## September 7, 2006

### Connes on Spectral Geometry of the Standard Model, III

#### Posted by Urs Schreiber I’ll try to outline the main technical ingredients that are involved in the computation of a spectral action # from a given spectral triple #.

Recall that we are imagining something along these lines:

We have a particle with worldline supersymmetry whose dynamics is defined in terms of a generalized Dirac operator

(1)$D$

which is an odd-graded operator on a $\mathbb{Z}_2$-graded Hilbert space

(2)$H$

on which an algebra

(3)$A$

of observables is represented

(4)$\rho : A \to B(H)$

by means of bounded operators on $H$.

More precisely, we want the $\mathbb{Z}_2$-grading on $H$ to be induced by an involutive operator

(5)$\gamma$

which is self-adjoint, squares to the identity, commutes with $A$ and anti-commutes with $D$.

Furthermore, we want there to be a real structure on $H$, which is an operator

(6)$J$

squaring to $\pm 1$ and commuting and/or anticommuting with $D$ and $\gamma$.

In order to visualize this, you should

• think of $H$ as a Hilbert space of square integrable sections of a spinor bundle on a Riemannian spin manifold $X$;
• think of $X$ as the target space that our superparticle propagates on;
• think of $D$ as the Dirac operator of the superparticle – i.e. of $D^2$ as the operator measuring the energy of the superparticle;
• think of $A$ as the algebra of smooth functions on $X$ acting on $H$ by multiplication (this are our “position operators”)
• think of expressions $[D,a]$ with $a \in A$ as Clifford fields (e.g. $[D,a] = [\gamma^\mu \nabla_\mu,a ] = \gamma^\mu \nabla_\mu a$)
• think of $\gamma$ as what physicists call $\gamma^5$.

All taken together, we call this data a spectral triple,

(7)$(A,H,D),$

and the point is that we think of any such triple of data satisfying some more or less obvious conditions as describing a superparticle propagating on a target space which is a generalization of the Riemannian target space $X$ used above.

Notice that

• $A$ encodes the topology of $X$;
• $D$ encodes the metric geometry of $X$.

In fact, Connes goes as far as identifying $D$ with the line element $ds$ on $X$ by writing

(8)$ds = D^{-1}\;.$

We associate two notions of dimension to such a generalized target space, called the

• metric dimension $d_\mathrm{met}$ determined by the decay rate of the eigenvalues of the operator $D$;
• and the KO-dimension $d_\mathrm{KO} \in \mathbb{Z}/8\mathbb{Z}$ determined by the choice of signs in the definition of the operator $J$ above.

We can think of the metric dimension as counting the number of “macroscopic” dimensions of target space of the kind we would measure by ordinary notions of movement in space.

And, while I am not completely sure at this moment, I believe we may think of the KO-dimension as the number of ordinary dimensions (modulo 8) of an ordinary Riemannian manifold which comes close to “approximating” our generalized Riemannian target space in some sense.

We shall be interested in finding spectral triples describing effective generalized target spaces $X$ that are good candidate models for the universe that we observe. This will make us want to have $d_\mathrm{met} = 4$ in order to get the basic properties of space and time right that we observe, and will make us want to have $d_\mathrm{KO} = 4+6 \in \mathrm{mod} 8$ in order to get the basic properties of the forces and particles we observe right.

But so far we have just a single superparticle propagating happily and undisturbed through an effective world $X$. In order for this to yield anything interesting as far as physics is concerned, we need to allow this particle to interact with itself.

Once it interacts with itself, we can regard our superparticle as a quantum of some quantum field $\Phi$ on $X$, such that that the dynamics of that quantum field are determined by the Feynman diagrams that describe the interaction process of our particle.

Instead of trying to write down these interactions explcitly, we shall at this point instead close our eyes, keep our fingers crossed and argue as follows:

“A sensible choice of interactions should yield a nice effective action $S$ for the quantum field. “Nice” means that it has a neat canonical expression in terms of just the data provided by our spectral triple. Moreover, we want the action to be additive under disjoint union of target space $X = X_1 \sqcup X_2$, which corresponds to $D = D_1 \oplus D_2$.

