One-dimensional null space for a 2-form

Question

Consider the following 2-form on $\mathbb R^{2n+1}$ with coordinates $x_1...x_n;y_1...y_n;t$: $$\omega^2=\sum dx_i \land dy_i-\omega^1 \land dt$$ where $\omega^1$ is any 1-form on $\mathbb R^{2n+1}$. How can I show that the null vector space of such form is 1-dimensional?

Answer 1

A $1$-form on $\mathbb{R}^{2n+1}$ can be thought of as a row vector. A $2$-form is, in particular, a $(0,2)$-tensor, and can thus be thought of as a square matrix. Moreover, if the row vectors $\varphi_1$ and $\varphi_2$ represent the $1$-forms $\alpha_1$ and $\alpha_2$, respectively, then the wedge product $\alpha_1\wedge\alpha_2$ is represented by the matrix $$\varphi_1^T\varphi_2-\varphi_2^T\varphi_1.$$

The $2$-form $\sum dx^i\wedge dy^i$ corresponds to the matrix $$J=\left(\begin{array}{ccc}0&I&0\\-I&0&0\\0&0&0\end{array}\right),$$ where $I$ denotes the identity matrix of size $n\times n$. Let the row vector $$\varphi=(\varphi_1,\ldots,\varphi_{2n+1})$$ represent the $1$-form $\omega^1$. Then the $2$-form $\omega^1\wedge dt$ corresponds to the matrix $$\begin{align}H&=\varphi^T(0,\ldots,0,1)-(0,\ldots,0,1)^T\varphi\\&=\left(\begin{array}{cccc}0&\cdots&0&\varphi_1\\\vdots&\;&\vdots&\vdots\\0&\cdots&0&\varphi_{2n+1}\end{array}\right)-\left(\begin{array}{ccc}0&\cdots&0\\\vdots&\;&\vdots\\0&\cdots&0\\\varphi_1&\cdots&\varphi_{2n+1}\end{array}\right)\\&=\left(\begin{array}{cccc}0&\cdots&0&\varphi_1\\\vdots&\;&\vdots&\vdots\\0&\cdots&0&\varphi_{2n}\\-\varphi_1&\cdots&-\varphi_{2n}&0\end{array}\right).\end{align}$$ The $2$-form in question is represented by the matrix $J-H$, which has $2n$ linearly independent rows.