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.
Best Answer
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.)