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September 12, 2008


Posted by John Baez

My final talk in Glasgow will be about the number 24.

This has long been my favorite number, since it shows up throughout math and physics in mysterious, shocking and even monstrous ways… which seem to fit together in a gargantuan conspiracy, as if pure mathematics itself were the work of an insane god — or at least one with a devilish sense of humor.

You can see the transparencies for this talk by clicking on the title below. If you see mistakes in it, please let me know!

Abstract: The numbers 12 and 24 play a central role in mathematics thanks to a series of “coincidences” that is just beginning to be understood. One of the first hints of this fact was Euler’s bizarre “proof” that

1+2+3+4+=1/121 + 2 + 3 + 4 + \cdots = -1/12

which he obtained before Abel declared that “divergent series are the invention of the devil”. Euler’s formula can now be understood rigorously in terms of the Riemann zeta function, and in physics it explains why bosonic strings work best in 26=24+2 dimensions. The fact that

1 2+2 2+3 2+24 21^2 + 2^2 + 3^2 + \cdots 24^2

is a perfect square then sets up a curious link between string theory, the Leech lattice (the densest known way of packing spheres in 24 dimensions) and a group called the Monster. A better-known but closely related fact is the period-12 phenomenon in the theory of “modular forms”. We shall do our best to demystify some of these deep mysteries.

I’ve been writing about these things for years in This Week’s Finds. The Glasgow Mathematical Trust promised to help me write up my talks as papers, and I hope to use this opportunity to assemble my scattered thoughts into something nice and coherent.

You may wonder why my very first and simplest talk, on the number 5, is the last one to show up here. That’s because I think more people will come to that one, including some schoolkids of ages 16 and 17. So, I want it to be really fun.

Note added later: you can now see a streaming video of this talk.

Posted at September 12, 2008 12:41 AM UTC

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23 Comments & 0 Trackbacks

Re: 24

To comment on only the most important aspect: I see you’ve switched from the nickel to the brass half of Home Depot’s excellent SoftCurve 4 inch numeral range. The talk posters stick to nickel.

Posted by: Tom Leinster on September 12, 2008 2:54 AM | Permalink | Reply to this

Re: 24

I love the talk posters — did you make ‘em?

I hope no disaster will be caused by me switching from ‘silver’ to ‘gold’ upon reaching the number 24. That was supposed to indicate that it’s the most interesting of the three. I’m sorry it doesn’t quite match your posters.

One great thing about the bibliography of this talk is that I get to use the phrase ‘Home Depot, op. cit.

Posted by: John Baez on September 12, 2008 5:17 AM | Permalink | Reply to this

Re: 24

Oh, and yes, I made ‘em. Glad you like ‘em.

Posted by: Tom Leinster on September 13, 2008 12:44 AM | Permalink | Reply to this

Re: 24

John wrote:

I hope no disaster will be caused

I reckon we’ll cope… actually, I think the “gold” would have looked better against the colours of the eggs on the 24 poster, and briefly considered changing it. But then I realized I’d have to make the momentous decision of how to divide two metals among three posters, and chickened out.

Posted by: Tom Leinster on September 12, 2008 1:41 PM | Permalink | Reply to this

Re: 24

Typo on page 6:

x + 2x + 3x2 + …

should presumably start with 1.

Posted by: Theo on September 12, 2008 4:17 AM | Permalink | Reply to this

Re: 24

Fixed! Thanks!

Posted by: John Baez on September 12, 2008 5:18 AM | Permalink | Reply to this

Re: 24

Still there is a typo: derivative of 11x\frac{1}{1-x} is 1(1x) 2\frac{1}{{(1-x)}^2} and the next sum (which is equal to 14-\frac{1}{4}) is not needed.
Posted by: Anton on September 12, 2008 10:50 PM | Permalink | Reply to this

Re: 24

Aargh — thanks! Fixed.

As you can see, the actual equations are the last thing on my mind when I’m making these slides… I’m more worried about layout and telling a fun story. But it would not be good to have them think I can’t do calculus. They might demand their money back.

Posted by: John Baez on September 13, 2008 12:35 AM | Permalink | Reply to this

Re: 24

Hi John,

I’m sorry I’m not in England now, otherwise, I would have bugged you to come down to London for a while. In fact, Yunhyong has been working in Glasgow for some years, but she’s also in Greece at the moment!

