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 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

## Leave a Reply