Report-Back on BMC

April 22, 2007

Posted by Urs Schreiber

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– guest post by Bruce Bartlett –

I was born in a large Welsh industrial town at the beginning of the Great War: an ugly, lovely town (or so it was, and is, to me), crawling, sprawling, slummed,unplanned, jerry-villa’d, and smug-suburbed by the side of a long and splendid-curving shore…

Thus described Dylan Thomas his childhood home of Swansea, Wales - the venue of the British Mathematics Colloquium this year :

Inspired by John’s blurb about the higher categories workshop at Fields earlier this year, I thought I’d send Urs a report-back of the (admittedly less glamorous) “BMC” , and mention a few things possibly of interest to $n$ -café patrons.

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Alain Connes was one of the plenary speakers. He spoke about his recent work with Ali Chamseddine and Matilde Marcolli on getting the Standard Model (together with gravity) out of an elegant noncommutative geometry framework.

In the beginning of the talk, he showed a slide with the lagrangian for the Standard Model - written out in gory detail :

Gulp! It takes up the whole slide - and that’s apparantly just the lagrangian on a flat background!

In Connes’ picture, spacetime is described as $X = M \times F$ where $M$ is an ordinary (`commutative’) space (it’s a manifold with a spin structure), and $F$ is a finite `noncommutative space’. In fact, the algebra representing $F$ is $F = \mathbb{C} \oplus \mathbb{H} \oplus M_3 (\mathbb{C}).$ So its a copy of the complex numbers, the quaternions, and the $3 \times 3$ matrices.

It’s really cool, at least to a newcomer like me. Physicists have often expressed their gut feeling that somehow spacetime has a discrete texture to it at small length scales… Connes’ picture makes that precise in a really simple and elegant way.

Moreover, apparantly one can turn the “noncommutative geometry crank” on the spacetime $X = M \times F$ and out pops the Standard Model - together with ‘neutrino mixing’ (whatever that means) - in all its gory detail! It’s possible that Urs has gone over all this stuff before… but I think I missed it.

[Yes, we had a series of posts on that: I, II, III, IV – Urs ]

Another quasi-plenary speaker was Tom Leinster, who gave a cool talk on “New perspectives on Euler characteristic.” Some of this stuff has been discussed on this blog before,

but I was fascinated to see how one can calculate the Euler characteristic of fractals like the Julia set associated to a rational complex-valued function:

We also had a nice talk on “Ricci flow and geometrization of three-manifolds” by Huai-Dong Cao who I think was (back in the day) one of Shing-Tung Yau’s students. Indeed Ricci flow has also been discussed on this blog before - and how it relates to the dilaton field in string theory!

But for me, the thing I was really looking forward to in Swansea was an opportunity to see the legendary “Annie’s Place” restaurant. As I understand it, it was at this restaurant in 1988, at the International Congress of Mathematical Physics, that Witten, Segal and Atiyah hatched their evil plan to combine geometry with quantum field theory and thus revolutionize mathematics. It was during dinner that Witten came up with the idea of Chern-Simons theory to explain the Jones polynomial, and decided the next day to not give the talk he had planned, but to talk about this new theory born only the night before!

Sadly Annie’s restaurant is no longer - it’s now a French restaurant called “Didier and Stephanie” :

Tough noogies.

Posted at April 22, 2007 8:01 PM UTC

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Re: Report-Back on BMC

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Hi Bruce!

Thanks a lot for the nice report.

Yes, the standard model Lagrangian written out in full detail looks intimidating.

It is important to realize that all the terms are of just a handful of different types, with all the remaining difference being a difference in realization of the same underlying pattern.

First there are the gauge-kinetic terms of the form $F \wedge \star F \,,$ where $F$ is the curvature 2-form of a bundle with connection. Terms of this form describe the dynamics of the “gauge fields”, the “force fields” of nature.

All other terms are bilinears in “fermions”, usually denoted $\psi$ , which are sections of spinor bundles associated to the bundles with connection from above.

First there the kinetic terms of the fermions. In the component and coordinate-ridden way these are usually displayed in this context, they look like $\bar \psi \gamma^\mu \frac{\partial}{\partial x^\mu} \psi \,.$

Then there are the “minimal coupling” terms which encode the interaction between the fermions and the gauge bosons from before. With $A$ being the connection 1-form (not only is the background metric usually assumed to be flat here, but also all bundles are assumed to be trivial) these read $\bar \psi \gamma^\mu A_\mu \psi \,.$

It is clear that together these two terms really should thought of as $\langle \psi | D \psi \rangle \,,$ where $D$ is the canonical Dirac operator on the spinor bundles with connection associated to our gauge bundles.