Taken together, this suggests that we set

(9)$S : (D,\psi) \mapsto \mathrm{Tr}(f(D)) + \langle \psi , D\psi \rangle \,,$

where $\psi \in H$ is a (generalized) spinor field and $f$ is some function (and we use functional calculus to apply it to $D$).”

Keep your fingers crossed.

From the superparticle-with-interactions point of view that I have adopted here, the above is an attempt to guess the right form of the effective target space action without actually writing down the interactions. I feel motivated to adopt this point of view, because Ali Chamseddine has, long ago, made some necessary consistency checks which show that it has a chance of being justified.

Namely, if we think of our superparticle as being the point-particle limit of a superstring, then it’s interactions are fixed, and the effective target space action is known - and Chamseddine checked # #, to lowest order, that, indeed, it coincides with the expression $S$ above.

Be that as it may, we call $S$ the spectral action associated to the spectral triple. We want to Taylor-expand it and find spectral triples $(A,H,D)$ such that the first terms of this expansion reproduce the action functionals of Einstein-gravity coupled to the standard model of particle physics.

And we want to do so by the algebraic analog of a Kaluza-Klein compactification.

Given any effective target space encoded by a spectral triple as above, we say $X$ is a compactification on a $d_\mathrm{KO}$-dimensional internal space $F$ if

• the Dirac operator on $X$ is a tensor product $D_X = D_Y \otimes D_F$
• such that the spectral triple $(A_F,H_F, D_F)$ has vanishing metric dimension $d_{\mathrm{metric},F} = 0 \;;$
• and has the specified KO-dimension.

We address this as a “compactification”, because we imagine that there is an ordinary Riemannian manifold of dimension $d_\mathrm{ext} + d_\mathrm{int}$ such that its spectral triple is smoothly connected, in some sense, along a path through the moduli space of all spectral triples, somehow, with the generalized spectral triple described above.

So we imagine taking $d_\mathrm{int}$ of the original dimensions and making them gradually ever smaller and possibly ever more “non-commutative” in order to get from the higher-dimensional ordinary Riemannian space to the lower-dimensional exotic one.

Let $D_\mathrm{ext}\otimes D_\mathrm{int}$ describe a compactified effective target space.

We are interested in the case where $D_\mathrm{ext}$ is itself a combination of

• the ordinary Dirac operator $D_0$ of the ordinary Riemannian manifold that remains after the compactification;
• plus a part $\delta D$ that is a measure for the mixed terms of the metric of the internal and the external space .

Then, in total, the metric on our compactified effective target space will be determined by

• the standard metric on the external space encoded in $D_0$;
• the metric on the internal, compact space encoded in $D_\mathrm{int}$;
• the “inner fluctuations” of the metric, encoded in $\delta D$.

We will see in the end that

• $D_0$ determines the gravitational field on the effective compactified target space;
• $\delta D$ determines the gauge fields on that effective target space;
• and $D_\mathrm{int}$ encodes the metric on the internal compact space, which will manifest itself in terms of the coupling constants of the gauge fields

This is essentially precisely as in Kaluza-Klein theory for ordinary Riemannian spaces.

(Hm, and it looks as if I needed yet another entry to go into the details of the compactifications that we want to consider…)

Posted at September 7, 2006 3:27 PM UTC

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### Re: Connes on Spectral Geometry of the Standard Model, III

It would be great if you could go ahead to the point of describing the specific target space Connes uses for the Standard Model!

You know, it’s funny how everyone is adding 4+6 modulo 8 and getting 10 instead of 2.

