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 . If R is the tautological subbundle on Gr(r-1, n), then is a vector bundle over Gr(r-1, n) and hence is irreducible. But the image of Z under the projection 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 where 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 given by , where is the standard basis for . 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 for all y. Then the form restricted to any complementary subspace L’ of L inside of will be nondegenerate. So for any nonzero x in L’, we can find y in L’ such that . Repeating this on the orthogonal complement of L’ inside of L, we can find a basis for L’ such that and is the Kronecker delta. Representing M in this basis gives the desired form. Note that if B is the matrix whose columns are , then the matrix representing the bilinear form with respect to this new basis is instead of .

This says three things: (1) the rank of M is always even, and (2) is either 0 or 1, so 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 for , so that , and let K be the function field . By the above, the determinant of M is a square of something in , which we denote by Pf(M). But the determinant of M is a polynomial in , so by the fact that is a UFD and Gauss' lemma, this square root actually lives in . 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 dimensional projective space by representing a subspace by an r x n matrix and then sending it to the -tuple of its r x r minors. The homogeneous ideal which defines the image is a Plücker ideal . Of course, we could forget that this ideal is homogeneous, and then it would define a subvariety (the affine cone) of dimensional affine space. In the case that r=2, the elements of dimensional affine space can be identified with skew-symmetric n x n matrices (there are 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 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

In this post I’d like to say what a P-partition is, what a Gorenstein ring is, and plan to discuss a chain of topics which will lead from one to the other. Roughly the first half of this post can be found in Section 4.5 of Richard Stanley’s Enumerative Combinatorics, Vol 1.

First, let’s start with P-partitions. Here P is a finite poset with p elements. The standard definition of partition is of course a way of decomposing a positive number into a sum of positive numbers, i.e., 5 = 1 + 3 + 1. Since we don’t care about the order, we just list them in descending order, so the example becomes (3,1,1). Another way to interpret this is in terms of posets: let [m] be the usual ordering on the set {1, 2, …, m}. Then a partition of n using at most m parts is the same as an order-reversing map , where is the natural numbers (including 0) under the usual order, such that . Now replace [m] with an arbitrary partition P and we can talk about P-partitions. Also, we’ll say that a P-partition is strict if from before is strictly order-reversing.

Given a poset P, let a(n), resp. , be the number of P-partitions, resp. strict P-partitions of n, and define the generating functions and . I’ll note two facts: as rational functions, we have

where is a polynomial of degree strictly less than , so that is a quasi-polynomial. Second, we have a reciprocity law:

,

and this implies that , where a(-n) makes sense since it is a quasi-polynomial. There is a refinement of this reciprocity which I won’t mention.

Here’s another way to think of P-partitions: think of them as points inside of . But not just any points: the order-reversing requirement gives some linear inequalities, so they live in some rational cone. Then a(n) is the number of lattice points of this cone with the intersection of the hyperplane . These intersections will be polytopes, but in general not ones with integer vertices.

But there is a way to get integral polytopes. Instead of P-partitions of n, let be the number of P-partitions such that for all x in P. This is the order polynomial of P (it’s not that bad to show that it is indeed a polynomial. Similarly, let be the number of strict P-partitions such that for all x in P. Then we can interpret as the number of lattice points in with inequalities and for all inequalities in P. This is a polytope O(P,n) (called the order polytope of P), and in fact, it will have integral vertices. We can even describe what they are: define a filter F of P to be subset such that if and y is in F, then so is x. Then each vertex is $\sum_{x \in F} x$ where F is a filter. Then the integer points in O(P,n) are in bijection with order-preserving maps , which are in turn in bijection with order-reversing maps. In this case, the order polynomial is a special case of the Ehrhart polynomial of a polytope O(P) = O(P,2). Since counts the interior points of O(P,n+1), reciprocity for Ehrhart polynomials implies that .

Well now that we have an integral polytope, we also get a toric variety. This is the projective variety corresponding to the Ehrhart ring, which is defined as follows. Given an integral polytope and a field K, let K[Q] be the K-vector space with basis elements corresponding to integer points , where dP is the dth dilation of Q. These are multiplied just like monomials are multiplied, and we grade K[Q] by the degree of z. Then by definition, the Hilbert function of K[Q] is the Ehrhart polynomial of Q.

