
This week I'll deviate from my plan of discussing number theory, and instead say a bit about something else that's been on my mind lately: structure types. But, you'll see my fascination with Galois theory lurking beneath the surface.
Andre Joyal invented structure types in 1981  he called them "espèces de structure", and lots of people call them "species". Basically, a structure type is just any sort of structure we can put on finite sets: an ordering, a coloring, a partition, or whatever. In combinatorics we count such structures using "generating functions". A generating function is a power series where the coefficient of x^{n} keeps track of how many structures of the given kind we can put on an nelement set. By playing around with these functions, we can often figure out the coefficients and get explicit formulas  or at least asymptotic formulas  that count the structures in question.
The reason this works is that operations on generating functions come from operations on structure types. For example, in "week190", I described how addition, multiplication and composition of generating functions correspond to different ways to get new structure types from old.
Joyal's great contribution was to give structure types a rigorous definition, and use this to show that many calculations involving generating functions can be done directly with structure types. It turns out that just as generating functions form a set equipped with various operations, structure types form a category with a bunch of completely analogous operations. This means that instead of merely proving equations between generating functions, we can construct isomorphisms between their underlying structure types  which imply such equations, but are worth much more. It's like the difference between knowing two things are equal and knowing a specific reason WHY they're equal!
Of course, this business of replacing equations by isomorphisms is called "categorification". In this lingo, structure types are categorified power series, just as finite sets are categorified natural numbers.
A while back, James Dolan and I noticed that since you can use power series to describe states of the quantum harmonic oscillator, you can think of structure types as states of a categorified version of this physical system! This gives new insights into the combinatorial underpinnings of quantum physics.
For example, the discrete spectrum of the harmonic oscillator Hamiltonian can be traced back to the discreteness of finite sets! The commutation relations between annihilation and creation operators boil down to a very simple fact: there's one more way to put a ball in a box and then take one out, than to take one out and then put one in. Even better, the whole theory of Feynman diagrams gets a simple combinatorial interpretation. But for this, one really needs to go beyond structure types and work with a generalization called "stuff types".
I've been thinking about this business for a while now, so last fall I decided to start giving a yearlong course on categorification and quantization. The idea is to explain bunches of quantum theory, quantum field theory and combinatorics all from this new point of view. It's fun! Derek Wise has been scanning in his notes, and a bunch of people have been putting their homework online. So, you can follow along if you want:
1) John Baez and Derek Wise, Quantization and Categorification.
Fall 2003 notes: http://math.ucr.edu/home/baez/qgfall2003
Winter 2004 notes: http://math.ucr.edu/home/baez/qgwinter2004/
Spring 2004 notes: http://math.ucr.edu/home/baez/qgspring2004/
I'd like to give you a little taste of this subject now. But, instead of explaining it in detail, I'll just give some examples of how structure types yield some farout generalizations of the concept of "cardinality". This stuff is a continuation of some themes developed in "week144", "week147", "week185", "week190", so I'll start with a review.
Suppose F is a structure type. Let F_{n} be the set of ways we can put this structure on a nelement set, and let F_{n} be the number of ways to do it. In combinatorics, people take all these numbers F_{n} and pack them into a single power series. It's called the generating function of F, and it's defined like this:
F_{n} F(x) = ∑  x^{n} n!It may not converge, so in general it's just a "formal" power series  but for interesting structure types it often converges to an interesting function.
What's good about generating functions is that simple operations on them correspond to simple operations on structure types. We can use this to count structures on finite sets. Let me remind you how it works for binary trees!
There's a structure type T where a Tstructure on a set is a way of making it into the leaves of a binary tree drawn in the plane. For example, here's one Tstructure on the set {a,b,c,d}:
b d a c \ \ / / \ \/ / \ / / \/ / \ / \/Thanks to the choice of different orderings, the number of Tstructures on an nelement set is n! times the number of binary trees with n leaves. Annoyingly, the latter number is traditionally called the (n1)st Catalan number, C_{n1}. So, we have:
T(x) = ∑ C_{n1} x^{n}where the sum starts at n = 1.