The “mod 8” in KO theory is all about Bott periodicity. And, over the years, I’ve often emphasized how

10 = 2 + 8

hints that the target space and the worldsheet in string theory are related via Bott periodicity. The strongest evidence that it’s not just a numerological coincidence is how the 8 transverse degrees of freedom of a string in 10-space are neatly encoded using octonions. Properties of the octonions, or triality, then “explain” the supersymmetry. But, precisely this math is what gives rise to Bott periodicity!

To quote Robert Helling from “week104”:

Let me add a technical remark that I extract from Green, Schwarz, and Witten, Vol 1, Appendix 4A.

The appearance of dimensions 3,4,6, and 10 can most easily been seen when one tries to write down a supersymmetric gauge theory in arbitrary dimension. This means we’re looking for a way to throw in some spinors to the Lagrangian of a pure gauge theory:

-1/4 F2

in a way that the new Lagrangian is invariant (up to a total derivative) under some infinitesimal variations. These describe supersymmetry if their commutator is a derivative (a generator of spacetime translations). As usual, we parameterize this variation by a parameter epsilon, but now epsilon is a spinor.

From people that have been doing this for their whole life we learn that the following Ansatz is common:

δAm = i/2 $\overline{\epsilon}$ Γm ψ

δ ψ = -1/4 Fmn Γmn $\epsilon$

Here A is the connection, F its field strength and ψ a spinor of a type to be determined. I suppressed group indices on all these fields. They are all in the adjoint representation. Γ are the generators of the Clifford algebra described by John Baez before.

For the Lagrangian we try the usual Yang-Mills term and add a minimally coupled kinetic term for the fermions:

-1/4 F2 + ig/2 ψ Γm Dm ψ

Here Dm is the gauge covariant derivative and g is some number that we can tune to to make this vanish under the above variations. When we vary the first term we find g = 1. In fact everything cancels without considering a special dimension except for the term that is trilinear in ψ that comes from varying the connection in the covariant derivative in the fermionic term. This reads something like

fabc $\overline{\epsilon}$ Γm ψa ψb Γm ψc

where I put in the group indices and the structure constants fabc. This has to vanish for other reasons since there is no other trilinear term in the fermions available. And indeed, after you’ve written out the antisymmetry of f explicitly and take out the spinors since this should vanish for all choices of ψ and ε. We are left with an expression that is only made of gammas. And in fact, this expression exactly vanishes in dimensions 3, 4, 6, and 10 due to a Fierz identity. (Sorry, I don’t have time to work this out more explicitly.)

This is related to the division algebra as follows (as explained in the papers pointed out by John Baez): Take for concreteness d = 10. Here we go to a light-cone frame by using coordinates

x+ = x0 + x9

and

x- = x0 - x1.

Then we write the Γm as block matrices where Γ+ and Γ-

have the +/- unit matrix as blocks and the others have γi as blocks where γi are the SO(8) Dirac matrices (i=1,…,9). But they are intimately related to the octonions. Remember there is triality in SO(8) which means that we can treat left-handed spinors, right-handed spinors and vectors on an equal basis (see week61, week90, week91). Now I write out all three indices of γi. Because of triality I can use i,j,k for spinor, dotted spinor and vector indices. Then it is known that

γijk = cijk for i,j,k < 8

γijk = δij for k=8 (and ijk permuted)

γijk = 0 for more than 2 of ijk equal 8.

is a representation of Cliff(8) if cijk are the structure constants of the octonions (i.e. ei ej = cijk ek for the 7 roots of -1 in the octonions).

When plug this representation of the Γ’s in the above mentioned gamma expression you will will find that it vanishes due to the antisymmetry of the associator

[a,b,c] = a(bc) - (ab)c

in the division algebras. This is my understanding of the relation of supersymmetry to the divison algebras.

Robert

I’ve also said many times that there should be a relation like “2d conformal field theories are to 3d topological quantum field theories as 10d string theory is to 11d M-theory”.

What’s interesting is that maybe some of the same numerology is showing up from a completely different viewpoint in Connes’ analysis of the Standard Model.

