A while back, Sam wrote a post about du Val singularities and gave explicit calculations for the blowups of the and singularities. One neat characterization of du Val singularities is the concept of absolutely isolated double point: they can be resolved by successive blowups, where each blowup is at an isolated double point. In particular, this means that when we blowup a du Val singularity, the result is either nonsingular, or the singularities are also of du Val type. In this post I will discuss this in further detail. These are mainly rough notes for myself, so apologies in advance if it’s not so enlightening to anyone else. We work over the complex numbers.

These are related to Dynkin diagrams in the following way. Let X be a du Val singularity, and let X’ be its minimal resolution, so we have a proper birational map and X’ is nonsingular. The preimage of the origin will be a tree of projective lines, and any two either intersect transversely in a single point, or are disjoint. Furthermore, any point of intersection contains only two lines. We draw a graph whose nodes are the projective lines, and connect two nodes with an edge if the corresponding lines intersect. Then this graph will be a Dynkin diagram of type ADE. Furthermore, this configuration determines the singularity up to analytic isomorphism, so we can give them names using Dynkin diagram language. Due to the recursive nature of these resolutions, we should be able to see how these trees are “built up” from successive blowups, and that’s what I want to investigate. For the E case, I’ll just give some remarks.

One thing I’ve been thinking about lately is the tensor product multiplicities for the classical groups. The case of the general (special) linear group is the Littlewood-Richardson rule, so I wanted to discuss how to use these numbers to gain some information for the orthogonal and symplectic groups.

When we count some family of objects, sometimes the counting function is a polynomial, and sometimes it is only a polynomial after we ignore finitely many “bad cases.” There’s a clean way of packaging that together using generating functions. For instance, let f(n) be some function that takes the nonnegative integers to integers. Then the formal power series is a rational function of the form if and only if there exists N and a polynomial g(n) of degree d such that g(n) = f(n) for . There’s a natural followup question here: how is the polynomial g(n) related to my original counting problem? And is there any meaning to plugging in negative values into g?

1. Hilbert functions and Hilbert polynomials

I’d like to discuss these two questions for Hilbert functions. We’ll let be the polynomial ring in d variables over some field K, which we think of as a graded ring in the standard way using the total degree. M will denote a finitely generated graded module over A which is generated in degree 1. Relevant definitions can be found here. The Hilbert function of M is defined by where is the nth graded part of M. Then there is a polynomial for which for all n sufficiently large. This is the Hilbert polynomial of M, and for me it was a little mysterious why there was an error term in the first place. The reasonable answer seemed to be that one can always add extra junk terms to the low degrees of a module without changing its higher degree terms. But there’s a better answer! It involves local cohomology, for which I’ll give the relevant definitions.

My girlfriend is taking an abstract algebra course this year, and they had the following “longterm project” to think about for the term: Show that every group of size less than 60 is either Abelian or not simple. If we assume Burnside’s theorem that a group of order (p and q are primes) is solvable, this gives every case except for 30 and 42. (A proof of Burnside’s theorem can be found here on p.19, Exercise 6 of one my old solution manuals, but you’ll need Serre’s book to see what the cited theorems are). But since I had never done this exercise, I wanted to write a post about how to do this using only techniques from a first course in algebra (i.e., Sylow theorems).

Let’s recall what the Sylow theorems say. First, let G be a finite group of order n. Let p be a prime and write where p does not divide m. A maximal subgroup of G whose order is a power of p is a Sylow p-subgroup. Let be the number of Sylow p-subgroups. Then we have the following theorem.

Theorem. 1. The size of a Sylow p-subgroup is .
2. All Sylow p-subgroups are conjugate to one another via inner automorphisms of G.
3. The index of the normalizer of any Sylow p-subgroup is . In particular, divides m and is congruent to 1 modulo p.

The third statement is usually enough. If we can show that for some p, then the corresponding Sylow p-subgroup (assuming p divides n) must be normal since any conjugate of it must have the same size. There were three general facts that have relatively short proofs that take care of most of the cases. Then the rest was a (quick) case-by-case analysis. If the reader knows of another quick general fact that makes the problem more efficient, leave a comment!

This puzzle was presented by Tanya Khovanova at a casual combinatorial meeting, apparently from some Russian olympiad for middle/high-schoolers. I really like this type of problem since I haven’t seen it before (also, it starts in a way that makes most people think “oh I’ve seen this before…” only to be wrong). Before I present the problem, we will look at a “baby version” to see how the problem works (this is also another excuse to make a Matryoshka problem).

We have 100 coins. Both Cecil and Lisa know the fact that “there are either 1 or 2 counterfeit coins.” Counterfeit coins have the same weight and are lighter than real coins. Given that Lisa knows that there are actually 2 counterfeit coins instead of 1 (and knows exactly which coins they are), find a way for Lisa to prove this fact to Cecil, aided with just a balance, such that at the end of the proof Cecil now knows there are 2 counterfeit coins but does not know the identity of any particular coin.

