# Irreducible Representations

1. Jul 20, 2014

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let V have dimension 3 and consider P_1(V ) = P(1,0,0) = span of {x,y,z}.Let I denote the subspace of all polynomials in P_1 of the form {rx+ry +rz|r any scalar}.Let W denote the subspace of all polynomials in P_1 of the form {rx+sy+tz|r+s+t = 0}. I and W are S_3 invariant subspaces

Show that I and W are subspaces that are irreducible. Find a basis for the subspaces I and W. Show that P1 = I⊕W.

3. The attempt at a solution
Math newb here, going to need a lot of help on this.

P_1(V) is a representation of S_3, which means that S_3 performed it's group action of composition on a vector space V, and the resulting representation was a polynomial with degree 1 (P_1). P_1 is a reducible representation and can be decomposed into two non-irreducible sub-representations I and W.

My first job is to show that I and W are irreducible. This would mean that there are no G-invariant subspaces for representations I and W?

My second job is to find the basis for the subspaces I and W. Is a basis a vector that is able to span the vector space? So I'm looking for vectors that span the representations I and W? Are representations still considered vector spaces?

Finally I must show and I+W = P_1. P_1 is a sort of abstract notion right? It's a three dimensional vector that has been acted upon by the S_3 symmetric group, which means that the representation P_1 contains different permutations of the basis vectors of V that include (1), (1,2), (1,3), (2,3) (1,2,3), (1,3,2) in cycle notation.

Is anything I typed remotely on the right track here? Trying to wrap my head around this stuff.

2. Jul 20, 2014

### pasmith

$S_3$ is acting linearly on $V$ in such a way as to permute the members of some basis set for $V$. The components of a vector in $V$ with respect to that basis are $x$, $y$ and $z$.

$P_1$ is the set of polynomials of the form $ax + by + cz$ where $a$, $b$ and $c$ are arbitrary scalars and $x$, $y$ and $z$ are as above. It is a vector space under the operations of pointwise addition and scalar multiplication.

The action of $\sigma$ on $V$ induces an action on $P_1$ by $\sigma(p) = p \circ \sigma^{-1}$ for $p \in P_1$ and $\sigma \in S_3$ (the inverse is necessary in order that $\rho(\sigma(p)) = (\rho\sigma)(p)$). Since $P_1$ consists of linear polynomials the action of $S_3$ on $P_1$ is linear, and so a representation.

Yes. When looking for invariant subspaces it doesn't matter whether you ask if $p \circ \sigma^{-1} \in I$ or $p \circ \sigma \in I$ since every element is the inverse of some element (and the only elements of $S_3$ that you need to check are the transpositions, which are all self-inverse anyway).

Yes. Strictly the representation is the homomorphism from $G$ to the general linear group on the vector space, although (confusingly in my view) it is apparently common to refer to the space as the representation if the homomorphism is clear.

The claim is that every $p \in P_1$ can be written in the form $p = q + w$ where $q \in I$ and $w \in W$.