This post will be an exercise in algebraic (differential) geometry: I’ll calculate the tangent bundle of the Grassmannian of r-dimensional subspaces of a fixed vector space E defined over a field k. We’ll deduce from this that Grassmannians are Fano varieties.

The two methods will be similar, but working in the differential category requires less machinery, so we’ll assume k is the real numbers first. In this case, we’ll use the definition of tangent spaces at a point x as the tangent vectors of curves starting at x. We’ll need the tautological bundle R on X, which is the subbundle of E x X given by . The quotient bundle will be denoted Q. Recall that the Plücker embedding is a map sends a subspace W of E to its rth exterior power . Under this embedding, the line bundle on P restricts to since is the tautological subbundle of .

At the risk of embarrassing myself, here’s a linear algebra question that has turned up in a knot theory project I’m working on. The post might be embarrassing because the answer might be “obviously not,” and I just don’t have enough linear algebra intuition to see it. On the other hand, this would contradict some things which seem to make sense from a knot theory perspective.

Say we have an matrix . We can assume throughout that is invertible, though maybe it doesn’t matter. Its determinant has the expansion

where is the symmetric group on objects. One nice way to think of this (which might help for the problem), is that a permutation corresponds to a “choice” of a column of by each row of (so row chooses column ), amounting to a particular bijection between rows and columns. Then, in the summand corresponding to in the above expansion, we see the element of each row which lies in the column it chose. (I’ve just learned that this way of looking at the expansion is exactly how Kauffman came up with his states-sum model for the Alexander polynomial. Would anyone be interested in an intro-post on the Alexander polynomial?)

Now suppose we have a matrix whose entries are only 0’s and 1’s. Suppose, furthermore, that all non-zero summands in the above determinant expansion are , i.e., the corresponding permutation is even. Call such a matrix special. Then the set of bijections between rows and columns which contribute a non-zero amount to the expansion has cardinality the determinant; call this set S1. First question: how to characterize special matrices in some less convoluted way (this is a side-question, but if there were some nice geometric characterization, it might help with the other question).

Now, another set which has cardinality is the set S2 of solutions to , where is some vector in . To see this, just consider —it contains exactly integral lattice points, and their preimages under are the solutions in question.

So, my big question: is there some natural bijection between S1 and S2, some way to go from a bijection between rows and columns, to a particular solution to the system of equations over ? As I said at the beginning, I’m not sure whether there should be. It seems to me that maybe one just has to sit down and explicitly translate from the algebraic to the geometric definition of determinant. But I told myself I would abandon this project until the end of the summer, so maybe someone else wants to have a go? I’d appreciate any insights! Another option is that maybe there is some kind of “affine” bijection: we can get from one permutation to another by transpositions, and maybe there is some corresponding way to go between solutions?

In a later post, maybe I’ll talk about the knot theory part of this. On one side we have binary-dihedral representations of a knot group, and on the other side we have Kauffman states, or spanning trees of a certain graph, if you like, which happen to generate both the Khovanov and knot Floer complexes. The “Alexander matrices” which show up are special, in the above sense, in the case of alternating knots.

Thanks for any help!

Last time I gave the definition of -rings and tried to motivate them from the perspective of K-theory. I’d like to continue with how to define abstract Chern classes for -rings and explain Riemann–Roch formalism. We’ll assume all -rings R have an augmentation, are equipped with a positive structure, and have an involution. I’ll once again use K-theory as motivation for the definitions.

In topological K-theory, if we have a space X with a vector bundle E, there exists a map such that the pullback decomposes into a direct sum of line bundles, and the map on cohomology rings is injective. The corresponding statement for -rings is that we can embed any -ring R into a larger one such that every element can be written as a sum of elements with augmentation .

(guest post by Joel Lewis, another of my office-mates. -Y)

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I frequent the Art of Problem Solving forum (also knows as MathLinks where at some point I stumbled across the following problem (advertised (.pdf) as appearing on the 1996 Iranian Mathematical Olympiad, a claim I have no reason to doubt but no way to verify):

Consider a chessboard in which the opposite edges have been identified to yield a torus. What is the maximal number of knights that can be placed on this board so that no two attack each other?

There are several ways to attack this problem. If you’ve spent some time thinking about chess-related mathematics, you’re probably familiar with the existence of a knight’s tour on a (regular) chessboard: it’s possible to start with a knight on any square of the chessboard and make a sequence of 64 moves so that it visits every square exactly once before returning to its original position. It follows that we can place at most 32 knights on the board so that no two attack each other: no two squares visited consecutively in the knight’s tour may both be occupied, so at most half of the board may be covered. Moreover, it’s easy to find a set of 32 squares on which we can place the knights — for example, the 32 black squares do nicely, since a knight on a black square attacks only white squares. (In fact, one can extend this argument to show that this and the 32 white squares are the only sets of 32 squares on which we can place nonattacking knights.)

That was a regular chessboard — we still haven’t dealt with the torus. Note that when we identify edges, we can’t possibly increase the number of knights that fit on the board. Also, it is still true on the torus that knights on black squares attack only black squares. Thus 32 squares are still maximal, and we’ve answered our question.

To summarize what we’ve done using the terminology of graph theory, we’ve used the Hamiltonicity of the knight’s graph on an chessboard, the fact that the knight’s graph is bipartite, and the fact that adding extra edges to a graph can only decrease its independence number. This Hamiltonicity result is reasonably high-powered, though. Let’s try to avoid using it. One way to do this might be to start with a non-Hamiltonian board, for example, the board.

Consider a chessboard in which the opposite edges have been identified to yield a torus. What is the maximal number of knights that can be placed on this board so that no two attack each other?

In this post, I want to discuss Grothendieck’s -rings and how they provide an abstract setting for Riemann–Roch formalism. The references I’ll be using are

• Donald Knutson, Lambda-Rings and the Representation Theory of the Symmetric Group
• William Fulton and Serge Lang, Riemann–Roch Algebra

The definition of a -ring is a bit technical, but it starts with a commutative ring R together with operations for all nonnegative integers i such that and for all r in R together with some axioms. In particular, we should say what these lambda operations do to sums and products, and we might also want to know what compositions of them look like. To motivate these axioms, we’ll look at K-theory (where it originates).

I spent three weeks of June and July at a summer school and conference on geometric representation theory at the University of Ottawa. This conference is already mentioned on the mathematical blogosphere (I refuse to talk about the “blathosphere” or the “blathyscape”), at the Secret Blogging Seminar.

While I could talk about things I learned in Ottawa, that would be contrary to my recent practice on Concrete Nonsense — namely, not posting about math!

In this post, I want to give an example of a Lie group which is not isomorphic to a subgroup of a matrix group. This contrasts with the algebraic picture, where every affine algebraic group can be realized as a subgroup of a matrix group. I’ll just sketch the details.

The group we’ll use is the universal cover G of . One can realize as a fiber bundle with fiber over via the map which sends a matrix to its first row, so its fundamental group is . Similarly, we can use this map to show that is a fiber bundle with fiber over and hence is simply connected.

Now let be any smooth map. It induces a map on Lie algebras . We can use this to define a map by . It’s straightforward to check that f preserves the Lie bracket. Since is simply connected, f is the derivative of some smooth map .

To show that r cannot be injective, we will show that the composition is not injective, where the second map is the usual inclusion. I claim that this composition is the same map as the composition , where is the covering space map. By construction, these two compositions induce the same differentials on their Lie algebras. Now using the fact that both G and are connected, this implies that the map on the level of Lie groups must be equal as well. But is not injective: we saw above that the fundamental group for is nontrivial, so we are done.

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 .