I noticed while looking at your slides that you nowhere mention that the equation 1 2+2 2++m 2=n 21^2+2^2+\cdots+m^2=n^2 is also an elliptic curve x(x+1)(2x+1)=6y 2,x(x+1)(2x+1)=6y^2, which is a bit odd, given that elliptic curves figure so prominently in your talk. Perhaps it would be more confusing than amusing to throw that in?

Posted by: Minhyong Kim on September 12, 2008 4:52 AM | Permalink | Reply to this

Re: 24

More confusing than amusing, since I’m carefully avoiding saying anything about how elliptic curves come from cubic equations. To people attending this talk, an ‘elliptic curve’ is just a goofy name for what you get by taking a parallelogram and folding it into a torus. That’s supposed make elliptic curves seem pathetically simple.

But maybe I can just mutter that elliptic curves were used to solve Lucas’ puzzle…

And these talks are just an elaborate warmup for writing some papers where I explain these issues in a bit more detail. So, I can say something about this ‘strange loop’ of ideas there.

There are also probably a bunch of cool facts about the number 24 that I’m forgetting.

It would be fun to see you in London, but I couldn’t pop down there even if you were around, since classes start here shortly after I get back.

I hadn’t known Yunhyong was in Glasgow!

Posted by: John Baez on September 12, 2008 5:27 AM | Permalink | Reply to this

Re: 24

A very silly question, probably easily looked up in Silverman or Lang, but is there an obvious link between n-torsion subgroups of SL(2,Z) and the Shimura complex multiplication action on families of Jacobians of elliptic curves with enlarged endomorphism rings?

It seems “unexpectedly large symmetries through cyclotomic actions” (slides 35/36/37) show up in both but I’m not completely clear on the link, if any.

Sorry, this is a vague question, and probably nonsense.

Posted by: curious on September 12, 2008 5:43 AM | Permalink | Reply to this

Re: 24

I bet Minhyong could answer this better than I could.

Posted by: John Baez on September 12, 2008 7:16 PM | Permalink | Reply to this

Re: 24

Wow. If that’s one of your very silly questions, I can’t wait to see your quite sensible ones.

Posted by: David Corfield on September 13, 2008 9:48 AM | Permalink | Reply to this

Re: 24

It’s a stupid question in the sense that the motivation for it is nothing other than a vague analogy between two areas of mathematics connected with elliptic curves.

I have no reason other than a gut feeling to suspect there might be a connection - no calculations to back anything up! And bitter experience has shown my gut to be somewhat off the mark…

Posted by: curious on September 13, 2008 4:50 PM | Permalink | Reply to this

Re: 24

Well, to live up to John’s vote of confidence, I wanted to wait for a quiet moment to reply, but it seems it won’t be coming soon. So here’s a sketch:

The general principle is this. Suppose you are interested in a moduli space M M of some objects. Then, it often happens that there is a surjective map NMN \rightarrow M from another moduli space NN of the same kind of objects with some extra structure, and so that M=N/GM=N/G for some group GG that serves to get rid the redundancy. Now suppose xNx\in N and [x]M[x]\in M is the point lying under it (obtained by forgetting the extra structure). Then it often happens that the isotropy group G xG_x of xNx\in N acts as automorphisms of the object that [x]M[x]\in M represents. Of course I don’t intend this as a theorem, although perhaps one could turn it into one with appropriate assumptions.

