# The Unapologetic Mathematician

## Group Actions and Representations

From the module perspective, we’re led back to the concept of a group action. This is like a $G$-module, but “discrete”. Let’s just write down the axioms for easy reference: we’ve got a set $S$ and a function $A:G\times S\to S$ such that

• $A$ preserves the identity: $A(e,s)=s$.
• $A$ preserves the group operation: $A(g,A(h,s))=A(gh,s)$.

Notice how this looks almost exactly like the axioms for a $G$-module, except since $S$ is just a set we don’t have any sense of “linearity”.

Now, from a group action on a finite set $S$ we can get a finite-dimensional representation. We let $V=\mathbb{C}S$ — the vector space defined to have $S$ as a basis. That is, vectors in $\mathbb{C}S$ are of the form $\displaystyle\sum\limits_{s\in S}c_s\mathbf{s}$

for some complex coefficients $c_s$. We get a $G$-module by extending $A$ to a bilinear function $\mathbb{C}[G]\times\mathbb{C}S\to\mathbb{C}S$. We already know how it behaves on the basis of the form $(\mathbf{g},\mathbf{s})$, and the extension to a bilinear function is uniquely defined. We call $\mathbb{C}S$ the “permutation representation” associated to $S$, and the elements $\mathbf{s}$ for $s\in S$ we call the “standard basis”.

As an example, the group $S_n$ is defined from the very beginning by the fact that it acts on the set $\{1,2,\dots,n\}$ by shuffling the numbers around. And so we get a representation from this action, which we call the “defining representation”. By definition, it has dimension $n$, since it has a basis given by $\{\mathbf{1},\dots,\mathbf{n}\}$. To be even more explicit, let me write out the defining matrix representation for $S_3$. Technically, going from an abstract representation to a matrix representation requires not just a basis, but an ordered basis, but the order should be pretty clear in this case. And so, with no further ado: \displaystyle\begin{aligned}\rho\left(e\right)&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\rho\left((1\,2)\right)&=\begin{pmatrix}{0}&1&0\\1&0&0\\{0}&0&1\end{pmatrix}\\\rho\left((1\,3)\right)&=\begin{pmatrix}{0}&0&1\\{0}&1&0\\1&0&0\end{pmatrix}\\\rho\left((2\,3)\right)&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\\\rho\left((1\,2\,3)\right)&=\begin{pmatrix}{0}&0&1\\1&0&0\\{0}&1&0\end{pmatrix}\\\rho\left((1\,3\,2)\right)&=\begin{pmatrix}{0}&1&0\\{0}&0&1\\1&0&0\end{pmatrix}\end{aligned}

To see how this works, note that the permutation $(1\,2\,3)$ sends $3$ to $1$. Similarly, we find that $\left[\rho\left((1\,2\,3)\right)\right](\mathbf{3})=\mathbf{1}$. That is: $\displaystyle\begin{pmatrix}{0}&0&1\\1&0&0\\{0}&1&0\end{pmatrix}\begin{pmatrix}{0}\\{0}\\1\end{pmatrix}=\begin{pmatrix}1\\{0}\\{0}\end{pmatrix}$

We also see that composition of permutations turns into matrix multiplication. For example, $(1\,2\,3)(2\,3)=(1\,2)$. In terms of the matrices we calculate: $\displaystyle\begin{pmatrix}{0}&0&1\\1&0&0\\{0}&1&0\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}=\begin{pmatrix}{0}&1&0\\1&0&0\\{0}&0&1\end{pmatrix}$

You can check for yourself all the other cases that you care to.

Notice that in general the matrices are index by two elements of $S$, and the matrix element $\rho(g)_s^t$ — the one in the $s$th row and $t$th column — is $\delta_{A(g,t),s}$. That is, it’s $1$ if $A(g,t)=s$ — if the action of $g$ sends $s$ to $t$ — and $0$ otherwise. This guarantees that every entry will be either $0$ or $1$, and that each row and each column will have exactly one $1$. Such a matrix we call a “permutation matrix”, and we see that the matrices that occur in permutation representations are permutation matrices.

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September 16, 2010