Principal failure of canonical quantum gravity (Witten)

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  • #1
tom.stoer
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In the thread https://www.physicsforums.com/showthread.php?t=485800&page=17 posz #257 fzero mentions that
all known approaches to use gauge theory to describe canonical gravity fail nonperturbatively (Witten in http://arxiv.org/abs/0706.3359 "Three-Dimensional Gravity Revisited"):
- gauge theories always include solutions where the vierbein is not invertible (not visible perturbatively)
- gauge theory only describes diffeomorphisms which are connected to the identity
fzero concludes that any attempt to formulate quantum gravity as a gauge theory in which there is a canonical map between degrees of freedom is incomplete

I would like to ask if and how this affects LQG (both in the old-fashioned approach where one uses Ashtekar and loop variables for quantization and in the new approach where one omits quantization in the sense of a construction and postulates spin networks as the kinematical basis for LQG)

In addition I would like to ask how this is different for "ordinary" gauge theory (e.g. in the canonical or BRST approach) as we know that these formulations have similar problems, e.g. singular gauge field confugurations due to Gribov copies etc., visible only non-perturbatively.

Then I would like to ask if this is a problem at all as we typically see such singular configurations already in ordinary quantum mechanics, but usually the system cures this via "repulsive potentials" and "boundary conditions" (orbital angular momentum in hydrogen atom, QCD in canonical formulation with "repulsive" terms due to Jacobian = Fadeev-Popov determinant, ...).
 

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  • #2
fzero
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From the gauge theory perspective, these configurations are not singular: [tex]A=e=\omega=0[/tex] is the simplest classical solution that corresponds to a singular geometry. The problem solutions are completely natural in gauge theory, but very bad once we try to translate back to gravity.
 
  • #3
tom.stoer
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The problem solutions are completely natural in gauge theory ...
agreed!

... but very bad once we try to translate back to gravity.
so what? the metric is a bad choice in these cases, but why should we bother about it?
 
  • #4
Physics Monkey
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I always wondered about this myself. One might imagine that the ability to describe an "uncondensed" phase with [tex] ds^2 = 0 [/tex] or a "phase of nothing" would be a good thing. Perhaps one could then understand how something comes about dynamically.
 
  • #5
fzero
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so what? the metric is a bad choice in these cases, but why should we bother about it?
These are classical configurations which have only a partial or no geometric interpretation at all. They're not solutions to Einstein's equations, since neither the Levi-Civita or curvatures are defined for them ([tex]g_{\mu\nu} =\eta_{\mu\nu} [/tex] is the trivial solution, not [tex]g_{\mu\nu}=0[/tex]). It seems odd to include them in a sum over geometries or to have tunneling between geometric universes and one of these singular configurations.
 
  • #6
tom.stoer
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There is no reason why a sum over geometries or geometry (metric) should be the correct starting point. For standard classical configurations both tetrad- and metric-formulations are strictly equivalent, for the quantum case it seems (for many reasons) that metric variables are useless.

Therefore the conclusion is that the tetrad is the preferred choice

(there is no reason why it should be possible to quantize a classical theory at all; there are many different quantum theories with the same classical limit; but if there is a classical theory w/o quantization then my conclusion is that one should change this classical starting point i.e. get rid of the metric at all)

There are many people out there preferring Einstein-Cartan instead of Einstein-Riemann.
 
  • #7
fzero
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The comment about sum over geometries may be extraneous. But so are the comments about what degrees of freedom are relevant for the quantum case. The fact remains that we are talking about classical solutions of the Chern-Simons theory (or appropriate gauge theory for the 4d case) that are not solutions of GR. Whatever the correct quantum degrees of freedom we use, the classical solutions should agree with GR.
 
  • #8
tom.stoer
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Whatever the correct quantum degrees of freedom we use, the classical solutions should agree with GR.
Of course.

