# Is There a Basis Free Definition of the Pfaffian

$\DeclareMathOperator{\pf}{pf}$ 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.

• Is there a basis free definition of the determinant? I don't think so... Jul 28 '15 at 10:21
• @darijgrinberg If $T:V\to V$ is a linear operator on an $n$-dimensional vector space $V$, then we know that $\bigwedge^n T:\bigwedge^n V\to \bigwedge^n V$ satisfies $\bigwedge^n T= cI$ for some constant $c$. This constant is defined to be the determinant of the operator $T$. Jul 28 '15 at 10:24
• Ah! I thought the proof of welldefinedness should also be basis-free. Jul 28 '15 at 10:25
• @darijgrinberg Proof of well-definedness of the determinant? If one goes by defining the determinant of $T$ by first forming the matrix representation $M$ of $T$ and then writing $\det T=\det M$, then yes, there is a well-definedness issue. Of course, in the exterior power approach there is no such problem. Jul 28 '15 at 10:28
• In the exterior power approach the problem is to verify that $\wedge^n V$ is nonzero. Aug 1 '16 at 9:54

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.
• However, it is well-defined up to a choice of basis of $\Omega^{2n}(V)$ (the space of alternating forms on $V$) - essentially an "orientation" on $V$: you can identify skew-symmetric or skew-adjoint maps $T$ with alternating $2$-forms via $\omega(v,w) := \langle v, Tw \rangle$, where $\langle -,- \rangle$ is a fixed bilinear form. Then the Pfaffian becomes the map $\Omega^2(V) \rightarrow \Omega^{2n}(V)$, $\omega \mapsto \frac{1}{n!} \omega^n$; essentially as you mentioned in the question. Jul 29 '16 at 17:08
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.