagate will not change the expression above. thus, if our two elements are not conjugate, we obtain that $\sum_\chi\chi(g)\overline{\chi(g')}=0$ . On the other hand, if $g\sim g'$ , then

in the sum above, which reduced to the expression we desired. $\qedsymbol$

A special case of this result, applied to

is that $\vert G\vert=\sum_{\chi}\chi(1)^2$ , that is, the sum of the squares of the dimensions of the irreducible representations of any finite group is the order of the group.