Let $R$ be a commutative ring (zero characteristic). Take a skew-symmetric matrix $A\in Mat^{skew-sym}(n,R)$.

  1. If $n$ is even, then $\det(A)=Pf^2(A)$ and there exists the "Pfaffian adjugate/adjoint" matrix satisfying: $A\times adj^{Pf}(A)=Pf(A)Id$. What is the standard notation for this $adj^{Pf}(A)$? (And the standard reference? preferably some linear-algebra book)

  2. If $n$ is odd then $\det(A)=0=Pf(A)$. Still, one would like to distinguish between the generic matrices (of corank=1) and more degenerate (of corank$\ge2$). There are various ways to distinguish these two sets, i.e. various scheme structures on the stratum of corank$\ge2$ matrices, i.e. various ideals that measure the degeneracy. One such ideal is: $\sum\limits_i \det(A_{ii})$, where $A_{ii}$ is the block obtained by erasing the i'th row and column. Note that each $A_{ii}$ is skew-symmetric and of even size. Thus one might use the refined ideal: $\sum\limits_i Pf(A_{ii})$. (The two ideals have the same radical.) Any other related ideals/scheme structures?

There is no direct "Pfaffian-adjugate" in the odd case. In my case I consider the submodule $Span(UA+AU^T)\subseteq Mat(n,R)$, where $U$ runs over all the possible matrices (not necessarily skew-symmetric). I need the largest ideal $J$ satisfying: $Span(UA+AU^T)\supseteq Mat^{skew-sym}(n,J)$. For $n=3$ one easily sees: $J=\sum\limits_i Pf(A_{ii})$. But for $n\ge5$ I can only prove: $J\supseteq\sum\limits_i Pf^2(A_{ii})$.

Any better bounds? (And the references? Probably this is well known.)

ps. To the moderators: maybe it's good to add the tag "Pfaffians"?

pps. Regarding the last question, indeed for odd $n$ there holds: $Span(UA+AU^T)\supseteq Mat^{skew-sym}(n,Pf_{n-1}(A))$, where $Pf_{n-1}(A)=\sum\limits^n_{i=1} Pf_{n-1}(A_{ii})$, the sum of Pfaffian ideals of skew-symmetric 'central' blocks of size (n-1). This can be proved 'directly' (many thanks to Vijay Kodiyalam) or deduced from the much deeper statement of Eisenbud-Buchsbaum, 1977 about the Gorenstein ideals of height 3)


1 Answer 1


These questions are all answered in terms of the exterior algebra over the ring $R$ of the free module $M=R^n$. Namely, the skew-symmetric matrices naturally live in $\Lambda^2(M) \simeq R^N$ where $N=\tfrac12n(n{-}1)$. The top exterior power is $\Lambda^n(M)\simeq R$.

If $n=2m$, and $A\in\Lambda^2(M)$ is given, then one finds that $A^m = m!\, \mathrm{Pf}(A)\,\omega$ where $\omega$ is the generator of $\Lambda^{2m}(M) = \Lambda^n(M)\simeq R$. Moreover, what you are calling the Pfaffian adjugate is just the element $A^{m-1}/(m{-}1)!\in \Lambda^{2m-2}(M) = \Lambda^{n-2}(M)$, which is naturally dual to $\Lambda^2(M)$.

If $n=2m-1$, then $A^{m-1}/(m{-}1)!\in \Lambda^{2m-2}(R) = \Lambda^{n-1}(R)$. Since $\Lambda^{n-1}(M)$ is naturally paired with $\Lambda^1(M) = M = R^{2m-1}$, it follows that the entries of $A^{m-1}/(m{-}1)!$ constitute $2m{-}1$ polynomials in the entries of $A$ that are homogeneous of degree $m{-}1$, and their vanishing measures the 'first degeneracy' of $A$, i.e., their vanishing is how you detect when $A$ has corank more than $1$. Call the ideal they generate $I_1$. More generally, you can look at the coefficients of $A^{m-k}/(m{-}k)!\in \Lambda^{2m-2k}(R)$ which are polynomials of degree $m{-}k$ in the entries of $A$ and these generate an ideal $I_k$. You have the decreasing sequence of ideals $I_{m-1}\supset I_{m-2}\supset \cdots\supset I_1$, which cut out the nested 'rank' conditions.

All of this should be found in any good treatment of exterior algebra.

  • $\begingroup$ Excellent, thanks! Still: what about the biggest ideal $J$, satisfying $Span(UA+AU^T)\supseteq Mat^{skew−sym}(n,J)$? Is this also written down somewhere? $\endgroup$ Commented Jan 15, 2016 at 18:47

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