Is There a Basis Free Definition of the Pfaffian

Question

$\DeclareMathOperator{\pf}{pf}$ I recently came across a delightful fact that:

The determinant of a $2n\times 2n$ skew-symmetric matrix is a the square of a certain polynomial called the pfaffian.

I was looking for a "conceptual proof" of the above. So naturally I first wanted to understand pfaffians. The description of pfaffian I have seen (here) is not very satisfactory to me.

Question. Is there a notion of the pfaffian of a linear operator?

A promising description of the Pfaffian is available on the above mentioned article: Assume for simplicity that the entries of $M$ are complex numbers, and the $ij$-th entry be written as $a_{ij}$. Let $e_1, \ldots, e_{2n}$ be the standard basis of $\mathbf C^{2n}$. To $M$ we associate a bivector $\omega=\sum_{i<j}a_{ij}\ e_i\wedge e_j$ and let $\omega^n$ denote the wedging of $\omega$ with itself $n$ times. Then $$\frac{1}{n!}\omega^n= \pf(M)e_1\wedge \cdots \wedge e_{2n}$$

If you know a nice proof of the fact mentioned above then please share it.

Answer 1

The Pfaffian only makes sense for skew-symmetric matrices in even dimensions, so there's no hope of generalizing it as far as an arbitrary linear operator.

One coordinate-free description of the input to the Pfaffian is that it is an element of the orthogonal Lie algebra $\mathfrak{so}(2n)$; in this incarnation the Pfaffian appears as an invariant polynomial on $\mathfrak{so}(2n)$ (so it is invariant under orthogonal, rather than arbitrary, changes of coordinates). It is in fact the polynomial which corresponds via Chern-Weil theory to the Euler class; this is more or less the content of the Chern-Gauss-Bonnet theorem. The Pfaffian squaring to the determinant corresponds via Chern-Weil theory to the Euler class squaring to the top Pontryagin class.

It should be possible to give a conceptual proof that the Pfaffian squares to the determinant using the wedge product definition. As in user357105's comment to his answer, the setting to work in is a finite-dimensional real (for simplicity) inner product space $V$; here you can identify skew-symmetric linear operators $T : V \to V$ with elements of $\wedge^2 V$ (both of which can in turn be identified with the Lie algebra $\mathfrak{so}(V)$), and you use this identification to relate the determinant and the Pfaffian. I haven't worked through the details, though.

Answer 2

The Pfaffian is an invariant of matrices, but not of underlying linear operators. Two skew-symmetric matrices that are similar can have different Pfaffians: for example, $$\mathrm{Pf}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 1, \; \; \mathrm{Pf} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = -1.$$ In other words, the Pfaffian depends not only on the linear operator, but also on a choice of basis - so there can be no basis-free definition.

Answer 3

The Pfaffian $Pf(T)$ of a skew-adjoint operator $T$ of an oriented Euclidean vector space $(V,\langle\cdot,\cdot \rangle)$ of dimension $2n$ is by definition $$Pf(T)\Omega:=\frac{\beta_T^n}{n!} \in \bigwedge { }^{2n} V^*$$ where $\Omega$ is the normalized volume form and $\beta_T\in \bigwedge^2 V^*$ is the alternating bilinear form on $V$ associated to $M$ via the inner product, thus $\beta_T(v,w)=\langle Tv,w\rangle$ for all $v,w \in V$.

If a basis is chosen and $T$ is represented by a matrix, then clearly $Pf(T)$ is a homogeneous polynomial of degree $n$ in the coefficients of $T$.

We have $$Pf(C^tTC)=\det(C)Pf(T)$$ since $\beta_{C^tTC}=C^*\beta_T$ and $(C^* \beta_T)^n=\det(C)\beta_T^n$.

Claim: $\det(T)=Pf(T)^2.$

To prove this claim, let $\omega$ be a symplectic form on $V$. Then there exists $B\in \text{End}(V)$ such that $B^*\omega=\beta_T$. If $\beta_T$ is nondegenerate (thus symplectic), this amounts to the fact that any two linear symplectic forms on $V$ are equivalent.

The symplectic form $\omega$ on $V$ can be chosen to satisfy $\frac{\omega^n}{n!}=\Omega$ and $\det(T_\omega)=1$. In this situation we show below that $Pf(T)=\det(B)$ and $\det(T)=\det(B)^2$, thus yielding the claim.

Indeed, we have $Pf(T)=\det(B)$ since $Pf(T)\Omega=\frac{(B^*\omega)^n}{n!}=\det(B)\Omega$ and $\frac{\omega^n}{n!}=\Omega$.

On the other hand, we have $T=T_{B^*\omega}=B^tT_\omega B$ where $T_b$ is the skew-adjoint operator associated to an alternating bilinear form $b$ via the inner product (so $T_b$ is such that $\langle T_b v,w\rangle=b(v,w)$ for all $v,w\in V$).

Thus $\det(T)=\det(B^tTB)=\det(B)^2\det(T_\omega)=\det(B)^2$ since $\det(T_\omega)=1.$