The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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October 21, 2010 - Posted by | Algebra, Group theory, Representation Theory

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 Irreducible Characters are Orthogonal « The Unapologetic Mathematician | October 22, 2010 | Reply

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

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


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