Secret Blogging Seminar http://sbseminar.wordpress.com Representation theory, geometry and whatever else we decide is worth writing about today. Fri, 20 Nov 2009 20:47:43 +0000 http://wordpress.com/ en hourly 1 http://www.gravatar.com/blavatar/8a36546006b3c842cdbb641e20a9b06c?s=96&d=http://s.wordpress.com/i/buttonw-com.png Secret Blogging Seminar http://sbseminar.wordpress.com The diamond lemma http://sbseminar.wordpress.com/2009/11/20/the-diamond-lemma/ http://sbseminar.wordpress.com/2009/11/20/the-diamond-lemma/#comments Fri, 20 Nov 2009 20:47:43 +0000 David Speyer http://sbseminar.wordpress.com/?p=2205 ]]>

A few results

1 (Bjorner, Eidelman and Ziegler) Suppose we have a finite collection of great circles on a sphere, none of them through the north or nouth pole. Let R be the set of regions in the complement of these circles, and suppose that every region is a triangle. Put a partial order on R by x \leq y if x is south of every circle that y is south of. Show that, for x and y \in R, there is some z \in R such that w \leq z if and only if w \leq x and w \leq y.

2 (Mozes, see also IMO 1986.3) Let G be a finite graph, and let r be a real valued function on the vertices of G. Consider the following (solitaire) game: find a vertex i for which r_i is negative. Replace r_i by - r_i and, for every vertex j that neighbors i, decrease r_j by -r_i. The game ends if all of the r_i are nonnegative. You and I start playing with the same graph and the same r. Show that, if my game ends in N moves at position z, then your game will end in the same position, in the same number of moves.

3 (Poincare, Birkhoff and Witt) Define U to be the ring generated by E, F and G, subject to the relations FE=EF+G, GE=EG+F and GF=FG+E. Show that any element of R can be expressed uniquely as a sum of elements of the form E^i F^j G^k. (Uniqueness is up to rearranging the sum and combining like terms.)

4 (Jordan and Holder) Let G be a finite group. Let G = G_0 \supsetneq G_1 \supsetneq G_2 \supsetneq \cdots \supsetneq G_r = \{ e \}
G = H_0 \supsetneq H_1 \supsetneq H_2 \supsetneq \cdots \supsetneq H_s = \{ e \}
by two sequences of subgroups such that G_{i+1} is normal in G_i, with G_{i+1}/G_i simple, and the same is true for the H’s. Then r=s and the quotients H_{i+1}/H_i are a permutation of the quotients G_{i+1}/G_i.

What do all of these have in common? You can remember all of their solutions by drawing the same figure — the diamond!

Solution 1

We say that z is a meet of x and y if z has the required properties.

For any region r, let \ell(r) be the number of circles below r. We will prove the following statement by induction on \ell(r):

Inductive Claim Suppose that r \geq x and r \geq y. Then x and y have a meet.

This establishes the result, as we can take r to be the region containing the north pole and the hypotheses on r become trivial. In the other hand, the base case is trivial because, when \ell(r)=0, then r must be the region containing the north pole, we have r=x=y, and r is a meet of x and y.

Now for the inductive part. Let r \to \cdots \to x and r \to \cdots \to y be southward traveling paths from r to x and y. Let the first steps on these paths be r \to s and r \to t, crossing lines i and j. If s=t, or equivalently i=j, then we can take s as a new r and we are done by induction. Otherwise, let u be the region due south of the crossing of i and j. It is easy to see that u is a meet of s and t.

Applying the inductive hypothesis to (s, x, u), let z' be a meet of x and u. Applying the inductive hypothesis to (t, u, y), let z'' be a meet of u and y. Applying the inductive hypothesis to (u, z', z''), let z be a meet of z' and z''. We leave it to the reader to check that z is a meet of x and y.

Solution 2

The proof is by induction on N. Say my game begins r \to s \to \cdots \to z and your game begins r \to t \to \cdots. (We don’t know yet that your game ends.) Let my first move be at vertex i and yours at vertex j. The case i=j is an immediate induction, so let’s assume i \neq j. We know that r_i and r_j <0.

My brother and your sister come to join us. Here is how they play. If i and j are not adjacent, my brother starts off playing at i, then at j and your sister starts of at j and then at i. So, after two moves, both he and she reach the same configuration u. If, on the other hand, i is adjacent to j, then my brother starts with iji and your sister starts jij. In three moves, they have again reached the same configuration. (Exercise! Remember to check that all vertices which are played are in fact negative at the time.) Call this configuration u.

After this, my brother plays in any manner he wishes. By induction, after he moves to s, he will make N-1 more moves and end at z. Your sister also plays in any manner she wishes. By induction between her and my brother, after she gets to u, she will make either N-2 or N-3 more moves and end at z, having made N moves in total. Finally, we apply the induction hypothesis to you and your sister. After you both move to t, you will make N-1 more moves, ending at z. So we all get to z in N moves.

Generalizing

I hope it is clear that Solutions 1 and 2 have the same inductive structure, although the details differ. The general strategy goes as follows: suppose we have some process which goes from one state to another. We have two different paths, r \to s \to \cdots \to x and r \to t \to \cdots \to y that start at the same state, and we want to show that, in some sense, they come together again. In the figure above, we are presented with the black structure, and we need to show that the paths rejoin. We will do this by adding the blue structure.

