Posted by: Steven Sam | June 24, 2009

Pfaffians and Plücker ideals

In this post, I want to discuss Pfaffians, a topic which I wish I had learned about as an undergraduate. I’m very interested in syzygies of ideals and such, and every now and then Pfaffians come up, so if only I knew what they were! Now that I know, I want to explain what they are and how they’re related to Plücker ideals.

Everything will be over the field K. If an n x n matrix has rank < r, then this can be checked by showing that all of the r x r submatrices of it have determinant 0. In particular, since these r x r minors are polynomials in the entries of the matrix, this says that the set of all matrices of rank < r is an algebraic subset Y of the space of all matrices. That it's irreducible can be seen by the following argument: let X be the space of n x n matrices, and let Gr(r-1, n) be the Grassmannian of r-1 planes in n-dimensional affine space. Then consider the subset Z of Gr(r-1, n) x X given by \{(W, f) \mid \text{image}(f) \subseteq W \}. If R is the tautological subbundle on Gr(r-1, n), then Z = \mathrm{Hom}(K^n, R) is a vector bundle over Gr(r-1, n) and hence is irreducible. But the image of Z under the projection \text{Gr}(r-1, n) \times X \to X is Y, so Y is also irreducible.

It's not so clear that the ideal generated by the r x r minors of a generic (= entries are algebraically independent variables over K) n x n matrix is radical, but this turns out to be true (one way to show that this is true is to find an explicit Gröbner basis for it).

But what if we only care about skew-symmetric matrices? To check if a matrix has rank < r, we could do the same as above, but the ideal generated by the r x r minors of a generic skew-symmetric matrix will NOT be radical. One problem already is that the determinant of any skew-symmetric matrix is always a perfect square in the field K, and hence our ideal should contain these square roots, which are called Pfaffians.

First I’d like to prove this last statement. Say M is a skew-symmetric n x n matrix. We want to find a number r such that M can be transformed isometrically into the matrix \left[ \begin{matrix} 0 & I_r & 0 \\ -I_r & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] where I_r is the r x r identity matrix (r could be 0, in which case these rows don’t appear), and the bottom row consists of n-2r rows of zeroes. First, M represents a bilinear form \langle\ ,\ \rangle \colon K^n \times K^n \to K given by \langle e_i, e_j \rangle = m_{i,j}, where e_1, \dots, e_n is the standard basis for K^n. The skew-symmetry of M means that this form is skew-symmetric. Let L be the kernel of this form, i.e., the set of all x such that \langle x, y \rangle = 0 for all y. Then the form restricted to any complementary subspace L’ of L inside of K^n will be nondegenerate. So for any nonzero x in L’, we can find y in L’ such that \langle x, y \rangle = 1. Repeating this on the orthogonal complement of L’ inside of L, we can find a basis x_1, \dots, x_r, y_1, \dots, y_r for L’ such that \langle x_i, x_j \rangle = \langle y_i, y_j \rangle = 0 and \langle x_i, y_j \rangle = \delta_{ij} is the Kronecker delta. Representing M in this basis gives the desired form. Note that if B is the matrix whose columns are x_1, \dots, x_r, y_1, \dots, y_r, then the matrix representing the bilinear form \langle\ ,\ \rangle with respect to this new basis is B^t M B instead of B^{-1} M B.

This says three things: (1) the rank of M is always even, and (2) \det(B^t) \det(M) \det(B) = \det(B)^2 \det(M) is either 0 or 1, so \det(M) is the square of something, which we call the Pfaffian of M, and (3) if we want to test the condition that rank M < r, we need only check if the principal r x r minors (i.e., when the row indices and column indices are the same) are 0. Since these submatrices are themselves skew-symmetric, they also have Pfaffians, so it's equivalent to check if they are 0.

Do these give polynomial equations? Let M now be a generic skew-symmetric matrix with algebraically independent entries x_{ij} for 1 \le i < j \le n, so that x_{ji} = -x_{ij}, and let K be the function field {\bf Q}(x_{ij}). By the above, the determinant of M is a square of something in {\bf Q}(x_{ij}), which we denote by Pf(M). But the determinant of M is a polynomial in {\bf Z}[x_{ij}], so by the fact that {\bf Z}[x_{ij}] is a UFD and Gauss' lemma, this square root actually lives in {\bf Z}[x_{ij}]. There's some ambiguity up to sign regarding which root to take, so we just pick one. So we see that the Pfaffians are polynomials in the entries of M, and hence the locus of skew-symmetric matrices of rank < r is also a subvariety of the space of skew-symmetric matrices. Since the rank of M must always be even, we should assume that r is odd. It turns out that the ideal generated by the Pfaffians of the (r+1) x (r+1) principal minors of M (just call these the (r+1) x (r+1) Pfaffians of M) will be radical.

Now that we've done that, how are Pfaffians related to Plücker ideals? First I'll say what those are: given the Grassmannian Gr(r, n), we can embed it inside of \binom{n}{r}-1 dimensional projective space by representing a subspace by an r x n matrix and then sending it to the \binom{n}{r}-tuple of its r x r minors. The homogeneous ideal which defines the image is a Plücker ideal I_{r,n}. Of course, we could forget that this ideal is homogeneous, and then it would define a subvariety (the affine cone) of \binom{n}{r} dimensional affine space. In the case that r=2, the elements of \binom{n}{2} dimensional affine space can be identified with skew-symmetric n x n matrices (there are \binom{n}{2} elements above the diagonal), and the image of the Grassmannian corresponds to those which have rank at most 2 (because they are minors of a 2-dimensional subspaces). Hence the Plücker ideal I_{2,n} is generated by the 4 x 4 Pfaffians of a generic n x n skew-symmetric matrix.

Why is this relevant? It's known how to write down the minimal free resolutions for the Pfaffians of a generic skew-symmetric matrix (you can find this in Section 6.4 of Jerzy Weyman's book Cohomology of Vector Bundles and Syzygies), so this gives for free the case of the Plücker ideals when r=2. The general case r>2 is not known, however. The case r=3 and n=6 can be done in Macaulay 2 in a few seconds (use the command res Grassmannian(2,5)), but I think any larger examples are computationally too expensive. In principal, they can be written down by hand using the techniques of Section 7.3 of the above book. There, the difficulties arise in trying to solve some plethysm problems for the general linear group. So it’s likely that a closed form solution will never be attainable.

-Steven

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Responses

  1. Dear Steven,
    this is a very interesting post. Do you know how to prove that if k>2 the affine cone over Gr(k,n) is not isomorphic to a determinantal variety?
    Thank you!


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