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

$\displaystyle\begin{aligned}\chi(g)=\mathrm{Tr}\left(X(g)\right)&=\sum\limits_ix_i^i(g)\\\psi(g)=\mathrm{Tr}\left(Y(g)\right)&=\sum\limits_jy_j^j(g)\end{aligned}$

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.

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October 22, 2010 - Posted by John Armstrong | Algebra, Group theory, Representation Theory

3 Comments »

[...] We have some immediate consequences of the orthonormality relations. [...]

Pingback by Consequences of Orthogonality « The Unapologetic Mathematician | October 25, 2010 | Reply
[...] let’s take the consequences we just worked out of the orthonormality relations and finish the [...]

Pingback by One Complete Character Table (part 1) « The Unapologetic Mathematician | October 26, 2010 | Reply
[...] we make use of our orthonormality relations. That is, if we use the regular dot product on the rows of the modified character table (considered [...]

Pingback by The Character Table as Change of Basis « The Unapologetic Mathematician | November 22, 2010 | Reply

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