In the case at hand, MM is the moduli space of elliptic curves, or lattices in CC modulo homothety. NN, on the other hand, is the space of (L,B)(L,B), where BB is an oriented basis for LL, again modulo homothety. Of course, SL 2(Z)SL_2(Z) acts on NN, so that the orbits are the different oriented bases for the same lattice. Now, note that NN can be realized as the upper half plane HH. This is by mapping (L,B=(z 1,z 2))(L, B=(z_1,z_2)) to (L/z 1,(1,z 2/z 1))(L/z_1, (1,z_2/z_1)) (It should be clear at this point what I meant by an oriented basis.) Now, what does it mean that τH\tau \in H is fixed by an element ((a,b),(c,d))SL 2(Z)?((a,b),(c,d)) \in SL_2(Z)? If you think about it, this is exactly saying that (L,(1,τ))(L, (1,\tau)) is homothetic to (L,(a+bτ,c+dτ)),(L, (a+b\tau, c+d\tau)), by some zCz\in C. But then, of course, we must have z=a+bτz=a+b\tau and $$ z L=L.Thatis, That is, zmustbeanautomorphismof must be an automorphism of L.Unknown character/pUnknown characterUnknown characterpUnknown characterEvenwithoutfurtherdetails,Ihopethismakesclearwhynontrivialisotropyin. </p> <p>Even without further details, I hope this makes clear why non-trivial isotropy in SL_2(Z)fora for a \tauimpartstothecorrespondingellipticcurveextraautomorphism.Inparticular,whenevertheisotropygrouphasorder imparts to the corresponding elliptic curve extra automorphism. In particular, whenever the isotropy group has order \geq 3,theellipticcurvemusthavecomplexmultiplication,which,inturn,mustbebythefields, the elliptic curve must have complex multiplication, which, in turn, must be by the fields Q(i)Q(i)or or Q(ζ 3),Q(\zeta_3),sincethesearetheonlyimaginaryquadraticfieldswithmorethantworootsofunity.Furthermore,inthiscase,theendomorphismringmustbe since these are the only imaginary quadratic fields with more than two roots of unity. Furthermore, in this case, the endomorphism ring must be Z[i]and and Z[\zeta_3](exercise),andthelatticemustbe (exercise), and the lattice must be [1, i]or or [1,\zeta_3].Unknown character/pUnknown characterUnknown characterpUnknown characterNow,IpresumeJohnhasalreadyexplainedsomewherethelinkbetweentorsionsubgroupsof.</p> <p>Now, I presume John has already explained somewhere the link between torsion subgroups of SL_2(Z)andtheisotropygroupsofpointsin and the isotropy groups of points in H.Thatistosay,Unknown character/pUnknown characterUnknown characterpUnknown charactertorsionoforder. That is to say,</p> <p>torsion of order \geq 3\rightarrowisotropygroupsoforder isotropy groups of order \geq 3forsomepointsin for some points in H\rightarrow$ two specific elliptic curves with complex multiplication.

Posted by: Minhyong Kim on September 14, 2008 6:35 PM | Permalink | Reply to this

Re: 24

Thanks, awesome explanation!

Posted by: curious on September 14, 2008 10:42 PM | Permalink | Reply to this

Re: 24

Great. enjoyed it a lot.

slide 19: i know you don’t want to get into it, but the attentive but otherwise lay reader may wonder here which rule you use to determine whether two tori are “the same”

slide 30: “appendix my talk”

Posted by: Urs Schreiber on September 12, 2008 7:32 AM | Permalink | Reply to this

Re: 24


I’m adding a bit more stuff in the endnotes about the definition of elliptic curve and when two elliptic curves count as isomorphic.

I cleverly set it up so that in the talk I only need to explicitly consider metric-preserving isomorphisms. But of course we secretly need to consider the rest as well — for example, to see why it’s sufficient to consider tori formed by parallelograms with a fixed ‘width’.

How much I actually talk about this will depend on how much time I have, how well people seem to be following, etc. I’m trying to zip from the harmonic oscillator to string theory in 10 minutes for an audience who won’t know any physics, so I have to cut corners. But, it’s good to have explanations and references lurking in the endnotes.

Posted by: John Baez on September 12, 2008 5:00 PM | Permalink | Reply to this

Re: 24

Small comments:

Misspelling on page 4: should be ‘Édouard’. Ditto on p.38, reference 3.

Display on page 16: should it be Z(t)Z(t) rather than Z(τ)Z(\tau) on the left hand side? (Compare the formula on the previous page.)

Posted by: Tom Leinster on September 12, 2008 1:52 PM | Permalink | Reply to this

Re: 24

Thanks — fixed!

I think my fingers are used to typing ‘edu’, as in .edu.

Posted by: John Baez on September 12, 2008 5:02 PM | Permalink | Reply to this

Re: 24

I’m very tempted to make the 400 mile car journey to see this talk but will probably settle for watching the video.

I know I am not the only one who has become fascinated by this story of the number 24 from all the stuff said about it in TWF. There has to be some marvellous structure that explains the “gargantuan conspiracy” but I fear I would never understand it even if I had 1300 years to think about it. However I did find another place where the number twenty four turns up which seems to be connected, so I’ll explain what it is as briefly as possible.

I aquired an interest in hyperdetermiants a few years ago when they came up in a number theory problem I was looking at. Hyperdeterminants are a generalisation of determinants from matrices to higher rank tensors (see wikipedia). They are polynomials in the components of the tensor and they get big very quickly as the rank and dimensions of the tensor go up. It is non-trivial to even work out what the polynomial degree of the hyperdetermiants are but for a rank four tensor of dimension 2x2x2x2 the hyperdetermiant is of degree 24.

