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

How many elements of 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 doesn’t seem to work for n > 2.

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 of objects, a subset of the maps . (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, can be concretized by the functor which sends a set to the set of subsets of , and sends a map of sets to the preimage map .

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 be a category and a functor from to . Then is a concretization if, for any objects and , and any morphisms and from to , we have only if .

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

Yoneda’s lemma tells us that, if and induce the same map from to for every , then . The proof is simply to take .

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 , which act as if the objects of 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 of test objects, so that it is enough to see whether on for .

There are several great theorems of this form: In the category of varieties of finite type over , the Nullstellansatz tells us that a map is determined by its values on — 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 , for all spheres . 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 has a small concretization if there is a set of objects of such that 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 were a concretization. For simplicity, I’ll assume that is the set with a single element. For any connected space , let be the element of corresponding to the unique homotopy class of map .

Freyd constructs a poset , with cardinality greater than , two sequences of connected spaces and indexed by , and maps . These have the property that the composite does not factor through but , with , does. Since is so big, there must be some such that and map to the same subset of . Call this subset .

Then factors through , so the map sends to . But then also sends to , contradicting that does not factor through a point.

To give a little more detail, one first constructs a sequence of groups , with nonzero maps , such that, for any map with , the composite would be zero. Let be a . Note that is a , so we have functorial maps . Then is the suspension , the map is the suspension of , and is the mapping cone of . If you want more detail than this, you should read Freyd’s paper.

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.

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 , , , , and .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 for the number of such matchings.

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

If , then the hexagon degenerates to a parallelogram, and there is only one tiling. If , then the horizontal edge has length . 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 .

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

If , 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 -tuples of noncrossing paths, and you wind up computing determinants. I’m going to skip over these and go straight to a wonderful formula due to MacMahon.

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 , to which we’ll be returning throughout this post.

           .

Starting with the initial condition , we can recursively use to compute for all , then and , then and so forth.

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

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 : of the superpositions come from the first term, and the remaining 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 , and . So, as well as , we have the variants of obtained by permuting the coordinates. If we restrict , and to lie between and , this gives variables , subject to 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:

So, here is my question: What are the other solutions to this trio of recurrences?

I’ve noticed a few easy things. You can rescale by for any . You can translate any
solution through three-space to give another solution. Also, I found one more peculiar solution: .

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

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.

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.

There’s been a very interesting dicussion in comments on Scott and Anton’s post about the strengths and weaknesses of MathOverflow and nLab, and I thought it might be good to divert to a new post (and use my position as blogger to top-post my thoughts). (edited since the original post. Look below for more)

Now just a moment ago, I followed a link, and started reading the n-Forum, and as a result, discovered that nLab has no page on knot invariants. And I thought, “hmmm, someone should fix that, and I guess that someone should be me.” So, I clicked a link, ended up with a blank page on knot invariants, and immediately realized that that person shouldn’t have been me at all.

What it comes down to for me, is that when I try to edit nLab, I always end up with a blank page, and no idea what I want to say. Andrew’s right that one should keep in mind that one is not writing Wikipedia (I think in my previous nLab forays, I lapsed into Wikipedian without even thinking about it). But still on some fundamental level, I had no idea what I wanted to tell people about knot invariants, or what they wanted to know. And “What knot invariants are.” is my standard explaining my research to a non-mathematician spiel; if there’s one thing I should be able to effortlessly write an nLab page for, that was it. And yet, it felt forced, and it was definitely not fun.

Whereas in my experience, MathOverflow is instantly absorbing and fun. Several other regular users have already told me they’ve found it an addictive time sink. I think the important difference is the level of feedback and interaction. On MO, one really feels as though other users are engaged with your questions and answers, whereas in nLab, I feel like I am talking into empty space. I think the other point is that the hardest part of mathematics is knowing which mathematics to do, so having a lot of concrete suggestions, not just for topics to write on, but questions to answer overcomes a real psychological barrier.