Now let’s give some properties of the order polynomial of P which might tell us some information about the Ehrhart ring K[O(P)] (and its toric variety). First, let L be the length of the longest chain of P. Then for . We say that P is graded if for any given elements x and y, all maximal chains between x and y have the same length. Then P is graded if and only if for all m. We can see this as saying that the number of interior integer points of O(P, L+m) is the same as the number of integer points of O(P,m) for all m. Combined with the vanishing statements, this is also equivalent to saying that when we write the generating function as a rational function, then the numerator polynomial has palindromic coefficients. Since this is the Hilbert series of K[O(P,2)], we now mention

Theorem (Stanley). If R is a Cohen–Macaulay domain over a field K, then R is Gorenstein if and only if the numerator polynomial of its Hilbert series has palindromic coefficients.

A proof can be found in Stanley’s paper Hilbert functions of graded algebras, along with other interplay between numerical conditions of Hilbert series and properties of the ring.

So I’ve fulfilled my promise of connecting the two topics in the title, but I need to say what a Gorenstein ring is! The significance comes from Serre duality. We’ll give a geometric definition first. For a variety X of dimension n, X is Gorenstein if the canonical bundle is a line bundle, where is the cotangent bundle of X. Of course, X nonsingular implies that X is Gorenstein. For a finitely generated domain over K, we say it is Gorenstein if its corresponding affine variety is Gorenstein. Chasing through all of the above, we have conclude that if P is a graded poset, then the affine cone of the toric variety associated to its order polytope (this means the affine variety of K[O(P)] forgetting that its graded) is Gorenstein (what a mouthful!). In other words, the toric variety is arithmetically Gorenstein (this implies that it’s Gorenstein also).

Now an algebraic definition says that a local K-algebra R of dimension n is Gorenstein if and only if is 0 for , and is K for i=n. And then a ring is Gorenstein if all of its localizations are Gorenstein. There’s a lot more equivalent definitions, which you can find on the wiki article.

Finally, here’s something which connects to my previous posts on Boij–Söderberg theory. If is Gorenstein where I is a homogeneous ideal, then the graded Betti table exhibits symmetry (we may have to assume that I is prime, but I’m not sure). To be more precise, if we write the table with the convention that is the rank of the ith syzygy module in degree -i-j, and the table has c+1 columns and r+1 rows, then (assuming that the top left corner is entry (0,0)). In fact, this is another characterization.

And now we have a lot of examples of symmetric Betti tables!

-Steven

It’s been a while since the last post, but I want to give an indication of how the Eisenbud and Schreyer proved the Boij–Söderberg conjectures. This fell into roughly 3 steps. There are some constructions that are involved which I will not mention, but may mention in a future post. In particular, I will take it as a given that for any given degree sequence d, there exists a Cohen–Macaulay module M whose Betti diagram is pure with degree sequence d. Recall the setup from the previous post: we picked degree sequences and to restrict our attention to a finite-dimensional space of Betti tables.

Step 1: Identify the exterior facets of the cone spanned by the pure diagrams.

This step was done by Boij and Söderberg in their original paper. Recall that we put a partial ordering on the set of degree sequences given by pointwise comparison and that the cone spanned by the pure diagrams offers a geometric realization of this poset. In particular, the facets are given by chains of degree sequences which are obtained uniquely by removing a single element from a maximal chain . The uniqueness part means that there do not exist two different maximal chains such that one can remove a single element from each to get the same resulting chain. Let i denote the index of the degree sequence that was removed. They classified the exterior facets into three types:

Case 1: We remove the maximal or minimal element.

Case 2: The degree sequences and differ in exactly one entry.

Case 3: The degree sequences and differ in exactly two entries.

In both cases 1 and 2, there is a unique coordinate x of the Betti table that is nonzero for but zero for the other degree sequences in the chain. In this case, x gives a linear functional which is positive on the set of all Betti tables and is zero on the facet in question. The nontrivial case of course is case 3. The first thing Eisenbud and Schreyer show in this case is that there is a hyperplane which contains this facet and those degree sequences which are greater than or equal to , and furthermore that this hyperplane is unique. Similarly, one can obtain a hyperplane which contains all degree sequences less than or equal to . The construction of the linear functionals is inductive and we won’t repeat it here. The point is that given this uniqueness, one can hope to reconstruct these linear functionals in a different way which illucidates that they are nonnegative on the set of Betti tables. We sketch this in the next section.

Step 2: Define a bilinear pairing between Betti tables and cohomology tables.