This is clear. One thing that Connes added to this was to observe that also the fourth kind of terms, namely the Yukawa couplings, which read $\bar \psi \phi \psi$ for $\phi$ the Higgs field, can be absorbed in a term simply of the form $\langle \psi | D \psi \rangle$ if only we allow the Dirac operator to be that associated to a noncommutative algebra.

That’s the main point of Connes’ approach as far as pure particle physics is concerned.

The other observation he made is that also the gauge kinitic terms $F \wedge \star F$ may be unified with the Einstein-Hilbert term $R$ by extracting both from a heat kernel expansion of somewthing like the exponential of the trace of the squared Dirac operator.

Given these two observations, the remaining task is to carefully identify precisely the right noncommutative geometry (a spectral triple), such that the associated Dirac operator reproduces all the gory details of the standard model by just expanding $\mathrm{tr}(f(D/\Lambda)) + \langle \psi | D \psi\rangle \,.$ That’s what people have done for almost 20 years now.

For all that time, the noncommutative geometry that had been found yielded something awefully close, but not quite coinciding with the standard model.

Interestingly, all that was missing was a tiny modification in one of the gradings that enter the full definition of a real spectral triple. With that modified grading, suddenly everything nicely falls into place.

The resulting spectral triple describes a noncommutative space whose algebra of functions looks like that on $\mathbb{R}^4$ times the noncommutative algebra you mentioned.

The ordinary spectral dimension of the “internal” noncommutative factor is 0, as it should be for a “compactified” space. Curiously, the dimension of this compact space as seen by K-theory is 6 modulo 8.

Posted by: urs on April 22, 2007 8:43 PM | Permalink | Reply to this

Euler Characteristic of a Julia Set

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Could anyone give a hint on how we think of a Julia set as a category, such as to associate an Euler characteristic with it?

Posted by: urs on April 23, 2007 7:21 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

Tom wrote about fractals here, which links to two more detailed papers.

Posted by: David Corfield on April 23, 2007 7:46 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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Tom wrote about fractals here, which links to two more detailed papers.

Thanks! Now I recall having briefly seen this before.

On p. 4 it says that a self-similarity system is a functor $M : A^{\mathrm{op}} \times A \to \mathrm{Set}$ satisfying some conditions ( $A$ being small).

So, is the Euler characteristic of a Julia set then the Euler characteristic of the corresponding category $A$ ?

Posted by: urs on April 23, 2007 7:57 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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Hi Urs. Just a quickie as I have to run…

The first thing to say is that this hasn’t been worked out rigorously, which is why it hasn’t been written up properly either.

The second thing is that you’re not meant to think of the space concerned (e.g. that Julia set) as corresponding to a category. In fact, the notion of Euler characteristic of a category doesn’t come into this at all, as far as I know.

The idea is more like the following. Suppose you wanted to know the Euler characteristic of the Cantor set $C$ . Since $C = C + C$ (i.e. $C$ is homeomorphic to the disjoint union of two copies of itself), we’d like to deduce that $\chi(C) = \chi(C) + \chi(C)$ , and so $\chi(C) = 0$ .

Gotta dash… to be continued another time

Posted by: Tom Leinster on April 24, 2007 5:07 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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The second thing is that you’re not meant to think of the space concerned (e.g. that Julia set) as corresponding to a category. In fact, the notion of Euler characteristic of a category doesn’t come into this at all, as far as I know.

Oh, I see. I was assuming that whenever you consider an Euler characteristic and a category, you are considering the Euler characteristic of that category. :-)

I was about to ask what the Euler characteristic of the infinite binary tree $\array{ &\swarrow &&& \searrow \\ \swarrow & \searrow & & & \swarrow & \searrow }$ – regarded as a fractal – would be, and how that compared to its Leinster-measure (roughly as discussed here and here) but maybe that doesn’t make that much sense, then.

