
Last week I posed this puzzle: to find a "Golden Object".
A couple days ago I got a wonderful solution from Robin Houston, a computer science grad student at the University of Manchester. So, I want to say a bit more about the golden number, then describe his solution, and then describe how he found it.
Supposedly the Greeks thought the most beautiful rectangle was one such that when you chop a square off one end, you're left with a rectangle of the same shape. If your original rectangle was 1 unit across and G units long, after you chop a 1by1 square off the end you're left with a rectangle that's G1 units across and 1 unit long:
G ......................... . . . . . . . . . . . . 1 . . . 1 . . . . . . . 1 . G1 . .........................So, to make the proportions of the little rectangle the same as those of the big one, you want
"1 is to G as G1 is to 1"or in other words:
1/G = G  1or after a little algebra,
G^{2} = G + 1so that
G = (1 + sqrt(5))/2 = 1.618033988749894848204586834365...while
1/G = 0.618033988749894848204586834365...and
G^{2 } = 2.618033988749894848204586834365...(At this point I usually tell my undergraduates that the pattern continues like this, with G^{3} = 3.618... and so on  just to see if they'll believe anything I say.)
These days, the number G is called the Golden Number, the Golden Ratio, or the Golden Section. It's often denoted by the Greek letter Φ, after the Greek sculptor Phidias. Phidias helped design the Parthenon  and supposedly packed it full of golden rectangles, to make it as beautiful as possible.
The golden number is a great favorite among amateur mathematicians, because it has a flashy sort of charm. You can find it all over the place if you look hard enough  and if you look too hard, you'll find it even in places where it's not. It's the ratio of the diagonal to the side of a regular pentagon! If you like the number 5, you'll be glad to know that
5 + sqrt(5) G = sqrt[] 5  sqrt(5)If you don't, maybe you'd prefer this:
G = exp(arcsinh(1/2))My favorite formulas for the golden number are
G = sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + ...and the continued fraction:
1 G = 1 +  1 + 1  1 + 1  1 + 1  1 + 1  1 + 1  . . .These follow from the equations G^{2} = G + 1 and G = 1 + 1/G, respectively. If you chop off the continued fraction for G at any point, you'll see that G is also the limit of the ratios of successive Fibonacci numbers. See "week190" for a very different proof of this fact.
However, don't be fooled! The charm of the golden number tends to attract kooks and the gullible  hence the term "fool's gold". You have to be careful about anything you read about this number. In particular, if you think ancient Greeks ran around in togas philosophizing about the "golden ratio" and calling it "Φ", you're wrong. This number was named Φ after Phidias only in 1914, in a book called The Curves of Life by the artist Theodore Cook. And, it was Cook who first started calling 1.618... the golden ratio. Before him, 0.618... was called the golden ratio! Cook dubbed this number "φ", the lowercase baby brother of Φ.
In fact, the whole "golden" terminology can only be traced back to 1826, when it showed up in a footnote to a book by one Martin Ohm, brother of Georg Ohm, the guy with the law about resistors. Before then, a lot of people called 1/G the "Divine Proportion". And the guy who started that was Luca Pacioli, a pal of Leonardo da Vinci who translated Euclid's Elements. In 1509, Pacioli published a 3volume text entitled Divina Proportione, advertising the virtues of this number. Some people think da Vinci used the divine proportion in the composition of his paintings. If so, perhaps he got the idea from Pacioli.
Maybe Pacioli is to blame for the modern fascination with the golden ratio; it seems hard to trace it back to Greece. These days you can buy books and magazines about "Elliot Wave Theory", a method for making money on the stock market using patterns related to the golden number. Or, if you're more spiritually inclined, you can go to workshops on "Sacred Geometry" featuring talks about the healing powers of the golden ratio. But Greek texts seem remarkably quiet about this number.
The first recorded hint of it is Proposition 11 in Book II of Euclid's Elements. It also shows up elsewhere in Euclid, especially Proposition 30 of Book VI, where the task is "to cut a given finite straight line in extreme and mean ratio", meaning a ratio A:B such that
A:B::(A+B):A (i.e., "A is to B as A+B is to A")This is later used in Proposition 17 of Book XIII to construct the pentagonal face of a regular dodecahedron.
Of course, Euclid wasn't the first to do all these things; he just wrote them up in a nice textbook. By now it's impossible to tell who discovered the golden ratio and just what the Greeks thought about it. For a sane and detailed look at the history of the golden ratio, try this:
1) J. J. O'Connor and E. F. Robertson, The Golden Ratio, http://wwwgap.dcs.stand.ac.uk/~history/HistTopics/Golden_ratio.html
While I'm at it, I should point out that Theodore Cook's book introducing the notation "Φ" is still in print:
2) Theodore A. Cook, The Curves of Life: Being an Account of Spiral Formations and Their Application to Growth in Nature, to Science, and to Art: with Special Reference to the Manuscripts of Leonardo da Vinci, Dover Publications, New York, 1979.