EDIT: Now that I think about it, there may be a good answer to this question. The sense I get from Andrew and Urs is that I haven’t been using nLab selfishly enough. I’ve been thinking about writing things that will be useful for other people, when perhaps I should be writing things that are useful for me to write, with some vague hope that other people will also find them useful and have input. Perhaps I will even try this at some point when I finish up a couple of current project and am in more of a reorganizing stage.

I think another point is that it would never occur to me to have an online lab book, because I don’t have a physical one. Some people (Scott M. and Aaron Lauda are examples that spring to mind) have physical lab books that they write lots of notes and calculations in. I don’t; I never take notes when reading papers, I write all calculations on random papers which I immediately proceed to lose (I’ve been using extra copies of a Number Theory exam recently), I never write anything down in any kind of structured form before I start drafting the paper. I’m sure I know how to use a lab book, which why I am probably not yet read to use nLab.

(written collaboratively by Anton Geraschenko and Scott Morrison, in google wave)

Math Overflow (MO) is a brand new mathematics questions and answers site. You should go give it a try! Several of the regular readers of this blog are already there. It’s much more fun if you actually ask a question, and the only way to get the full experience.

Math Overflow has lots of features that we think makes it really awesome:

• Questions and answers get voted up and down, making it easy to find the gems in a large pool of content.
• There is a reputation system. As your questions and answers get voted up, your reputation goes up. The more reputation you get, the more the system allows you to do. Above 10,000 reputation, there is little difference between between users and moderators.
• It’s sorta wiki. Once you reach 2000 reputation, you can edit other peoples’ posts to correct typos and make the presentation clearer. Questions and answers can also be marked “community wiki”, in which case all established users (100+ reputation) can edit them. (By the way, when you post material on the site, it is under the Creative Commons Attribution Share-alike license.

Now MO isn’t intended to fill the same niche as math blogging. In fact, we’re not really sure what niche(s) it’s going to fill; it’s an experiment! It will be much more focused on “smaller” and “more technical” questions, especially ones that have definite answers. This is of course both good and bad. On the bad side, some really interesting questions will be less suitable for MO, simply because it’s harder to develop answers incrementally, conversationally, or collaboratively. In particular the “polymath” style of problem solving probably isn’t going to happen on MO to the extent that it happens on a math blog. On the other hand, MO has already proved itself surprisingly effective, especially if you really want an answer to a particular question. Hopefully after a while google will start effectively indexing MO, and we’ll even see incoming searches to previous questions.

At present the focus is largely in algebraic geometry, although the past week has also seen lots of growth in TQFTs and homological algebra. This is in part just because of the current user population, and it will hopefully broaden as people other than our friends start using it! Even if it remains concentrated in certain areas it may still be successful. If you’d like to influence the development, and encourage good coverage of your field, go ask and answer some questions, and invite your colleagues! One worry, of course, is that it will be overwhelmed by calculus students. Before you worry too much about this, go and browse through the current questions, observe the one or two inappropriate questions and how we’ve dealt with them. For now it really doesn’t seem to be a problem.

We’re hoping that this comment thread (and perhaps new ones as necessary) will become the central place for “meta” discussion of Math Overflow. In particular, if you have comments or criticisms about the site, or even the very idea, please tell us here. Complaints about bugs, the community, or the overbearing manner of the moderators are all okay too. We’re even happy to hear that you don’t like the color scheme. Some aspects of the site are beyond our control, but we’d like to do whatever we can to make Math Overflow a great place to do math.