Last time, I discussed a bit about the relationship between the Chow groups of a nonsingular variety X and its K-theory. In this post, I want to specialize to the case when X is a Grassmannian and explain some of the combinatorics behind this relationship.

Set-valued tableaux

For a partition , we let denote its Young diagram, i.e., we draw boxes in the ith row (counting from top to bottom) all left-justified. Also denotes the sum of its parts. Given two partitions , with (i.e., for all i), we set denote its Young diagram.

Given two finite subsets A and B of positive integers, denote A<B if max(A) < min(B) and if . This is not meant to define a partial order.

A set-valued tableau of is an assignment T of a finite subset of positive integers to each cell in such a way that and using the notation for subsets defined above. Given a set-valued tableau T, we associate to it a monomial where is the number of times that the integer i appears in a cell of T. Also, let |T| denote the total degree of this monomial. We define the (single stable) Grothendieck polynomial to be

where the sum is over all set-valued tableaux of . Then is a symmetric function in the . This is not obvious, but one can prove it to be true purely combinatorially in the same way one can prove that Schur functions are symmetric purely combinatorially (it is enough to show that it is invariant under switching and , and this can be shown by directly switching the number of i’s and (i+1)’s within any given set-valued tableau).

Steven greeted me with a puzzle when I entered the office today. I got it after thinking a bit, though I definitely have seen this a long time ago:

Given 51 numbers between 1 and 100, show that you can find two of them that are relatively prime

Lately some of us at MIT have been thinking about counting -rational points on some classes of varieties related to linear algebra that provide natural q-analogues for various classes of permutations. One thing we came across was some classes that have the same counts over every finite field. Yan wanted me to post about the following, so I’ll delay my post on the K-theory of the Grassmannian until next time.

We’ll consider varieties defined over a fixed field K. Form the free Abelian group on the isomorphism classes of such varieties. If Z is a closed subvariety of X, then we impose the relation

.

We can put a product structure on this group via

though it will not be relevant for this post. Related to this product structure is a paper by Bjorn Poonen which shows that if the characteristic is 0, then this ring is not an integral domain. And presumably the result is true over positive characteristic also, but the paper uses the existence of resolution of singularities. This is the Grothendieck ring of varieties. This is at least one way to make sense of statements of the form: .

I want to write a post about the set-valued tableaux of Buch and how they are related to Schur polynomials. But since these two things are related to the K-theory and Chow ring, respectively, of the Grassmannian, I thought I would write a post explaining some basic generalities between the Chow ring and K-theory. Let X be a variety over an algebraically closed field K.

First let’s define the Chow groups. We first form the k-cycles to be the free Abelian group spanned by the k-dimensional subvarieties of X. Let [V] be the basis element corresponding to a subvariety V. Pick a subvariety W of X of dimension k+1, and a nonzero rational function f/g defined on W. If V is a codimension 1 subvariety of W, let be the ring obtained by taking the ring of polynomial functions on W and inverting all polynomial functions which are not identically zero on V. We define the order to be , where the dimension is as K-vector spaces. The divisor of f/g is given by . We say these divisors are rationally equivalent to 0, and define the Chow group to be the group of k-cycles modulo rational equivalence.

Let K be a field of characteristic 0, and let V be a vector space over K of dimension n, and pick k < n. Let X be the Grassmannian Grass(k, V). We’ll briefly explore the (bounded) derived category of coherent sheaves of X, denoted .

1. Derived category review

For those unfamiliar with derived categories, here’s a quick summary. If A is any Abelian category, set K(A) to be the category of (co)chain complexes of A with the morphisms being chain maps modulo homotopy equivalence. Chain maps which induce isomorphisms are formally inverted, and the result is the derived category of A. Usually we only want to consider bounded complexes, or at least complexes with finitely many nonzero (co)homology groups, and in this case we denote the category . The category is equipped with a shift functor, which just shifts the degrees of a given complex.

One thing we can do is reformulate derived functors. Given a left exact functor , we define its right derived functor as follows. Given an object X in A, an injective resolution of X becomes an isomorphism in (considering X as a complex with one nonzero term), so we define to be the complex obtained by applying F to . Actually, we don’t need an injective resolution, we only need a resolution consisting of F-acyclic objects (i.e., the usual right derived functors of F vanish for them). To define on a general complex , we need to find a double complex which is term by term an injective resolution for each (these are called Cartan–Eilenberg resolutions). Then we apply F to the total complex of . A similar story is true for right exact functors G, so we get left derived functors . For notation, the left derived functor of the tensor product is denoted .