If you want to see what Euclid said about the golden ratio, you can also pick up a cheap copy of the Elements from Dover  but it's probably quicker to go online. There are a number of good places to find Euclid's Elements online these days.
Topologists know David Joyce as the inventor of the "quandle"  an algebraic structure that captures most of the information in a knot. Now he's writing a hightech edition of Euclid, complete with Java applets:
3) David E. Joyce's edition of Euclid's Elements, http://aleph0.clarku.edu/~djoyce/java/elements/toc.html
Joyce is carrying on a noble tradition: back in 1847, Oliver Byrne did a wonderful edition of Euclid complete with lots of beautiful color pictures and even some popup models. You can see this online at the Digital Mathematics Archive:
4) Oliver Byrne's edition of Euclid's Elements, online at the Digital Mathematics Archive, http://www.sunsite.ubc.ca/DigitalMathArchive/
The most famous scholarly English translation of Euclid was done by Sir Thomas Heath in 1908. You can find it together with an edition in Greek and a nearly infinite supply of other classical texts at the Perseus Digital Library:
5) Thomas L. Heath's edition of Euclid's Elements, online at The Perseus Digital Library, http://www.perseus.tufts.edu/
But I'm digressing! My main point was that while G = (1 + sqrt(5))/2 is a neat number, it's a lot easier to find nuts raving about it on the net than to find truly interesting mathematics associated with it  or even interesting references to it in Greek mathematics! The cynic might conclude that the charm of this number is purely superficial. However, that would be premature.
First of all, there's a certain sense in which G is "the most irrational number". To get the best rational approximations to a number you use its continued fraction expansion. For G, this converges as slowly as possible, since it's made of all 1's:
1 G = 1 +  1 + 1  1 + 1  1 + 1  1 + 1  1 + 1  . . .We can make this more precise. For any number x there's a constant c(x) that says how hard it is to approximate x by rational numbers, given by
lim inf x  p/q = c(x)/q^{2} q → ∞where q ranges over integers, and p is an integer chosen to minimize x  p/q. This constant is as big as possible when x is the golden ratio!
It'd be ironic if the famously "rational" Greeks, who according to legend even drowned the guy who proved sqrt(2) was irrational, chose the most irrational number as the proportions of their most beautiful rectangle! But, it wouldn't be a coincidence. Their obsession with ratios and proportions led them to ponder the situation where A:B::(A+B):A, and this proportion instantly implies that A and B are incommensurable, since if you assume A and B are integers and try to find their greatest common divisor using Euclid's algorithm, you get stuck in an infinite loop. Euclid even mentions this idea in Proposition 2 of Book X:
If, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable.He doesn't explicitly come out and apply it to what we now call the golden ratio  but how could he not have made the connection? For more info on the Greek use of continued fractions and the Euclidean algorithm, check out the chapter on "antihyphairetic ratio theory" in this book:
6) D. H. Fowler, The Mathematics of Plato's Academy: A New Reconstruction, Oxford U. Press, Oxford, 1987.
Anyway, it's actually important in physics that the golden number is so poorly approximated by rationals. This fact shows up in the Kolmogorov ArnoldMoser theorem, or "KAM theorem", which deals with small perturbations of completely integrable Hamiltonian systems. Crudely speaking, these are classical mechanics problems that have as many conserved quantities as possible. These are the ones that tend to show up in textbooks, like the harmonic oscillator and the gravitational 2body problem. The reason is that you can solve such problems if you can do a bunch of integrals  hence the term "completely integrable".
The cool thing about a completely integrable system is that time evolution carries states of the system along paths that wrap around tori. Suppose it takes n numbers to describe the position of your system. Then it also takes n numbers to describe its momentum, so the space of states is 2ndimensional. But if the system has n different conserved quantities  that's basically the maximum allowed  the space of states will be foliated by ndimensional tori. Any state that starts on one of these tori will stay on it forever! It will march round and round, tracing out a kind of spiral path that may or may not ever get back to where it started.
Things are pretty simple when n = 1, since a 1dimensional torus is a circle, so the state has to loop around to where it started. For example, when you have a pendulum swinging back and forth, its position and momentum trace out a loop as time passes.
When n is bigger, things get trickier. For example, when you have n pendulums swinging back and forth, their motion is periodic if the ratios of their frequencies are rational numbers.
