## Can you “Wedge” two representations?

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.

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I think you've answered your own question. You cannot. The Wedge is just the alternation (if that is word) of the tensor product, which is the sign representation of the symmetric group $S_2$ acting on $V \otimes V$. There is no action of the symmetric group on $V \otimes W$. – José Figueroa-O'Farrill Mar 7 2010 at 0:21
Not all Schur functors can be obtained just from symmetric products and wedge products; rather Schur functors generalize the definition of Sym^k V and /\^k V as the elements in Tensor^k V which transform in a certain way under permutation of the factors. – Tom Church Mar 7 2010 at 0:36
What are you trying to achieve by the putative wedge product? Answering that first is the first step in order to see if there is an operation which does what you want... – Mariano Suárez-Alvarez Mar 7 2010 at 14:59

$V \wedge V$ 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 $V\wedge V \ne \wedge^2 V.$

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

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

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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 $S_n$.

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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".

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 In the case of the dihedral group, there are some non-isomorphic as representations which are isomorphic as vector spaces. Perhaps the construction I need is not functorial... – John Mangual Mar 8 2010 at 15:32