## June 4, 2006

### Around the Blogs

I thought I’d take some time out for a little tour around the physics blogosphere. While there’s plenty of sturm und drang to be found in the usual places, there is also also some really nifty and worthwhile stuff out there

Georg, over at Life on the Lattice has a nice post on NRQCD on the lattice. In studying heavy quark bound states on the lattice, it is often very convenient to replace the Wilson (or Kogut-Susskind) action for the fermions with its non-relativistic counterpart. Obviously, halving the number of degrees of freedom simplifies the task of computing a fermi determinant. But that’s not where the saving really lies. It’s really in removing the rest-energy contribution to the Hamiltonian, (which, otherwise would dominate the decay of Euclidean correlation functions, $C(t,0) \sim e^{-m t}$) that makes the nonrelativistic approximation worthwhile.

I’d love to hear more about the tradeoffs in tuning the higher-order terms in the NRQCD Lagrangian and what the added complexity buys you, in terms of better numerical stability and faster convergence to the continuum.

Urs has a series of posts on Mathai’s approach to the “topological” aspects of T-duality. How T-duality acts on the K-theory of $M\times T^n$ is more-or-less understood. Mathai would like to bootstrap this to some statement about the algebra of functions $C(M\times T^n)$ (in which the K-theory is the algebraic K-theory of modules over this algebra).

Personally, I’m rather dubious of the approach, in that T-duality is more naturally described in terms of the loop space of $X$, rather than $X$ itself (remember, it exchanges momenta and windings). It’s neat, but somehow seemingly accidental, that it reduces to a simple mapping of the K-theories of $X$ and its T-dual.

Travis Stewart reports that the LHC’s ATLAS detector has seen cosmic ray events, an excellent sign that things are working as they should.

Mike Schmitt and Tomasso Dorigo have some posts about techniques for tagging b-jets.

Finally, on a more elmentary level, Dmitri Terryn has a nice little post about the elastic electromagnetic scattering cross-section, from its nonrelativistic, classical expression (Rutherford scattering) to its more complicated, fully quantum-mechanical variations.

Posted by distler at June 4, 2006 3:37 AM

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### Re: Around the Blogs

Jacques Distler wrote:

Personally, I’m rather dubious of the approach, in that T-duality is more naturally described in terms of the loop space of $X$, rather than $X$ itself (remember, it exchanges momenta and windings). It’s neat, but somehow seemingly accidental, that it reduces to a simple mapping of the K-theories of $X$ and its T-dual.

Since, as you say, the approach to T-duality as followed by Mathai (and others) knows nothing about loop space, let alone about CFT (which is really where the duality lives), it is clearly just some sort of “shadow” of the full thing in terms of topological notions on target space.

(The question about the apparent “continuous orbifolding” which I mentioned is probably related to that, since clarifying it seems to call for a a setup that knows about the T-dual CFTs.)

I am not sure, though, how accidental the mapping between K-theories on T-dual target spaces is. Since these classes are where the RR-charges live, it seems necessary that such an isomorphism exists.

But that’s more a gut feeling than a precise statement.

It is no secret that “topological T-duality theorists” (like for instance also Ulrich Bunke math.GT/0501487) pretty mich just take the K-theory setup as given and are happy doing their math in that framework, without actively worrying about the connection to physics/CFT. The “abstract T-duality” definition which I mention at the end of part III certainly certifies this.

I would like to see the connection of the topological T-duality technology to CFT. As I tried to indicate before, there is an interesting analogy, which might point in the right direction.

So on the one hand side we see in the topological T-duality approach the Fourier-Mukai transformation on the correspondence space of the torus and its dual. In as far as the sheaves on $M \times \mathbf{T}$ that we are dealing with are modules for the structure sheaf, the Fourier-Mukai correspondence is a bimodule, and its operation on these sheaves by pullback can be understood alternatively in terms of the tensor product of modules.

Now, this is rather reminiscent of the way T-duality is realized in the Frobenius-algebraic approch to (R)CFT ($\to$).

As will be detailed in upcoming work that has been announced in cond-mat/0404051, T-duality (like other dualities) is induced by the action of certain bimodules internal to the representation category of the chiral vertex algebra.

I am hoping that this can be used to make the connection between T-duality in full CFT and the topological T-duality on target space studied by Mathai, Bunke et. al.

