# The Unapologetic Mathematician

## Inner Products in the Character Table

As we try to fill in the character table, it will help us to note another slight variation of our inner product formula:

$\displaystyle\langle\chi,\psi\rangle=\frac{1}{\lvert G\rvert}\sum\limits_K\lvert K\rvert\overline{\chi_K}\psi_K$

where our sum runs over all conjugacy classes $K\subseteq G$, and where $\chi_K$ is the common value $\chi_K=\chi(k)$ for all $k$ in the conjugacy class $K$ (and similarly for $\psi_K$). The idea is that every $k$ in a given conjugacy class gives the same summand. Instead of adding it up over and over again, we just multiply by the number of elements in the class.

As an example, consider again the start of the character table of $S_3$:

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

Here we index the rows by irreducible characters, and the columns by representatives of the conjugacy classes. We can calculate inner products of rows by multiplying corresponding entries, but we don’t just sum up these products; we multiply each one by the size of the conjugacy class, and at the end we divide the whole thing by the size of the whole group:

$\displaystyle\begin{array}{cclccclcc}\langle\chi^\mathrm{triv},\chi^\mathrm{triv}\rangle&=(&1\cdot1\cdot1&+&3\cdot1\cdot1&+&2\cdot1\cdot1&)/6=&1\\\langle\chi^\mathrm{triv},\mathrm{sgn}\rangle&=(&1\cdot1\cdot1&+&3\cdot1\cdot-1&+&2\cdot1\cdot1&)/6=&0\\\langle\mathrm{sgn},\mathrm{sgn}\rangle&=(&1\cdot1\cdot1&+&3\cdot-1\cdot-1&+&2\cdot1\cdot1&)/6=&1\end{array}$

We find that when we take the inner product of each character with itself we get $1$, while taking the inner product of the two different characters gives $0$. This is no coincidence; for any finite group $G$ irreducible characters are orthonormal. That is, different irreducible characters have inner product $0$, while any irreducible character has inner product $1$ with itself. This is what we will prove next time.

## 2 Comments	»

1. [...] Characters are Orthogonal Today we prove the assertion that we made last time: that irreducible characters are orthogonal. That is, if and are -modules with characters and , [...]

Pingback by | October 22, 2010 | Reply

2. [...] dealing with lines in the character table, we found that we can write our inner product [...]

Pingback by | November 22, 2010 | Reply

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