This is how it works for any completely integrable system. For any torus, there's an ntuple of numbers describing the frequency with which paths on this torus wind around in each of the n directions. If the ratios of these frequencies are all rational, paths on this torus trace out periodic orbits. Otherwise, they don't!
KAM theory says what happens when you perturb such a system a little. It won't usually be completely integrable anymore. Interestingly, the tori with rational frequency ratios tend to fall apart due to resonance effects. Instead of periodic orbits, we get chaotic motions instead. But the "irrational" tori are more stable. And, the "more irrational" the frequency ratios for a torus are, the bigger a perturbation it takes to disrupt it! Thus, the most stable tori tend to have frequency ratios involving the golden number. As we increase the perturbation, the last torus to die is called a "golden torus".
You can actually watch tori breaking into chaotic dust if you check out the applet illustrating the "standard map" on this website:
7) Takashi Kanamaru and J. Michael T. Thompson, Introduction to Chaos and Nonlinear Dynamics, standard map applet, http://brain.cc.kogakuin.ac.jp/~kanamaru/Chaos/e/Standard/
The "standard map" is a certain dynamical system that's good for illustrating this effect. You won't actually see 2d tori, just 1d crosssections of them  but it's pretty fun. For more details, try this:
8) M. Tabor, Chaos and Integrability in Nonlinear Dynamics: An Introduction, Wiley, New York, 1989.
In short, the golden number is the best frequency ratio for avoiding resonance!
Some audiophiles even say this means the best shaped room for listening to music is one with proportions 1:G:G^{2}. I leave it to you to find the flaw in this claim. For more dubious claims, check out the ad for expensive speaker cables at the end of this article.
KAM theory is definitely cool, but we shouldn't rest content with this when skeptics ask if the golden number is all it's cracked up to be. I figure it's part of our job as mathematicians to keep on discovering mindblowing facts about the golden number. A small part, but part: we shouldn't give up the field to amateurs!
Penrose has done his share. His "Penrose tiles" take crucial advantage of the selfsimilarity embodied by the golden number to create nonperiodic tilings of the plane. This helped spawn a nice little industry, the study of "quasicrystals" with 5fold symmetry. Here's a good introduction for mathematicians:
9) Andre Katz, A short introduction to quasicrystallography, in From Number Theory to Physics, eds. M. Waldschmit et al, Springer, Berlin, 1992, pp. 496537.
By the way, this same book has some nice stuff on the role of the golden number in KAM theory and the theory of iterated maps from the circle to itself:
10) Predrag Cvitanovic, Circle maps: irrationally winding, in From Number Theory to Physics, eds. M. Waldschmit et al, Springer, Berlin, 1992, pp. 631658.
11) JeanChristophe Yoccoz, Introduction to small divisors problems, in From Number Theory to Physics, eds. M. Waldschmit et al, Springer, Berlin, 1992, pp. 659679.
Conway and Sloane are also pulling their weight. Starting from the relation between the golden ratio, the isosahedron, and the 4dimensional big brother of the icosahedron (the "600cell"), they've described how to construct the coolest lattices in 8 and 24 dimensions using "icosians"  which are certain quaternions built using the golden ratio. I discussed this circle of ideas in "week20", "week59" and "week155".
But if you want some really scary formulas involving the golden ratio, Ramanujan is the one to go to. Check these out:
1  1 + exp(2π)  1 + exp(4π) = exp(2π/5) [sqrt(G sqrt(5))  G]  1 + exp(6π)  1 + exp(8π)  . . .and
1 + exp(2π sqrt(5))  1 + exp(4π sqrt(5)) sqrt(5)  = exp(2π/5) [   G] 1 + exp(6π sqrt(5)) 1 + [5^{3/4} (G  1)^{5/2}  1]^{1/5}  1 + exp(8π sqrt(5))  . . .These are special cases of a monstrosity called the RogersRamanujan continued fraction, which is a kind of "qdeformation" of the continued fraction for the golden ratio. For details, start here:
12) Eric W. Weisstein, RogersRamanujan Continued Fraction, http://mathworld.wolfram.com/RogersRamanujanContinuedFraction.html
It's these two formulas, and one other like it, that led Hardy to write the famous lines:
A single look at them is enough to show that they could only be written down by a mathematician of the highest class. They must be true because, if they were not true, no one would have had the imagination to invent them.
For more by Hardy on these continued fractions, see section 1 and section 6.17 of his book:
13) G. H. Hardy, Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work, Chelsea Publishing Co., New York, 1959.