There's a nice recursive definition of T:
"To put a Tstructure on a set, either note that it has one element, in which case there's just one Tstructure on it, or chop it into two subsets and put a Tstructure on each one."In other words, any binary tree is either a degenerate tree, with just one leaf:
Xor a pair of binary trees stuck together at the root:
       T   T        \ / \ / \ / \/We can write this symbolically as
T ≅ X + T^{2}Here's why: X is a structure type called "being the oneelement set", + means "exclusive or", and squaring a structure type means you chop your set in two parts and put that structure on each part. (I explained these rules more carefully in "week190".)
Note that we only have an isomorphism between structure types here, not an equation. But if we take the generating function of both sides we get an actual equation, and the notation is set up to make this really easy:
T = x + T^{2}In "week144" I showed how you can solve this using the quadratic equation:
T = (1  sqrt(1  4x))/2.and then do a Taylor expansion to get
T = x + x^{2} + 2x^{3} + 5x^{4} + 14x^{5} + 42x^{6} + ...Lo and behold! The coefficient of x^{n} is the number of binary trees with n leaves!
There's also another approach where we work directly with the structure types themselves, instead of taking generating functions. This is harder because we can't subtract structure types, or divide them by 2, or take square roots of them  at least, not without stretching the rules of this game. All we can do is use the isomorphism
T ≅ X + T^{2}and the basic rules of category theory. It's not as efficient, but it's illuminating. It's also incredibly simple: we just keep sticking in "X + T^{2}" wherever we see "T" on the righthand side, over and over again. Like this:
T ≅ X + T^{2} T ≅ X + (X + T^{2})^{2} T ≅ X + (X + (X + T^{2})^{2})^{2}and so on. You might not think we're getting anywhere, but if you stop at the nth stage and expand out what we've got, you'll get the first n terms of the Taylor expansion we had before! At least, you will if you count "stages" and "terms" correctly.
I won't actually do this, because it's better if you do it yourself. When you do, you'll see it captures the recursive process of building a binary tree from lots of smaller binary trees. Each time you see a "T" and replace it with an "X + T^{2}", you're really taking a little binary tree:
    T    and replacing it with either a degenerate tree with just a single leaf:
Xor a pair of binary trees:
       T   T        \ / \ / \ / \/So, each term in the final result actually corresponds to a specific tree! This is a good example of categorification: when we calculate the coefficient of x^{n} this way, we're not just getting the number of binary planar trees with n leaves  we're getting an actual explicit description of the set of such trees.
Now, what happens if we take the generating function T(x) and evaluate it at x = 1? On the one hand, we get a divergent series:
T(1) = 1 + 1 + 2 + 5 + 14 + 42 + ...This is the sum of all Catalan numbers  or in other words, the number of binary planar trees. On the other hand, we can use the formula
T = (1  sqrt(1  4x))/2to get
T(1) = (1  sqrt(3))/2It may seem insane to conclude
1 + 1 + 2 + 5 + 14 + 42 + ... = (1  sqrt(3))/2but Lawvere noticed that there's a kind of strange sense to it.
The trick is to work not with generating function T but with the structure type T itself. Since T(1) is equal to the number of planar binary trees, T(1) should be naturally isomorphic to the set of planar binary trees. And it is  it's obvious, once you think about what it really means.