But that’s just me.

Posted by: urs on June 4, 2006 11:02 AM | Permalink | Reply to this

### Accidental

I am not sure, though, how accidental the mapping between K-theories on T-dual target spaces is. Since these classes are where the RR-charges live, it seems necessary that such an isomorphism exists.

I meant “accidental” in a very specific sense. $K^\bullet(X)$ is the algebraic K-theory of $C(X)$; $K^\bullet(\hat{X})$ is the algebraic K-theory of $C(\hat{X})$. They are exchanged under T-duality. But $C(X)$ and $C(\hat{X})$ (the algebras whose algebraic K-theories are exchanged) are not so-exchanged.

Maybe replacing $C(X)$ by some appropriate crossed-product algebra produces a correct statement, but I’m not quite smart enough to see it.

Posted by: Jacques Distler on June 4, 2006 11:55 AM | Permalink | PGP Sig | Reply to this

### Re: Accidental

Maybe replacing $C(X)$ by some appropriate crossed-product algebra produces a correct statement

Isn’t that the statement “1) the algebraic action of T-duality” which I mentioned in part III?

Posted by: urs on June 4, 2006 12:45 PM | Permalink | Reply to this

### Re: Accidental

Umh, yeah, that’s the statement. But is it … correct?

Posted by: Jacques Distler on June 4, 2006 1:12 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental

is it … correct?

It seems to be a direct consequence of that second version of Green’s theorem.

But unless I sit down and learn the details I will have to trust that Mathai did not get the basics for his theory wrong.

Posted by: urs on June 4, 2006 1:20 PM | Permalink | Reply to this

### Re: Accidental

The stated equivalence may indeed be correct. But there remain the questions

1. Is there any nontrivial content to it? (As you, yourself, point out, topologically, $M\times T^n$ and $M\times \hat{T}^n$ are the same, and Morita equivalence is a very weak statement of equivalence.
2. Is it, in any sense, related to T-duality?

We know a rather precise statement of T-duality, and how it acts on the closed string states on $M\times T^n$. Is there any relation between this cross-product algebra and the space of closed string states on $M\times T^n$? If so, is the stated action equivalent to the known action on closed string states?

Finally, I’m a little puzzled by the alleged labelling of this as a “topological” version of T-duality. $C(X)$ carries much more information than the topology of $X$. Indeed, it’s really the same information as “$X$” itself. Passing to the algebraic K-theory gives us something topological. Is there something intermediate (cross product algebra up to Morita equivalence, or some such thing) that is also befitting of the appelation, “topological”?

Posted by: Jacques Distler on June 4, 2006 1:50 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental

$C(X)$ carries much more information than the topology of X

Ah, here is a misunderstanding. What is meant is that $C(X)$ characterizes a (locally compact Hausdorff) topological space, so it gives you $X$ (as a set of points) together with a topology on it (in the sense of systems of open sets).

So as long as everything is formulated only in terms of (commutative or not) $C^*$ algebras, the setup is topological in that we know target space as a topological space, but not as A Riemannian manifold.

Usually, we’d say that T-duality is something that involves spaces with a metric, because otherwise it does not make sense to say that the radius of some circle fiber is inverted.

In order to incorporate such metric information into Mathai’s algebraic formulation of T-duality one would probably have to equip the $C^*$-algebra with a representation by bounded operators on a graded Hilbert space, together with a Dirac operator on that Hilbert space (i.e. a spectral triple).

Is there any nontrivial content to it?

Indeed, that’s a question that arises. The question has been raised in both talks that I have heard by Mathai, and he says there is. I’ll ask him about more details when I visit him next week at the ESI.

If so, is the stated action equivalent to the known action on closed string states?

The “topological T-duality” setup certainly contains less information. What it does reproduce is the action of T-duality on

a) the topology of spacetime

b) the $H$-flux.

So, for instance, given an $H$-flux on a trivial circle bundle, the T-dual circle bundle is that whose Chern class is the integral of $H$ over the original $S^1$ fiber (if I recall correctly). So you know the Chern class of the T-dual spacetime (regarded as a circle bundle) and the T-dual H-flux. Hence the “topological” information.

This is reproduced, I gather, by Mathai’s approach.