First, make a careful analysis of the case where the two paths have length 1, and prove your result in this case.

Now, using your analysis from the first step, build paths r \to s \to \cdots \to u and r \to t \to \cdots \to u which come together in the required way. Inductively, bring the paths s \to x and s \to u together at some z'; bring the paths t \to u and t \to y together at some z''; finally, bring u \to z' and u \to z'' together at some z.

Often, as in Solution 1, the starting point r is trivial in the most important application. But bringing it into the problem allows us to apply this inductive structure.

One of the main difficulties is figuring out what to induct on. Roughly, one wants to use the length of the paths, but this may not be precisely right.

There are many good papers on the diamond lemma, but they tend to focus on applications to one particular field, and call it by different names. Here are some examples of the Diamond Lemma in Grobner bases, noncommutative rings, braid groups (chapter 14, see figure 12!), lattices (lemma 2.1), anti-matroids (lemma 2.6).

Solution 3 (a sketch)

Let’s talk about Puzzle 3. This is typical of applications to ring theory, and there are many subtleties which are particular to this context. I would like to refer to Bergmann’s superb paper for the details but, sadly, it is not publicly available online. For those without academic access, the best reference I can find online is Wenfeng Ge’s masters thesis.

Let’s call E^i F^j G^k a standard monomial, and a sum of standard monomials a standard polynomial. Let the states of our system be formal noncommutative polynomials in E, F and G. Let our operations be finding a term of the form FE, GE or GF, and replacing it by EF+G, EG-F or FG+E respectively. So the states where we can perform no operations are precisely the standard polynomials. We want to show that, starting with any polynomial r, this process will terminate and, if we perform the process in two different ways, r \to \cdots \to x and r \to \cdots \to y, then x=y. I’ll ignore termination, which is the easier question, to focus on the latter issue.

A better way to phrase our claim is that if we have any two paths r \to \cdots \to x and r \to \cdots \to y, which may not end at standard polynomials, then we can join them together into two paths r \to \cdots x \to \cdots \to z and
r \to \cdots y \to \cdots \to z which have a common endpoint. This trick is frequently useful: by phrasing our claim to apply to more paths, we make induction easier.

We start with a careful analysis of the case of paths of length 1. Say we have r \to s and r \to t. If the two changes happen in different monomials, or if they happen in nonoverlapping parts of the same monomial, then we can just make both changes in the two possible orders to get to the same destination in two steps.

So the only interesting case is two overlapping changes. For example, GFE \to (FG+E)E and GFE \to G(EF+G). In this case, we make the following moves

The reader familiar with Lie algebras might like to see how this computation works in a general universal enveloping algebra; it’s a bit more complicated because the quadratic terms may not be standard.

The structure of the proof is now the same. Of course, we have to figure out what to induct on, and that’s a little subtle. But the worse issue is the following: Suppose our starting state is GFE + FGE and our first move was to E^2 on one path, and to GEF + G^2 + FGE on the other. According to the rubric above, we should move E^2 \to FEG - F^2 + E^2 - FGE. But, as we’ve defined things, this isn’t a legal move, because there is no GE to replace in E^2. This possibility of terms cancelling is a major nuisance; I leave it to Bergmann to explain how to fix it.

Solution 4

You do it!

]]> http://sbseminar.wordpress.com/2009/11/20/the-diamond-lemma/feed/ 5 davidspeyer Diamond Diamond3 Quantum mechanics and geometry http://sbseminar.wordpress.com/2009/11/16/quantum-mechanics-and-geometry/ http://sbseminar.wordpress.com/2009/11/16/quantum-mechanics-and-geometry/#comments Mon, 16 Nov 2009 23:02:53 +0000 Scott Morrison http://sbseminar.wordpress.com/?p=2634 ]]>

Here’s a nice little story about quantum mechanics, which surprisingly few mathematicians seem to know about. The essential idea is “quantum mechanics on the projective space looks remarkably like classical mechanics”! Everything I say here comes from two papers Geometrical Formulation of Quantum Mechanics (gr-qc/9706069), Ashtekar and Schilling, and Geometry of stochastic state vector reduction (#), Hughston. If you’re interested in more details, I’d encourage you to read these papers — they’re well written and contain many further references.

As you’ll recall, quantum mechanics says that systems are described by Hilbert spaces, with states given by vectors. I’ll stick with finite-dimensional systems (e.g. particles with spin) for simplicity, but this isn’t essential for what follows. A particular self-adjoint operator H, called the Hamiltonian, governs the dynamics of the system via the Schrodinger equation \frac{d}{dt} \psi = i H \psi. Quantum mechanics also says something about measurement, which we’ll come to in a moment.

Now the Schrodinger equation defines a one parameter flow via U(t) = exp(i H t). This preserves the unit sphere in our Hilbert space, and descends to a flow on the projective space. The projective space is naturally a Kahler manifold, and in particular a symplectic manifold, so we immediately ask if this flow is Hamiltonian. The answer is unsurprising but underappreciated: yes, the flow is Hamiltonian, and the Hamiltonian function is just the expectation value of the Hamiltonian operator \langle \psi, H \psi \rangle.