Some known issues (and preemptive responses):

• “MO is not a discussion site”. There was some talk in the comments here about whether MO is amenable to discussions. The Q&A format encourages people to ask particular questions which can be given definite answers. Some discussion will take place in the comments, in particular request for clarification or pointing out errors. To keep things on-topic, comments are limited to 600 characters. Also keep in mind that answers will often change order as later ones get voted up. All of this means that the framework forces any discussion to remain fairly focused. If you want to discuss something more broadly, well, come back the Secret Blogging Seminar. Remember that MO is not meant to fill the same niche a blog does.
• There’s currently no LaTeX support. Obviously, this is the top-priority feature. This definitely won’t be implemented until the Stack Exchange software is out of beta, but it will probably be implemented shortly thereafter. In the meantime, do your best. You’re welcome to type raw LaTeX, use basic html (e.g. Ω and x₁), or whatever you would normally write in email.
• A few users have reported problems with their Google-issued OpenID. Occasionally, for reasons unknown, Google will issue an OpenID different from the one it issued before, so the site will not recognize you. If you have this problem, just create a new user and email Anton (geraschenko@mathoverflow.net). He’ll merge the two users and add the other OpenID as an alternate, so your account will respond to both. If it keeps being a problem, the easiest thing is to use a non-Google OpenID (like myopenid.com).

Finally, some statistics and a little background: Math Overflow is only 2 weeks old, it already has about 50 active users, 150 questions, and over 300 answers. It runs the same software (Stack Exchange) as the very successful stackoverflow.com for programmers. The hosting costs have been generously funded by Ravi Vakil at Stanford.

Hi folks,

Here are the latest problems from the 20 Questions seminar for the query-hungry :)

====1 Sune====

[read/reply on wiki.]

Is it possible to explicitly define (without AC) a total ordering on a set with cardinality greater than ?

====2 Yuhao====

[read/reply on wiki.]

Let be the function field of an algebraic curve . Then by the
birational nature of the genus, it determines the genus of the
curve. How can we see directly from the field?

(One way is to PICK a function in and we get a ramified morphism , then use Riemann-Hurwitz to compute the genus; However, I
think there should be a more “intrinsic” way to see that, i.e. without
picking a function in an arbitrary manner. The genus is a well-defined
invariant for any extension of the ground field of tr.deg. 1, right?)

====3 Anon====

[read/reply on wiki.]

Determine explicitly a partition of the plane into two sets A and B such that neither A nor B contains (the image of) a non-constant continuous curve.

====4 An H====

[read/reply on wiki.]

Let denote the cyclotomic field obtained by adjoining th roots of unity to , where is prime. Let be a prime element in the ring of
integers of which is coprime to . Let denote the Kummer extension of
by adjoining a th root of to .

Question: is it always true that divides the exponent of the
prime ideal in the relative discriminant ? (as usual,
denotes a primitive pth root of unity)

====5 Critch====

[read/reply on wiki.]

In the categories Set and AbGrp (and I believe any topos), finte limits (e.g. kernels, products, equalizers…) commute with directed colimits (aka direct limits). Give an example category where this fails.

====6 Anton====

[read/reply on wiki.]

Can a coequalizer in the category of Schemes fail to be surjective? (Note: it must hit all closed points in the target, because otherwise the closed point could be removed to make the coequalizer smaller.)

See the mathoverflow.net question: [http://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective]

====7 Darsh====

[read/reply on wiki.]

Let be rational.

In what generalized notions of convergence does the sequence converge?

====8 Harold====

[read/reply on wiki.]

In the category of smooth manifolds, when does the fibre product of two maps exist?

A) Necessary conditions

B) Sufficient conditions (e.g. that each map is a fibration).

====9 James T====

[read/reply on wiki.]

Which bounded linear maps on a Hilbert space are the exponentials of other maps? I.e., what is the image of the map ?

====10 Pablo====

[read/reply on wiki.]

Give a simple example of a (necessarily infinte dimensional) Lie algebra that is not the lie algebra of any (necessarily infinite dimensional) lie group.

====11 Theo====

[read/reply on wiki.]

[http://mathoverflow.net/questions/134 See mathoverflow.net].

****

Cheers!
-Critch

Vaughan Jones often quips at the beginning of talks on Planar Algebras (see these lectures, for example) that the worst thing you can say about Planar Algebras is that they have not yet yielded a proof of the 4-color theorem. In this post I’ll sketch how a common “evaluation algorithm” (used by Greg Kuperberg and by Emily Peters, for example) almost proves the 4-color theorem. I believe this (failed) argument is due to Penrose, though I’m taking it from an article of Chmutov, Duzhin, and Kaishev and some notes of John Baez’s. There are some more elaborate attacks (by Kauffman, Saleur, Bar Natan, and probably others) that I won’t discuss at all. This is the second of what hopefully will be a short series of posts on “evaluation algorithms” (the first was on the Jellyfish algorithm).