The number of binary planar trees is not very interesting, but the set of them is. In particular, if we take the isomorphism
T ≅ X + T^{2}and set X = 1, we get an isomorphism
T(1) ≅ 1 + T(1)^{2}which says
"a planar binary tree is either the tree with one leaf or a pair of planar binary trees."Starting from this, we can derive lots of other isomorphisms involving the set T(1), which turn out to be categorified versions of equations satisfied by the number
T(1) = (1  sqrt(3))/2For example, this number is a sixth root of unity. While there's no onetoone correspondence between 6tuples of trees and the 1 element set, which would categorify the formula
T(1)^{6} = 1there is a very nice onetocorrespondence between 7tuples of trees and trees, which categorifies the formula
T(1)^{7} = T(1)Of course the set of binary trees is countably infinite, and so is the set of 7tuples of binary trees, so they can be placed in onetoone correspondence  but that's boring. When I say "very nice", I mean something more interesting: starting with the isomorphism
T ≅ x + T^{2}we get a onetoone correspondence
T(1) ≅ 1 + T(1)^{2}which says that any binary planar tree is either degenerate or a pair of binary planar trees... and using this we can construct a onetoone correspondence
T(1)^{7} ≅ T(1)The construction is remarkably complicated. Even if you do it as efficiently as possible, I think it takes 18 steps, like this:
T(1)^{7 } ≅ T(1)^{6} + T(1)^{8} ≅ T(1)^{5} + T(1)^{7} + T(1)^{8} . . . ≅ 1 + T(1) + T(1)^{2} + T(1)^{4} ≅ 1 + T(1) + T(1)^{3} ≅ 1 + T(1)^{2} ≅ T(1)I'll let you fill in the missing steps  it's actually quite fun if you like puzzles.
If you get stuck, you can find the answer online in a couple of different places:
2) Andreas Blass, Seven trees in one, Jour. Pure Appl. Alg. 103 (1995), 121. Also available at http://www.math.lsa.umich.edu/~ablass/cat.html
3) Marcelo Fiore, Isomorphisms of generic recursive polynomial types, to appear in 31st Symposium on Principles of Programming Languages (POPL04). Also available at http://www.cl.cam.ac.uk/~mpf23/papers/Types/recisos.ps.gz
Or, take a peek at the "Addenda" down below.
Robbie Gates, Marcelo Fiore and Tom Leinster have also proved some very general theorems about this sort of thing. Gates focused on "distributive categories" (categories with with products and coproducts, the former distributing over the latter), while the work of Fiore and Leinster applies to more general "rig categories":
4) Robbie Gates, On the generic solution to P(X) = X in distributive categories, Jour. Pure Appl. Alg. 125 (1998), 191212.
5) Marcelo Fiore and Tom Leinster, Objects of categories as complex numbers, available as math.CT/0212377.
A rig category is basically the most general sort of category in which we can "add" and "multiply" as we do in a ring  but without negatives, hence the missing letter "n". It turns out that whenever we have an object Z in a rig category and it's equipped with an isomorphism
Z = P(Z)where P is a polynomial with natural number coefficients, we can associate to it a "cardinality" Z, namely any complex solution of the equation
Z = P(Z)Which solution should we use? Well, for simplicity let's consider the case where P has degree at least 2 and the relevant Galois group acts transitively on the solutions of this equation, so "all roots are created equal". Then we can pick any solution as the cardinality Z. Any polynomial equation with natural number coefficients satisfied by one solution will be satisfied by all solutions, so it won't matter which one we choose.
Now suppose the cardinality Z satisfies such an equation:
Q(Z) = R(Z)where neither Q nor R is constant. Then the results of Fiore and Leinster say we can construct an isomorphism
Q(Z) = R(Z)!In other words, a bunch of equations satisfied by the object's cardinality automatically come from isomorphisms involving the object itself.
This explains why the set T(1) of binary trees acts like it has cardinality
T(1) = (1  sqrt(3))/2or equally well,
T(1) = (1 + sqrt(3))/2(Since the relevant Galois group interchanges these two numbers, we can use either one.) More generally, the set T(n) consisting of binary trees with ncolored leaves acts a lot like the number T(n).
