# Is there a definition of the Pfaffian parallel to the definition of the determinant in terms of the exterior algebra?

If $V$ is a finite-dimensional vector space, we have some $A \in \operatorname{End}(V)$ then we can extend $A$ to act on the exterior algebra $\bigwedge(A)$ by setting $$A \cdot (v_1 \wedge \cdots \wedge v_k) := (A v_1) \wedge \cdots \wedge (A v_k).$$ If $V$ is $n$-dimensional then $\bigwedge\nolimits^n V$ is one-dimensional, so the action of $A$ on this space is just scalar multiplication. Of course, we call this scalar the determinant.

If we have a skew-symmetric bilinear form $Q$ on $V$, we can regard it as an element of $(V \wedge V)^\ast$, and given the superficial similarity between the definitions of the Pfaffian and the determinant it seems as though we should be able to give a definition of the Pfaffian that involves lifting $Q$ to some larger algebra, maybe $\bigoplus_i \bigwedge\nolimits^{2i} V \subseteq \bigwedge(V)$. But I can't see a reasonable way of doing this.

• What's wrong with using the $\subseteq$ you gave? – blue Jul 9 '14 at 23:08
• Well, I guess the natural thing would be to try to define, say, $Q \cdot v_1 \wedge v_2 \wedge v_3 \wedge v_4$ as $Q(v_1 \wedge v_2) Q(v_3 \wedge v_4)$ but that's not well-defined. – Daniel McLaury Jul 9 '14 at 23:45
• Seems to me we can use exterior products to define the Pfaffian via $$\frac{1}{n!}{\bigwedge}^n\left(\frac{1}{2}\sum_{i,j}Q(v_i,v_j)v_i\wedge v_j\right)={\rm pf}(A)(v_1\wedge \cdots\wedge v_{2n}).$$ This is what's given on Wikipedia, essentially. – blue Jul 9 '14 at 23:59
• Hmmm... unless I'm mistaken, that gives a $k$-linear map from $\bigoplus_i \bigwedge^{2i} V$ to $k$ but not a $k$-algebra homomorphism. The analogous construction for the determinant gave a $k$-algebra automorphism of $\bigwedge(V)$. Maybe the lesson here is that that was just a coincidence, though. – Daniel McLaury Jul 10 '14 at 0:04