As far as I understand, he can also reproduce, and go beyond, known examples where T-duals are noncommutative.

Posted by: urs on June 4, 2006 3:12 PM | Permalink | Reply to this

### Re: Accidental

I never meant to imply that $C(X)$ carries metric information about $X$. But, depending on exactly what you mean by “$C(X)$”, it carries more than topological information. For instance, if you take it to mean the ring of smooth functions on $X$ (that’s the case that leads to the topological K-theory of $X$, and so was my best guess for what you meant), then $C(X)$ encodes the differentiable structure of $X$. If you take it to be the ring of (local) holomorphic functions, that encodes the complex or algebraic structure of $X$.

In both cases, it encodes more than purely topological information.

Maybe examples with H-flux and nontrivial circle bundles give something nontrivial. But the case of $M\times T^n$ is a yawner.

So, here’s an example: let $S^3\to S^2$ be the Hopf fibration. Put $n$-units of H-flux on the $S^3$ (perhaps including the case $n=0$). What does Mathai’s formalism tell you about the T-dual of this system.

Try not to “cheat” and use known physics to guess the answer.

Posted by: Jacques Distler on June 4, 2006 8:34 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental?

Well we will have 1 unit of dual $H$-flux and the $S^1$ bundle with Chern/Euler class $n$ times the generator of $H^2(S^2,Z) = Z$. It’s not an interesting’ example in the sense that the $H$-flux can have only one leg along the fibres. Without looking back at the Bouwknegt-Mathai paper, I vaguely recall the T-dual bundle should be the lens space gotten by dividing the appropriate sphere by $Z_n$. (Possibly modulo signs on classes)

You say $M \times T^n$ is uninteresting, and without H-flux it is, but try to dualise $M \times T^3$ with the $H$-flux on the torus and things fall apart’.

Not being a physicist, Morita equivalence doesn’t seem too weak, since it is the natural equivalence one uses considering presentations of stacks. This shows up all the time in physics in coordinate transforms, change of atlas, gauge transformations even. It’s just all the non-geometric stuff (to borrow a phrase) which appears strange.

DM Roberts

Posted by: David Roberts on June 5, 2006 12:54 AM | Permalink | Reply to this

### Re: Accidental?

Yes, the $S^1$ bundle over $S^2$ with Euler class “$n$” is the Lens space, $S^3/\mathbb{Z}_n$.

So do we agree, then, that the T-dual of $S^3$ with $n$ units of H-flux is $S^3/\mathbb{Z}_n$ with 1 unit of H-flux (which is the same thing as taking the original configuration of $S^3$ with $n$ units of H-flux and quotienting by the freely-acting $\mathbb{Z}_n$ symmetry)?

Posted by: Jacques Distler on June 5, 2006 8:46 AM | Permalink | PGP Sig | Reply to this

### Re: Accidental?

I do agree, but I find it hard to imagine how one generalises such a description to non-geometric backgrounds. Given a Lens space $S^3/Z_n$ with $m$ units of H-flux, I’m genuinely curious how you would describe taking the T-dual.

I add to the general discussion that there is also a formula for the T-dual RR fields in the topological approach’.

DM Roberts

Posted by: David Roberts on June 5, 2006 7:58 PM | Permalink | Reply to this

### Re: Accidental?

I think I can answer that. But, first, we need to agree on what the question means.

To me, “T-duality” is an isomorphism between two conformal field theories. These conformal field theories are formulated as nonlinear $\sigma$-models, with target spaces, $X$ and $X'$, which happen to be torus bundles.

What might a “topological” version of “T-duality” mean?

Well, given two such Riemannian manifolds, $X$, $X'$, we could forget about all the geometrical data (that, in particular, made the corresponding $\sigma$-models conformally-invariant), and retain only topological information (the topology of $X$, the cohomology class of $H$, etc).

We would then declare these two topological spaces (with, perhaps, that extra topological data) to be T-dual.

We might examine the cases we’ve studied, and then find some “general rules” for T-dualizing other topological spaces.

But we’d better be sure that, whatever our definition, it had better agree with the CFT one, whenever there exists a choice of geometrical data on $X$, $X'$ which turns them into CFT backgrounds.

If we find such a definition, and it survives these checks, then we can be confident that our definition of “topological T-duality” is a good one.