This has gotten me interested in trying to find a nice example of a "Golden Object": an object G in some rig category that's equipped with an isomorphism
G^{2} = G + 1The Golden Object doesn't fit into Fiore and Leinster's formalism, since this isomorphism is not of the form G = P(G) where P has natural number coefficients. But, it still seems that such an object deserves to have a "cardinality" equal to the golden number:
G = (1 + sqrt(5))/2 = 1.618033988749894848204586834365...James Propp came up with an interesting idea related to the Golden Object: consider what happens when we evaluate the generating function for binary trees at 1. On the one hand we get an alternating sum of Catalan numbers:
T(1) = 1 + 1  2 + 5  14 + 42 + ...On the other hand, we can use the formula
T = (1  sqrt(1  4x))/2to get
T(1) = (1  sqrt(5))/2which is 1 divided by the golden number. Of course, it's possible we should use the other sign of the square root, and get
T(1) = (1 + sqrt(5))/2which is just the golden number! Galois theory says these two roots are created equal. Either way, we get a bizarre and fascinating formula:
 1 + 1  2 + 5  14 + 42 + ... = (1 + sqrt(5))/2Can we fit this into some clear and rigorous framework, or is it just nuts? We'd like some generalization of cardinality for which "the set of binary trees with 1colored leaves" has cardinality equal to the golden number.
James Propp suggested one avenue. A while back, Steve Schanuel made an incredibly provocative observation: if we treat "Euler measure" as a generalization of cardinality, it makes sense to treat the real line as a "space of cardinality 1":
6) Stephen H. Schanuel, What is the length of a potato?: an introduction to geometric measure theory, in Categories in Continuum Physics, Spring Lecture Notes in Mathematics 1174, Springer, Berlin, 1986, pp. 118126.
7) Stephen H. Schanuel, Negative sets have Euler characteristic and dimension, Lecture Notes in Mathematics 1488, Springer Verlag, Berlin, 1991, pp. 379385.
James Propp has developed this idea in a couple of fascinating papers:
8) James Propp, Euler measure as generalized cardinality, available as arXiv:math/0203289.
9) James Propp, Exponentiation and Euler measure, available as arXiv:math/0204009.
Using this idea, it seems reasonable to consider the space of binary trees with leaves labelled by real numbers as a rigorous version of "the set of binary trees with 1colored leaves". So, we just need to figure out what generalization of Euler characteristic gives this space an Euler characteristic equal to the golden number. It would be great if we could make this space into a Golden Object in some rig category, but that may be asking too much.
Whew! There's obviously a lot of work left to be done here. Here's something easier: a riddle. What's this sequence?
un, dos, tres, quatre, cinc, sis, set, vuit, nou, deu,...The answer is at the end of this article.
Now I'd like to mention some important papers on ncategories. You may think I'd lost interest in this topic, because I've been talking about other things. But it's not true!
Most importantly, Tom Leinster has come out with a big book on ncategories and operads:
10) Tom Leinster, Higher Operads, Higher Categories, Cambridge U. Press, Cambridge, 2003. Also available as arXiv:math.CT/0305049.
As you'll note, he managed to talk the press into letting him keep his book freely available online! We should all do this. Nobody will ever make much cash writing esoteric scientific tomes  it takes so long, you could earn more per hour digging ditches. The only financial benefit of writing such a book is that people will read it, think you're smart, and want to hire you, promote you, or invite you to give talks in cool places. So, maximize your chances of having people read your books by keeping them free online! People will still buy the paper version if it's any good....
And indeed, Leinster's book has many virtues besides being free. He gracefully leads the reader from the very basics of category theory straight to the current battle front of weak ncategories, emphasizing throughout how operads automatically take care of the otherwise mindnumbing thicket of "coherence laws" that inevitably infest the subject. He doesn't take wellestablished notions like "monoidal category" and "bicategory" for granted  instead, he dives in, takes their definitions apart, and compares alternatives to see what makes these concepts tick. It's this sort of careful thinking that we desperately need if we're ever going to reach the dream of a clear and powerful theory of higherdimensional algebra. He does a similar careful analysis of "operads" and "multicategories" before presenting a generalized theory of operads that's powerful enough to support various different approaches to weak ncategories. And then he describes and compares some of these different approaches!