Posted by: urs on April 24, 2007 5:34 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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OK, continuing. Let’s do a slightly more complicated example: $[0, 1]$ . Since $[0, 1] = [0, 1/2] \cup [1/2, 1]$ and $[0, 1/2] \cap [1/2, 1] = \{ pt \}$ , we’d like to deduce that $\chi([0, 1]) = \chi([0, 1/2]) + \chi([1/2, 1]) - \chi(pt).$ Why? Because Euler characteristic is meant to behave like cardinality. And since $\chi$ is obviously meant to be homeomorphism-invariant, we can solve to conclude that $\chi([0, 1]) = 1$ .

Here we’ve used the ‘inclusion-exclusion formula’ for the cardinality of the union of two spaces. A union is a pushout (along injections). In my Euler characteristic paper, there’s a generalization of the inclusion-exclusion formula to arbitrary colimit shapes. The result is this: if $X: A \to FinSet$ is a suitably ‘nondegenerate’ functor then $| colim_a X(a) | = \sum_a k^a |X(a)|$ where $k^\cdot$ is a weighting on $A$ .

Another example: suppose we have two spaces $X$ and $Y$ known to satisfy $X = X + X + Y, Y = Y + Y.$ Then we’d like to deduce $\chi(X) = \chi(X) + \chi(X) + \chi(Y), \chi(Y) = \chi(Y) + \chi(Y)$ and so $\chi(X) = \chi(Y) = 0$ .

Now let’s go to the general setting: you have a family of spaces each described as a colimit of other members of the family. (By “other” I don’t exclude the possibility of self-reference.) This is a system of simultaneous equations of some kind. We know what the cardinality of a colimit is supposed to be - that’s the result above. So we can apply $\chi$ everywhere to obtain an ordinary system of simultaneous linear equations involving the Euler characteristics of the spaces. With luck, we can then solve to find what those Euler characteristics are.

Let me repeat that this is not fully worked out. In particular - crucially! - there is a question of consistency: could this procedure give two different answers for the Euler characteristic of a space? There are examples telling us that we have to be careful, but I believe we’ll be able to find a way of making it work.

Posted by: Tom Leinster on April 24, 2007 5:47 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

Tom, you may be aware of this, but I’ll throw it out there anyway: the consistency of Euler characteristic (in the sense of an integer-valued invariant valuation) has been worked out for “tame” subsets of Euclidean space. Here “tame” means “belongs to an o-minimal structure”, such as the collection of semi-algebraic sets. See “Tame Topology and O-minimal Structures” by Lou van den Dries, Proposition 4.2.2.

Sadly, fractals fall well outside anything that could ever pass for “tame” in set-ups like this.

(In case anyone is wondering, an o-minimal structure is, roughly speaking, a collection of subsets of finite-dimensional Euclidean spaces which, considered as a collection of “definable” predicates, is closed under all operations of first-order logic with equality, and which contains some basic relations such as the order relation on the reals, and any real constant. The o- or order-minimality condition is that the only definable subsets of the reals are finite unions of points and intervals.)

Posted by: Todd Trimble on April 24, 2007 9:28 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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Thanks Todd - I wasn’t aware of this.

Here’s the warning example I had in mind. Take the universal solution of the equation $X = 2X + 1$ , i.e. the terminal coalgebra of the endofunctor $Top \to Top, X \mapsto 2X + 1.$ Then it’s not too hard to see that, in fact, there is a (‘bad’) homeomorphism $X \to X + 1$ .

What this suggests to me is that we should be thinking about spaces that carry their recursive structure around with them, in some suitable sense. Let’s call such a thing a ‘recursive space’, for the purposes of this discussion. The bad homeomorphism wouldn’t be an isomorphism of recursive spaces.

The hope would be that spaces such as finite cell complexes carry a unique recursive space structure, which would explain why you can define the Euler characteristic of such spaces without being explicitly told a recursive structure. Perhaps the same would be true of tame Euclidean spaces too.

Posted by: Tom Leinster on April 25, 2007 10:41 AM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

Is the terminal coalgebra of the functor F(X) = 2X+1 just the union of the usual Cantor set with the midpoints of the “middle thirds” intervals (which were deleted in forming the Cantor set)? Since the midpoint set is countable and discrete, it’s easy to see that this coalgebra is a fixed point of F(X) = X+1, as you say.

By the way, can the Euler characteristics for these fractal sets can take on rational (non-integer) values? (I’m guessing they can.)

Posted by: Todd Trimble on April 27, 2007 1:29 AM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

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Re your description of the terminal coalgebra: yes, I think that’s right. (It’s not exactly how I’d thought of it.) But it’s not a disjoint union, i.e. it’s not a coproduct in Top, because you can have a sequence of midpoints converging to something in the Cantor set itself. So I don’t think it’s completely obvious that it satisfies $X = X + 1$ .