The example you should have in mind at this point is a simple spin 1/2 system in a magnetic field, whose Hilbert space is \mathbb{C}^2, Hamiltonian \begin{pmatrix}1 &0 \\ 0 & -1\end{pmatrix}. The projective space is \mathbb{CP}^1 and the Hamiltonian function we get as the expectation value is just the usual z coordinate of the standard embedding of \mathbb{CP}^1 = S^2 in R^3. The Hamiltonian flow rotates points along lines of latitude, completing each orbit in \pi units of time (go calculate the unitary).

Eigenvectors for the Hamiltonian operator correspond to critical points for the Hamiltonian function, and in particular fixed points of the flow. (That’s the north and south poles in the example above.) The flow described above is just a rigid rotation of the sphere, and in fact this is generally true: the flow on projective space coming from a self-adjoint operator is Killing, that is, it preserves the metric. This is the first appearance of the metric, but it’s really essential, because the converse of this statement is true — Hamiltonian functions whose corresponding flows preserve the metric are precisely those which arise as expectation values of self-adjoint operators on the Hilbert space.

That’s not all the metric is good for! Quantum mechanics also tells us something about what happens during “measurement”. This is that when a “measurement” (yes, I’m going to keep using scare quotes, so you’re not allowed to argue with me about what measurement means) occurs, the system jumps discontinuously to one of the eigenvector of the Hamiltonian, and the probabilities of reaching the the various different eigenvectors  are given by the absolute value squared of the inner product of the current state and the eigenvector. This probability is exactly \cos^2 \theta, where \theta is the metric distance between the current state and the corresponding fixed point. (In the spin 1/2 example, let’s normalise this metric so it just measures angles between points on S^2.)

It gets even better, but at this point I’m going to stop talking about the conventional description of quantum mechanics, and begin describing a proposed modification of quantum mechanics. Physicists have already thought a lot about whether modifications like this are reasonable, but I’ll postpone that for now. At this point, if you’re reading the actual articles, we’re switching from the Ashtekar/Schilling paper to the Hughston one.

So what is this proposed modification? Well, let’s imagine the symplectic flow as some differential equations describing the trajectory of our state. We now want to add in a stochastic term, in particular an isotropic Brownian motion term with an amplitude that depends on the position in the projective space. This amplitude will be (some simple function of?) the energy uncertainty, namely the quantity \langle \psi, H^2 \psi \rangle - \langle \psi, H \psi \rangle^2. In fact, this energy uncertainty is exactly the squared velocity of the symplectic flow with respect to the metric. In our spin 1/2 example this velocity is \sqrt{1-z^2} (remember we have rigid rotation) and since H^2=1, \langle \psi, H^2 \psi \rangle - \langle \psi, H \psi \rangle^2 = 1-z^2. What happens? Well, at the fixed points it’s easy to see that the energy uncertainty is zero, so we might expect that the Brownian motion term drives the state away from areas with high energy uncertainty, towards the eigenstates — just like what is supposed to happen during “measurement”. This is precisely what happens: Hughston does a lot of financial mathematics, and he knows his stochastic calculus. His Proposition 5 says the energy uncertainty in this model is a supermartingale, that is, an on average decreasing function. As time passes, you expect to end up at one of the fixed points, each with various probabilities. Note that these are honest, stochastic probabilities, not just numbers we’ve declared to be interpreted as probabilities as in the naive set up. (ED: see below for Greg’s comment on this.) His next result, of course, is that these probabilities match up with what we want, namely that they are given simply by metric distances on the projective space.

I think this is a beautiful picture. The measurement process is now something more concrete, a stochastic term in the governing equation, and we can resume thinking probabilistically about quantum mechanical probabilities.Very roughly, you’re meant to think that in an “isolated quantum system” the stochastic term is extremely small, and symplectic flow dominates. On the other hand, during a “measurement”, presumably when the system is coupled with the macroscopic world, the scale of energy uncertainties becomes extremely large and the stochastic terms dominates, and the system is quickly driven to a fixed point of the symplectic flow.

You have to think hard, however, about where this stochastic terms comes from, and what it means. Hughston has some ideas about quantum gravity, but I’m not so sure I like them! There are also lots of no-go theorems ruling out stochastic variations on quantum mechanics, and I have to admit to not being clear about whether these results affect Hughston’s model.

A final idea for further thought, from the Ashtekar/Schilling paper: we can fully describe quantum mechanics solely in terms of the Kahler manifold structure of the projective space, so why not drop the requirement that it’s a projective space? That is, can we imagine systems on other Kahler manifolds? It seems that all we lose is the fact that on \mathbb{CP}^n any two points have a canonical \mathbb{CP}^1 through them — i.e. that we’re allowed to form linear superpositions of states. Is this really essential? Where might we look for finite dimensional systems described by “exotic” Kahler manifolds? And all you quantum topologist gallium-arsenide engineers out there — how might we try to make one?

