The Unapologetic Mathematician

Mathematics for the interested outsider

The Character Table is Square

We’ve defined the character table of a group, and we’ve seen that it must be finite. Specifically, it cannot have any more rows — G cannot have any more irreducible representations — than there are conjugacy classes in G. Now we can show that there are always exactly as many irreducible representations as there are conjugacy classes in G.

We recall that for any representation V the center of the endomorphism algebra Z_{\mathrm{End}_G(V)} is equal to the number of irreducible representations that show up in V. In particular, since we know that every irreducible representation shows up in the left regular representation \mathbb{C}[G], the number of irreducible representations is k=\dim\left(Z_{\mathrm{End}_G(\mathbb{C}[G])}\right). Thus to calculate this number k, we must understand the structure of the endomorphism algebra and its center.

But we just saw that \mathrm{End}_G(\mathbb{C}[G]) is anti-isomorphic to \mathbb{C}[G] as algebras, and this anti-isomorphism induces an anti-isomorphism on their centers. In particular, their centers have the same dimension. That is:

\displaystyle k=\dim\left(Z_{\mathrm{End}_G(\mathbb{C}[G])}\right)=\dim\left(Z_{\mathbb{C}[G]}\right)

So what does a central element of the group algebra look like? Let z be such a central element and write it out as

\displaystyle z=\sum\limits_{g\in G}c_gg

Now since z is central, it must commute with every other element of the group algebra. In particular, for every h\in G we have zh=hz, or z=hzh^{-1}. That is:

\displaystyle\sum\limits_{g\in G}c_gg=z=hzh^{-1}=\sum\limits_{g\in G}c_ghgh^{-1}

Since z is invariant, the coefficients c_g and c_{hgh^{-1}} must be the same. But as h runs over G, hgh^{-1} runs over the conjugacy class of g, so the coefficients must be the same for all elements in the conjugacy class. That is, we have exactly as many free parameters when building z as there are conjugacy classes in G — one for each of them.

So we’ve established that the center of the group algebra has dimension equal to the number of conjugacy classes in G. We also know that this is the same as the dimension of the center of the endomorphism algebra of the left regular representation. Finally, we know that this is the same as the number of distinct irreducible representations that show up in the decomposition of the left regular representation. And so we conclude that any finite group G must have exactly as many irreducible representations as it has conjugacy classes. Since the conjugacy classes index the columns of the character table of G, and the irreducible characters index the rows, we conclude that the character table is always square.

As a quick corollary, we find that the irreducible characters span a subspace of the space of class functions with dimension equal to the number of conjugacy classes in G. Since this is the dimension of the whole space of class functions, the irreducible characters must form an orthonormal basis of this space.

November 19, 2010 - Posted by | Algebra, Group theory, Representation Theory

2 Comments »

  1. […] 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 […]

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

  2. […] of finite group representations. But our goal is to talk about symmetric groups in particular. Now, we’ve seen that the character table of a finite group is square, meaning there are as many irreducible […]

    Pingback by The Road Forward « The Unapologetic Mathematician | December 7, 2010 | Reply


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