The number theory problem I was interested in showed a link between hyperdeterminants and elliptic curves. Here is one way you can see such a link. Consider a fixed 2x2x2x2 tensor contracted over three vectors each with 2 ineteger components. I require that the resulting vector is zero. If you try and solve this as a diophantine problem you will find that it reduces to looking for rational points on an elliptic curve whose coefficients are polynomials in the components of the tensor. You will also discover that the discriminant of the elliptic curve is the hyperdeterminant of the tensor. This establishes a tenuous link between the degree 24 of the hyperdeterminant and the fact (mentioned on the slides) that the discriminant is the 24th power of the Dedekind eta function.

Posted by: PhilG on September 12, 2008 3:52 PM | Permalink | Reply to this

Re: 24

As usual I seem to have failed to get anyone interested in hyperdeterminants, but I know that lots of people are interested in E_8 so I am going to have one last go by trying to convince them that there might be a relationship between the invariants of E_8 and the 2x2x2x2 hyperdeterminant.

Start with quant-ph/0609227 by Duff and Ferrara which shows a correspondance between Cartan’s quartic invariant of E_7 (which acts on the adjoint representation of dimension 56) and Cayley’s hyperdetrminant which is the hyperdeterminant for a 2x2x2 tensor. You get the hyperdeterminant from Cartan’s invariant when you limit the symmetry to SL(2)^3 corresponding to any of the lines in the Fano plane. This is an inevitable result since the hyperdeterminant is the only SL(2)^3 invariant of degree 4.

So how will this generalise to E_8 where the adjoint representation of degree 248 is also the fundamental representation? In this case I believe we can look at subgroups isomorphic to SL(2)^4 so that any polynomial invariant of E_8 will have to reduce to invariants of SL(2)^4. The 2x2x2x2 hyperdeterimant is one such invariant but there are others.

A full analysis of the invariants of SL(2)^4 can be found in quant-ph/0212069 by Luque and Thibon. There is a quadratic invariant H and others of degree 4 (L,M and N) and 6 (D_xx, D_xy, D_xz). These generate the full ring of invariants. To get the hyperdeterminant they construct an invariant of degree 8 called S and an invariant of degree 12 called T, then the hyperdeterminant is the degree 24 invariant given by D = S^3 - 27T^2. Although the hyperdeterminant is not itself a fundamental generator of the ring of invariants it is still a very special invariant because of its role as a discriminant for the quadrilinear form given by the 2x2x2x2 tensor.

So what do we know about the invariants on the fundamental rep of E_8? Firstly there is the killing form which is of degree 2 so it must reduce to H on the SL(2)^4 subgroups. For some info about higher order invariants look at hep-th/0702024 by Cederwell and Palmkvist which concentrates on an octic invariant but mentions others of degree 12, 14, etc. My guess would be that the octic invariant reduces to S and another of degree 12 will reduce to T. In that case we could form an invariant of degree 24 which will reduce to the 2x2x2x2 hyperdeterminant by uplifting the relationship D = S^3 - 27T^2.

If that is correct then perhaps this invariant of E_8 has some special properties in the same way as the hyperdeterminant has special properties. For example it might be a discriminant for something. You could even use these ideas to construct a rational invariant of E_8 corresponding to the J-invariant via the relationship between the hyperdeterminant and elliptic curves. I.e. using J = S^3/D

I’ll stop there but if anyone thinks this might be of interest please let me know.

Posted by: PhilG on September 14, 2008 5:55 PM | Permalink | Reply to this

Hyperdeterminants, Bond, Clones; Re: 24

I’ve been told that, before Cayley, a good chunk of Hyperdeterminants were invented by William Spottiswoode FRS [11 January 1825, London - 27 June 1883 London], English mathematician and physicist.

Can anyone tell my more than is in the wikipedia page on him, and as to how far he went to inventing tensors?

The current major film director/writer is indeed a descendent. Roger Spottiswoode [5 January 1945 in Ottawa, Ontario] began his career as an editor in the 1970s. He’s directed a number of notable films and television productions, including Under Fire (1983) and the 1997 James Bond film Tomorrow Never Dies starring Pierce Brosnan. Spottiswoode was one of the writing team responsible for the delightful 48 Hrs. starring Eddie Murphy and Nick Nolte. In 2000 he directed the Sci-fi action thriller The 6th Day starring Governor Arnold Schwarzenegger.

One of those John/Joan Baez families that disproves “The Two Cultures” hypothesis.

Posted by: Jonathan Vos Post on September 14, 2008 7:14 PM | Permalink | Reply to this

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