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 $\displaystyle S_3$ and I'm not quite sure how it's arrived at.

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

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

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

Thanks

Re: Representation of symmetric groups

believe it or not, the way this comes about is starting with a three-dimensional representation:

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

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

one can then ask: what subspace in the standard basis {e_{1},e_{2},e_{3}} is invariant under (all six of) these mappings?

one answer is the hyperplane {x_{1}e_{1} + x_{2}e_{2} + x_{3}e_{3} : x_{1}+x_{2}+x_{3} = 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 S_{3} 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 e_{1} - e_{2} = (1,-1,0), and 1 + (-1) + 0 = 1, so this is one possible basis vector. another possible basis vector is e_{1} - e_{3} = (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 = {e_{1}-e_{2},e_{1}-e_{3}}.

so let's look at what the 3x3 matrices do to e_{1} - e_{2}, and e_{1} - e_{3}.

the 3x3 matrix for (1 2) sends e_{1} - e_{2} to e_{2} - e_{1}. so it sends the vector [1,0]_{B} to [-1,0]_{B}.

the 3x3 matrix for (1 2) sends e_{1} - e_{3} to e_{2} - e_{3} = -(e_{1} - e_{2}) + e_{1} - e_{3},

so it sends the vector [0,1]_{B} to [-1,1]_{B}. thus in the basis B, the induced representation of S_{3} on our hyperplane gives for (1 2) the matrix:

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

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

e_{1} - e_{2} gets sent to e_{2} - e_{3} = -(e_{1} - e_{2}) + e_{1} - e_{3}, so (1 2 3) sends [1,0]_{B} to [-1,1]_{B}

e_{1} - e_{3} gets sent to e_{2} - e_{1} = -(e_{1} - e_{2}) 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:

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

Re: Representation of symmetric groups

Quote:

Originally Posted by

**Deveno** so let's look at what the 3x3 matrices do to e_{1} - e_{2}, and e_{1} - e_{3}.

the 3x3 matrix for (1 2) sends e_{1} - e_{2} to e_{2} - e_{1}. so it sends the vector [1,0]_{B} to [-1,0]_{B}.

the 3x3 matrix for (1 2) sends e_{1} - e_{3} to e_{2} - e_{3} = -(e_{1} - e_{2}) + e_{1} - e_{3},

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?

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 S_{3} 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 C^{3} (say U) such that ρ(U) ⊆ U, for every ρ in S_{3}?

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 C^{3} = <(1,1,1)> ⊕ W?

how about the vectors in C^{3} 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 S_{3}.

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

for any vector v = (x,y,z) in C^{3}, we can write it uniquely as:

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

now T_{ρ}(v) = T_{ρ}(v) = T_{ρ}(a(1,1,1)) + T_{ρ}(w) = a(1,1,1) + T_{ρ}(w).

so the (1,1,1) part doesn't tell us anything, T_{ρ} 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 S_{3} out in this basis. i'll get you started: the "change of basis matrix" is:

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

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

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

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

(in the course of doing this, i realized i had the wrong matrix for $\displaystyle \rho_{(123)}$ 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:

$\displaystyle \begin{bmatrix}I&0\\0&B \end{bmatrix}$, where I is the 1x1 identity matrix, and B is one of the 2x2 matrices in the "two-dimensional" version of S_{3}.