]]> http://sbseminar.wordpress.com/2009/11/16/quantum-mechanics-and-geometry/feed/ 19 semorrison Why is physical intuition possible? http://sbseminar.wordpress.com/2009/11/16/why-is-physical-intuition-possible/ http://sbseminar.wordpress.com/2009/11/16/why-is-physical-intuition-possible/#comments Mon, 16 Nov 2009 17:16:59 +0000 Noah Snyder http://sbseminar.wordpress.com/?p=2627 ]]>

This post is based on a conversation I had with Allan Adams at Mathcamp a few summers ago, and I was reminded of it by an aside in Mike Freedman’s talk in Scott’s backyard on Friday. As usual with blog posts based on other people’s talks, all good ideas in this post should be attributed to Allan and Mike and all mistakes to me. Furthermore I think everything I say here is obvious to people who actually know physics.

My basic confusion was how physical intuition (in particular in quantum field theory) could be applied to so many mathematical settings when there’s only one physical world so there’s no reason to think any intuition built up within that single example would apply any more generally than that one example. What Allan pointed out to me is that it’s not true that physicists are only studying one example. Although there may only be one fundamental theory of physics, by looking at various particular physical systems the limiting behavior becomes its own theory. The physics at the surface of a black hole can be thought of as its own example; the physics of superconductors is its own example; etc. Because all of these examples are physical (they involve minimizing actions, they’re quantum, etc.) they have a lot of attributes in common, so intuition and general techniques can be developed by understanding their commonalities.

Mike made two comments in his talk (on K-theory and superconductors) that flesh out this idea further. He was discussing the BCS superconductor and explained that when physicists refer to a theory by initials they’re not just being polite, what they mean is that you’re studying the mathematical model rather than any particularly instantation of it. In particular, the model doesn’t care if there are exactly 10^9 electron pairs or the exact composition of the material, it is studying the abstract setting that appears in the limit. By calling it the “BCS superconductor” they mean that in some sense they’re studying the physics of a different world. In particular, in the BCS setting since you’re assuming that there’s a huge sea of electron pairs the “vacuum” consists of this huge sea. This explains how physicists can develop intuition for more general notions of vacuum: they’re not always studying the absolute vacuum, they’re also studying other systems with states that have the properties of being a “vacuum.” This particular vacuum has a delightfully strange property. Since a new electron pair doesn’t change the underlying vacuum, in this “world” electric charge isn’t preserved!

]]> http://sbseminar.wordpress.com/2009/11/16/why-is-physical-intuition-possible/feed/ 3 nsnyder Choosing problems for grad. students http://sbseminar.wordpress.com/2009/11/11/choosing-problems-for-grad-students/ http://sbseminar.wordpress.com/2009/11/11/choosing-problems-for-grad-students/#comments Wed, 11 Nov 2009 18:12:23 +0000 David Speyer http://sbseminar.wordpress.com/?p=2596 ]]>

I am coming to the point in my career where I will be expected to take graduate students, and I’d like some advice about finding problems for them. How responsible am I for making sure that a problem is solvable and not already under attack elsewhere? I have a (private) list of problems that might be suitable for attack with tropical methods, or using cluster algebras. In most cases, the reason that I have not worked on these problems myself is that I would have to do a fair bit of research to find out the current state of the field and make sure that I wasn’t missing something stupid. Is it fair to pass this sort of thing off to a graduate student?

]]> http://sbseminar.wordpress.com/2009/11/11/choosing-problems-for-grad-students/feed/ 28 davidspeyer Remarks on career advice http://sbseminar.wordpress.com/2009/11/11/remarks-on-career-advice/ http://sbseminar.wordpress.com/2009/11/11/remarks-on-career-advice/#comments Wed, 11 Nov 2009 11:06:35 +0000 Scott Carnahan http://sbseminar.wordpress.com/?p=2592 ]]>

There have been a few questions about the job application process on MathOverflow, and I’d like to make a few remarks in an open forum.

First of all, I think there have been some really good questions, and really good answers. I found it especially illuminating when mathematicians who have been on hiring committees weighed in on what they thought was important in an application. Depending on your social circle and who your advisor is, it can be difficult to get accurate information when you are a graduate student (or a postdoc – I recently learned that my research statement was too long by a factor of 2 or 3). So, hats off to the people who give well-informed advice. Please keep it up.

This brings us to the other side of the discussion, which is uninformed job advice. I hear and see it in person and on the internet (including MathOverflow, where fortunately, it has been relatively easy to detect). Some of it is quite obnoxious, with a cynical, hypercompetitive perspective of the mathematical community. I’m not prepared to discuss the psychological foundations of this sort of attitude, but I’ve seen bad advice cause a lot of unnecessary stress in people who receive it. As far as I can tell, the job application process is already stressful, where small strategic errors can mean big differences in outcome, so perhaps we should treat baseless speculation more severely in this context than in the realm of normal mathematical discourse.

I’ve been asked to include one more remark about applications that may be obvious to some of you, and it is that different types of schools look for different features in applications. As a basic example, Research-I schools tend to focus less on teaching statements than four-year colleges. Community colleges apparently have their own priorities, which are somewhat different from the other two classes. As a piece of meta-advice, if you’re surrounded by research faculty (quite likely in grad school), but you’re planning to teach at a four-year or community college, you might benefit from seeking some application and career advice from outside your program.