The outline of the post is as follows. First I’ll explain a standard reduction of the 4-color theorem to a question about 3-coloring edges of trivalent graphs. Second I’ll explain why 3-colorings of edges is a question about finding a positive evaluation algorithm for a certain planar algebra. Third, I’ll discuss “Euler characteristic” evaluation algorithms. Fourth I’ll explain how this technique almost answers the 4-color theorem.

Suppose you have a planar graph and you want to 4-color the faces (note, I said faces, not vertices). First notice that you can reduce to the case of 3-valent planar graphs by “adding a new country at every higher valency vertex” (see Week 22 in Mathematical Physics for a delightful ASCII drawing). This observation is from the early days of the 4-color problem and due to Cayley and Kempe. The next reduction also comes from the 19th century, and is due to Tait. There is a bijection between “4-colorings of faces” and “4-coloring one distinguished face and then 3-coloring the edges.” The way to understand this bijection is to label the edges with “rules for changing face colorings.” Namely there are 3 ways to assign a pairing of the four colors. If you’ve chosen such a rule for every edge then you can propogate your single face coloring to the whole graph. It’s a simple exercise to check that this gives a bijection between valid colorings.

The invariant of planar graphs “number of 3-colorings of edges” extends to a planar algebra/TQFT/tensor category by the recipe similar to that Ben outlined here. Any planar graph with boundary can be thought of as a functional on the space Span{labelings of the boundary}. Extending by linearity any linear combination of planar graphs with boundary can also be thought of as such a functional. Mod out by the equivalence setting two diagrams equal to each other if they give the same functional. The key fact is that these relations play well with gluing together diagrams and hence gives a planar algebra (or tensor category together with a chosen generating object).

What explicitly do these relations look like? You can remove a bigon for a multiplicative cost of 2, a triangle for a multiplicative cost of 1, and a circle with no trivalent vertices for a multiplicative cost of 3. Furthermore there is a version of the I = H relation (I strongly encourage you to check yourself that this relation between number of 3-colorings of edges holds for any coloring of the boundary):

So what do we need to do to prove the 4-color theorem? We need to give a manifestly positive algorithm for evaluating closed diagrams in this planar algebra! That’s it.

One common technique for evaluating closed planar diagrams is to concentrate on the faces. Notice above we know that we can remove all bigons and triangles. What about bigger faces? Well the beautiful thing is that by Euler characteristic arguments all you need to do is prove that you can remove squares and pentagons because any planar diagram has a pentagonal or smaller face. In fact this technique and small variations on it have been remarkably successful at answer planar algebraic questions (once for quantum groups and once for an exotic subfactor).

So let’s try that. Here’s the square (finding this formula using the above relations is kinda fun):

Now there’s just one case left, the pentagon. Just find a positive formula there and you’d be done!

Rereading Noah’s graduate school advice post, I realized I’d forgotten to stir up trouble at the time about his comments on REUs. In part, one should understand this post as an attempt to goad him into explaining.

First, a little personal history. I’m basically the poster-child for REUs; doing an REU in the year between my junior and senior years was the only organized math outside of school I participated in before grad school, and was the only experience with research I had before grad school, essentially. I had an excellent relationship with my advisor from the REU, met several people I liked a lot, did good enough research to turn it into a solo paper (admittedly, several years later), and generally had a really excellent experience. I’ve always recommended REUs to students, especially if they were considering grad school.

This has always seemed like a no brainer to me. If nothing else, since an REU is essentially the only real chance that an undergrad has to test drive grad school before committing years of their life to it. So, I’ll admit, I was a little surprised to find out that “REU-hater” is a category of person that exists. And now I’m curious; are there any more of you out there?