This may seem to stray from our goal, but in fact, this is the crucial observation to showing that the case 3 linear functionals in Step 1 are in fact positive on all Betti tables. The point will be that it is easy to show positivity for this pairing, and then one need only pick the right cohomology tables to get the desired facet-defining equations. Here of course I will sketch only the constructions and give some ideas, since there are far too many details to discuss.

First, let be a graded Betti table, and let be a graded cohomology table (i.e., the degrees of some cochain complex). Eisenbud and Schreyer introduce a bilinear functional . In the case that comes from a free resolution F and comes from a cochain complex E, we can write this sum as . Here denotes the truncation of F to degrees at least j, and   denotes the usual Euler characteristic. The next step is to examine the double complex where we think of F as having cohomological indices, i.e., . We assume that E is bounded and consists of free modules since we only need this case. Since F is a resolution, the total cohomology of this complex is zero in negative cohomological degrees (seen by using the spectral sequence that starts by computing cohomology of the rows ). But we’re really interested in some truncation of this double complex, so we go to the next page of the spectral sequence starting in the other direction (along columns) and then remove the things with positive cohomological index (they don’t appear in our bilinear form). Then in fact our bilinear form can be identified with the Euler characteristic of the total cohomology of this truncation. But we know that the total cohomology vanishes in negative degrees, and we’ve removed the positive part, so in fact our bilinear form is just the dimension of the degree 0 part of the total cohomology of the truncated double complex, and hence nonnegative. Doing some more analysis, one can determine when this total cohomology vanishes, but we’ll skip that.

Unfortunately, this isn’t quite good enough. For one thing, we have not used the fact that F should be a minimal free resolution. To get around this, Eisenbud and Schreyer introduce a second bilinear form which is some kind of augmentation of the one above. Then using nonnegativity of the original bilinear form, they show that the augmented one is nonnegative when F is a minimal free resolution, and this is form is the correct thing to proceed with.

Step 3: Find the appropriate cochain complexes.

What is needed are cochain complexes which have the right vanishing conditions (which I didn’t explain) subject to the above bilinear form. Briefly I can say where these come from. The first way to get interesting cochain complexes is via linear monads for vector bundles on (n-1)-dimensional projective space (here n is the number of variables of our polynomial ring). What this means precisely is if is a vector bundle on such that its dual bundle is a-regular (a-regular means that for i>0), then there exists a linear complex E such that (here A is our polynomial ring, and the are just some integers) whose cohomology is close to the cohomology of . To be more precise, for i < n-1, and . Without going too much deeper, I will just end by saying then that the vector bundles whose linear monads give the desired linear functionals are those with supernatural cohomology. To be more precise, this means that 1) each twist of the vector bundle has at most one nonzero cohomology group, and 2) the roots of its Hilbert polynomial are all integers. Eisenbud and Schreyer show that given any prescribed set of integers, there indeed exists a supernatural vector bundle whose roots are precisely those integers (this is in some sense, dual to the problem of construction pure CM modules with a given degree sequence). To get the ones that give our linear functionals, the roots are chosen according to the chain of degree sequences that the facet is defined by. As an unexpected bonus, one can use the bilinear functionals to show that every cohomology table of a vector bundle on is a positive linear combination of the supernatural cohomology tables.

And this is about as much as I want to say. Sorry if this post seemed much more dense than the others, but hopefully this gives some kind of picture of what goes behind this particular theorem. The interersted reader should of course consult Eisenbud and Schreyer’s paper.

I may say some more about the existence of supernatural vector bundles and pure CM modules in a later post. They exist over any arbitrary field, but in characteristic 0, one can get -equivariant sheaves and modules using the Borel–Weil–Bott theorem, and I’ll try to say something about the equivariant constructions later.

-Steven

Last time in this post, I gave an introduction to minimal resolutions over polynomial rings and stated a theorem of Eisenbud and Schreyer. This time I want to describe the significance of the Cohen–Macaulay property, and in part III, I will start explaining the proof of the Boij–Söderberg conjectures.

The first point to address is the notion of a Cohen–Macaulay module. Let’s first assume that M is a module over a local ring R with maximal ideal P (either this is an actual local ring, or R is graded, and it has a unique homogeneous maximal ideal). An M-sequence satisfies 1) multiplication by is an injective function where and for i>0, and 2) , and the depth of M, depth(P,M), is the longest length of an M-sequence. The dimension of M, denoted dim(M), is the Krull dimension of R / ann(M) where ann(M) denotes the annihilator ideal of M. In general, the inequality holds, and we say that M is Cohen–Macaulay (CM from now on) in case of equality. The ring R is CM if it is CM as a module over itself, and we extend these definitions to the global case by saying that a ring / module is CM if its localization at every maximal ideal is CM. But actually, since we will be dealing with graded modules, we will think of the polynomial ring as a “local graded ring” because it has a unique homogeneous maximal ideal.