But we know that Einstein-Cartan agrees with GR - up to the spin-density and the induced torsion which are extremely small within matter and exactly zero outside as torsion does not propagate. Therefore Einstein-Cartan is fully compliant with all (experimentally) known GR phenomena and is valid as classical theory of geometry / gravity. And of course we expect that a quantization of Einstein-Cartan reproduces Einstein-Cartan (and therefore GR) in the classical limit. That's why I see no problem in using Einstein-Cartan (i.e. the full tetrad formalism) instead of applying the restricted metric formalism.

Does that mean that Wittens reminds us of an pseudo-problem?
 
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How would you deal with coupling spinors without introducing a tetrad?
 
  • #10
tom.stoer
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I wouldn't do that; the tetrad formalism is perfectly valid, including (and especially for) the coupling spinors. Again: there is no problem with Einstein-Cartan and tetrads.
 
  • #11
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Yes, I agree, my comment wasn't really a response to your post. I was more suggesting that actually the tetrad formalism is necessary to couple spinors to gravity. Also in that sense one might want to take seriously the extra solutions corresponding to degenerate metrics that have no analogue in metric gravity.

Edit: However, I'd be interested to learn more about Witten's (or atyy's) second objection. I heard it for the first time.
 
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  • #12
tom.stoer
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I don' think that the second objection is critical as there are similar (but solvable) problems in gauge theory where the Gauss law generates only "small" gauge transformations.
 
  • #13
fzero
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I don' think that the second objection is critical as there are similar (but solvable) problems in gauge theory where the Gauss law generates only "small" gauge transformations.
It's more complicated than that. The Gauss law constraint is local and never generates large gauge transformations. But the problem in gravity is that the diffeomorphism group is neither locally compact (it's infinite dimensional) nor generated by it's Lie algebra. So it's not clear that it would even be sufficient to consider superselection sectors on the gauge theory side.

As Witten points, out, perturbative string theory uses 2d gravity on the worldsheet in a fundamental way. Both of the issues arise when considering the consistency of the theory. First, one needs a nondegenerate worldsheet metric to perform the integration over the moduli space of Riemann surfaces. Furthermore, modular invariance (one class of "large" diffeomorphisms) is crucial to avoid gravitational anomalies. So it's pretty clear that in 2d gravity only considering diffeomorphisms connected to the identity leads to an inconsistent quantum theory.

I was looking at Witten's '88 paper and he refers to a paper of Regge about degenerate metrics: http://inspirebeta.net/record/167638?ln=en I don't have access to the paper, but it might shed some light on consistency.
 
  • #14
tom.stoer
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What are the generators or the Lie algebra of diffeomorphisms in general (I mean not the special cases for space-time foliations a la Wheeler-deWitt or Ashtekar)? Why should the diffeomorphism group be infinite dimensional? I agree that it's the case in 2d, but why for d>2?

Are you referring to the Cr diffeomorphisms which are (in a certain topology) locally homeomorphic to Cr vector fields which can be seen as a Banach space? That would mean that the infinite dimensional Banach space induces an infinite dimensional diffeomorphism group; is it that what you are saying? But what about the following: certain (smooth) functions on S³ are of course members of an infinite dimensional function space, nevertheless SU(2)~S³ is a finite-dimensional Lie group. I do not see how the dimension of the function space and the dimension of the group can be related directly.
 
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  • #15
fzero
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What are the generators or the Lie algebra of diffeomorphisms in general (I mean not the special cases for space-time foliations a la Wheeler-deWitt or Ashtekar)?
The infinitesimal diffeomorphisms are associated with vector fields. The Lie algebra is identified with the space of all vector fields on the manifold. This is an infinite dimensional space.

