The Unapologetic Mathematician

Mathematics for the interested outsider

Irreducible Characters are Orthogonal

Today we prove the assertion that we made last time: that irreducible characters are orthogonal. That is, if V and W are G-modules with characters \chi and \psi, respectively, then their inner product is 1 if V and W are equivalent and 0 otherwise. Strap in, ’cause it’s a bit of a long one.

Let’s pick a basis of each of V and W to get matrix representations X and Y of degrees m and n, respectively. Further, let A be any m\times n matrix with entries a_i^j. Now we can construct the m\times n matrix

\displaystyle B=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(g)AY\left(g^{-1}\right)

Now I claim that B intertwines the matrix representations X and Y. Indeed, for any h\in G we calculate

\displaystyle\begin{aligned}X(h)B&=X(h)\left(\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(g)AY\left(g^{-1}\right)\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(h)X(g)AY\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(hg)AY\left(g^{-1}h^{-1}h\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(hg)AY\left((hg)^{-1}h\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X(f)AY\left(f^{-1}\right)Y(h)\\&=\left(\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X(f)AY\left(f^{-1}\right)\right)Y(h)\\&=BY(h)\end{aligned}

At this point, Schur’s lemma kicks in to tell us that if X\not\cong Y then B is the zero matrix, while if X\cong Y then B is a scalar times the identity matrix.

First we consider the case where X\not\cong Y (equivalently, V\not\cong W). Since B is the zero matrix, each entry must be zero. In particular, we get the equations

\displaystyle\frac{1}{\lvert G\rvert}\sum\limits_{k,l}\sum\limits_{g\in G}x_i^k(g)a_k^ly_l^j\left(g^{-1}\right)=0

But the left side isn’t just any expression, it’s a linear function of the a_k^l. Since this equation must hold no matter what the a_k^l are, the coefficients of the function must all be zero! That is, we have the equations

\displaystyle\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}x_i^k(g)y_l^j\left(g^{-1}\right)=0

But now we can recognize the left hand side as our alternate expression for the inner product of characters of G. If the functions x_i^k and y_l^j were characters, this would be an inner product, but in general we’ll write

\displaystyle\langle x_i^k,y_l^j\rangle'=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}x_i^k(g)y_l^j\left(g^{-1}\right)=0

Okay, but we actually do have some characters floating around: \chi and \psi. And we can write them out in terms of these matrix elements as


And now we can use the fact that for characters our two bilinear forms are the same to calculate

\displaystyle\begin{aligned}\langle\chi,\psi\rangle&=\langle\chi,\psi\rangle'\\&=\left\langle\sum\limits_ix_i^i,\sum\limits_jy_j^j\right\rangle'\\&=\sum\limits_{i,j}\langle x_i^i,y_j^j\rangle'\\&=0\end{aligned}

So there: if V and W are inequivalent irreps, then their characters are orthogonal!

Now if V\cong W we can pick bases so that the matrix representations are both X. Schur’s lemma tells us that there is some c\in\mathbb{C} so that b_i^j=c\delta_i^j. Our argument above goes through just the same as before to show that

\displaystyle\langle a_i^k,a_l^j\rangle'=0

so long as i\neq j. To handle the case where i=j, we consider our equation

\displaystyle\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}X(g)AX\left(g^{-1}\right)=cI_d

We take the trace of both sides:

\displaystyle\begin{aligned}cd&=\mathrm{Tr}(cI_d)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\mathrm{Tr}\left(X(g)AX\left(g^{-1}\right)\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\mathrm{Tr}(A)\\&=\mathrm{Tr}(A)\end{aligned}

and thus we conclude that b_i^i=c=\frac{1}{d}\mathrm{Tr}(A). And so we can write

\displaystyle\frac{1}{\lvert G\rvert}\sum\limits_{k,l}\sum\limits_{g\in G}x_i^k(g)a_k^lx_l^i\left(g^{-1}\right)=\frac{1}{d}\left(\sum\limits_ja_j^j\right)

Equating coefficients on both sides we find

\displaystyle\langle x_i^k,x_l^i\rangle'=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}x_i^k(g)x_l^i\left(g^{-1}\right)=\frac{1}{d}\delta_l^k

And finally we can calculate

\displaystyle\begin{aligned}\langle\chi,\chi\rangle'&=\sum\limits_{i,j}\langle x_i^i,x_j^j\rangle'\\&=\sum\limits_i\langle x_i^i,x_i^i\rangle'\\&=\sum\limits_i\frac{1}{d}\\&=1\end{aligned}

exactly as we asserted.

Incidentally, this establishes what we suspected when setting up the character table: if V and W are inequivalent irreps then their characters \chi and \psi must be unequal. Indeed, since they’re inequivalent we must have \langle\chi,\psi\rangle=0. But if the characters were the same we would have to have \langle\chi,\psi\rangle=\langle\chi,\chi\rangle=1. So since inequivalent irreps have unequal characters we can replace all the irreps labeling rows in the character table by their corresponding irreducible characters.

October 22, 2010 Posted by | Algebra, Group theory, Representation Theory | 3 Comments