Geometrcal proof of "Pfaffian squared is determinant"

Question

Let $(V,g)$ be an oriented Euclidian vector space of dimension $2n$, with a non-singular operator $A\in End(V)$ anti-symmetric, then $w_A(u,v)=g(u,Av)$ defines a linear symplecitc form. Define its Pfaffian by $\frac{w_A^n}{n!}=Pf(A)\mathrm{vol}$, where $\mathrm{vol}$ denotes the volume form of $V$.

You can pull back the relation $\frac{w_A^n}{n!}=Pf(A)\mathrm{vol}$ by a $B\in End(V)$. You will get

$$Pf(B^TAB)=Pf(A)\det(B). $$

Use the normal form theorem in linear symplectic geometry, we may assume $V=\mathbb{R}^{2n}$ and $(w_0,g_0,J_0)$ denotes the standard compatible triple and there's $P:(V,w_A)\to (V,w_0)$ symplectomorphism. $$w_A(u,v)=g_0(u,Av)=P^*w_0(u,v)=w_0(Pu,Pv)=g_0(J_0Pu,Pv)=g_0(P^TJ_0Pu,v)=g_0(-Au,v).$$ So, $A=P^TJ_0^{-1}P$, and $Pf(A)^2=\det(A)$.

But this proof uses the normal form theorem. Is there a more geometrical one without citing this theorem?

Answer 1

Here is another idea of how to prove it geometrically, but still using a basis theorem. $\newcommand{\innprod}[1]{\left\langle #1 \right\rangle}$

Theorem: Let $V$ be a finite-dimensional real vector space, equipped with an inner product $\innprod{-, -} \colon V \times V \to \mathbb{R}$ and a symplectic form $\omega \colon V \times V \to \mathbb{R}$. Then there exists an orthonormal basis $e_1, \ldots, e_n, f_1, \ldots, f_n$ and scalars $\lambda_1 \geq \cdots \geq \lambda_n > 0$ such that the pairing $\omega$ on the basis is given by $$ \omega(e_i, e_j) = \omega(f_i, f_j) = 0, \quad \omega(e_i, f_j) = \lambda_i.$$ Furthermore, the set of scalars is canonically determined, and when $A_\omega \colon V \to V$ is the skew-symmetric isomorphism associated to $\omega$, it acts on the basis by $A_\omega e_i = \lambda_i\;\!\;\! f_i$ and $A_\omega f_i = - \lambda_i e_i$.

Perhaps you have come across this theorem, or perhaps in a slightly different form. It is not difficult to prove, I can outline the steps if you like.

But now it is simple to check that $\det A = \lambda_1^2 \cdots \lambda_n^2$, and that $$ \omega^n(e_1, \ldots, e_n, f_1, \ldots, f_n) = n! \, \lambda_1 \ldots \lambda_n \operatorname{vol}(e_1, \ldots, e_n, f_1, \ldots, f_n)$$ hence $\operatorname{Pf}(\omega) = \pm \lambda_1 \ldots \lambda_n$ and indeed we have $\operatorname{Pf}(\omega)^2 = \det(A)$.

I would count this proof as very geometric, since the actual definition of Pfaffian (of a linear operator) depends on understanding the interaction between the orthogonal structure $\innprod{-, -}$ and the induced symplectic structure $\omega(-, -)$.

Answer 2

Here is an idea for a proof without the normal form theorem, but of unknown geometric interpretation. For a skew-symmetric $2n \times 2n$ matrix $A$ consider the following simultaneoues row and column operations. $$ C := \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix} \rightsquigarrow \begin{pmatrix} A & iA \\ iA & 0 \end{pmatrix} \rightsquigarrow \begin{pmatrix} 0 & iA \\ iA & 0 \end{pmatrix} = B^T \Omega B $$ where $\Omega = \begin{pmatrix} 0 & I_{2n} \\ -I_{2n} & 0 \end{pmatrix}$ and $B := \begin{pmatrix} iA & 0 \\ 0 & I_{2n} \end{pmatrix}$ which is again skew-symmetric. Or in other words $$ C = T^T B^T \Omega B T \qquad\text{with}\qquad T = \begin{pmatrix} I_{2n} & 0 \\ -\frac{i}{2} I_{2n} & I_{2n} \end{pmatrix} \begin{pmatrix} I_{2n} & -i I_{2n} \\ 0 & I_{2n} \end{pmatrix} $$ Then the transformation property of the Pfaffian under base change gives $$ \newcommand{Pf}{\operatorname{Pf}} \Pf(A)^2 = \Pf(C) %= \Pf(T^T B^T \Omega B T) = \Pf(\Omega) \det(B) \det(T) = \det(A) $$ where we used $\Pf(\Omega) = (-1)^\frac{2n(2n-1)}{2} = (-1)^n$ and $\det(B) = \det(iA) = (-1)^n \det(A)$.