The golden number also shows up in the theory of quantum groups. I talked about this in "week22" so I won't explain it again here. But, I can't resist mentioning that Freedman, Larsen and Wang have subsequently shown that a certain topological quantum field theory called ChernSimons theory, built using the quantum group SU_{q}(2), can serve as a universal quantum computer when the parameter q is a fifth root of unity. And, this is exactly the case where the spin1/2 representation of SU_{q}(2) has quantum dimension equal to the golden number!
14) Michael Freedman, Michael Larsen, Zhenghan Wang, A modular functor which is universal for quantum computation, available at quantph/0001108.
But don't get the wrong idea: it's not that some magic feature of the golden number is required to build a universal quantum computer! It's just that the 5 seems to be the smallest number n such that SU_{q}(2) ChernSimons theory is computationally universal when q is an nth root of 1.
That's pretty much everything I know about the golden number. So now, what about this "Golden Object" puzzle?
Basically, the problem was to find an object that acts like the golden number. The golden number has G^{2} = G + 1, so we want to find a object G equipped with a nice isomorphism between G^{2} and G + 1.
If G is just a set, this means we want a nice onetoone correspondence between pairs of elements of G, and elements of G together with one other thing. It doesn't matter what that other thing is, so let's call it "@".
(You may be wondering about the word "nice". The point is, the problem is too easy if we don't demand that the solution be nice in some way  some way that I don't feel like making precise.)
Here's Robin Houston's answer:
Define a "bit" to be either 0 or 1. Define a "golden tree" to be a (planar) binary tree with leaves labelled by 0, 1, or *, where every node has at most one bitchild. For example:
/\ /\ 1 0 *is a golden tree, but
/\ /\ * 0 1is not. Let G be the set of golden trees. We define an isomorphism
f: G^{2} → G + {@}as follows. First we define f(X, Y) when both X and Y are golden trees with just one node, this node being labelled by a bit. We can identify such a tree with a bit, and doing this we set
f(0, 0) = 0 f(0, 1) = 1 f(1, 0) = * f(1, 1) = @In the remaining case, where the golden trees X and Y are not just bits, we set
f(X, Y) = /\ X YThere are different ways to show this function f is a onetoone correspondence, but the best way is to see how Houston came up with this answer! He didn't just pull it out of a hat; he tackled the problem systematically, and that's why his solution counts as "nice".
It's easy to find a set S equipped with an isomorphism
S = P(S)where P is some polynomial with natural number coefficients. You just use the fixedpoint principle described in "week108". Namely, you start with the empty set, keep hitting it with P forever, and take a kind of limit. This is how I built the set of binary trees last week, as a solution of T = T^{2} + 1.
The problem is that the isomorphism we seek now:
G^{2} = G + 1 (1)is not of this form. So, what Houston does is to make a substitution:
G = H + 2Given this, we'd get (1) if we had
H^{2} + 4H + 4 = H + 3 (2)and we'd get (2) if we had
H^{2} + 4H + 1 = H (3)which is of the desired form.
We can rewrite (3) as
H = 1 + H^{2} + 2H + H2and in English this says "an element of H is either a *, or a pair consisting of two guys that are either bits or elements of H  but not both bits". So, a guy in H is a golden tree! But, if it has just one node, that node can only be labelled by a *, not a 0 or 1. This means there are precisely 2 golden trees not in H. So, G = H + 2 is the set of all golden trees, and our calculation above gives an isomorphism G^{2} = G + 1.
Voila!
Note that to derive (3) from (1) we need to subtract, which in general is not allowed in this game. Here we are subtracting constants, and Houston says that's allowed by the "GarsiaMilne involution theorem". I don't know this theorem, so I'll make a note to myself to learn it. But luckily, we don't really need it here: we only need to derive (1) from (3), and that involves addition, so it's fine.
Part of what makes Houston's solution "nice" is that it suggests a general method for turning polynomial equations into recursive definitions of the form S = P(S). Another nice thing is that his trick delivers a structure type G(X) that reduces to G when X = 1. To get this, first use the fixedpoint method to construct a structure type H(X) with an isomorphism
H(X) = (H(X) + X)^{2} + 2H(X)Then, define
G(X) = H(X) + X + 1and note that this gives
G(X)^{2} = G(X) + Xwhich reduces to G^{2} = G + 1 when X = 1.
As if this weren't enough, Houston also gave another solution to the puzzle. He showed that James Propp's proposed Golden Object, described last week, really is a Golden Object! Maybe Propp already knew this, but I sure didn't.
