# Representation of symmetric groups

• Nov 24th 2012, 07:26 AM
Conn
Representation of symmetric groups
Hi, just working my way through the beginnings of some material for a group representations course I'll be taking next semester. The notes give an example of a representation of $S_3$ and I'm not quite sure how it's arrived at.

Let $\rho: S_3 \to GL_2\mathbb{C}$ be specified on the generators $(12)$ and $(123)$ by

$\rho_{(12)} =\begin{pmatrix}-1 & -1 \\0 & 1\\\end{pmatrix}$ and $\rho_{(123)} =\begin{pmatrix}-1 & -1 \\1 & 0\\\end{pmatrix}$

I'm just not entirely sure how these matrices are obtained, could anyone please enlighten me?

Thanks
• Nov 24th 2012, 12:33 PM
Deveno
Re: Representation of symmetric groups
believe it or not, the way this comes about is starting with a three-dimensional representation:

$\rho_{(12)} = \begin{bmatrix}0&1&0\\1&0&0\\0&0&1 \end{bmatrix},\ \rho_{(123)} = \begin{bmatrix}0&1&0\\0&0&1\\1&0&0 \end{bmatrix}$

(the other 4 matrices can be obtained in the usual way from these generators).

one can then ask: what subspace in the standard basis {e1,e2,e3} is invariant under (all six of) these mappings?

one answer is the hyperplane {x1e1 + x2e2 + x3e3 : x1+x2+x3 = 0}.

note that the defining equation of the hyperplane is the (elementary) symmetric polynomial of degree 1 in 3 variables.

since we have an invariant subspace, we might consider the induced action of S3 on this subspace.

to come up with matrices we, of course, need to pick a BASIS for this subspace. there are various ways to do this (each one yielding different matrices).

well e1 - e2 = (1,-1,0), and 1 + (-1) + 0 = 1, so this is one possible basis vector. another possible basis vector is e1 - e3 = (1,0,-1).

since a hyperplane in a n-dimensional space has dimension n-1, these are all we need. now let's look at which 2x2 matrices we get in the basis B = {e1-e2,e1-e3}.

so let's look at what the 3x3 matrices do to e1 - e2, and e1 - e3.

the 3x3 matrix for (1 2) sends e1 - e2 to e2 - e1. so it sends the vector [1,0]B to [-1,0]B.

the 3x3 matrix for (1 2) sends e1 - e3 to e2 - e3 = -(e1 - e2) + e1 - e3,

so it sends the vector [0,1]B to [-1,1]B. thus in the basis B, the induced representation of S3 on our hyperplane gives for (1 2) the matrix:

$\begin{bmatrix}-1&-1\\0&1 \end{bmatrix}$.

now let's do the same thing for (1 2 3):

e1 - e2 gets sent to e2 - e3 = -(e1 - e2) + e1 - e3, so (1 2 3) sends [1,0]B to [-1,1]B

e1 - e3 gets sent to e2 - e1 = -(e1 - e2) so (1 2 3) sends [0,1]B to [-1,0]B,

which gives us for (1 2 3) in the basis B the matrix:

$\begin{bmatrix}-1&-1\\1&0 \end{bmatrix}$.
• Nov 26th 2012, 05:18 PM
Conn
Re: Representation of symmetric groups
Quote:

Originally Posted by Deveno
so let's look at what the 3x3 matrices do to e1 - e2, and e1 - e3.

the 3x3 matrix for (1 2) sends e1 - e2 to e2 - e1. so it sends the vector [1,0]B to [-1,0]B.

the 3x3 matrix for (1 2) sends e1 - e3 to e2 - e3 = -(e1 - e2) + e1 - e3,

so it sends the vector [0,1]B to [-1,1]B.

Sorry, I've never reduced a matrix from 3d to 2d before and I just don't understand where these results are coming from, like I understand why the basis vectors in the new basis formulae are being switched around, but I don't really see how this leads to [-2,0] being added to what I assume are just the standard basis vectors for 2x2 matrices?
• Nov 27th 2012, 10:22 AM
Deveno
Re: Representation of symmetric groups
ok, think of an element in a 3-dimensional vector space (over a field, in this case, the complex numbers) as being just "three things", its coordinates.

so we can have S3 act on such vectors by "permuting the coordinates". so the permutation (1 2) for example, sends:

(1,0,0) to (0,1,0)
(0,1,0) to (1,0,0)
(0,0,1) to itself, and we "extend by linearity".

it's natural to ask: are there any subspaces of C3 (say U) such that ρ(U) ⊆ U, for every ρ in S3?

well, an obvious one is the subspace generated by (1,1,1), switching around the coordinates of a vector whose coordinates are all the same won't change the vector at all.

can we find another subspace W with C3 = <(1,1,1)> ⊕ W?

how about the vectors in C3 perpendicular to (1,1,1)? which vectors are these?

if <(x,y,z),(1,1,1)> = 0, then x+y+z = 0. the is called the hyperplane defined by the line t(1,1,1).

now let's look at what this means for our permutations ρ in S3.

remember, every ρ gives us a linear transformation Tρ in GL(3,C). and since C3 = <(1,1,1)> ⊕ W,

for any vector v = (x,y,z) in C3, we can write it uniquely as:

v = a(1,1,1) + w (where w is orthogonal to (1,1,1)).

now Tρ(v) = Tρ(v) = Tρ(a(1,1,1)) + Tρ(w) = a(1,1,1) + Tρ(w).

so the (1,1,1) part doesn't tell us anything, Tρ is the identity on that subspace.

check it out, instead of using the standard basis for the 3x3 matrices, use this basis instead:

B = {(1,1,1),(1,-1,0),(1,0,-1)}

write all 6 matrices for S3 out in this basis. i'll get you started: the "change of basis matrix" is:

$P = \begin{bmatrix}1&1&1\\1&-1&0\\1&0&-1 \end{bmatrix}$

so the matrix for (1 2 3) in the basis B is:

$\begin{bmatrix}\frac{1}{3}&\frac{1}{3}&\frac{1}{3} \\ \frac{1}{3}&\frac{-2}{3}&\frac{1}{3}\\ \frac{1}{3}&\frac{1}{3}&\frac{-2}{3} \end{bmatrix} \begin{bmatrix} 0&0&1\\1&0&0\\0&1&0 \end{bmatrix} \begin{bmatrix}1&1&1\\1&-1&0\\1&0&-1 \end{bmatrix}$

$= \begin{bmatrix}1&0&0\\0&-1&-1\\0&1&0 \end{bmatrix}$

(in the course of doing this, i realized i had the wrong matrix for $\rho_{(123)}$ in my earlier post, it should be the transpose of that matrix. oh bother!).

convince yourself that in "this new basis" all of the matrices have the block form:

$\begin{bmatrix}I&0\\0&B \end{bmatrix}$, where I is the 1x1 identity matrix, and B is one of the 2x2 matrices in the "two-dimensional" version of S3.