I made my own attempt to see the Standard Model as 10-dimensional; I have no idea if there’s a mathematical relation to Connes’ work.

Posted by: John Baez on September 8, 2006 4:40 AM | Permalink | Reply to this

### dimensions - internal and external

You know, it’s funny how everyone is adding 4+6 modulo 8 and getting 10 instead of 2.

I see your point. But in the present context it is actually justified - if one assumes that KO-dimension is ordinary dimension mod 8 “after compactification”.

From this point of view, what one should be worrying about is if 4 + 6 is maybe 4 + 14 or 4 + 22, and so on.

You possibly feel very different about this, but the way I feel about it is like this:

For a long time Connes had demonstrated that the standard model action may be encoded neatly into something operator-theoretic. Nice, but so what? Now there is for the first time a hint that this might point to a UV-completion.

Or so I think. Maybe I am hallucinating.

Posted by: urs on September 8, 2006 11:39 AM | Permalink | Reply to this

### bulk and boundary

there should be a relation like “2d conformal field theories are to 3d topological quantum field theories as 10d string theory is to 11d M-theory”.

The best I can say about it is this observation:

An $E_8$ WZW 2D CFT ($E_8$ current algebra as in the heterotic string) arises on the boundary of an $E_8$-Chern-Simons theory, which again is part of the M2-brane action.

So, this seems to indicate there is a connection of the sort you indicate.

But maybe I am wrong.

Posted by: urs on September 8, 2006 11:51 AM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

I made my own attempt to see the Standard Model as 10-dimensional;

What I cannot quite integrate about this observation into the rest of my world view is that I am not sure which physics models would turn the holonomy group of a space into the observed gauge group.

It looks like this would have to be a theory where a configuration is represented by a path in a 10d Calabi-Yau, with gauge transformations corresponding to adding loops to the ends of the path.

So a gauge invariant state would be represented by a path up to such loops. Hm, so maybe just by the endpoints?

In any case, I don’t think I have ever seen a model that would encode physical configurations this way. But maybe that’s just me.

Posted by: urs on September 8, 2006 12:11 PM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

I think someone told that part of the objection against Calabi Yau manifolds was that one whants to have a complete diffeomorfism group and the restriction to 4D+CY do not let to implement it fully.

Posted by: Alejandro Rivero on September 10, 2006 4:47 PM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

[…] one whants to have a complete diffeomorfism group […]

In general, yes, but not for every vacuum that we pick.

In particular, we want to fix the compactification metric. It fixes the Yukawa couplings.

Posted by: urs on September 11, 2006 12:02 PM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

Hmm the finite spectral triple of Connes is now six-dimensional, and we add the four dimensions of space time via tensor product. On the contrary, your model (Baez his) seems to be 10 dimensional already at the Standard Model level, before adding any dimensions of space time nor Poincare Group. So it is more of a 14 dimensional model, isnt it? Or is there a deep trick so that the SU(3) force relates to the hidden dimensions while SU(2)xU(1) meets the usual space time?

Posted by: Alejandro Rivero on September 10, 2006 5:02 PM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

I suppose it is pointed out in some of the weeks, but it could be worthy here to remember the paper

Supersymmetry and the division algebras
, by Kugo and Townsend

Posted by: Alejandro Rivero on October 28, 2006 4:14 AM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

You know, it’s funny how everyone is adding 4+6 modulo 8 and getting 10 instead of 2.

Actually, Connes writes:

But 10 is also 2 modulo 8 which might be related to the observations of [Lauscher&Reuter] about gravity.

His “internal space”, by the way, is the algebra

$A = \mathbb{C} \oplus \mathbb{H} \oplus M_3(\mathbb{C})$

That is: the complex numbers for the U(1) of hypercharge, plus the quaternions for the SU(2) of the weak force, plus 3×3 complex matrices for the SU(3) of the strong force. This has been in his work for a long time.

