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.

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.

In this post, I want to discuss to what extent a group’s character table determines it up to isomorphism.

First, let’s do the easy case of Abelian groups. The set of characters of an Abelian group G is itself a group called given by pointwise multiplication. In fact, G is isomorphic to (though not canonically) as follows: we can write where the are prime powers. Fix generators for each cyclic summand. For each j, the function given by sending the generator of to gives an element of order in , and the map given by is injective. Since the number of characters of G is equal to the order of G, we conclude that it is an isomorphism. To do this canonically (without picking the direct sum decomposition), we can just do it twice (picking the decomposition twice ends up cancelling the fact that we made a choice), and we get an isomorphism by sending x to the character of defined by evaluation: . This is a special instance of Pontryagin duality, which holds more generally for any locally compact Abelian group. So we know that the characters determine the group up to isomorphism in this case.

Now let’s look at the noncommutative case. The first value of n for which there exist two nonisomorphic noncommutative groups of order n is 8, in which case we have the dihedral group which is the symmetries of the square, and the quaternion group .

While I was working on my final paper for the course on quivers that I’m taking this semester, I came across the following result (the notation means the finite field with elements, and is the field of complex numbers):

Theorem. Let X be a variety defined over Z and assume for some prime p, and all r>0, that the function is obtained by plugging in into some polynomial P(t). Then P(t) has integral coefficients, and P(1) is the Euler characteristic of (computed using cohomology with compact support).

If you’re not comfortable with varieties, just think of the solution set of some collection of polynomials in multiple variables, and if you don’t know what the rest of the words mean, that won’t matter much for the stuff starting with the next paragraph. Here I’m using the notation X(F) to denote the solutions for X over F, and I’m thinking of as a complex analytic space. The proof uses -adic cohomology (with compact support) and the Grothendieck–Lefschetz trace formula. I won’t get into that, but I’d like to use this as an excuse to talk about the q-analogues of the natural numbers.

In connection with my last post on the Boij–Söderberg conjectures, I mentioned constructing equivariant supernatural vector bundles and equivariant pure Cohen–Macaulay modules using the Borel–Weil–Bott theorem. So in this post, I’d like to say something about what this theorem says and next time discuss how it can be used. I learned the stuff on Bott’s theorem from Jerzy Weyman’s book Cohomology of Vector Bundles and Syzygies [warning: there are some mistakes in the statement of Bott's theorem for general reductive groups]. Bott’s theorem is usually stated for a reductive group, but for concreteness we’ll stick with the general linear group, since that’s all we’ll need.

The setup is as follows: k denotes some field of characteristic 0, , B is the subgroup of upper triangular matrices, and T is the subgroup of diagonal matrices. Then G/B is the complete flag variety whose k-points correspond to maximal flags where . We are interested in realizing representations of G as cohomology groups of line bundles over G/B. But first we’ll state the relative version of the theorem.

A not often-mentioned skill that I have been trying to develop is the evaluation of the “usefulness” of statements. Given my horrible memory and degrading mental RAM, I must prioritize which facts to consciously absorb and which ones to skim, since I’m sure I forget at least one theorem every time I learn a new one. Luckily, books aid to an extent by labelling the important things “Theorems” and “Lemmas,” though sometimes the latter are just as important as (if not much more than) the former, and sometimes even more important items appear in the Exercises section. Something about Exercise 7.70 in Richard Stanley’s Enumerative Combinatorics 2 really struck me as “useful.” The statement is:

where the are sets of variables, the product of the hook-lengths, gives the cycle structure, and are permutations of size .

Two things make the intuition of “useful” more concrete:

One, the statement has a lot of flexibility; By the symmetry of the symmetric group and the fact that cycles are preserved under inverses, you can sum over, say, products of all ordered -tuples of by rewriting the sum on the right as over , for example, or sum over pairs which give the same product, say . Furthermore, having the choice over the variables makes a lot of specialization techniques useful. For example, you can “pick out” cycles by specializing , where is an -th root of unity, because the power sums then give unless .