Also, one final request: if you’re answering a career-advice question on the internet, it would be great if you could add background context including the type of school (e.g., “I have done hiring at a four-year college in the US”). I see some people doing this already, and it’s great. The career questions at MathOverflow have so far leaned toward research-intensive tracks, but it would help in the future in case more general questions come up.

Update, Nov. 15: Tom Leinster has pointed out that I’ve used some US-specific vocabulary (e.g., Research-I), so I’ll try to give a brief explanation for the benefit of people who haven’t had extensive contact with university administrators in the US. The terms essentially come from an outdated version of The Carnegie Classification of Institutions of Higher EducationTM. As I understand it, Research-I was a designation for doctoral-granting schools for which research performance played the primary role in faculty hiring and promotion decisions, while Research-II denoted doctoral-granting schools where research was less of a focus. There was a revision in 2000 that turned the Roman numeral I into a letter standing for “Intensive” and II was changed to X for “eXtensive”. I think this was done to make it appear less judgmental (although many people still pronounce it “one”). There was a further revision in 2006, (see the bottom of this page for a full list), breaking the two groups into 3: RU/VH (research university/very high activity), RU/H, and DRU. The list of classes I linked also gives a list of schools that fall into each class, and you can search for your favorite examples. Four-year colleges are institutions whose graduate programs are either very small or don’t exist (classified as “Baccalaureate Colleges”). Community colleges are usually schools that offer two-year programs (classified as “Associate’s Colleges”). That latter two classes tend to make hiring and promotion decisions based on considerations other than research performance, although there is some variation.

]]> http://sbseminar.wordpress.com/2009/11/11/remarks-on-career-advice/feed/ 13 Scott Carnahan Classified problem http://sbseminar.wordpress.com/2009/11/04/classified-problem/ http://sbseminar.wordpress.com/2009/11/04/classified-problem/#comments Thu, 05 Nov 2009 07:26:49 +0000 Scott Carnahan http://sbseminar.wordpress.com/?p=2575 ]]>

Today at tea, some grad students were discussing the following enumeration problem:

How many elements of GL_n(\mathbb{F}_q) have zeroes in all diagonal entries?

I think they [Redacted]. The answer is apparently known but classified. It’s a sort of q-analog of derangements (i.e., permutations without fixed points), but if you take the derangement formula and add q-numbers in the naive way, the formula (q-1)^n \sum_{k=0}^n (-1)^k \frac{[n]!}{[k]!} doesn’t seem to work for n > 2.

]]> http://sbseminar.wordpress.com/2009/11/04/classified-problem/feed/ 27 Scott Carnahan Concrete Categories http://sbseminar.wordpress.com/2009/10/26/concrete-categories/ http://sbseminar.wordpress.com/2009/10/26/concrete-categories/#comments Mon, 26 Oct 2009 18:01:46 +0000 David Speyer http://sbseminar.wordpress.com/?p=2540 ]]>

In many introductions to category theory, you first learn the notion of a concrete category: A concrete category is a collection of sets, called the objects of the category and, for each pair (X,Y) of objects, a subset of the maps X \to Y. (There are, of course, axioms that these things must obey.) In a concrete category, the objects are sets, and the morphisms are maps that obey certain conditions. So the category of groups is concrete: a map of groups is just a map of the underlying sets such that multiplication is preserved. So are the category of vector spaces, topologicial spaces, smooth manifolds and most of the other most intuitive examples of categories.

Using terminology from a discussion at MO, I’ll call a category concretizable if it is isomorphic to a concrete category. For example, \mathrm{Set}^{op} can be concretized by the functor which sends a set X to the set 2^X of subsets of X, and sends a map of sets f:X \to Y to the preimage map f^{-1}: 2^Y \to 2^X.

At one point, I learned of a result of Freyd: The category of topological spaces, with maps up to homotopy, is not concretizable. I thought this was an amazing reflection of how subtle homotopy is. But now I think this result is sort of a cheat. As I’ll explain in this post, if you are the sort of person who ignores details of set theory, then you might as well treat all categories as concrete. My view now is that specific concretizations are very interesting; but the question of whether a category has a concretization is not. I’ll also say a few words about small concretizations, and Freyd’s proof.

Let me start by saying exactly what you need to check to see whether a functor is a concretization. Let C be a category and F a functor from C to \mathrm{Set}. Then F is a concretization if, for any objects X and Y, and any morphisms f and g from X to Y, we have F(f)=F(g) only if f=g.

Now, Yoneda’s lemma almost gives us such a functor in every case. Define

F(X) := \bigsqcup_{S \in \mathrm{Ob}(C)} \mathrm{Hom}_C(S,X).

Yoneda’s lemma tells us that, if f and g induce the same map from \mathrm{Hom}_C(S,X) to \mathrm{Hom}_C(S,Y) for every S, then f=g. The proof is simply to take S=X.

So, why doesn’t Yoneda’s lemma tell us that all categories are concretizable? Because the collection of objects of our category may not be a set. I assume that you have at some point been introduced to Russell’s paradox, which is resolved by declaring that the collections of all sets is “too big” to be a set. Similarly, you run into trouble if you try to talk about “the set of all vector spaces” or “the set of all groups.” And since people do consider the category of all vector spaces; category theorists have learned to be careful with expressions like \bigsqcup_{S \in \mathrm{Ob}(C)}, which act as if the objects of C are a set.