I can’t yet tell if his “dimension 10 mod 8” is mathematically related to my crazy idea relating the Standard Model to 10 dimensions. In my idea, the number 10 comes from the fact that $SO(10)$ is the smallest $SO(n)$ into which

$\mathbf{G} = (U(1) \times SU(2) \times SU(3))/\mathbb{Z}_6$

the true gauge group of the Standard Model, embeds. To get a single generation of the Standard Model fermions, we take the 32-dimensional Dirac spinor representation of $Spin(10)$ and restrict it to $\mathbf{G} \subset SO(10)$ (on which it becomes single-valued).

This 32-dimensional space is the bimodule $\mathcal{M}_F$ which Connes mentions on page 5!

He also considers a 90-dimensional Hilbert space $\mathcal{H}_F$, but I don’t see how he gets this. He seems to say in Def. 2.3 that

$\mathcal{H}_F = \mathcal{M}_F \otimes \mathbb{C}^3$

which makes sense: we’re tripling the number of fermions, since we have 3 generations.

That gives me

$\mathrm{dim}(\mathcal{H}_F) = 32 \times 3 = 96,$

not 90.

But Connes is better at math than I am, so I’m sure he’s right. Posted by: John Baez on September 8, 2006 6:24 AM | Permalink | Reply to this

### Generation

To get a single generation of the Standard Model fermions, we take the 32-dimensional Dirac spinor representation of $Spin(10)$ and restrict it to $G\subset SO(10)$ (on which it becomes single-valued).

Hmmm?

To get a single generation of Standard Model fermions, we take the 16-dimensional Weyl spinor of representation of $Spin(10)$.

And that gives us an extra 16th fermion, the $G$-singlet sterile neutrino. And this guy must have a very large mass.

There are various mechanisms for giving it a mass, the most parsimonious of which involves a “Higgs” scalar transforming in the 126-dimensional anti-self-dual 5-form representation of $SO(10)$.

What mechanism does Connes invoke to give it a large mass?

Posted by: Jacques Distler on September 8, 2006 6:42 AM | Permalink | PGP Sig | Reply to this

### Re: Generation

What mechanism does Connes invoke to give it a large mass?

I know essentially nothing about this issue, but it seems that Connes does address it.

On the top of p. 7 he says that $M$ will be set to order Planck mass by the equations of motion.

Then there is a more detailed discussion on the top of p. 14.

Posted by: urs on September 8, 2006 3:21 PM | Permalink | Reply to this

### Re: Generation

On the top of p. 7 he says that $M$ will be set to order Planck mass by the equations of motion.

If true, then this theory is falsified already. That’s 4 orders of magnitude too large to give the observed neutrino masses.

Posted by: Jacques Distler on September 8, 2006 3:53 PM | Permalink | PGP Sig | Reply to this

### Re: Generation

That’s 4 orders of magnitude too large to give the observed neutrino masses.

That mass is $\sim \frac{m^2}{M}$ (the smaller of the two eigenvalues), right?

At the moment I don’t see what Connes gets for $m$. All I see is a footnote saying that he has no argument for it having any particular value.

Posted by: urs on September 8, 2006 4:40 PM | Permalink | Reply to this

### Re: Generation

It is not so clear that it implies a mass in the Planck range. The GUT range is equally valid, and moreover the estimate of the top mass is done in this paper by running down from GUT scale as usual. Besides, Connes made some amusing remarks about/against Planck scale and Newton’s constant.

Posted by: Alejandro Rivero on September 9, 2006 3:22 PM | Permalink | Reply to this

### Numerology

He also considers a 90-dimensional Hilbert space $ℋ_F$, but I don’t see how he gets this.

I assume that $90= 3\times 15\times 2$. because there are 3 generations, each of which consists of 15 fermions which, in turn, transform as the $\mathbf{2}$ of $SL(2,\mathbb{C})$.