The idea of the proof is pretty general. Suppose we're in some category that's a "2rig" in the sense of "week191". And, suppose we've got an object X equipped with an isomorphism
X = 1 + 2X (4)so that X acts like "1". For example, following Schanuel and Propp, we can take the category of "σpolytopes" and let X be the open interval: then isomorphism (4) says
(0,1) = (0,1/2) + {1/2} + (1/2,1)Or, following Houston, we can take the category of sets and let X be the set of finite bitstrings. Then (4) says "a finite bitstring is either the empty bitstring, or a bit followed by a finite bitstring". The relation between these two examples is puzzling to me  if anyone understands it, let me know! But anyway, either one works.
Now let G be the object of "binary trees with Xlabelled leaves":
G = X + X^{2} + 2X^{3} + 5X^{4} + 14X^{5} + 42X^{6} + ...where the coefficients are Catalan numbers. Let's show that G is a Golden Object. To do this, we'll use (4) and this isomorphism:
G = G^{2} + X (5)which says "a binary tree with Xlabelled leaves is a pair of such trees, or a degenerate tree with just one Xlabelled node". The formula for G involving Catalan numbers is really just the fixedpoint solution to this!
Here is Houston's fiendishly clever argument. Suppose Z is any type equipped with an isomorphism
Z = Z' + Xfor some Z'. Then
Z + X + 1 = Z' + 2X + 1 = Z' + X = ZThis applies to Z = G^{2}, since
G^{2} = (X + G^{2})^{2} = (2X + 1 + G^{2})^{2}has a X term in it when you multiply it out, so it's of the form Z' + X. Therefore we have an isomorphism
G^{2} = G^{2} + X + 1But we also have an isomorphism G + 1 = G^{2} + X + 1 by (5). Composing these, we get our isomorphism
G^{2} = G + 1.Golden! I'll stop here.
Addendum: The computer scientist Sebastiano Vigna pointed out this paper:
15) Paolo Boldi, Massimo Santini, and Sebastiano Vigna, Measuring with jugs, or: what if mathematicians were asked to defuse bombs?, Theoret. Comput. Sci. 2 (2002). Also available at http://vigna.dsi.unimi.it/papers.php
which shows that if you want to approximately measure an arbitrary amount of water using only two jugs, it's best if they have capacity 1 and G. This paper cites a a charming result by Swierczkowski which picks up where a famous theorem due to Dedekind leaves off. Dedekind showed that if x is any irrational number, the numbers nx mod 1 are uniformly distributed in the interval [0,1]. But if x = 1/G, these numbers have an especially nice property: each new point in the sequence (nx mod 1) lands in one of the longest intervals not containing a previous point! And, it chops this interval in a golden way.
Stephen Schanuel said some things about "week203" on the category theory mailing list, so I'll include his post here along with various replies, concluding with my own.
From: Stephen Schanuel Subject: categories: mystification and categorification Date: Thu, 4 Mar 2004 00:44:46 0500 I was unable to understand John Baez' golden object problem, nor his description of the solutions. He refuses to tell us what 'nice' means, but let me at least propose that to be 'tolerable' a solution must be an object in a category, and John doesn't tell us what category is involved in either of the solutions; at least I couldn't find a specification of the objects, nor the maps, so I found the descriptions 'intolerable', in the technical sense defined above. He is very generous, allowing one to use a category with both plus and times as extra monoidal structures. (Does anyone know an example of interest in which the plus is not coproduct?) This freedom is unnecessary; a little algebra plus Robbie Gates' theorem provides a solution G to G^2=G+1 which satisfies no additional equations, in an extensive category (with coproduct as plus, cartesian product as times). Briefly, here it is. A primitive fifth root of unity z is a root of the polynomial 1+X+X^2+X^3+X^4, hence satisfies 1+z+z^2+z^3+z^4+z=z, which is of the 'fixed point' form p(z)=z with p in N[X] and p(0) not 0. Gates' theorem then says that the free distributive category containing an object Z and an isomorphism from p(Z) to Z is extensive, and its Burnside rig B (of isomorphism classes of objects) is, as one would hope, N[X]/(p(X)=X); that is, Z satisfies no unexpected equations. Since the degree of p is greater than 1, an easy general theorem tells us (from the joint injectivity of the Euler and dimension homomorphisms) that two polynomials agree at the object Z if and only if either they are the same polynomial or both are nonconstant and they agree at the number z.Now the 'algebra': the golden number is 1+z+z^4. So G=1+Z+Z^4 satisfies G^2=G+1, as desired. It satisfies no unexpected equations, because the relation X^2=X+1 reduces any polynomial in N[X] to a linear polynomial, and these reduced forms have distinct Euler characteristics, i.e. differ at z. Thus the homomorphism from N[X]/(X^2=X+1) to B (sending X to G) is injective, and that's all I wanted. Since in the category of sets, any nasty old infinite set satisfies the golden equation and many others, I have taken the liberty of interpreting 'nice' to mean at least 'satisfying no unexpected equations'. One could ask for more; the construction above has produced a distributive, but not extensive, category whose Burnside rig is N[X]/(X^2=X+1), the full subcategory with objects polynomials in G. (If it were extensive, it would be closed under taking summands, but every object in the larger category