If, like me, you don’t care about this sort of set theoretic issue, then you might as well think that all categories are concretizable. But you should still object to the Yoneda method of concretization. When will you ever be able to check something for all the objects of a category? What you want is some reasonable collection T \subset \mathrm{Ob}(C) of test objects, so that it is enough to see whether f_*=g_* on X(S) for S \in T.

There are several great theorems of this form: In the category of varieties of finite type over \mathbb{C}, the Nullstellansatz tells us that a map is determined by its values on X(\mathbb{C}) — so we can think of a variety as made up of its points. In the category of finite CW complexes, with maps up to homotopy, Whitehead’s theorem tells us that it is enough to study \mathrm{Hom}(S^n, X), for all spheres S^n. The preceeding statement is completely false; thanks to Eric Wofsey for a counter-example. I’ll try to find a better, and more true, example.

My mathematical aesthetic would be to adopt a subjective standard here: the goal of concretization is to find a “nice” set of test objects, and the term “nice” is defined by the judgement of the mathematical community. The choice of a single point, in the Nullstellansatz example, is very nice. The choice of, for example, all Artinian rings, would still be nice, but less so. (PARAGRAPH REWRITTEN due to error in the preceding paragraph.)

For those who seek an objective criterion, Tom Leinster proposes saying that C has a small concretization if there is a set T of objects of C such that X \mapsto \bigsqcup_{S \in T} \mathrm{Hom}_C(S,X) is a concretization. Cautionary exercise: the category whose objects are sets, and whose morphisms are surjections, is a concrete category but has no small concretization.

I don’t want to close the post without saying something about how it is proved that the category of topological spaces, with maps modulo homotopy, is not concretizable. Even though I don’t find concretization interesting, the idea that it can be proved impossible is interesting to me. This is a result of Peter Freyd, whose explanation of the technical points is about as good as it could be, so I’ll leave the details to him.

Suppose that F were a concretization. For simplicity, I’ll assume that F(\{ \mathrm{point} \}) is the set with a single element. For any connected space X, let z(X) be the element of F(X) corresponding to the unique homotopy class of map \{ \mathrm{point} \} \to X.

Freyd constructs a poset P, with cardinality greater than 2^{F(S^2)}, two sequences of connected spaces A_p and B_p indexed by P, and maps A_p \to S^2 \to B_p. These have the property that the composite A_p \to B_p does not factor through \{ \mathrm{point} \} but A_p \to B_q, with q<p, does. Since P is so big, there must be some q<p such that F(A_p) and F(A_q) map to the same subset of F(S^2). Call this subset I.

Then A_p \to S^2 \to B_q factors through \{\mathrm{point}\}, so the map S^2 \to B_q sends I to z(B_q) . But then A_q \to S^2 \to B_q also sends F(A_q) to z(B_q), contradicting that A_q \to B_q does not factor through a point.

To give a little more detail, one first constructs a sequence of groups G_p, with nonzero maps \mathbb{Z} \to G_p, such that, for any map G_p \to G_q with q<p, the composite \mathbb{Z} \to G_q would be zero. Let C_p be a K(G_p,1). Note that S^1 is a K(\mathbb{Z}, 1), so we have functorial maps S^1 \to C_p. Then B_p is the suspension \Sigma(C_p), the map S^2 \to B_p is the suspension of S^1 \to C_p, and A_p is the mapping cone of S^1 \to C_p. If you want more detail than this, you should read Freyd’s paper.

]]> http://sbseminar.wordpress.com/2009/10/26/concrete-categories/feed/ 12 davidspeyer MSC vs. ArXiv (and some interesting info on mathjobs) http://sbseminar.wordpress.com/2009/10/25/msc-vs-arxiv-and-some-interesting-info-on-mathjobs/ http://sbseminar.wordpress.com/2009/10/25/msc-vs-arxiv-and-some-interesting-info-on-mathjobs/#comments Mon, 26 Oct 2009 03:29:17 +0000 Noah Snyder http://sbseminar.wordpress.com/?p=2535 ]]>

One of my pet peeves is how annoyingly the AMS’s math subject classification is for people working in quantum algebra and quantum topology. The MSC has 97 different major subjects and my field is not one of them, and instead appears many times a subheading. In the new 2009 classification there’s at least the following: 16T, 17B37, 18D10, 20G42, 33D80, 57R56, 58B32, 81R50, and 81T45. Here I’m only counting things that are obviously quantum algebra and quantum topology (for example I didn’t list subfactors, quantum computation, knot invariants, etc.) By way of contrast, on the ArXiv there are only 32 categories, yet one of them (math.QA) contains the vast majority of work in my field (of course, many of those are cross-posted).

