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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.
Here is an elaboration of Qiaochu's comment above:
A $2n\times 2n$ matrix $A$ induces a pairing (say on column vectors), namely $$\langle v,w \rangle := v^T A w.$$ Thus we can think of $A$ as being an element of $(V\otimes V)^*$ (which is the space of all bilinear pairings on $V$), where $V$ is the space of $2n$-dimensional column vectors.
If $A$ is skew-symmetric, then this pairing is anti-symmetric, and so we can actually regard $A$ as an element of $\wedge^2 V^*$. We can then take the $n$th exterior power of $A$, so as to obtain an element of $\wedge^{2n} V^*$. This latter space is $1$-dimensional, and so if we fix some appropriately normalized basis for it, the $n$th exterior power of $A$ can be thought of just as a number. This is the Pfaffian of $A$ (provided we chose the right basis for $\wedge^{2n} V^*$).
How does this compare to the usual description of determinants via exterior powers:
For this, we regard $A$ as an endomorphism $V \to V$, which induces an endomorphism $\wedge^{2n} V \to \wedge^{2n} V$, which is a scalar (being an endomorphism of a $1$-dimensional space); this is $\det A$.
So now we see where the formula $\det(A) =$ Pf$(A)^2$ comes from: computing the determinant involves taking a $2n$th exterior power of $A$, while computing the Pfaffian involves only taking an $n$th exterior power (because we use the skew-symmetry of $A$ to get an exterior square "for free", so to speak).
The sorting out the details of all this should be a fun exercise.
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).$\Box$
--
$^1$ 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 $n$.
Let $(V,\Omega)$ be a vector space of dimension $2n$ with a nonzero $\Omega\in \bigwedge^{2n} V^*$.
Let $A$ be an alternating bilinear form on $V$. Then $A\in \bigwedge^2 V^*$ and $\Omega$ spans $\bigwedge^{2n} V^*$.
Define the Pfaffian $Pf(A)$ of $A$ w.r.t. $\Omega$ by $$Pf(A)\Omega:=\frac{A^{\wedge n}}{n!}\in \bigwedge { }^{2n} V^*$$
If a basis is chosen and $A$ is represented by a matrix, then clearly $Pf(A)$ is a homogeneous polynomial of degree $n$ in the coefficients of $A$.
Let now $\omega$ be a symplectic form on $V$. Then there exists $B\in \text{End}(V)$ such that $B^*\omega=A$. If $A$ is nondegenerate (thus also symplectic), this amounts to the fact that any two linear symplectic forms on $V$ are equivalent.
Since $A^{\wedge n}=(B^*\omega)^{\wedge n}=B^*\omega^{\wedge n}=\det(B)\omega^{\wedge n}$ we have $Pf(A)=\det(B)$, if $\omega$ is chosen such that $\omega^{\wedge n}/n!=\Omega$.
Fix an inner product on $V$. Then $\omega$ can be viewed as an endomorphism of $V$. Since $B^T\omega B=A$, we have $\det(A)=\det(B)^2\det(\omega)$ and thus $$ \det(A)=Pf(A)^2\det(\omega).$$ If the inner product is such that $\det (\omega)=1$, this reduces to $\det(A)=Pf(A)^2$.
By continuity, we can assume that $A$ can always be reduced to block diagonal form with blocks $$ \left(\begin{matrix} 0 &\lambda_i\cr -\lambda_i &0 \end{matrix} \right) $$ on the diagonal. In this case computing the determinant gives $\prod \lambda^2_i$ and computing the Pfaffian gives $\prod \lambda_i$, so the determinant is the square of the Pfaffian.
