TITLE: Decomposition of product of exterior products
QUESTION [7 upvotes]: Suppose $V$ is a $n$-dimensional vector space.
What is the kernel of
$$\bigwedge^p V \otimes \bigwedge^q V\longrightarrow \bigwedge^{p+q} V$$
here $p+q \le n$.

REPLY [5 votes]: This is a long comment rather than a complete answer. [Updated with a complete answer below.] 
$\wedge^k V$ is a simple example of a Schur functor. You might know of the classification of irreducible representations of the symmetric group in terms of partitions. Namely, given a partition $\lambda=(\lambda_1,\cdots,\lambda_k)$ where $\lambda_1\geq \lambda_2\cdots\geq\lambda_k\geq 1$ where $\sum \lambda_i=n$, there is an associated irreducible representation of the symmetric group $\Sigma_n$ denoted $F_\lambda$. Now $\mathbb S_\lambda$ denotes a functor called a Schur functor defined on vector spaces by the formula $\mathbb S_\lambda (V)= F_\lambda\otimes_{\Sigma_n}V^{\otimes n}$. There are nice formulas for computing the dimension of these vector spaces. Also $\mathbb S_{(1,\ldots,1)}(V)=\wedge^n(V)$ and $\mathbb S_{n}(V)$ is the symmetric power $S^n(V)$. It is a fundamental fact that any functor on vector spaces decomposes as a direct sum of Schur functors, so that we can definitely say that $\wedge^p V\otimes\wedge^q V$ decomposes as a direct sum of such functors, associated to partitions and you can even show that the partitions have to be of $p+q$. Looking at the kernel of the map you gave will correspond to dropping one copy of $\mathbb S_{(1,\ldots,1)}(V)$ from the decomposition. For any given $i$ and $j$ you can use the dimension formulas to calculate exactly what the decomposition is, but I'm not sure if there is a general formula for the result, since I'm not an expert on this stuff.
So for example, when $p=q=1$, $V\otimes V=S^2V\oplus \wedge^2 V$ so the kernel is isomorphic to $S^2V$. A more complicated example is $p=q=2$. According to my calculations
$$(\wedge^2 V)\otimes(\wedge^2 V)\cong \wedge^4 V\oplus \mathbb S_{(2,2)}(V)\oplus \mathbb S_{(2,1,1)}(V),$$
so that the kernel is isomorphic to $\mathbb S_{(2,2)}(V)\oplus \mathbb S_{(2,1,1)}(V)$.
Update(July 2012): I mentioned that I wasn't an expert on this stuff but I am learning! In general one needs the Littlewood-Richardson rule to decompose a product of Schur functors. This is probably not the best place to expound on how this rule works. Wikipedia has a somewhat confusing explanation. I learned it from Fulton and Harris's classic text on representation theory. The case at hand is $\mathbb S_{(1^p)}(V)\otimes \mathbb S_{(1^q)}(V)$, and the Littlewood-Richardson rule gives an isomorphism
$$\mathbb S_{(1^p)}(V)\otimes \mathbb S_{(1^q)}(V)\cong \bigoplus_{n=0}^{\min\{p,q\}} \mathbb S_{(2^n,1^{p+q-n})}(V).$$ The OP's map is surjective, so the kernel eliminates the $\wedge^{p+q}V=\mathbb S_{(1^{p+q})}(V)$ direct summand. So we have a complete answer:
$$\ker(\wedge^pV\otimes\wedge^qV\to \wedge^{p+q}V)\cong \bigoplus_{n=1}^{\min\{p,q\}} \mathbb S_{(2^n,1^{p+q-n})}(V).$$