Does this seem like a reasonable game-plan to you? Do you have an alternate proposal?

Posted by: Jacques Distler on June 5, 2006 9:10 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental?

I do not think it possible to declare arbitrary spaces to be T-dual.

T-duality is a deeper relation, involving twisted K-theory (which I do not claim to understand in any great detail) - it is really an equivalence of deeper structures, which we happened to study using manifolds - I don’t know was this structure is, but I suspect people like Bunke do.

Really the original goal was to classify the change in topology under T-duality, which was known to occur from the physics (Nilmanifold, nilmanifold…). It is bolstered because it takes the (possibly twisted) K-theory classifying type IIA RR fields to the the (possibly twisted) K-theory classifying IIB fields. I leave the CFT to those who know.

DM Roberts

Posted by: David Roberts on June 5, 2006 10:55 PM | Permalink | Reply to this

### Re: Accidental?

Really the original goal was to classify the change in topology under T-duality, which was known to occur from the physics… I leave the CFT to those who know.

So does that mean that you do, or you don’t care whether the two definitions agree, in those instances where they both apply?

Posted by: Jacques Distler on June 5, 2006 11:30 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental?

you do, or you don’t care whether the two definitions agree

Certainly, one should care.

Still, the game currently being called “topological T-duality” is one where they call any operation an abstract T-duality which has the properties which I listed at the very end of partIII of that little summary.

Some of the people working on this, who I have talked to, said they know little more about the string/CFT-theoretic origin of T-duality than that it exists.

As I have tried to indicate in another comment, I believe there are one or two hints that indicate how what they do could make explicit connection to CFT.

And there are others who try to make the connection. As I mentioned towards the end of part I, there is an upcoming paper by Ellwood and Hashimoto which claims to rederive some of the more surprising results of topological T-duality using genuinely stringy analysis.

While nice, that still leaves room for a general conceptual understanding of why (if at all) topological T-duality actually corectly captures aspects of T-duality in full CFT.

Posted by: urs on June 6, 2006 1:38 PM | Permalink | Reply to this

### Re: Accidental?

Certainly, one should care.

I guess what I’m asking is: if (hypothetically) one found a counterexample, where topological T-duality and (“CFT”) T-duality disagreed, would anyone working on the subject say, “Whoops! We must have a bad definition of topological T-duality.”?

Or would they shrug their shoulders and say: “Conformal Field Theory, what’s that?”

Posted by: Jacques Distler on June 6, 2006 2:00 PM | Permalink | PGP Sig | Reply to this

### Re: Accidental?

Both, I think.

The definitions there are serve the purpose of interesting definitions in math: they lead to nontrivial theorems.

But of course after having a theorem one wants to have an application of it, hence something which satisfies the axioms given in the definitions.

To me the situation looks comparable to what happened with other parts of string theory. There are people studying stability conditions on triangulated categories without direct connection to SCFT motivations. For instance.

Posted by: urs on June 6, 2006 2:22 PM | Permalink | Reply to this

### Re: Accidental?

For the record, if the two definitions disagree, then I’d like to know why, and how we can generalise to fix’ things.

As you say there is more information around, which is why I think Bouwknegt is working on a generalised geometry version of T-duality - certainly there one has metric (and generalised metric) information.

If people manage to construct a counter-example to the topological T-duality using the CFT approach, then the mathematicians will have found enough interesting stuff to leave the string theory behind (as Urs has said). To fix it one would have to think about how one classifies various charges, since T-duality exchanges more than just winding numbers and momenta, but is supposed to be symmetry of the full quantum theory.

Personally I would be surprised at a counter-example, though.

DM Roberts

Posted by: David Roberts on June 7, 2006 12:58 AM | Permalink | Reply to this

### H-flux and noncommutativity

Hi David,

thanks for helping out with information. I now remember that Varghese Mathai has mentioned this, but I would not have properly recalled it.

Maybe you can also help me with something related:

We know from open strings that nontrivial B-fields lead to noncommutativity in the effective field theory.

Also, the $C^*$-algebra used in topological T-duality for the case of nontrivial $H$-flux is noncommutative.

What is the relation?

Maybe it’s obvious, but right now I do not quite see it.