This mini-rant of mine came up at dinner at an AMS meeting in Waco (more on the excellent “fusion categories” special session later). Someone pointed out an interesting side-effect of this issue that I hadn’t thought of. One of the awesome things about mathjobs is that rather than simply having a large paper stack of applications, the people on hiring committees can instead sort the applications automatically in many different ways. It makes a lot of sense that mathjobs has this feature, but none of us who were on the applying side of things had ever considered it. Here are a few examples of things you might want to search for: look at people applying from a specific school, find everyone who has a recommendation letter from Prof. X, and (relevant to this post) sort by AMS subject classification.

This means that choosing the right AMS subject classifications is actually somewhat important. If you choose poorly then someone who might be interested in hiring you might never actually find your application among the hundreds they’re looking through. So if you’re in a situation like mine it’s worth asking a professor or two which AMS subject classifications they’d be most likely to look through.

Since then I’ve been wondering whether it might be a useful for mathjobs that the data they ask for also include which arxiv classifications applicants have posted preprints under, as that’s the search that I would want to use if I were on a hiring committee. What do people think? Mathjobs is very responsive to requests, so if people think this makes sense I may send them an email.

]]> http://sbseminar.wordpress.com/2009/10/25/msc-vs-arxiv-and-some-interesting-info-on-mathjobs/feed/ 18 nsnyder Rhombus tilings and an over-constrained recurrence http://sbseminar.wordpress.com/2009/10/21/rhombus-tilings-and-an-over-constrained-recurrence/ http://sbseminar.wordpress.com/2009/10/21/rhombus-tilings-and-an-over-constrained-recurrence/#comments Thu, 22 Oct 2009 00:53:40 +0000 David Speyer http://sbseminar.wordpress.com/?p=2452 ]]>

I recently visited Robin Pemantle and his student Peter Du at UPenn. We talked about tilings of planar regions, generating functions and asymptotics. Towards the end, we talked about a bit about a very classical example, which is what I want to tell you about today.

In most planar tiling problems, the goal is an asymptotic analysis for tilings of large regions, because there isn’t enough structure to do better. This is the approach taken in the beautiful work of Kenyon, Okounkov, and collaborators.1, 2, 3 Sometimes, there is enough structure to give exact solutions with explicit generating functions. This is the situation with Aztec Diamonds, fortresses, and other several other examples.4,5 The central name here is Jim Propp 6, 7, who has developed this theory together with many undergraduate and graduate students (including me).

And then there is one case: rhombus tilings of a hexagon. These have almost too much structure; more structures than one would expect to be compatibly possible. In this post, I want to talk about this example. In particular, I want to ask you a question which I thought about a bit on the train ride back and see whether any of you have some thoughts.

A QUICK INTRODUCTION TO RHOMBUS TILINGS

Tile the plane with equilateral triangles, and cut out a hexagonal region where the opposite sides are parallel and of the same length. We’ll call the lengths of the sides r, s, t, r, s and t.This hexagon contains equally many up and down triangles, so we can hope to tile it with unit rhombi, each of which cover two adjacent triangles. We write M(r,s,t) for the number of such matchings.

Hexagon with (r,s,t)=(3,2,4)

Hexagon with (r,s,t)=(3,2,4)

If t=0, then the hexagon degenerates to a parallelogram, and there is only one tiling. If t=1, then the horizontal edge has length 1. In that case, the two types of horizontal rhombi form a path from the bottom of the hexagon to the top (colored blue). It is easy to see that there is one matching for each possible path, so M(r,s,1) = \binom{r+s}{r}.

One of the 6!/3!3! tilings of this hexagon

One of the 6!/3!3! tilings of this hexagon

If t=2, this turns into the problem of counting noncrossing pairs of paths; by a happy coincidence, this topic was recently discussed on Math Overflow. More generally, you need to count t-tuples of noncrossing paths, and you wind up computing t \times t determinants. I’m going to skip over these and go straight to a wonderful formula due to MacMahon.

M(r,s,t) = \prod_{i=1}^r \prod_{j=1}^s \prod_{k=1}^t \frac{i+j+k-1}{i+j+k-2}. \quad (*)

Of course, there is a lot of cancellation here, so this can be rewritten in many ways.

One can also give a bilinear recurrence for M(r,s,t), to which we’ll be returning throughout this post.

M(r,s,t) M(r-1,s-1,t) = M(r,s-1,t) M(r-1,s,t) +
           M(r,s,t-1) M(r-1,s-1,t+1) \quad (**).

Starting with the initial condition M(r,s,0) = M(r,0,t) = M(0,s,t)=1, we can recursively use (**) to compute M(1,1,t) for all t, then M(1,2,t) and M(2,1,t), then M(2,2,t) and so forth.

I believe it was Eric Kuo who first realized that (**) has a direct combinatorial meaning.

Kuo's interpretation of (**)

Kuo's interpretation of (**)

Each term in this “equation” involves two hexagons, plus some extra rhombi. The meaning is to take all the tilings of the red hexagon, and all the tilings of the blue hexagon, and superimpose them in all possible ways, together with the excess black rhombi. Then every superposition of rhombi appears with the same multiplicity on the left hand side and on the right.

Why is this important? Suppose you wanted to know the expectation of some statistic of a random tiling — for example, the probability that a particular tile would be used. Then you can just take expectations of both sides of Kuo’s relation. Of course, in order to do this, you need to know the relative sizes to the two right hand terms. This is easy, using the explicit formula (*): (s+t-1)/(r+s+t-1) of the superpositions come from the first term, and the remaining r/(s+t-1) come from the second.

