A non-zero quantity associated to an invertible skew-symmetric matrix of even order.

Question

Once again, I failed to make a concise post so feel free to skip to the emphasized parts.

In the context of symplectic and contact geometry, I would like to establish the following linear algebra fact:

Proposition. Let $E$ be a real vector space of dimension $2n$ and let $\omega\in\Lambda^2 E^*$, then $\omega$ is non-degenerate if and only if $\omega^n$ is a volume form of $E$.

Proof. There exists $(\omega_{i,j})_{1\leqslant i,j\leqslant 2n}$ a skew-symmetric $2n\times 2n$ matrix with real entries such that, one has: $$\omega=\sum_{i,j=1}^{2n}\omega_{i,j}\,\mathrm{d}x_i\wedge\mathrm{d}x_j.$$ A straightforward computation then leads to the following equality: $$\omega^n=\left(\sum_{\sigma\in\mathfrak{S}_{2n}}\varepsilon(\sigma)\prod_{i=1}^n\omega_{\sigma(2i-1),\sigma(2i)}\right)\mathrm{d}x_1\wedge\cdots\wedge\mathrm{d}x_{2n},$$ where $\mathfrak{S}_{2n}$ stands for the permutation group on the set $\{1,\ldots,2n\}$ letters and $\varepsilon(\sigma)$ is the signature of $\sigma$. Therefore, let us define the following quantity associated to $\omega$:

$$\operatorname{vol}(\omega):=\sum_{\sigma\in\mathfrak{S}_{2n}}\varepsilon(\sigma)\prod_{i=1}^n\omega_{\sigma(2i-1),\sigma(2i)}.$$

The proof boils down to show that $(\omega_{i,j})_{1\leqslant i,j\leqslant 2n}$ is invertible if and only if $\operatorname{vol}(\omega)$ is non-zero. $\Box$

Hence, my question arises:

Question. How can I show that $\operatorname{vol}(\omega)$ is non-zero if and only if $(\omega_{i,j})_{1\leqslant i,j\leqslant 2n}$ is invertible?

I am aware that $\operatorname{vol}(\omega)=2^nn!\operatorname{pf}(\omega)$, where $\operatorname{pf}(\omega)$ is the Pfaffian of $\omega$ and that one has: $$\det(\omega)=\operatorname{pf}(\omega)^2,\tag{1}$$ which immediately leads to the desired result, but $(1)$ is a pain to establish and I would like to avoid using it. If you are aware of a clever and rather elementary way to prove $(1)$, I will also be happy with it.

Any enlightenment will be greatly appreciated.

Answer 1

This answer is based on fredgoodman's suggestion in the comments.

In what follows, the key observation is that $\operatorname{vol}(\omega)$ is the unique scalar $\lambda$ such that $w^n=\lambda\,\mathrm{d}x_1\wedge\cdot\wedge dx_{2n}$.

Proposition. Let $M$ be a $2n\times 2n$ matrix with real entries, then one has the following equality: $$\textrm{vol}(M\omega{}^\intercal M)=\det(M)\operatorname{vol}(\omega).$$

Proof. Let us define $\omega':=M^*\omega$, then it is easily seen that on the level of matrices, one has: $$\omega'=M\omega{}^\intercal M.$$ Besides, one has the following equality: $${\omega'}^n=M^*\omega^n=M^*(\operatorname{vol}(\omega)\,\mathrm{d}x_1\wedge\ldots\wedge\mathrm{d}x_{2n})=\det(M)\operatorname{vol}(\omega)\,\mathrm{d}x_1\wedge\ldots\wedge\mathrm{d}x_{2n}.$$ Whence the result. $\Box$

Lemma. Let $\omega:=\operatorname{diag}\left(\begin{bmatrix}0&\lambda_1\\-\lambda_1&0\end{bmatrix},\ldots,\begin{bmatrix}0&\lambda_n\\-\lambda_n&0\end{bmatrix},0,\ldots,0\right)$, then one has the following: $$\operatorname{vol}(\omega)=2^nn!\lambda_1\ldots\lambda_n.$$ In particular, one has $\operatorname{vol}(\omega)^2=(2^nn!)^2\det(\omega)$.

Proof. By construction of the matrix, one has: $$\omega=2\sum_{i=1}^n\lambda_i\,\mathrm{d}x_{2i-1}\wedge\mathrm{d}x_{2i}.$$ Therefore, by a straightforward computation, one has: $$\begin{align}\omega^n&=2^n\sum_{i_1=1}^{n}\cdots\sum_{i_n=1}^{n}\lambda_{i_1}\ldots\lambda_{i_n}\,\mathrm{d}x_{2i_1-1}\wedge\mathrm{d}x_{2i_1}\wedge\ldots\wedge\mathrm{d}x_{2i_n-1}\wedge\mathrm{d}x_{2i_n},\\&=2^n\sum_{\sigma\in\mathfrak{S}_n}\lambda_{\sigma(1)}\ldots\lambda_{\sigma(n)}\,\mathrm{d}x_{2\sigma(1)-1}\wedge\mathrm{d}x_{2\sigma(1)}\wedge\cdots\wedge\mathrm{d}x_{2\sigma(n)-1}\wedge\mathrm{d}x_{2\sigma(n)},\notag\\&=2^n\sum_{\sigma\in\mathfrak{S}_n}(-1)^{2\varepsilon(\sigma)}\lambda_1\ldots\lambda_n\,\mathrm{d}x_1\wedge\cdots\wedge\mathrm{d}x_{2n},\\&=2^nn!\lambda_1\ldots\lambda_n\,\mathrm{d}x_1\wedge\ldots\wedge\mathrm{d}x_{2n}.\end{align}$$ Whence the result, since $\det(\omega)={\lambda_1}^2\ldots{\lambda_n}^2$. $\Box$

Theorem. One has $\det(\omega)=\operatorname{pf}(\omega)^2$.

Proof. Using the spectral theorem for skew-symmetric along with the proposition and the lemma, one has: $$\operatorname{vol}(\omega)^2=(2^nn!)^2\det(\omega).$$ Whence the result, since $\operatorname{vol}(\omega)=2^nn!\operatorname{pf}(\omega)$. $\Box$