The Unapologetic Mathematician

Mathematics for the interested outsider

One Complete Character Table (part 2)

Last time we wrote down the complete 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\\\chi^\perp&2&0&-1\end{array}

which is all well and good except we haven’t actually seen a representation with the last line as its character!

So where did we get the last line? We had the equation \chi^\mathrm{def}=\chi^\mathrm{triv}+\chi^\perp, which involves the characters of the defining representation V^\mathrm{def} and the trivial representation V^\mathrm{triv}. This equation should correspond to an isomorphism V^\mathrm{def}\cong V^\mathrm{triv}\oplus V^\perp.

We know that there’s a copy of the the trivial representation as a submodule of the defining representation. If we use the standard basis \{\mathbf{1},\mathbf{2},\mathbf{3}\} of V^\mathrm{def}, this submodule is the line spanned by the vector \mathbf{1}+\mathbf{2}+\mathbf{3}. We even worked out the defining representation in terms of the basis \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2},\mathbf{3}\} to show that it’s reducible.

But what we want is a complementary subspace which is also G-invariant. And we can find such a complement if we have a G-invariant inner product on our space. And, luckily enough, permutation representations admit a very nice invariant inner product! Indeed, just take the inner product that arises by declaring the standard basis to be orthonormal; it’s easy to see that this is invariant under the action of G.

So we need to take our basis \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2},\mathbf{3}\} and change the second and third members to make them orthogonal to the first one. Then they will span the orthogonal complement, which we will show to be G-invariant. The easiest way to do this is to use \{\mathbf{1}+\mathbf{2}+\mathbf{3},\mathbf{2}-\mathbf{1},\mathbf{3}-\mathbf{1}\}. Then we can calculate the action of each permutation in terms of this basis. For example:

\displaystyle\begin{aligned}\left[\rho((1\,2))\right](\mathbf{1}+\mathbf{2}+\mathbf{3})&=\mathbf{1}+\mathbf{2}+\mathbf{3}\\\left[\rho((1\,2))\right](\mathbf{2}-\mathbf{1})&=\mathbf{1}-\mathbf{2}=-(\mathbf{2}-\mathbf{1})\\\left[\rho((1\,2))\right](\mathbf{3}-\mathbf{1})&=\mathbf{3}-\mathbf{2}=-(\mathbf{2}-\mathbf{1})+(\mathbf{3}-\mathbf{1})\end{aligned}

and write out all the representing matrices in terms of this basis:

\displaystyle\begin{aligned}\rho(e)&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\rho((1\,2))&=\begin{pmatrix}1&0&0\\{0}&-1&-1\\{0}&0&1\end{pmatrix}\\\rho((1\,3))&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&-1&-1\end{pmatrix}\\\rho((2\,3))&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\\\rho((1\,2\,3))&=\begin{pmatrix}1&0&0\\{0}&-1&-1\\{0}&1&0\end{pmatrix}\\\rho((1\,3\,2))&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&-1&-1\end{pmatrix}\end{aligned}

These all have the required form:

\displaystyle\left(\begin{array}{c|cc}1&0&0\\\hline{0}&\ast&\ast\\{0}&\ast&\ast\end{array}\right)

where the 1 in the upper-left is the trivial representation and the 2\times 2 block in the lower right is exactly the other representation V^\perp we’ve been looking for! Indeed, we can check the values of the character:

\displaystyle\begin{aligned}\chi^\perp(e)&=\mathrm{Tr}\begin{pmatrix}1&0\\{0}&1\end{pmatrix}=2\\\chi^\perp((1\,2))&=\mathrm{Tr}\begin{pmatrix}-1&-1\\{0}&1\end{pmatrix}=0\\\chi^\perp((1\,2\,3))&=\mathrm{Tr}\begin{pmatrix}-1&-1\\1&0\end{pmatrix}=-1\end{aligned}

exactly as the character table predicted.

October 27, 2010 - Posted by | Algebra, Group theory, Representation Theory

2 Comments »

  1. […] Alternative Path It turns out that our efforts last time were somewhat unnecessary, although they were instructive. Actually, we already had a matrix […]

    Pingback by An Alternative Path « The Unapologetic Mathematician | October 28, 2010 | Reply

  2. […] a basis, we can pick . We recognize this pattern from when we calculated the invariant subspaces of the defining representation of . And indeed, is the defining […]

    Pingback by Examples of Specht Modules « The Unapologetic Mathematician | December 28, 2010 | Reply


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