# The Unapologetic Mathematician

## Consequences of Orthogonality

We have some immediate consequences of the orthonormality relations.

First of all, since irreducible characters are orthonormal any collection of them forms an orthonormal basis of the subspace it spans. Of course whatever subspace this is, it has to fit within the space of class functions, and so it can’t have any more basis elements than the dimension of this larger space. That is, there can be at most as many irreducible characters as there are conjugacy classes in $G$. And so we know that the character table must have only finitely many rows. For instance, since $S_3$ has three conjugacy classes, it can have at most three irreducible characters. We know two already, so there’s only room for one more, if there are any more at all.

For something a little more concrete, let $V^{(i)}$ be a collection of irreps with corresponding characters $\chi^{(i)}$. Then the representation $\displaystyle V=\bigoplus\limits_{i=1}^km_iV^{(i)}$

has character $\displaystyle\chi=\sum\limits_{i=1}^km_i\chi^{(i)}$

That is, direct sums of representations correspond to sums of characters. This is just the tip of a far-reaching correspondence between the high-level properties of the category of representations and the low-level properties of the algebra of characters.

Anyway, proving this relation is pretty straightforward. If $A$ and $B$ are two matrix representations then their direct sum is $\displaystyle\left[A\oplus B\right](g)=\left(\begin{array}{c|c}A(g)&0\\\hline0&B(g)\end{array}\right)$

It should be clear that the trace of the direct sum on the left is the sum of the traces on the right. This is all we need, since we can just split off one irreducible component after another to turn the direct sum on one side into a sum on the other.

Next we have a way of reading off the coefficients. Let $V$ be the same representation from above, with the same character. I say that the multiplicity $m_j=\langle\chi^{(j)},\chi\rangle$. Indeed, we can easily calculate \displaystyle\begin{aligned}\langle\chi^{(j)},\chi\rangle&=\left\langle\chi^{(j)},\sum\limits_{i=1}^km_i\chi^{(i)}\right\rangle\\&=\sum\limits_{i=1}^km_i\langle\chi^{(j)},\chi^{(i)}\rangle\\&=\sum\limits_{i=1}^km_i\delta_{j,i}\\&=m_j\end{aligned}

Notice that this is very similar to the result we showed at the end of calculating the dimensions of spaces of morphisms. This is not a coincidence.

More generally, if $V$ is the representation from above and $W$ is another representation that decomposes as $\displaystyle W=\bigoplus\limits_{j=1}^kn_jV^{(j)}$

then the character of $W$ is $\displaystyle\psi=\sum\limits_{j=1}^kn_j\chi^{(j)}$

and we calculate the inner product \displaystyle\begin{aligned}\langle\chi,\psi\rangle&=\left\langle\sum\limits_{i=1}^km_i\chi^{(i)},\sum\limits_{j=1}^kn_j\chi^{(j)}\right\rangle\\&=\sum\limits_{i=1}^k\sum\limits_{j=1}^km_in_j\langle\chi^{(i)},\chi^{(j)}\rangle\\&=\sum\limits_{i=1}^k\sum\limits_{j=1}^km_in_j\delta_{i,j}\\&=\sum\limits_{i=1}^km_in_i\end{aligned}

In particular, we see that $\displaystyle\langle\chi,\chi\rangle=\sum\limits_{i=1}^km_i^2$

We see that in all these cases $\langle\chi,\psi\rangle=\dim\hom_G(V,W)$. Just like sums of characters correspond to direct sums of representations, inner products of characters correspond to $\hom$-spaces between representations. We just have to pass from plain vector spaces to their dimensions when we pass from representations to their characters. Of course, this isn’t much of a stretch, since we saw that the character $\chi$ of a representation $V$ includes information about the dimension: $\chi(e)=\dim(V)$.

This goes even further: what happens when we swap the arguments to an inner product? We get the complex conjugate: $\langle\psi,\chi\rangle=\overline{\langle\chi,\psi\rangle}$. What happens when we swap the arguments to the $\hom$ functor? We get the dual space: $\displaystyle\hom_G(W,V)\cong\hom_G(V,W)^*$. Complex conjugation corresponds to passing to the dual space.

Finally, the character $\chi$ is irreducible if and only if $\langle\chi,\chi\rangle=1$. Indeed, if $\chi$ is itself irreducible then our decomposition only involves one nonzero coefficient, which is a $1$. The formula we just computed gives $\displaystyle\langle\chi,\chi\rangle=\sum\limits_{i=1}^km_i^2=1$

Conversely, if this formula holds then we have to write $1$ as a sum of squares. The only possibility is for all but one of the numbers $m_i$ to be $0$, and the remaining one to be $1$, in which case $\chi=\chi^{(i)}$, and is thus irreducble.

October 25, 2010 -

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