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 $\mathrm{Sym}^2\, V$ 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 D2m invariant isomorphism between these spaces but maybe using the Galois group one can get around it.
