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 D2m invariant isomorphism between these spaces but maybe using the Galois group one can get around it.
Best Answer
Nope.
Really, this is a linear algebra question. You can take tensor products of pairs of vector spaces, symmetric and exterior powers of a single vector space. These are all functorial, so extend to representations of a group.
Let $Vec$ be the category of (finite-dimensional) vector spaces and linear maps over a given field $k$. The tensor product can be thought of as a functor: $$\otimes: Vec \times Vec \rightarrow Vec.$$
The symmetric $n^{th}$ power can be thought of as a functor: $$Sym^n: Vec \rightarrow Vec.$$
There are various other linear algebraic functors, from $Vec^m \times (Vec^{op})^n$ to $Vec$, e.g., dual space, Schur functors, tensor products of such things, etc.. All such linear algebraic functors naturally yield functors on the category of representations of a given group.
But there is no "wedge product of two vector spaces" functor, like you are looking for.
There may be some interesting (not obvious) functors from $Vec \times Vec$ to $Vec$, perhaps depending on the characteristic of your ground field. But I don't know much about this -- I recommend searching for things like "polynomial functor" and "linear species".