is a summand of G.) I don't know whether there is an extensive category with N[X]/(X^2=X+1) as its full Burnside rig; perhaps one or both of the examples John mentioned would do, if I knew what they were. While I'm airing my confusions, can anyone tell me what 'categorification' means? I don't know any such process; the simplest exanple, 'categorifying' natural numbers to get finite sets, seems to me rather 'remembering the finite sets and maps which gave rise to natural numbers by the abstraction of passing to isomorphism classes'. Finally, a note to John: While you're trying to give your audience some feeling for the virtues of ncategories, couldn't you give them a little help with n=1, by being a little more precise about objects and maps? Greetings to all, and thanks for your patience while I got this stuff off my chest, Steve Schanuel
From: David Yetter Subject: categories: Re: mystification and categorification Date: Fri, 5 Mar 2004 10:55:26 0600 Categorification is a bit like quantization: it isn't a construction so much as a desideratum for a relationship between one thing and another (in the case of categorification an (n+1)categorical structure and an ncategorical structure; in the case of quantization a quantum mechanical system and a classical mechanical system). Categorification wants to find a higherdimensional categorical structure corresponding to a lowerdimensional one by weakening equations to natural isomorphisms and imposing new, sensible, coherence conditions. In general, for the original purpose for which it was proposedconstructions of TQFT's and models of quantum gravityone wants the highest categorical level to have a linear structure (hence Baez wanting tensor product and a sum it distributes over, rather than cartesian product and coproduct). Specific lowerdimensional categories with structure are 'categorified' by finding a higherdimensional category with the new structure which 'lies over' the lower dimensional one in the way an additive monoidal category lies over its Grothendieck rig. For instance any (klinear) monoidal category with monoid of isomorphism classes M is a categorification of M, and more generally (klinear) monoidal categories are a categorification of monoids. A simple example shows why it is not a construction: commutative monoids (as rather special categories with one object) admit two different categorifications: symmetric monoidal categories and braided monoidal categories (each regarded as a kind of bicategory with one object). There is a good argument for regarding braided monoidal categories as the 'correct' categorification: the EckmannHilton theorem ('a group in GROUPS is an abelian group' or, really as the proof shows, 'a monoid in MONOIDS is a commutative monoid') 'categorifies' to: A monoidal category in MONCAT is a braided monoidal category.
From: Vaughan Pratt Subject: categories: Re: mystification and categorification Date: Fri, 05 Mar 2004 22:49:56 0800 >While I'm airing my confusions, can anyone tell me what >'categorification' means? I don't know any such process; the simplest >exanple, 'categorifying' natural numbers to get finite sets, seems to me >rather 'remembering the finite sets and maps which gave rise to natural >numbers by the abstraction of passing to isomorphism classes'. A fair question. I attended John's Coimbra lectures on this stuff in 1999 but a lot of it leaked out afterwards. If I had to guess I'd say he was categorifying the free monoid on one generator to make it a monoidal category, but then how did the monoid end up as coproduct and the generator as the final object? One suspects some free association thereJohn, how do you make that connection? With regard to categorification in general, sets seem to play a central role in at least one development of category theory. The homobjects of "ordinary" categories are homsets. (In that sense I guess "ordinary" must entail "locally small.") 2categories are what you get if instead you let them be homcats, suitably elaborated. Going in the other direction, if you take homsets to be vacuous, not in the sense that they are empty but rather that they are all the same, then you get sets. One more step in that direction makes everything look the same, which may have something to do with the Maharishi Yogi hiring category theorists for the math dept. of his university in Fairfield, Iowa. (When I spoke last with the MY's "Minister of World Health," an MD who like Ross Street was a classmate of mine but eight years earlier starting in 1957, the entire conversation seemed to be largely a skirting of the minefield of the sameness of everything, which may subconsciously have been behind my obscure reply to Peter Freyd's posting a while back about unique existence going back to Descartes, where I tried to oneup him by claiming it went much further back.) Categorification isn't the only way to get to 2categories, which can be understood instead in terms of the interchange law as a twodimensional associativity principle. However John has got a lot of mileage out of the categorification approach, which one can't begrudge in an era where mileage and minutes are as integral to a balanced life as one's checkbook. (Q: How many minutes in a month? A: Depends on your plan.) >Since in the category of sets, any nasty old infinite set satisfies >the golden equation and many others, I have taken the liberty of >interpreting 'nice' to mean at least 'satisfying no unexpected >equations'. Quite right. I would add to this "and satisfying the expected equations." The "nasty sets" of which Steve speaks fail to satisy such expected equations as 2^2^X ~ X. (The power set of a set is a Boolean algebra, for heaven's sake. Why on earth forget that structure prior to taking the second exponentiation? Set theorists seem to think that they can simply forget structure without paying for it, but in the real world it costs kT/2 joules per element of X to forget that structure. If set theorists aren't willing to pay realworld prices in their modeling, why should we taxpayers pay them realworld salaries? Large cardinals are a figment of their overactive imaginations, and the solution to consistency concerns is not to go there.) Vaughan Pratt