Why should the diffeomorphism group be infinite dimensional? I agree that it's the case in 2d, but why for d>2?
As you say below, we're dealing with a function space that admits a Banach structure. Also note that the size of [tex]\text{Diff}(M)[/tex] doesn't really get smaller as we go higher in dimension. For example, it's obvious that [tex]\text{Diff}(T^n)[/tex] ([tex]T^n[/tex] being the n torus) contains [tex]\text{Diff}(S^1)[/tex]. We could similarly consider the subgroups of [tex]\text{Diff}(M)[/tex] that preserve some submanifold [tex]S[/tex]. We're not going to suddenly conclude that [tex]\text{Diff}(M)[/tex] suddenly becomes a finite-dimensional space.

Are you referring to the Cr diffeomorphisms which are (in a certain topology) locally homeomorphic to Cr vector fields which can be seen as a Banach space? That would mean that the infinite dimensional Banach space induces an infinite dimensional diffeomorphism group; is it that what you are saying? But what about the following: certain (smooth) functions on S³ are of course members of an infinite dimensional function space, nevertheless SU(2)~S³ is a finite-dimensional Lie group. I do not see how the dimension of the function space and the dimension of the group can be related directly.
[tex]\text{Diff}(M)[/tex] has a group structure of composition of functions. Any group structure on [tex]M[/tex] does not change the fact that [tex]\text{Diff}(M)[/tex] is infinite dimensional.

When [tex]M[/tex] is a group, we might be able to make some more precise statements. For example, if we denote the Lie algebra of [tex]\text{Diff}(S^1)[/tex] by [tex]\text{diff}(S^1)[/tex], then we can say that the elements of [tex]\text{diff}(S^1)/S^1[/tex] are linear combinations of [tex]f_m = \cos mt[/tex] and [tex]g_m = \sin mt[/tex]. However this is still an infinite dimensional space.
 
  • #16
tom.stoer
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I think my example with S³ was missleading; I'll try it again:

The dimension of the vector space of C-functions on S³ is infinite-dimensional. But if we look at the Lie group SU(2) we can write its elements as

[tex]U[f] = \exp\left\{i t^a f^a(x)\right\}[/tex]

The functions fa are members of an infinite dimensional vector space, but nevertheless we talk about a finite dimensional group b/c we have finitly many generators ta.

Looking at the Ashtekar formulation of gravity there are the Gauss-law generators Ga(x) and the spacelike diffeomorphism generators = vector-constraints Va(x). The transformations are generated via functionals like

[tex]G[f] = \int d^3x\,G^a(x)\,f^a(x);\quad V[f] = \int d^3x\,V^a(x)\,f^a(x);\quad[/tex]

Again we have finitly many generators even if the functions f belong to an infinite dimensional vector space.
 
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  • #17
fzero
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I think my example with S³ was missleading; I'll try it again:

The dimension of the vector space of C-functions on S³ is infinite-dimensional. But if we look at the Lie group SU(2) we can write its elements as

[tex]U[f] = \exp\left\{i t^a f^a(x)\right\}[/tex]

The functions fa are members of an infinite dimensional vector space, but nevertheless we talk about a finite dimensional group b/c we have finitly many generators ta.
Diffeomorphisms which are infinitesimally close to the identity look like gauge transformations. The problem is that infinitesimal diffeomorphisms do not generate the whole group.

Looking at the Ashtekar formulation of gravity there are the Gauss-law generators Ga(x) and the spacelike diffeomorphism generators = vector-constraints Va(x). The transformations are generated via functionals like

[tex]G[f] = \int d^3x\,G^a(x)\,f(x);\quad V[f] = \int d^3x\,V^a(x)\,f(x);\quad[/tex]

Again we have finitly many generators even if the functions f belong to an infinite dimensional vector space.
If there are a finite number of generators, we are not talking about the full diffeomorphism group, only the infinitesimal diffeomorphisms.
 
  • #18
tom.stoer
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Yes, I agree that these are only the "small" diffeomorphisms, but that is the same restriction as in gauge theory. I still do not understand why large gauge transformations are fine but large diffeomorphisms should be problematic.
 

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