The important point is that this is the right condition to be able to write down some nice equations. First we should point out that polynomial rings over fields are CM.  Now let M be a finitely generated module over a ring R with finite projective dimension, denoted pd(M) (this means that the shortest projective resolution we can find for M has length pd(M)). Furthermore, if either 1) R is a local ring and P is its maximal ideal, or 2) R is a finitely generated positively graded ring with a field, M is a graded R-module, and P is its maximal homogeneous ideal , then we have

Theorem (Auslander–Buchsbaum). pd(M) = depth(P,R) – depth(P,M).

In particular, if we let R be the polynomial ring in n variables over K, then depth(P,R) = n, and if in addition M is CM, this says that dim(M) = n – pd(M). How is this relevant? We’ll unfortunately need some more definitions. First, give M as above, define the Hilbert function (here we mean vector space dimension) and the Hilbert series .

Theorem (Hilbert–Serre). Let M be a finitely generated positively graded module over , and let d = dim(M). Then there exists a polynomial R(t) with integer coefficients such that

.

Now here’s where we can begin. First, we will assume that all modules we are dealing with are CM and of codimension c=n-d. Let , and write down the minimal resolution for a finitely generated graded A-module CM M (the Auslander–Buchsbaum formula tells us that the length of this resolution must be c):

.

Since the Hilbert function is additive on degree 0 exact sequences, we can write where . To get , we just note that , so we conclude that

.

Now since M is CM, the Hilbert–Serre theorem tells us that is a polynomial divided by . So this means that the polynomial must have 1 as a root with multiplicity c. This can be expressed as saying that the first c derivatives of have 1 as a root, and this gives us c linearly independent equations on the possible Betti numbers of graded CM modules of codimension c. Call these the HK (for Herzog and Kühl) equations. Furthermore, if we assume that M has a pure free resolution, then we can figure out what the Betti numbers have to be up to a multiple. I’ll spare you the details, but mention that if the degrees of i-th syzygy module is , i.e., if unless , then they must be of the form

(*)

for , and where r is some rational number.

Remember that the theorem of Eisenbud and Schreyer stated that every Betti table of a CM module (of a given codimension c) is a positive rational linear combination of pure Betti tables (of the same codimension). This last statement translates well into convex geometry where it says: in the vector space of all tables of rational numbers with c+1 columns and infinitely many rows, the cone spanned by the pure Betti tables contains all of the Betti tables of CM modules.

Of course, the space of all tables with c+1 columns is infinite-dimensional. To avoid doing infinite-dimensional convex geometry, we can always focus our attention on finite-dimensional subspaces. In particular, fix two degree sequences and . We’ll set to be the subspace of tables T such that unless and which satisfy the HK equations.

Also, define a partial ordering on the degree sequences by saying that if for all i. It is something which remains to be proved that for every degree sequence d, there exists a CM module with a pure free resolution of degree d, but we will postpone that. The key point is that one can deduce from (*) that Betti tables corresponding to degree sequences in a chain are linearly independent. The number of degree sequences in a maximal chain is since for each i, and can only differ in exactly one spot, and this difference is 1. The dimension of is (the -c comes from the fact that there are c HK equations, and they are linearly independent). So we conclude that every maximal chain forms a basis for .

Not too interesting yet. We know only that every Betti table is a linear combination of pure Betti tables at this point. But now let’s restrict to positive linear combinations. In this case, each maximal chain forms a simplicial cone, and in fact any two such cones intersect in a facet of both, so we get a simplicial fan. So what we must do is 1) identify the exterior facets of this fan, 2) find the facet-defining equations, and 3) show that these equations are nonnegative on all Betti tables.

Step 1) was done by Boij and Söderberg, and 2) can be done (although since we are working in a proper subspace of a Euclidean space, these functionals will not be uniquely determined). The key insight of Eisenbud and Schreyer was to construct the functionals in 2) in terms of a bilinear pairing between minimal free resolutions and cochain complexes. Then the facet defining equations come from very special cochain complexes: the linear monads of “supernatural” vector bundles on projective space .

I’ll explain this last paragraph and discuss what to do next in the next installment: part III.

-Steven