From: Tom Leinster Subject: Re: categories: mystification and categorification Date: 07 Mar 2004 20:50:39 +0000 Steve Schanuel wrote: > a category with both plus and times as extra monoidal structures. > (Does anyone know an example of interest in which the plus is not > coproduct?) Here are two examples that I've come across previously of rig categories in which the plus is not coproduct: (i) the category of finite sets and bijections, with + and x inherited from the category of sets; (ii) discrete rig categories, which are of course the same thing as rigs. > This freedom is unnecessary; a little algebra plus Robbie > Gates' theorem provides a solution G to G^2=G+1 which satisfies no > additional equations, in an extensive category (with coproduct as plus, > cartesian product as times). If you do allow yourself the freedom to use any rig category then an even simpler solution exists, also satisfying no additional equations: just take the rig freely generated by an element G satisfying G^2 = G + 1 and regard it as a discrete rig category. > Since in the category of sets, any nasty old infinite set satisfies > the golden equation and many others, I have taken the liberty of > interpreting 'nice' to mean at least 'satisfying no unexpected > equations'. I'd interpret "nice" differently. (Apart from anything else, the trivial example in my previous paragraph would otherwise make the golden object problem uninteresting.) "Nice" as I understand it is not a precise term  at least, I don't know how to make it precise. Maybe I can explain my interpretation by analogy with the equation T = 1 + T^2. A nice solution T would be the set of finite, binary, planar trees together with the usual isomorphism T ~> 1 + T^2; a nasty solution would be a random infinite set T with a random isomorphism to 1 + T^2. (Both these solutions are in the rig category Set with its standard + and x.) I regard the first solution as nice because I can see some combinatorial content to it (and maybe, at the back of my mind, because it has a constructive feel), and the second as nasty because I can't. I'm not certain what I think of the solution given by the set of notnecessarilyfinite binary planar trees (nice?), or by the set of binary planar trees of cardinality at most aleph_5 (probably nasty). Maybe the finding of a "nice" solution is similar in spirit to the finding of a "concrete interpretation" of a combinatorial identity. As an extremely simple example, consider the identity saying that each entry in Pascal's triangle is the sum of the two above it, (n+1 choose r) = (n choose r1) + (n choose r). This is a doddle to prove, but all the same you'd be missing something if you didn't know the standard "concrete interpretation": choosing r objects out of n+1 objects amounts to EITHER choosing the first one and then choosing r1 of the remaining n OR ... . Even if the challenge of finding a "nice solution" or "concrete interpretation" isn't made precise, I think there is a shared sense of what would count as an answer, and finding an answer is in general not straightforward. Best wishes, Tom
From: John Baez Subject: golden objects Date: Sun, 7 Mar 2004 12:50:29 0800 (PST) Dear Categorists  Sorry to take a while to respond. People at UCR have been unable to receive posts on the category theory mailing list, due to problems with our internet connection. I'd asked for some nice examples of an object G in a rig category equipped with an isomorphism from G^2 to G + 1. Steve Schanuel replied: >I was unable to understand John Baez' golden object problem, nor his >description of the solutions. He refuses to tell us what 'nice' means, [...] The problem was deliberately openended, but you seem to have understood it perfectly, since you've given a nice solution, including a precise specification of what you consider "nice". Let me repeat the two solutions given by Robin Houston: 1) The first solution works in any rig category having an object H equipped with an isomorphism to H^2 + 4H + 1. The solution is to take G = H + 2. I described how Houston uses the isomorphism H → H^2 + 4H + 1 to construct an isomorphism G^2 → G + 1. What's nice about this is that it reduces a problem that's not obviously of fixedpoint form to one that is. 2) Houston's second solution works in any monoidal cocomplete category, tensor product distributing over colimits, that contains an object X equipped with an isomorphism to 2X + 1. The solution is to let G be the object of "binary planar rooted trees with Xlabelled leaves", i.e. G = X + X^2 + 2X^3 + 5X^4 + 14X^5 + 42X^6 + ... where the coefficients are Catalan numbers. He uses the obvious isomorphism G → G^2 + X to