Rewriting a k-form as a wedge product with a 1-form

Question

I am trying to show that a general element of the kth exterior product $\Lambda^kV^*$ (of V an n-dimensional vector space) $$ \alpha = \sum_{i} \alpha_i e_i$$ (where the $\{e_i\}$, for $1\leq i\leq {n\choose k}$ are a basis of the vector space $\Lambda^kV^*$) can be written as : $$\alpha = f_j \wedge \beta_j, $$ where the $f_j$'s are at most $k$ linearly independent elements of $V^*$ and $\beta_j\in \Lambda^{k-1}V^*$ depends on $\alpha$ and on the possible $f_j$. I thought of using a recurrence on $k$ (it is obviously true when $k=1$ and $k=n$), but I don't just don't know how I could use the kth step to proceed up to the (k+1)th step...

The objective of this exercise, is to get a bound on the dimension of the kernel of : $$ x\mapsto x\wedge \alpha$$ when $x$ is a 1-form and $\alpha$ is a given $k$ form. I was able to show that the dimension of the kernel is equal to the number of $f_j$ of the preceding paragraph, and now I'm stuck.

I also tried a less general approach, consisting on counting the different basis elements $e_i$ of $\alpha$ (not including the elements appearing once in every $e_i$) that differ by just one element of the $\Lambda V^*$ basis (those that will give non-trivial equations for $x\wedge \alpha = 0$). This procedure looks however like a big mess.

If anyone could help, I would appreciate any clue.

Answer 1

This seems false. Consider $R^6$ with coordinates $a,b,c,x,y,z$. Then

$$ da \wedge db + dc \wedge dx + dy \wedge dz $$ which is a 2-form, can't be written (I think!) as a product of 1-forms $$ s \wedge \phi $$ where $s$.

At the very least, you can generalize the number of coordinates to something quite large; if you're restricting to just two $f_j$s, then it's gonna be tough.

By the way, do you mean to write $$ \alpha = \sum_j f_j \wedge \beta_j $$ perhaps? Or is every 2-form supposed to be either $f_1$ wedge something, OR $f_2$ wedge something, etc.?

I think that if you look up "indecomposable 2-forms", you'll see what I'm talking about. In fact, this question gives a criterion for a 2-form to be decomposable. :)

Answer 2

I find your first question pretty unclear as it is stated, so I will concentrate on the second half.

Consider the basis of $\Lambda^kV^*$ given by $$\{v_{i_1}\wedge\ldots v_{i_k}|1\le i_1<\ldots<i_k\le n\}.$$ We can write $\alpha$ as $$\alpha = \alpha^{i_1\ldots i_k}v_{i_1}\wedge\ldots v_{i_k}.$$ We have $$v_j\wedge\alpha = \left(\prod_{\ell=1}^k\delta_{ji_\ell}\right)(-1)^{\max\{\ell|i_\ell<j\}}\alpha^{i_1\ldots i_k}v_{i_1}\wedge\ldots\wedge v_j\wedge\ldots\wedge v_{i_k}.$$ This transforms your problem in pure linear algebra, namely to finding the dimension of the kernel of a matrix of dimension $k$ times $\binom{n}{k+1}$. Of course, a priori this is not an easy task to do by hand, but in many cases you are interested in $\alpha$'s giving you an easy form for your matrix.

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As an example, let $V^*=\mathbb{R}^4$ with basis $\{x,y,z,w\}$ and $\alpha = x\wedge y + z\wedge w$. Then \begin{align} x\wedge\alpha = & x\wedge z\wedge w\\ y\wedge\alpha = & y\wedge z\wedge w\\ z\wedge\alpha = & x\wedge y\wedge z\\ w\wedge\alpha = & x\wedge y\wedge w, \end{align} so that (after choosing an order of the basis of $\lambda^3V^*$) the associated matrix is $$\pmatrix{0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0},$$ which has maximal rank and is square, and thus has trivial kernel.