Decomposition of tensor product of two representations in $S_3$.

Question

Consider the group $S_3$. There are three irreducible representations, the trivial, $\varphi^{triv}$, the sign representation $\varphi^\epsilon$ (both 1-dimensional), and the two-dimensional one $\varphi^2$. If we take the tensor product representation of the two-dimensional representation $\varphi^2 \otimes \varphi^2$ (with $V$ a two-dimensional vector space), and decompose this new representation using characters and multiplicity of characters, we find that $$ \varphi^2 \otimes \varphi^2 \simeq \varphi^{triv} \oplus \varphi^\epsilon \oplus \varphi^2 $$ and hence $$ V \otimes V \simeq V_{triv} \oplus V_\epsilon \oplus V_2, $$ the $G$-invariant subspaces of $\varphi^{triv},\varphi^\epsilon,\varphi^2$ respectively (notice that $\dim V \otimes V = \dim(V)\dim(V) = 4$). On the other hand, we know that $$ V \otimes V \simeq \operatorname{Sym}^{(2)}(V) \oplus \operatorname{Alt}^{(2)}(V), $$ with $\dim \operatorname{Sym}^{(2)}(V) = 3$ and $\dim \operatorname{Alt}^{(2)}(V) = 1$.

Now my question: how do these two decompositions match up?

I suspect that $V_\epsilon \simeq Alt^{(2)}(V)$ and $V_{triv} \oplus V_2 \simeq Sym^{(2)}(V)$, which would match the dimensions, but I have no idea how to prove this.

What I started with is that $\{e_1 \otimes e_2 - e_2 \otimes e_1\}$ is a basis for $Alt^{(2)}(V)$ and $\{2(e_1 \otimes e_1), e_1 \otimes e_2 + e_2 \oplus e_1, 2(e_2 \otimes e_2\}$ is a basis for $Sym^{(2)}(V)$. I was thinking to map these basis elements to basis elements of $V_\epsilon, V_2, V_{triv}$, but I don't see what those basis elements are exactly.

Answer 1

You can get an explicit $2$-dimensional representation of $S_3$ by mapping $$ (123)\mapsto\begin{pmatrix}\omega & 0\\0 &\omega^2\end{pmatrix}, (12)\mapsto\begin{pmatrix}0 & 1\\1 &0\end{pmatrix} $$ where $\omega$ is a primitive cube root of unity.

You can then check that $$(123)\cdot e_1\wedge e_2= \omega e_1\wedge \omega^2 e_2=e_1\wedge e_2, \text{ and } (12)\cdot e_1\wedge e_2=e_2\wedge e_1= -e_1\wedge e_2 $$ so that this is indeed the sign-module.

For the symmetric module, it is easiest just to use quadratic polynomials in $e_1, e_2$. It is not difficult to see that $e_1e_2$ is a basis for the trivial module, and that $e_2^2, e_1^2$ a basis for the $2$-dimensional irreducible. (I write it this way to emphasise the swap of $\omega$ and $\omega^2$ when we square them.)

Answer 2

Given a basis $(E_1, E_2, E_3)$ of a $3$-dimensional vector space $W$ (say, over a field of characteristic not $2$ or $3$), $S_3$ acts by permutation on the basis. This action preserves $S := E_1 + E_2 + E_3$ and also the sum of coefficients of an element with respect to this basis, so $W = V + \langle S \rangle$, where $V := \{a E_1 + b E_2 + c E_3 : a + b + c = 0\}$. Checking shows that $V$ is irreducible. Transpositions act trivially on $V_{\operatorname{triv}}$ and by negation on $V_{\epsilon}$, so it's sufficient to compute whether, e.g., $(12)$ maps a nonzero element $\eta \in \operatorname{Alt}^{(2)} V$ to $+\eta$ or $-\eta$.

With respect to the standard basis $(F_1, F_2) := (E_1 - E_2, E_2 - E_3)$ of $V$, the permutation $(12)$ acts by \begin{align}(12) \cdot F_1 = (12) \cdot (E_1 - E_2) &= (E_2 - E_1) = -(E_1 - E_2) = -F_1, \\ (12) \cdot F_2 = (12) \cdot (E_2 - E_3) &= (E_1 - E_3) = F_1 + F_2 ,\end{align} and so the action on the spanning element $\eta := F_1 \wedge F_2 \in Alt^{(2)} V$ is $$(12) \cdot (F_1 \wedge F_2) = ((12) \cdot F_1) \wedge ((12) \cdot F_2) = (-F_1) \wedge (F_1 + F_2) = -F_1 \wedge F_2 = -\eta.$$ Thus, $V_{\epsilon} = \operatorname{Alt}^{(2)} V$.

Remark As you say, this implies that $\operatorname{Sym}^{(2)} V$ contains the trivial representation $V_{\operatorname{triv}}$, or equivalently, there is a nonzero $S_3$-invariant symmetric $2$-tensor on $V$, and it is unique up to scale.