Showing that a 2-form on an odd dimensional space is not degenerate

Question

On an odd-dimensional space $\mathbb R^{2n+1}$ with coordinates $x_1...x_n;y_1...y_n;t$ consider the following 2-form: $$\omega^2=\sum dx_i \land dy_i-\omega^1 \land dt$$ where $\omega^1$ is any 1-form on $\mathbb R^{2n+1}$. How to show that $\omega^2$ is non degenerate?

Answer 1

I'm pretty sure it isn't nondegenerate.

Let $e_{x_i}$, $e_{y_i}$, and $e_t$ denote the standard basis vectors in $\Bbb{R}^{2n+1}$. Define $v=e_t+\sum_i \omega^1(e_{x_i})e_{y_i}-\omega^1(e_{y_i})e_{x_i}$. Now we check that $\omega^2(v,e)=0$ for each standard basis vector $e$.

$$\omega^2(v,e_{x_i})=-v_{y_i}+\omega^1(e_{x_i})v_t=-\omega^1(e_{x_i})+\omega^1(e_{x_i})=0$$ for all $i$, similarly evaluating $\omega^2(v,e_{y_i})$ gives $$\omega^2(v,e_{y_i})=v_{x_i}+\omega^1(e_{y_i})v_t=-\omega^1(e_{y_i})+\omega^1(e_{y_i})=0.$$ Finally, evaluating at $e_t$, we get $\omega^2(v,e_t)=-\omega^1(v)+\omega^1(e_t)v_t$, and expanding $\omega^1(v)$, we get $$\omega^2(v,e_t)=-\omega^1(e_t)-\sum_i \left(\omega^1(e_{x_i})\omega^1(e_{y_i})-\omega^1(e_{y_i})\omega^1(e_{x_i})\right) +\omega^1(e_t)=0.$$

Hence $v$ exhibits the degeneracy of $\omega^2$. Anyway, it's been a while since I had to do concrete computations with differential forms, so do let me know if I've made any mistakes in here.

Edit:

By $v_{x_i}$, $v_{y_i}$, and $v_t$ I mean the real numbers such that $$v=v_te_t+\sum_i \left(v_{x_i}e_{x_i} + v_{y_i}e_{y_i}\right).$$ By construction of $v$, $v_{x_i}=-\omega^1(e_{y_i})$, $v_{y_i}=\omega^1(e_{x_i})$, and $v_t=1$.