On the one hand we have a Moyal star. On the other hand we have the algebra of sections of a bundle of compact operators, associated with the $PU(H)$-bundle representing the gerbe.

How are both related?

Posted by: urs on June 5, 2006 12:44 PM | Permalink | Reply to this

### Re: H-flux and noncommutativity

I know precious little about such things, but I answer with a question: Is the Moyal star a groupoid product?

I have been reminded to say that $C(X)$ is the continuous functions on $X$, so knows nothing about the differentiable structure. And that the smooth version of $C(X)$ isn’t a $C^*$-algebra, but is Frechet - it only has a family of seminorms. Mathai et al say nothing about such things.

DM Roberts

Posted by: David Roberts on June 5, 2006 7:50 PM | Permalink | Reply to this

### Re: H-flux and noncommutativity

Is the Moyal star a groupoid product?

Sorry, what precisely do you mean by groupoid product? The product in the category algebra of a groupoid, i.e. the convolution product on the algebra of functions on the groupoid’s morphisms?

Is the product of sections of a bundle of compact operators a groupoid product, then?

I should maybe add a comment to my question:

The noncommutative algebra constituted by the sections of a bundle of compacts associated to the $PU(H)$-bundle which represents the $U(1)-gerbe$ depends only on that gerbe without a connection and is noncommutative even if that gerbe is trivial. (But is it equivalent to the commutative algebra of continuous functions in that case?)

On the other hand, the Seiberg-Witten Moyal star product is determined by a nontrivial connection on a trivial gerbe (the constant $\theta \propto (g + B)^{-1}$, where $B$ is the constant $B$-field 2-form and $g$ the metric).

I always had the impression that both of these algebras had to be related, but it is apparently not at all obvious how.

This is a pity, because one would want to know the effective noncommutative field theory on a D-brane in, say, the presence of a nontrivial torsion gerbe, i.e. the effective noncommutative field theory on a gerbe module.

The only place that I am aware of where this is addressed at all is the abstract of hep-th/0206101. But I am not sure if the results of that paper really shed light on this question.

$C(X)$ is the continuous functions on X, so knows nothing about the differentiable structure

Yes, that’s the theorem, that commutative $C^*$-algebras are in bijection with (locally compact Hausdorff) topological spaces.

But of course Jacques is right that more specific function algebras may encode more than topological information, yet still less than metric information.

Posted by: urs on June 6, 2006 4:57 AM | Permalink | Reply to this

### Re: Accidental

then $C(X)$ encodes the differentiable structure of $X$

True, sorry for being dense.

Posted by: urs on June 5, 2006 8:51 AM | Permalink | Reply to this

### Re: Around the Blogs

One of the really nice things about NRQCD is that it gives you Schrodinger problem to work with. There is no computation of Fermi determinants needed at all. Given a set of gauge field configurations, you simply evolve the fermion forward in (Lattice) time via the Schrodinger equation. This is much faster than other approaches.

Posted by: Matthew on June 5, 2006 4:14 AM | Permalink | Reply to this

### Quenched

There is no computation of Fermi determinants needed at all.

Yah. I should have said (or been more clear) that, in the heavy quark limit, the quenched approximation is exact (loops of heavy quarks are negligible). What’s less obvious, at first glance, is the advantage of evolving via the the Schrœdinger versus the Dirac Hamiltonian.

Or maybe, if you’re smart enough, this is all obvious from the 'git go …

Posted by: Jacques Distler on June 5, 2006 9:01 AM | Permalink | PGP Sig | Reply to this

### Re: Around the Blogs

NRQCD isn’t actually used for dynamic simulations (the heavy quark loops are a small effect), so the size of the fermionic determinant doesn’t matter. When people talk of unquenched NRQCD results they mean that the configurations on which the NRQCD quarks were propagated were from an unquenched ensemble (with dynamical light quarks).

The heavy quark propagators in NRQCD can be computed using the Schroedinger equation as an explicit equation for G(t+1) in terms of G(t), which is nicer than having to invert a huge matrix.

Posted by: Georg on June 5, 2006 12:51 PM | Permalink | Reply to this
Read the post Mathai on T-Duality IV: Distler on CFT Checks
Weblog: The String Coffee Table
Excerpt: Jacques Distler comments on CFT checks of topological T-duality.
Tracked: June 10, 2006 6:55 AM

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