# The Unapologetic Mathematician

## 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 -

## 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 […]

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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 […]

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