In short: if you want to learn more about operads and ncategories, this is the book to read.
Leinster doesn't say too much about what ncategories are good for, except for a nice clear introduction entitled "Motivation for Topologists", where he sketches their relevance to homology theory, homotopy theory, and cobordism theory. But this is understandable, since a thorough treatment of their applications would vastly expand an already hefty 380page book, and diffuse its focus. It would also steal sales from my forthcoming book on higherdimensional algebra  which would be really bad, since I plan to retire on the fortune I'll make from this.
Secondly, Michael Batanin has worked out a beautiful extension of his ideas on ncategories which sheds new light on their applications to homotopy theory:
11) Michael A. Batanin, The EckmannHilton argument, higher operads and E_{n} spaces, available as arXiv:math.CT/0207281.
Michael A. Batanin, The combinatorics of iterated loop spaces, available as arXiv:math.CT/0301221.
Getting a manageable combinatorial understanding of the space of loops in the spaces of loops in the space of loops... in some space has always been part of the dream of higherdimensional algebra. These "kfold loop spaces" or have been important in homotopy theory since the 1970s  see the end of "week199" for a little bit about them. People know that kfold loop spaces have k different products that commute up to homotopy in a certain way that can be summarized by saying they are algebras of the E_{k} operad, also called the "little kcubes operad". However, their wealth of structure is still a bit mindboggling. James Dolan and I made some conjectures about their relation to ktuply monoidal categories in our paper "Categorification" (see "week121"), and now Batanin is making this more precise using his approach to ncategories  which is one of the ones described in Leinster's book.
There's also been a lot of work applying higherdimensional algebra to topological quantum field theory  that's what got me interested in ncategories in the first place, but a lot has happened since then. For a highly readable introduction to the subject, with tons of great pictures, try:
12) Joachim Kock, Frobenius Algebras and 2D Topological Quantum Field Theories, Cambridge U. Press, Cambridge, 2003.
This is mainly about 2d TQFTs, where the concept of "Frobenius algebra" reigns supreme, and everything is very easy to visualize.
When we go up to 3dimensional spacetime life gets harder, but also more interesting. This book isn't so easy, but it's packed with beautiful math and wonderfully drawn pictures:
13) Thomas Kerler and Volodymyr L. Lyubashenko, NonSemisimple Topological Quantum Field Theories for 3Manifolds with Corners, Lecture Notes in Mathematics 1765, Springer, Berlin, 2001.
The idea is that if we can extend the definition of a quantum field theory to spacetimes that have not just boundaries but corners, we can try to build up the theory for arbitrary spacetimes from its behavior on simple building blocks  since it's easier to chop manifolds up into a few basic shapes if we let those shapes have corners. However, it takes higherdimensional algebra to describe all the ways we can stick together manifolds with corners! Here Kerler and Lyubashenko make 3dimensional manifolds going between 2manifolds with boundary into a "double category"... and make a bunch of famous 3d TQFTs into "double functors".
Closely related is this paper by Kerler:
14) Thomas Kerler, Towards an algebraic characterization of 3dimensional cobordisms, Contemp. Math. 318 (2003) 141173. Also available as arXiv:math/0106253.
It relates the category whose objects are 2manifolds with a circle as boundary, and whose morphisms are 3manifolds with corners going between these, to a braided monoidal category "freely generated by a Hopf algebra object". (I'm leaving out some fine print here, but probably putting in more than most people want!) It comes close to showing these categories are the same, but suggests that they're not quite  so the perfect connection between topology and higher categories remains elusive in this important example.
Answer to the riddle: these are the Catalan numbers  i.e., the natural numbers as written in Catalan. This riddle was taken from the second volume of Stanley's book on enumerative combinatorics (see "week144").
