Starting with a representation $\rho:G \to \mathrm{GL}(V)$. Then we can build the tensor product of $V$ with itself by defining $g(v_1 \otimes v_2) = g(v_1) \otimes g(v_2)$. Then by saying $v_1v_2 = \frac{1}{2}(v_1 \otimes v_2 + v_2 \otimes v_1)$ and $v_1 \wedge v_2 = \frac{1}{2}(v_1 \otimes v_2 - v_2 \otimes v_1)$ we can define $\mathrm{Sym}^2V$ and $V \wedge V$. In fact, it looks like Schur functors are just combinations of symmetric and wedge product.

It is possible to tensor two different representations $V \otimes W$ by $g(v\otimes w) = g(v)\otimes g(w)$. In general (or in specific) is it possible to build wedge or symmetric product of two arbitrary representations? I'm betting it's not since $v \otimes w \in V \otimes W$ while $w \otimes v \in W \otimes V$. Then it's not clear how to add two elements in different space $v \otimes w + v \otimes v$. Can anyone help me out?

*@ Mariano*: For a friend, I was doing a write-up of the representations of the dihedral group, $D_{2m}$. There's Id, sgn and irredicible 2D representations for each root of unity (besides 1). I was supposed to also explain tensor products, symmetric and exterior powers, but I got caught up trying to define $W \wedge V$. I realize now it's not generally possible.

But even though you can't tensor arbitrary representations in general, there is a clear Galois action (i.e. $\mathrm{Gal}[\mathbb{Q}(\xi_m):\mathbb{Q}]$) on the roots of unity and therefore on the representations themselves. There is no D_{2m} invariant isomorphism between these spaces but maybe using the Galois group one can get around it.

thatfirst is the first step in order to see if there is an operation which does what you want... $\endgroup$ – Mariano Suárez-Álvarez Mar 7 '10 at 14:59