Yes, these Euler characteristics can be non-integers (if any of this makes any sense!), e.g. the terminal coalgebra for $X \mapsto 3X + 1$ should be $-1/2$ , as you’d guess. Also, the Euler characteristic of the classifying space $BG$ of a finite group $G$ should of course be $1/order(G)$ ; this isn’t a fractal, but I’d like to be able to handle it in the same framework.

Posted by: Tom Leinster on April 27, 2007 2:04 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

…the same framework

Sounds like what you might need for a self-similar group.

Posted by: David Corfield on April 27, 2007 2:28 PM | Permalink | Reply to this

Re: Euler Characteristic of a Julia Set

So I don’t think it’s completely obvious that it satisfies X=X+1 .

You’re right. Let me patch it up: the final coalgebra of F(X) = 2X+1 is (according to my description) the completion of a uniform structure on the set M of midpoints of deleted thirds; the uniform structure is given by the metric. Define a map on M which takes each element of the form 3^-n/2 to 3^-(n+1)/2, and every other point to itself. This map is uniformly continuous (bounded by Lipschitz constant 3, I think), hence extends uniquely to a map on the completion, and that map on the completion should do the trick.

As an aside, I noticed while writing the above that the midpoint set M is the initial algebra of the functor F(X) = 2X+1, which I thought was sort of intriguing. It got me thinking that there should be an interesting class of endofunctors where the final coalgebra is some sort of ‘uniform completion’ of the initial algebra. (I’ll ramble on this for a while. Probably something like this is well known, but I didn’t know it.)

A wild guess: suppose we take an endofunctor F on a category [A, Top] given by tensoring with a finite A-bimodule (satisfying technical conditions specified in your [Tom’s] paper), and add 1 to it. Could the final (1+F)-coalgebra be a completion of a uniform structure on the initial (1+F)-algebra?

Here’s my rough thinking: an element of the final F-coalgebra is something like a class of behaviors of an F-machine with feedback; by adding 1 to F, you add in the possibility of halting a behavior after finitely many steps. Then, I want to view these finite approximations to F-behaviors as elements of the initial algebra for 1+F: on the initial algebra side, the 1 would provide some ‘seed money’ or input to start the iterations with.

More precisely, since F is a left adjoint, the initial (1+F)-algebra should be 1 + F(1) + F²(1) + …

Well, I have to go now, but I’m conjecturing that once a suitable uniform structure has been specified on the free algebra, that the completion is the final (1+F)-coalgebra, where the ‘points at infinity’ form a final F-coalgebra.

Posted by: Todd Trimble on April 28, 2007 2:23 PM | Permalink | Reply to this

Re: Report-Back on BMC

This goes back to Bruce’s original posting where he refers to
`F is a finite `noncomm’ space.
It’s the word finite that causes me problems; perhaps just another example of a word that means something different in physics. The F in question is the direct some of the complex numbers, the quaternions
and 3 x 3 matrices over C. how in the world is that finite - unless finite means finite dimensional?

Posted by: jim stasheff on May 22, 2007 2:54 PM | Permalink | Reply to this

Re: Report-Back on BMC

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I believe Bruce was thinking of “finite volume” when saying “finite space”. This is a general idea in standard Kaluza-Klein models: one assumes that spacetime is a bundle with finite-volume fibers.

(But since the end of the last millenium people also consider Kaluza-Klein models where the fibers have infinite volume. But Connes’ model is not of this kind.)

The important point in Connes’ noncummutative geometry version of Kaluza-Klein compactification is in fact: this space $F$ (the fiber mentioned above) has vanishing spectral dimension. So it’s in fact 0-dimensional!

But one has to be careful here, since not all notions of dimension coincide for noncommutative spaces: the space $F$ also has a “K-theoretic” dimension, which knows about the nature of (noncommutative) vector bundles on it (i.e. modudles of the noncommutative algebra): this K-theoreitc dimension (really “KO dimension”) is 6 modulo 8.

This means, incidentally, that Alain Connes shows that the world we perceive has spectral dimension 4 and KO-dimension

4 + 6 mod 8.

Posted by: urs on May 22, 2007 8:18 PM | Permalink | Reply to this

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April 22, 2007