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Is string theory local?

+ 14 like - 0 dislike
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By locality I mean something like the Atiyah-Segal axioms for Riemannian cobordisms (see e.g. http://ncatlab.org/nlab/show/FQFT). I.e. to any (spacelike) hypersurface in the target we associate a Hilbert space and to any cobordism an S-matrix.

I am familiar with the S-matrix prescription for the target being $\mathbf{R}^n$ and the hypersurfaces being the asymptotic time infinities. Can one extend that to any cobordism?

Does locality appear only when we integrate over the worldsheet conformal structures and sum over all genera, or can we see it even for a fixed conformal structure?

I believe this is what string field theory is about, but why would one expect locality from the (perturbative) string theory point of view?


This post has been migrated from (A51.SE)

asked Oct 10, 2011 in Theoretical Physics by Pavel Safronov (1,115 points) [ revision history ]
edited Apr 20, 2014 by dimension10
@Urs Please read what I said. I did not say asking if string theory has a space-time S-matrix is not reasonable. It is reasonable, in fact string theory was essentially only a space-time S-matrix theory for the first 25 or so years of its existence. What I did say, however, is that Pavel's definition of "local" is not the standard definition of local. Thus, if his question found an answer, the answer would not necessarily have a bearing on the question as to if string theory is local.

This post has been migrated from (A51.SE)
@KellyDavis, I had an impression that the TQFT functor associates the time-evolution operator to a (Riemannian) cobordism, and then one can pair it with in/out states to compute the S-matrix. Also, Costello's definition would not apply to string theory directly, it only says something about the locality (of the sigma-model) on the worldsheet, while I am interested in the locality on the target.

This post has been migrated from (A51.SE)
You have to be careful. Axiomatic TQFT do not require a metric be present. Defining an S-Matrix requires some notion of an "infinite past" and and "infinite future" both of which require a metric to define. Also, an S-Matrix is unitary. Not all axiomatic TQFT are unitary. Costello's definition does apply to string theory, world-sheet and space-time, think of the equations the imposed on the space-time fields by conformal invariance. (See GSW section 3.4)

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@KellyDavis: Pavel is quite right, cobordism representations do not just serve to axiomatize TQFT, but also QFT with metric structure, notably CFT, as first emphasized by Graeme Segal. This is quite standard by now. References are here: http://ncatlab.org/nlab/show/conformal+field+theory#FQFTReferences . The morphism associated by such a metric QFT to a cobordism is indeed effectively the S-matrix. The question whether string field theory falls into this pattern is perfectly reasonable (even though the answer might be: no, it does not).

This post has been migrated from (A51.SE)
You realize your definition of locality is non-standard. (See for example, Definition 5.0.1 of Costello's book "Renormalization and Effective Field Theory" for a more standard definition.) Also, the Atiyah-Segal axioms don't say anything about an S-Matrix. (See for example, the entry by Turaev and Blanchet in the "Encyclopedia of Mathematical Physics", Elsevier for Axiomatic TQFT.)

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1 Answer

+ 16 like - 0 dislike

String theory as we know it admits only S-matrix as an observable. By its definition an S-matrix is a non-local object, it tells you about transition amplitudes between asymptotic states in past and future infinity. You cannot even ask local questions in spacetime, unless you somehow extend the formalism (which is the goal of string field theory, more about this below).

This is not (in my opinion) a quirk of the formalism. String theory is a quantum theory of gravity, and at long distances it coincides with General Relativity. GR also does not allow for local observables. Mathematically it is because there are no local diffeomorphism invariant quantities. Physically it is because there is no systematic way to locally probe the system without disturbing it — to construct a localized classical probe (a.k.a. measuring device) you’d want it to be very massive (i.e. has many degrees of freedom) to suppress quantum fluctuations. Alas, if it couples to gravity it then back-reacts on the geometry. If gravity couples weakly you can construct approximately localized probes, but this does not work in the fully quantum gravitational regime.

All of this does not mean the theory is not local, just that you have to be careful how to phrase the question and make sure it makes sense. There are a few indications that if you ask the question the right way string theory is local in some sense. Two such indications that come to mind:

  1. For a local QFT, the S-matrix obeys certain properties which follow from locality. Turns out string theory obeys those as well. This of course does not, strictly speaking, imply that string theory is local, but it is an indication that it is not obviously non-local.

  2. In extension of the formalism, like open SFT, in some sense interactions are local on the worldsheet - strings interact only when they touch in spacetime. Opinions vary on what this means, FWIW in my mind SFT is inherently perturbative, and for perturbative gravity perhaps it is not surprising one can construct quasi-localized probes. In any event, this is not (I don't think) a gauge invariant statement, so it cannot be made as sharp as one would have liked.

As for your specific question: In perturbative string theory, only after integrating over conformal structures you have a chance of getting objects which make sense physically, which are the S-matrix elements. If you fix the conformal structure you get objects which cannot be interpreted physically (e.g. they have "ghosts", negative norm states in the Hilbert space).

Specifically: probes of the theory are realized as punctures on the worldsheet, and conformal invariance (achieved by integration over conformal structures) pushes their location in spacetime to asymptotic null infinity. Heuristically, this is because a puncture is conformal to an infinitely long tube emanating from the worldsheet . Less heuristically, conformal invariance on the vertex operator inserted at the puncture (which expresses the specific probe of the theory) translates to mass-shell conditions in spacetime (which is the Fourier transform of the previous statement). Since you don't have all the Fourier modes of your probe, there is no way to localize it in spacetime.

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answered Oct 10, 2011 by Moshe (2,375 points) [ no revision ]
Added a comment, hope it helps.

This post has been migrated from (A51.SE)
Obvious caveat: I am a physicist, so when I mention locality I refer to the common usage of the term by theoretical physicists, which may well be different from what is meant by the term in other context, e.g. axiomatic topological QFT.

This post has been migrated from (A51.SE)
One more question. Do we have a Hilbert space only for something like R^9, or can one compute it for, say, T^9? Similarly, is there an "S-matrix" for the spacetime being T^9 x [0,1]?

This post has been migrated from (A51.SE)
You need asymptotic infinity to define an S- Matrix, so if space is compact no asymptotic Hilbert space is defined. What are the ST observables in such situation is a subtle issue. Of course you can partially comactify to leave room for some null infinity.

This post has been migrated from (A51.SE)
Thanks! Can you explain a little bit what you mean by your last sentence?

This post has been migrated from (A51.SE)

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