TITLE: Existence of the Pfaffian?
QUESTION [16 upvotes]: Consider a square skew-symmetric $n\times n$ matrix $A$. We know that $\det(A)=\det(A^T)=(-1)^n\det(A)$, so if $n$ is odd, the determinant vanishes.
If $n$ is even, my book claims that the determinant is the square of a polynomial function of the entries, and Wikipedia confirms this. The polynomial in question is called the Pfaffian.
I was wondering if there was an easy (clean, conceptual) way to show that this is the case, without mucking around with the symmetric group.

REPLY [4 votes]: Here is an approach using (possibly complex) Grassmann variables and Berezin integration$^1$ to prove the required relation $${\rm Det}(A)~=~{\rm Pf}(A)^2. \tag{1}$$ This approach isn't purely conceptional, but at least it is easy, we don't fudge the overall sign, we don't muck around much with the symmetric group, and Grassmann variables do implement exterior calculus.

Define the Pfaffian of a (possibly complex) antisymmetric matrix $A^{jk}=-A^{kj}$ (in $n$ dimensions) as$^2$
$$ \begin{align}
{\rm Pf}(A)&~:=~\int \!d\theta_n \ldots d\theta_1~ 
e^{\frac{1}{2}\theta_j A^{jk}\theta_k}
\cr &~=~(-1)^{\lfloor\frac{n}{2}\rfloor} \int \!d\theta_1 \ldots d\theta_n~ 
e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr &
\cr &~=~(-1)^{\frac{n}{2}} \int \!d\theta_1 \ldots d\theta_n~ 
e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr 
\cr &~=~i^n \int \!d\theta_1 \ldots d\theta_n~ 
e^{\frac{1}{2}\theta_j A^{jk}\theta_k}\cr 
&~=~ \int \!d\theta_1 \ldots d\theta_n~ 
e^{-\frac{1}{2}\theta_j A^{jk}\theta_k}.\end{align}
 \tag{2}$$
In the last equality of eq. (2), we rotated the Grassmann variables $\theta_k\to i\theta_k$ with the imaginary unit.  
Define the determinant as
$$ {\rm Det}(A)~:=~\int \!d\theta_1 ~d\widetilde{\theta}_1 \ldots d\theta_n ~d\widetilde{\theta}_n~ e^{\widetilde{\theta}_j A^{jk}\theta_k}
. \tag{3}$$
It is not hard to prove via coordinate substitution that eq. (3) indeed reproduces the standard definition of the determinant.
If we make a change of coordinates 
$$ \theta^{\pm}_k~=~ \frac{\theta_k\pm \widetilde{\theta}_k}{\sqrt{2}}, \qquad k~\in~\{1,\ldots,n\},\tag{4} $$
in eq. (3), the super-Jacobian becomes $(-1)^n$.
Therefore we calculate
$$\begin{align} {\rm Det}(A)&\stackrel{(3)+(4)}{=}~(-1)^n\int \!d\theta^+_1 ~d\theta^-_1 \ldots d\theta^+_n ~d\theta^-_n~
e^{\frac{1}{2}\theta^+_j A^{jk}\theta^+_k
-\frac{1}{2}\theta^-_j A^{jk}\theta^-_k}\cr
&~~=~\int \!d\theta^-_1 \ldots  d\theta^-_n~d\theta^+_n\ldots d\theta^+_1 ~~e^{\frac{1}{2}\theta^+_j A^{jk}\theta^+_k}
e^{-\frac{1}{2}\theta^-_j A^{jk}\theta^-_k}\cr
&~~\stackrel{(2)}{=}~{\rm Pf}(A)^2,  \end{align}\tag{5}$$
which proves eq. (1).$\Box$

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$^1$ We use the sign convention that Berezin integration $$\int d\theta_i~\equiv~\frac{\partial}{\partial \theta_i}\tag{6} $$ is the same as differentiation wrt. $\theta_i$ acting from left. See e.g. this Phys.SE post and this Math.SE post.
$^2$ The sign of the permutation $(1, \ldots, n)\mapsto(n, \ldots, 1)$ is given by $(-1)^{\frac{n(n-1)}{2}}=(-1)^{\lfloor\frac{n}{2}\rfloor}$, where $\lfloor\frac{n}{2}\rfloor$ denotes the integer part of $\frac{n}{2}$.  One may show that the Pfaffian (2) vanishes in odd dimensions $n$.