The Unapologetic Mathematician

The Character Table as Change of Basis

Now that we’ve seen that the character table is square, we know that irreducible characters form an orthonormal basis of the space of class functions. And we also know another orthonormal basis of this space, indexed by the conjugacy classes $K\subseteq G$:

$\displaystyle\left\{\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}f_K\right\}$

A line in the character table corresponds to an irreducible character $\chi^{(i)}$, and its entries $\chi_K^{(i)}$ tell us how to write it in terms of the basis $\{f_K\}$:

$\displaystyle\chi^{(i)}=\sum\limits_K\chi_K^{(i)}f_K$

That is, it’s a change of basis matrix from one to the other. In fact, we can modify it slightly to exploit the orthonormality as well.

When dealing with lines in the character table, we found that we can write our inner product as

$\displaystyle\langle\chi,\psi\rangle=\sum\limits_K\frac{\lvert K\rvert}{\lvert G\rvert}\overline{\chi_K}\psi_K$

So let’s modify the table to replace the entry $\chi_K^{(i)}$ with $\sqrt{\lvert K\rvert/\lvert G\rvert}\chi_K^{(i)}$. Then we have

$\displaystyle\sum\limits_K\overline{\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(i)}\right)}\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(j)}\right)=\langle\chi^{(i)},\chi^{(j)}\rangle=\delta_{i,j}$

where 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 as tuples of complex numbers) we find that they’re orthonormal. But this means that the modified table is a unitary matrix, and thus its columns are orthonormal as well. We conclude that

$\displaystyle\sum\limits_i\overline{\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_K^{(i)}\right)}\left(\sqrt{\frac{\lvert K\rvert}{\lvert G\rvert}}\chi_L^{(i)}\right)=\delta_{K,L}$

where now the sum is over a set indexing the irreducible characters. We rewrite these relations as

$\displaystyle\sum\limits_i\overline{\chi_K^{(i)}}\chi_L^{(i)}=\frac{\lvert G\rvert}{\lvert K\rvert}\delta_{K,L}$

We can use these relations to help fill out character tables. For instance, let’s consider the character table of $S_3$, starting from the first two rows:

$\displaystyle\begin{array}{c|ccc}&e&(1\,2)&(1\,2\,3)\\\hline\chi^\mathrm{triv}&1&1&1\\\mathrm{sgn}&1&-1&1\\\chi^{(3)}&a&b&c\end{array}$

where we know that the third row must exist for the character table to be square. Now our new orthogonality relations tell us on the first column that

$\displaystyle1^2+1^2+a^2=6$

Since $a=\chi^{(3)}(e)$, it is a dimension, and must be positive. That is, $a=2$. On the second column we see that

$\displaystyle1^2+1^2+b^2=\frac{6}{3}=2$

and so we must have $b=0$. Finally on the third column we see that

$\displaystyle1^2+1^2+c^2=\frac{6}{2}=3$

so $c=\pm1$.

To tell the difference, we can use the new orthogonality relations on the first and third or second and third columns, or the old ones on the first and third or second and third rows. Any of them will tell us that $c=-1$, and we’ve completed the character table without worrying about constructing any representations at all.

We should take note here that the conjugacy classes index one orthonormal basis of the space of class functions, and the irreducible representations index another. Since all bases of any given vector space have the same cardinality, the set of conjugacy classes and the set of irreducible representations have the same number of elements. However, there is no reason to believe that there is any particular correspondence between the elements of the two sets. And in general there isn’t any, but we will see that in the case of symmetric groups there is a way of making just such a correspondence.

November 22, 2010 -