Addenda: Long after this issue was written, we had a discussion on the nCategory Café about the "seven trees in one" problem. Let B be the set of binary planar trees  the set I was calling T(1) above. Starting from the isomorphism
B ≅ B^{2} + 1
we want to construct an isomorphism
B ≅ B^{7}
Here is the proof in Marcelo Fiore's paper:
At each step he either replaces B^{n} by B^{n1} + B^{n+1}, or the reverse. The underlined portion shows where this will be done. Over at the nCafé, Stuart Presnell made a beautiful picture of this proof:
He also made a picture of another proof, which is on page 29 of Pierre Ageron's book Logiques, Ensembles, Catégories: Le Point de Vue Constructif:
You can watch a movie of a proof here:
15) Dan Piponi, Arboreal isomorphisms from nuclear pennies, A Neighborhood of Infinity, September 30, 2007. Available at http://blog.sigfpe.com/2007/09/arborealisomorphismsfromnuclear.html.
It was in the ensuing discussion on this blog that George Bell came up with his more efficient proof. For a bit more discussion, see:
16) John Baez, Searching for a video proof of "seven trees in one", nCategory Café, July 16, 2009. Available at http://golem.ph.utexas.edu/category/2009/07/searching_for_a_video_proof_of.html.
Now, on to some older addenda!
My pal Squark pointed out that if we try to compute the generating function for binary trees by making an initial guess for T(x), say t, and repeatedly improving this guess via
t → x + t^{2}the guess will converge to the right answer if t is small  but the process will fail miserably, with t approaching ∞, if and only if the complex number x lies outside the Mandelbrot set!
After an earlier version of this Week appeared on the category theory mailing list, Steve Schanuel posted some corrections. I've tried to correct the text above as much as possible without making it too technical  for example, by citing the important work of Robbie Gates, and distinguishing more clearly between his work on distributive categories and the paper by Fiore and Leinster, which applies to rig categories. I tend to talk about 3 different sorts of ringlike categories in This Week's Finds:
17) M. Laplaza, Coherence for distributivity, Lecture Notes in Mathematics 281, Springer Verlag, Berlin, 1972, pp. 2972.
18) G. Kelly, Coherence theorems for lax algebras and distributive laws, Lecture Notes in Mathematics 420, Springer Verlag, Berlin, 1974, pp. 281375.
(These authors spoke of "ring categories", but the term "rig category" is more appropriate since, as in a rig, there need be no additive inverses.)
Dear colleagues,For those who read the most recent long discursion of John Baez, a few of the errors in the section on distributive categories merit correction:
(1) J. B. suggests that Blass published what Lawvere had already worked out. In fact, Lawvere (partly to counteract some incorrect uses of infinite series in analyses of 'data types' in computer science) had worked out the algebra of the rig presented by one generator X and one relation X=1+X^{2}, roughly by the method in (3) below, and conjectured that this rig could be realized as the isomorphism classes in a distributive (even extensive) category, which conjecture Blass then proved (and a bit more) in "Seven Trees...".
(2) The generalization of Blass's theorem to one generator ond one polynomial relation of the 'fixedpoint' form X=p(X), where p is a polynomial with natural number coefficients and nonzero constant term is not, as J. B. seems to suggest, due to Fiore and Leinster; it was part of the prizewinning doctoral thesis of Robbie Gates, who (using a calculus of fractions) described explicitly the free distributive category on one object X together with an isomorphism from p(X) to X, proving that this category is extensive and that its rig of isomorphism classes satisfies no further relations, i.e. is the rig R presented by one generator and the one relation above.
(3) If p is as in (2) and of degree at least 2, the algebra of the rig R is made by J. B. to seem mysterious. It is more easily understood in the way the X=2X+1 case was treated in my "Negative Sets..." paper; just show that the Euler and dimension homomorphisms, tensoring with Z and with 2 (the rig true/false) respectively, are jointly injective. In this case the dimension rig has only three elements, which explains why the Euler characteristic captures almost, but not quite, everything.
Greetings to all,
Steve Schanuel
© 2004 John Baez
baez@math.removethis.ucr.andthis.edu
