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

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.

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.)

• Thank you, just a quick question: what does the symbol $\wedge$ mean in this case? – Sigurd Oct 22 '18 at 18:14
• @Sigurd, surely this is part of the definition of the alternating module? Informally $u\wedge v$ is what $u\otimes v$ maps to when we factor out the relations that make the module "alternating". – ancient mathematician Oct 23 '18 at 6:52
• And I commend to you exploring what happens when you replace $3$ by any odd number $n$, and look at the decomposition of the tensor products of the various 2-dimensionaal irreducibles of $D_{2n}$ both in characteristic $0$ but also in characteristic $2$. – ancient mathematician Oct 23 '18 at 6:54

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.