A property of the wedge of the restriction of a $2$-form and a $1$-form to a codimension $1$ vector subspace

Question

Let $\omega \in \Lambda^2(V^*)$ be a $2$-form on a vector space $V$ of dimension $2k+2$. Suppose further we have a nonzero $1$-form $\eta$ on $V$ and a subspace $W$ of codimension $1$ with the property that $W = V_1 \oplus (\ker (\eta|_W))$ for some $V_1$ of dimension $1$. Is it true then that $\eta|_W \wedge (\omega|_W)^k \neq 0$ iff $\omega|_{\ker(\eta|_W)}$ is nondegenerate?

I can assume that $\omega$ is nondegenerate on $V$ if it is a necessary hypothesis somewhere.

I suspect the answer to be "Yes", because if $\omega|_{\ker(\eta|_W)}$ is nondegenerate, by the pointwise Darboux theorem we have that on $\ker(\eta|_W)$ we can write $\omega|_{\ker(\eta|_W)} = \sum_{i=1}^{2k} e_i^* \wedge f_i^*$ with $\omega(e_i, e_j) = 0 = \omega(f_i, f_j)$ and $\omega(e_i, f_j) = \delta_{i,j}$. Since on $W$ $\eta$ is nonzero, it seems plausible that when I write out the wedge product $\eta|_W \wedge (\omega|_W)^k$ I cannot get a $0$.

However, I don't know how to pass from the expression I have for $\omega|_{\ker(\eta|_W)}$ to an expression for $\omega|_W$, nor how to write the wedge product in a convenient way so that I can show it's nonzero.

The only true problem seems to be that I don't see how to use the expression of $\omega|_{\ker(\eta|_W)}$ to derive an expression for $\omega|_W$; if this were solved, I think the problem would also be solved just by the fact that $\omega|_{\ker(\eta|_W)}^k \neq 0$ iff $\omega|_{\ker(\eta|_W)}$ is nondegenerate (which follows from Darboux).

Answer 1

You can't in general derive an expression for $\omega|_W$ using the expression of $\omega|_{ker(\eta|_W)}$. Taking the basis $v_0, ..., v_{2k+1}$, and its dual basis $v_0^*, ..., v_{2k+1}^*$, where $v_0$ is the basis of $V_0$, $v_1$ is the basis of $V_1$, $(v_2, ..., v_{2k+1})$ is the basis of $ker(\eta|_W)$, and $V = V_0 \oplus W = V_0 \oplus V_1 \oplus ker(\eta|_W)$, you have:

$\eta = Av_0^* + Bv_1^*$

$\eta|_W = Bv_1^*$

$\omega = \sum_{0 \le i<j \le 2k+1} C_{i,j}v_i^*\wedge v_j^*$

$\omega|_W = \sum_{1 \le i<j \le 2k+1} C_{i,j}v_i^*\wedge v_j^*$

$\omega|_{ker(\eta|_W)} = \sum_{2 \le i<j \le 2k+1} C_{i,j}v_i^*\wedge v_j^*$

It's clear to see that having an expression for $\omega|_{ker(\eta|_W)}$ doesn't give you any information regarding $C_{i,j}$ for $i<2$.

Nonetheless, the answer to your question is yes. You showed that if $\omega|_{ker(\eta|_W)}$ is non-degenerate, then $\eta|_W \wedge \omega|_W \ne 0$. For the other direction, assume that $\omega|_{ker(\eta|_W)}$ is degenerate, and we'll show that $\eta|_W \wedge \omega|_W = 0$. The degeneracy gives that $(\omega|_{ker(\eta|_W)})^k = 0$. Therefore, all the terms of $(\omega|_W)^k$ (when written in the basis above), that have nonzero coefficients, must have $v_1^*$ in them. Hence, when wedged with $\eta|_W$, we get zero, which proves the other direction.

Answer 2

$$ (\omega|_W)^2 = \sum_{\substack{1 \leq i < j \leq 2k+1 \\1 \leq s< r \leq 2k+1 }} c_{i,j} c_{s,r} v_i^* \wedge v_j^* \wedge v_s^* \wedge v_r^* $$ Generalising, $$ (\omega|_W)^k = \sum_{\substack{1\leq i_s < j_s \leq 2k+1 \\ s=1}}^k c_{i_1, j_1} \cdots c_{i_k, j_k} v_{i_1}^* \wedge v_{j_1}^* \wedge \cdots \wedge v_{i_k}^* \wedge v_{j_k}^* $$

By the same reasoning: $$ (\omega|_K)^k = \sum_{\substack{2\leq i_s < j_s \leq 2k+1 \\ s=1}}^k c_{i_1, j_1} \cdots c_{i_k, j_k} v_{i_1}^* \wedge v_{j_1}^* \wedge \cdots \wedge v_{i_k}^* \wedge v_{j_k}^* $$

Now because $\eta|_W = B v_1^*$ we have: $$ (\eta|_W) \wedge (\omega|_W)^k = \sum_{\substack{1\leq i_s < j_s \leq 2k+1 \\ s=1}}^k c_{i_1, j_1} \cdots c_{i_k, j_k} B v_1^* \wedge v_{i_1}^* \wedge v_{j_1}^* \wedge \cdots \wedge v_{i_k}^* \wedge v_{j_k}^* $$

Since in every wedge product we have $e_1^*$, if some $i_s = 1$ then the product disappears. Therefore we are left with: $$ (\eta|_W) \wedge (\omega|_W)^k = \omega|_{\ker(\eta|_W)} $$