Posted by: Jacques Distler on September 8, 2006 6:48 AM | Permalink | PGP Sig | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

John wrote:

To get a single generation of the Standard Model fermions, we take the 32-dimensional Dirac spinor representation of $Spin(10)$ and restrict it to $\mathbf{G} \subset SO(10)$ (on which it becomes single-valued).

Jacques writes:

Hmmm?

To get a single generation of Standard Model fermions, we take the 16-dimensional Weyl spinor of representation of $Spin(10)$.

I was implicitly including the antiparticles of the fermions.

And that gives us an extra 16th fermion, the $G$-singlet sterile neutrino. And this guy must have a very large mass.

[…]

What mechanism does Connes invoke to give it a large mass?

Both Barrett and Connes invoke a version of the see-saw mechanism - see for example the bottom of page 6 and top of page 7 on Connes’ paper, and page 13 of Barrett’s. However, I don’t understand the details.

John wrote:

He also considers a 90-dimensional Hilbert space $ℋ_F$, but I don’t see how he gets this.

Jacques wrote:

I assume that $90= 3\times 15\times 2$ because there are 3 generations, each of which consists of 15 fermions which, in turn, transform as the $\mathbf{2}$ of $SL(2,\mathbb{C})$.

I guess that’s it! But, he really does seem to start with $\mathbb{C}^3 \otimes \mathbb{C}^{32} \cong \mathbb{C}^{96}$. Maybe at some point he just decides to ignore the sterile neutrino for certain purposes, taking him down from 96 to 90.

Hmm, yes - Barrett considers two models, one with a sterile neutrino, one without. The small one has a 90-dimensional fermionic Hilbert space $\mathcal{H}_F$; the big one has a 96-dimensional $\mathcal{H}_F$.

Posted by: John Baez on September 8, 2006 10:42 AM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

His “internal space”, by the way, is the algebra

(1)$A = \mathbb{C} \oplus \mathbb{H} \oplus M_3(\mathbb{C})$

That is: the complex numbers for the $U(1)$ of hypercharge, plus the quaternions for the $\mathrm{SU}(2)$ of the weak force, plus $3\times 3$ complex matrices for the $\mathrm{SU}(3)$ of the strong force. This has been in his work for a long time.

Yes, this is over ten years old by now.

But it didn’t quite work for over ten years.

The new ingredient that seems to save the day is a rather tiny modification of the original proposal: the $\mathbb{Z}_2$-grading $\gamma_F$ of the Hilbert space $H_F$ that $A$ is represented on is switched in the anti-particle sector.

It’s that choice which leads to the “signature” $d_\mathrm{KO} = 6$ for the internal space, as opposed to 0.

John Barrett gives a nice discussion (first few pages of hep-th/0608221) of how this now allows to project down the spinors to Weyl spinors, as it should be.

Posted by: urs on September 8, 2006 11:20 AM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

I guess that’s it!

Yes, that’s it. It’s spelled out in detail on p. 23 of hep-th/9603053.

Posted by: urs on September 8, 2006 11:24 AM | Permalink | Reply to this

### Re: Connes on Spectral Geometry of the Standard Model, III

What mechanism does Connes invoke to give it a large mass? Both Barrett and Connes invoke a version of the see-saw mechanism

The see-saw mechanism is the consequence (not the cause) of the sterile neutrino having a large mass. In the $2\times 2$ mass matrix, $\begin{pmatrix}M&m\\ m& 0\end{pmatrix}$ the “$0$” is a consequence of $G$-gauge invariance and the “$m$” is easily generated by a coupling to the standard ($SU(2)$ doublet) Higgs. It is the large entry, “$M$”, that needs explanation.

This has to be very high (nearly GUT scale) and generating it requires some degrees of freedom beyond the ones list (the Standard model particles + the sterile neutrino).

That’s because the theory without additional degrees of freedom has an accidental global symmetry that forbids $M\neq 0$.

Posted by: Jacques Distler on September 8, 2006 1:07 PM | Permalink | PGP Sig | Reply to this
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