Can you "Wedge" two representations?

Question

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.

Answer 1

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".

Answer 2

Not an answer, rather an attempt to hijack the question...

Some time ago I have also been wondering how to wedge two vector spaces and came up with the following construction:

Let $f:U\to V$ and $g:U\to W$ be two vector space morphisms. We define the vector space $V\wedge_U W$ (of course, this depends not only on $U$, $V$ and $W$, but also on $f$ and $g$, but we silently leave these out of the notation - just as in the case of fibered products) as the quotient of the tensor product $V\otimes W$ by the subspace spanned by all tensors of the form $f\left(u\right)\otimes g\left(u\right)$ for $u\in U$.

This is functorial, but does anyone know any use for it? Any results about the structure of $V\wedge_U W$ as a representation, if $U$, $V$ and $W$ are representations? How does this $\wedge_U$ operation "look like" in the representation ring (for instance, the usual wedge operations look like the lambda operations $\lambda^1$, $\lambda^2$, ...).

EDIT: In characteristic $\neq 2$, we have $V\wedge_U W=\left(V\otimes W\right)\diagup \left(\left(\left(f\otimes g\right)\circ\left(\mathrm{id}+\tau\right)\right)\left(U\otimes U\right)\right)$, where $\tau$ is the transposition of the two tensorands. But it is still interesting to find out what exactly is factored out in classical cases, e. g. in representation theory of $S_n$.

Answer 3

$V \wedge V$ is bad notation when $V$ is a representation, just as $V~ Sym ~V$ would be. $\wedge^2 V$ is less misleading.

You could try to define $V \wedge W = \wedge^2 (V \oplus W)$. For $v\in V, w\in W$, we can naturally identify $v\wedge w$ and $w \wedge v$ with elements of $\wedge^2 (V \oplus W)$, and $v\wedge w = - w \wedge v$.This has some nice properties, perhaps too trivially, but be careful that $V\wedge V \ne \wedge^2 V.$

$\wedge^2(V\oplus W) = \wedge^2V \oplus \wedge^2 W \oplus V \otimes W.$

$Sym^2(V\oplus W) = Sym^2 V \oplus Sym^2 W \oplus V \otimes W.$

Answer 4

This is a construction in some sense dual to the proposal of Darij Grinberg. Lets consider only $G=GL(n,\mathbb{C})$ for simplicity and take $V$ and $W$ to be subrepresentations of $\bigotimes^k \mathbb{C}^n$. (Which we can do without loss of generality for any two finite-dimensional representations.)

Then $V \otimes W$ sits in $\bigotimes^k \mathbb{C}^n \otimes \bigotimes^k \mathbb{C}^n$ and you can define $V \wedge W$ to be the projection of $V \otimes W$ to $\bigotimes^k \mathbb{C}^n \wedge \bigotimes^k \mathbb{C}^n$.