construct an isomorphism G^2 → G + 1. What's nice about this is that it shows Propp's originally proposed golden object really is one: just take the category of σpolytopes with its cartesian product, and let X be the open interval! And, it makes precise the sense in which the alternating sum of Catalan numbers equals the golden ratio. Steve writes: >I don't know whether there is an extensive category with N[X]/(X^2=X+1) >as its full Burnside rig; perhaps one or both of the examples John >mentioned would do, if I knew what they were. I think example 1) does the job if we take the free distributive category on an object H equipped with an isomorphism to H^2 + 4H + 1. Right? Steve also writes: >He is very generous, allowing one to use a category with both plus >and times as extra monoidal structures. (Does anyone know an example >of interest in which the plus is not coproduct?) This freedom is >unnecessary [...] It's unnecessary, but handy: I think there's also an golden object in the rig category of reps of quantum SU(2) at a suitable value of q. Here the tensor product is not cartesian. In the lingo of quantum group theory, this object has "quantum dimension" equal to the golden number. It's interesting how such nonintegral but algebraic "dimensions" show up naturally in quantum group theory, just as nonintegral but algebraic "cardinalities" show up in the theory of distributive categories. I don't know any golden objects in rig categories where the plus is not coproduct, and I agree that such rig categories arise less often than those where times is not product. But, if you use the obvious way of making the groupoid of finite sets into a rig category, + isn't coproduct, nor is x product. > While I'm airing my confusions, can anyone tell me what > 'categorification' means? I don't know any such process; the simplest > exanple, 'categorifying' natural numbers to get finite sets, seems to me > rather 'remembering the finite sets and maps which gave rise to natural > numbers by the abstraction of passing to isomorphism classes'. You're right: categorification is not a systematic process! I explained this idea back in week121, and also in my paper "Categorification", at http://www.arXiv.org/abs/math.QA/9802029. Here's what I said: If one studies categorification one soon discovers an amazing fact: many deepsounding results in mathematics are just categorifications of facts we learned in high school! There is a good reason for this. All along, we have been unwittingly `decategorifying' mathematics by pretending that categories are just sets. We `decategorify' a category by forgetting about the morphisms and pretending that isomorphic objects are equal. We are left with a mere set: the set of isomorphism classes of objects. To understand this, the following parable may be useful. Long ago, when shepherds wanted to see if two herds of sheep were isomorphic, they would look for an explicit isomorphism. In other words, they would line up both herds and try to match each sheep in one herd with a sheep in the other. But one day, along came a shepherd who invented decategorification. She realized one could take each herd and `count' it, setting up an isomorphism between it and some set of `numbers', which were nonsense words like `one, two, three, ...' specially designed for this purpose. By comparing the resulting numbers, she could show that two herds were isomorphic without explicitly establishing an isomorphism! In short, by decategorifying the category of finite sets, the set of natural numbers was invented. According to this parable, decategorification started out as a stroke of mathematical genius. Only later did it become a matter of dumb habit, which we are now struggling to overcome by means of categorification. While the historical reality is far more complicated, categorification really has led to tremendous progress in mathematics during the 20th century. For example, Noether revolutionized algebraic topology by emphasizing the importance of homology groups. Previous work had focused on Betti numbers, which are just the dimensions of the rational homology groups. As with taking the cardinality of a set, taking the dimension of a vector space is a process of decategorification, since two vector spaces are isomorphic if and only if they have the same dimension. Noether noted that if we work with homology groups rather than Betti numbers, we can solve more problems, because we obtain invariants not only of spaces, but also of maps. In modern language, the nth rational homology is a functor defined on the category of topological spaces, while the nth Betti number is a mere function, defined on the set of isomorphism classes of topological spaces. Of course, this way of stating Noether's insight is anachronistic, since it came before category theory. Indeed, it was in Eilenberg and Mac Lane's subsequent work on homology that category theory was born! Decategorification is a straightforward process which typically destroys information about the situation at hand. Categorification, being an attempt to recover this lost information, is inevitably fraught with difficulties. >Finally, a note to John: While you're trying to give your audience >some feeling for the virtues of ncategories, couldn't you give them a >little help with n=1, by being a little more precise about objects and >maps? I hope it's clearer now. Best, jb
© 2004 John Baez
baez@math.removethis.ucr.andthis.edu