Two, the statement has a lot of power; the proof of this fact has a “Stanley difficulty” of [3] (out of 5) means it contains some nontrivial usage of the symmetric function machinery already. Thus, by using it we know we are backed with some complex stuff in the background. Note when , we get the amazing Hook-length formula directly; when , by observing that summing over is really summing over a single (a special case of our discussion above on “flexibility”), we get the familiar “inner product” identity. These are two nontrivial statements without an obvious connection with each other, a pleasant surprise.

This warm “finding a hidden gem” feeling is very nice.

-Y

P.S. Well, there is a third (and less glamorous) cue, which is that at least two problems in the problem sets came with the hint: “Look at Exercise 7.70.” The power of hindsight in all its glory, I suppose. However, I liked this statement too much that I needed an excuse to write it down, so shhh.

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.

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.

I plan to continue my discussion of Boij–Söderberg theory from last time, but I’d like to make a quick detour first. This came up today in the office when I was talking with yanzhang.

Question: If I randomly pick numbers (with a uniform distribution) from the unit interval [0,1], what is the expected number of numbers that I need to pick before their sum is at least 1?

The answer for some strange reason is the number e, and I’ll illustrate this with two approaches. Both give different expressions of e, so this problem will show an equivalence between two definitions of e.

The first approach is to use calculus. For , let f(r) be the expected number of randomly selected numbers in [0,1] so that their sum is at least r, and define f(r) = 0 if r is negative. Then we have the following integral equation

.

To see this, pick a number t randomly in [0,1], and then add f(r-t). Since we need to account for all such t, we just integrate over [0,r] (I could integrate over [0,1], but remember that f(r-t) = 0 if t>r), and divide this integral by the length of [0,1], which is 1. The second inequality follows from a change of variables s=r-t. Taking derivatives of both sides, we get by the fundamental theorem of calculus, so for some constant c. Of course, f(0) = 1, so c=1. Our original question was for r=1, in which case f(1) = e, as we promised.

The second approach is the same, except we discretize everything. Let’s ask an analogous question. First fix k. If I randomly pick integers uniformly from [0,k], what is the expected number that I need before their sum is at least n, where ? Let g(n) be this expected value. Then again, we have

by the same reasoning as before. Let be the first difference operator, i.e., . Now apply to the above equation to get

.

Of course, g(-1) = 0, and rewriting this last equation gives

assuming that n>0. Since g(0) = 1, this means that . Now set k=n: If we rephrase this question, we also see that g(n) is the expected number of random selections of rational numbers with denominator n from [0,1] such that their sum is at least 1. Letting n go to infinity, we should get the answer from the previous method since the rationals form a dense subset of the reals.

So these two approaches show the equivalence of two definitions of e: one as the unique number such that , and the other as the limit .

Do you know of any other solutions to the original question?

-Steven

I wanted to say something about Boij–Söderberg theory. But before I can even explain what that means, I have to give some definitions in homological commutative algebra. I might as well also use this as an excuse to talk about some results in that area.

Fix a field and let be a polynomial ring in variables. Let be a module over . We say that a chain complex

is a free resolution of if each is a free module, and if the complex is exact in each degree, i.e., the image of each map is equal to the kernel of the one following it. It is easy to see that every module has a free resolution: pick some set of generators for , and let be the free module generated by them, so that we have a surjection . Now pick a set of generators for the kernel of this map, and let be the free module generated by them, so that we have a map which is exact at , and repeat as necessary (possibly forever).

But we won’t really be interested in this unless is graded. The polynomial ring has a natural grading where is the -vector space generated by degree monomials. By grading, I just mean that for all and . So a (positively) graded module will be one with a decomposition such that for all and . Elements in for some will be called homogeneous. The right notion of a homomorphism of graded modules will be degree 0 maps: those that send to for all . We could also talk about degree maps, but instead we will use the notation to denote a grading shift: . So what we might call a degree map is just a degree 0 map . So now that our modules are graded, we will be interested in graded free resolutions. In general, these will look like

(we will see later why I use instead of , where the will denote the number of generators in each graded piece (most of these numbers will be 0 of course). The existence of graded free resolutions works as before: instead of just picking random generators each time, we restrict our attention to homogeneous ones.

Okay, so how about some examples.