The Unapologetic Mathematician

Mathematics for the interested outsider

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 sth row and tth 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 Posted by | Algebra, Group Actions, Group theory, Representation Theory, Representations of Symmetric Groups | 9 Comments