Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p+q \le n$.

share|improve this question
    
You can typeset mathematics here using $ and $$ just as in LaTeX. –  Zhen Lin Aug 16 '11 at 11:57

2 Answers 2

I don't know how to describe the whole kernel, but I do know how to describe the generators of the kernel.

Recall that $e_1\wedge e_2\dots\wedge e_k=0$ if and only if $\{e_1,\dots,e_k\}$ is a linearly dependent set of vectors. Furthermore, if $\{e_1,\dots e_k\}$ and $\{e'_1,\dots,e'_k\}$ are bases for the same subspace, then $e_1\wedge e_2\dots\wedge e_k$ and $e'_1\wedge e'_2\dots\wedge e_k'$ are scalar multiples of each other (in fact the coefficient by which you multiply the former to obtain the latter is precisely the determinant of the linear map on the subspaces that sends $e_i\to e'_i$).

The above implies that you can think of the generators of $\bigwedge^k V$ as the $k$-dimensional subspaces of $V$. Hence, the generators of $\bigwedge^p V\otimes\bigwedge^q V$ can be though of as pairings of a $p$- and $q$-dimensional subspaces, and the map to $\bigwedge^{p+q} V$ is a map from pairings of $p$- and $q$-dimensional subspaces to $p+q$-dimensional subspaces, which takes the direct sum of the pair. Hence, the kernel of the map is generated by pairings of $p$- and $q$-dimensional subspaces, whose direct sum is not $p+q$-dimensional, or, in other words, the kernel is generated by pairings of subspaces with non-trivial intersection.

share|improve this answer

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: $$\ker(\wedge^pV\otimes\wedge^qV\to \wedge^{p+q}V)\cong \bigoplus_{n=1}^{\min\{p,q\}} \mathbb S_{(2^n,1^{p+q-n})}(V).$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.