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

\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.

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

3 Comments »

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

    Pingback by Consequences of Orthogonality « The Unapologetic Mathematician | October 25, 2010 | Reply

  2. […] 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

  3. […] 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 […]

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