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.

September 16, 2010 - Posted by John Armstrong | Algebra, Group Actions, Group theory, Representation Theory, Representations of Symmetric Groups

9 Comments »

[…] as with any group action on a finite set, we get a finite-dimensional permutation representation. The representing space has a standard basis corresponding to the elements of . That is, to every […]

Pingback by The (Left) Regular Representation « The Unapologetic Mathematician | September 17, 2010 | Reply
[…] around the cosets. That is, we have a group action of on the quotient set , and this gives us a permutation representation of […]

Pingback by Coset Representations « The Unapologetic Mathematician | September 20, 2010 | Reply
[…] an example, consider the defining representation of , which is a permutation representation arising from the action of on the set . This representation comes with the standard basis , and […]

Pingback by Reducibility « The Unapologetic Mathematician | September 23, 2010 | Reply
[…] take to be a permutation representation coming from a group action on a finite set that we’ll also call . It’s straightforward […]

Pingback by Characters of Permutation Representations « The Unapologetic Mathematician | October 19, 2010 | Reply
[…] trivial representation sends every group element to the identity matrix, whose trace is . We also know that every character’s value on the identity element is the degree of the corresponding […]

Pingback by The Character Table of a Group « The Unapologetic Mathematician | October 20, 2010 | Reply
[…] we have any other representations of to work with? Well, there’s the defining representation. This has a character we can specify by the three […]

Pingback by One Complete Character Table (part 1) « The Unapologetic Mathematician | October 26, 2010 | Reply
[…] can find such a complement if we have a -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 […]

Pingback by One Complete Character Table (part 2) « The Unapologetic Mathematician | October 27, 2010 | Reply
[…] step is to compute the character of as a left -module. The nice thing here is that it’s a permutation representation, and that means we have a shortcut to calculating its character: is the number of fixed point of […]

Pingback by Decomposing the Left Regular Representation « The Unapologetic Mathematician | November 17, 2010 | Reply
[…] that we have an action of on the Young tabloids of shape , we can consider the permutation representation that corresponds to it. Let’s consider a few […]

Pingback by Permutation Representations from Partitions « The Unapologetic Mathematician | December 14, 2010 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects

Subjects
Archives

September 2010

M T W T F S S

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22 23 24 25 26

27 28 29 30

« Aug Oct »
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider