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Given an explicit polynomial, is there any kind of trick/algorithm to check whether it is a pfaffian of a matrix with linear entries?

The pfaffian can be defined as $\sqrt{{\rm det}(A) } $ when $A$ is skew symmetric, or explicitly $${\rm pf}(A) = \frac{1}{2^n n!}\sum\limits_{\sigma \in S_{2n}}{\rm sgn} (\sigma)\prod\limits_{j=1}^n a_{\sigma (2j -1 ),\sigma(2j)}.$$

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What is a Pfaffian polynomial? – Qiaochu Yuan Jun 25 '11 at 17:26
Please try to improve the question! As it stands, it might (and probably should) get closed for lack of clarity, motivation etc. – Gjergji Zaimi Jun 26 '11 at 3:48
As motivation for related questions, Mumford gives, in his first paper on Prym varieties, an explicit Pfaffian of linear forms, which defines the tangent cone to the theta divisor of a Prym variety precisely when that Pfaffian is not identically zero. When this occurs is more complicated however and was only settled some 25 years later. – roy smith Jun 26 '11 at 15:51
I am sorry for the sloppyness of the question. I assumed, without stating, that the matrix had linear entries. Now I've edited it. – IMeasy Jun 26 '11 at 17:43
The Pfaffian of a skew symmetric matrix is a polynomial in the entries of the matrix, whose square is the determinant of the matrix, as given in your explicit formula. But the Pfaffian of a real matrix can be positive or negative, so the Pfaffian is not the square root, but only one of the two square roots, of the determinant. – Ben McKay Apr 19 at 16:09
up vote 16 down vote accepted

As Bruce Westubury noticed, the answer to this question is trivial as it is stated.

Surprisingly enough, however, the situation becomes very interesting when one considers representations of homogeneous polynomials as pfaffians of matrices with linear entries.

More precisely, let us consider the following version of the question:

Question. Let $F \in k[x_0, \ldots, x_n ]$ be a homogeneous polynomial of degree $d$. Does there exist a symmetric (resp. antisymmetric) matrix $M$, whose entries are linear forms, such that $$\det(M)=F \quad (\textrm{resp}. \ \textrm{Pf}(M)=F)?$$

This problem is studied in detail in Beauville's paper [Symmetric determinantal hypersurfaces, Michigan Mathematical Journal 48 (2000)], where the existence of a symmetric or pfaffian representation is related to the existence of certain vector bundles on the projective variety $X \subset \mathbb{P}^n_k$ defined by $F=0$.

Among other things, Beauville proves the following results (when $k= \mathbb{C}$):

$\bullet$ A general polynomial of degree $d$ admits a determinantal representation if and only if $$n=2 \quad \textrm{or}$$ $$n=3 \quad \textrm{and} \quad d\leq 3.$$

$\bullet$ A general polynomial of degree $d$ admits a pfaffian representation if and only if $$n=2,$$ $$n=3 \quad \textrm{and} \quad d \leq 15 \quad \textrm{or}$$ $$n=4 \quad \textrm{and} \quad d \leq 5.$$

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The Pfaffian of $\begin{pmatrix} 0 & A \\\\ -A & 0\end{pmatrix}$ is $\det A$. Since any polynomial is a determinant this means any polynomial is a Pfaffian.

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Now that the question makes sense, let me mention the following result (you can find it in Eisenbud, Exercise 20.17) which hinted at what Francesco wrote.

Let $R = k[[x,y,z]]$ and $f\in m=(x,y,z)$. Then $f$ is a determinant of a matrix (of size at least $2$) with entries in $m$ iff $R/(f)$ is not a UFD!

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Something is wrong here. Take $f=z$. Then $R/(f) = k[x,y]$ is a UFD, and $f = \det \ (z)$. – David Speyer Jun 26 '11 at 19:52
David, thanks. Of course the size of the matrix has to be bigger than one, I fixed it now. – Hailong Dao Jun 26 '11 at 20:05

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