This gives you a linear recurrence, a reasonably nice generating function and, we hope, a target to which we can apply Robin’s machinery. But that’s not what I want to talk about, because Peter and Robin are already thinking about it. Instead, I want to explain what I meant when I said that this example had more structure than you would think possible.

MUSINGS ABOUT THE RECURRENCE

In most tiling problems, you can’t find a simple recurrence like (**) at all. In the nice examples I talked about earlier, like Aztec diamonds, most of the above ideas work: you can reinterpret matchings as noncrossing paths, get a determinant, get some sort of product formula, and get a recurrence. It’s hard to recommend just one paper on this subject, but I like Greg Kuperberg’s exposition. (See Section 7 for the case considered here.)

But, in our case, (**) overdetermines the number of tilings! More precisely, the entire situation is symmetric in r, s and t. So, as well as (**), we have the variants of (**) obtained by permuting the coordinates. If we restrict r, s and t to lie between 0 and N, this gives N^3 variables M(r,s,t), subject to \approx 3 N^3 relations. If you were handed this set of quadratic equations, without being told where it came from, you would expect there to be no solutions! Similarly, running the trick I described above, we get more linear equations for the probability of using a given tile than we have variables.

For easy reference, I’ll write the variants obtained by permuting the variables:

F(r,s,t) F(r-1,s-1,t) = F(r,s-1,t) F(r-1,s,t)
          + F(r,s,t-1) F(r-1,s-1,t+1)

F(r,s,t) F(r-1,s,t-1) = F(r,s,t-1) F(r-1,s,t)
          + F(r,s-1,t) F(r-1,s+1,t-1)

F(r,s,t) F(r,s-1,t-1) = F(r,s,t-1) F(r,s-1,t)
           +F(r-1,s,t) F(r+1,s-1,t-1)

So, here is my question: What are the other solutions F(r,s,t) to this trio of recurrences?

I’ve noticed a few easy things. You can rescale F(r,s,t) by w x^r y^s z^t for any (w,x,y,z). You can translate any
solution through three-space to give another solution. Also, I found one more peculiar solution: F(r,s,t) = 2^{\binom{r+s+t}{2}}.

I currently have no idea how much freedom I have in choosing a solution. Here is a way to make that question precise: Let X_N be the set of nonzero* solutions to these equations, restricted to the range 0 \leq r,s,t \leq N. How does the dimension of X_N grow with N?

So, any ideas?

*I look at nonzero solutions for technical reasons: the variety of all solutions to the equations above has many trivial components, where we take enough of the variables to be zero in order to make all the relations trivially true. By restricting to nonzero solutions, I make sure we are studying the main component.

]]> http://sbseminar.wordpress.com/2009/10/21/rhombus-tilings-and-an-over-constrained-recurrence/feed/ 12 davidspeyer Rhombus1a Rhombus2a Rhombus3a Polymath projects on StackExchange/mathoverflow? http://sbseminar.wordpress.com/2009/10/20/polymath-projects-on-stackexchangemathoverflow/ http://sbseminar.wordpress.com/2009/10/20/polymath-projects-on-stackexchangemathoverflow/#comments Tue, 20 Oct 2009 18:48:53 +0000 Scott Morrison http://sbseminar.wordpress.com/?p=2506 ]]>

I’ve been thinking a bit about whether the StackExchange software (which mathoverflow is running on) could be used to host a polymath project.

I’d imagine it involving many many questions and answers, with links between them, modelling the division of the “big question” into its constituent chunks.

There are some big advantages — in particular, it’s easier to pay attention to individual parts, because there’s more structure than in blog comments.

As a first approximation, you might start out like this: Terry asks the polymath7 problem, linking elsewhere for motivation and background. Tim posts a first ‘answer’: “Could we attack this by proving Lemmas X, Y and then generalising the approach of Theorem Z?” and at the same time creates questions corresponding the Lemmas X and Y and a more open question about Theorem Z. Other participants can then go to those questions to give their thoughts. Answers don’t have to be “answers” in the convention sense — they’re just meant to correspond to “ideas”, and should often link to a new question if it’s obvious that the idea needs further development. The StackExchange software allows for comments on answers, which would allow short responses to previous answers.

The big disadvantages of StackExchange are that
* at this point, there’s no LaTeX support, although this will hopefully change.
* the reputation system inhibits new participants, at least at first (they can still ask and answer questions, but commenting and upvoting are limited).
* it may end up harder to understand the “big picture” than in a blog thread.

The solution to the first two of these may be to try a polymath project at mathoverflow.net itself, rather than a new installation. Many participants will already have reputation (and on an established site it’s very easy to gain enough reputation to comment and upvote, because any decent question will quickly garner reputation). It’s easy to filter questions by tags, so I think you could ignore everything else happening on mathoverflow.net if you wanted to.

The last problem might be addressed by having a “community wiki” answer at the top of each question, summarising progress so far, as well as regular progress reports on blogs.

]]> http://sbseminar.wordpress.com/2009/10/20/polymath-projects-on-stackexchangemathoverflow/feed/ 6 semorrison