# LQG and diffeomorphism group cocycles

51 views

Mar 9, 2004, 5:42:29 PM3/9/04
to
,
the signature Urs (Schreiber?) suggests that LQGists should
look for 2-cocycles of the diffeomorphism group. More precisely,
he writes

"Here by quantization of the group I of course mean a set of
operators U(phi) that satisfy the classical group algebra up
to quantum corrections, something like U(phi)U(psi) = U(phi
o psi)V, where V is a quantum correction. For instance for
the quantization of the Virasoro algebra V would be a phase
factor that comes from the anomaly in the Virasoro algebra.

But of course there is a big problem: So far in the LQG-like
quantization of gravity people have simply set V=1
identically (for the spatial diffeomorphisms or for all
group elements in Thiemann's LQG-string). I am convinced
that this cannot possibly be the right choice because it
amounts to eliminating all quantum effects whatsoever.
Thiemann's string shows how very different this choice is
from the usual V=phasefactor choice.

I believe that LQG-like quantization of gravity could make
sense for some V=complicated correction. Maybe.

But how should we find this V? And is anyone looking for it?

[...]

Ok, so it seems that I am proposing a new LQG program:

Find a V=something quantum correction to the diffeomorphism
group such that a sensible semiclassical limit is obtained. "

The condition U(phi)U(psi) = U(phi o psi)V means by definition
that V is a group 2-cocycle; the cocycle condition expresses
associativity, or the Jacobi identity on the Lie algebra level.
One place where one may find (all) candidate V's is in Askar
cocycles of the diffeomorphism group:

A. Dzhumadil'daev, Virasoro type Lie algebras and deformations,
Z. Phys. C 72 (1996) 509.

I spent a considerable time to understand this paper and
translate it into a formalism that I could understand. Since
this was a considerable effort, I wrote it up and put it on the
web: http://www.arxiv.org/abs/math-ph/0002016 .

To integrate the Lie algebra cocycle to a group cocycle is a
secondary problem, which I believe has been successfully
completed in some cases. Since all group cocycles give rise to
Lie algebra cocycles for phi and psi close to the identity, it

### Urs Schreiber

Mar 10, 2004, 7:13:02 AM3/10/04
to
"Thomas Larsson" <thomas_l...@hotmail.com> schrieb im Newsbeitrag

> the signature Urs (Schreiber?)

I'll admit or deny authorship after having seen the reaction of the
experts... ;-)

> suggests that LQGists should
> look for 2-cocycles of the diffeomorphism group.

So is this exactly what you are concerned with?

> But of course there is a big problem: So far in the LQG-like
> quantization of gravity people have simply set V=1
> identically

I think that there is also another big problem: Just knowing the symmetry
group doesn't tell you all the physics. The actions

S1 = int sqrt(g)

and

S2 = int sqrt(g) + int B

and

S3 = int sqrt(g) R

have the diffeo group as symmetry group, but S1 describes a brane in a
gravitational background, S2 a brane in a gravitational+ 2-form (B)
background and S3 describes GR.

Claiming that all these systems can be 'quantized' by just finding some
operator rep of a quantization of the symmetry group means ignoring
essentially all physical content.

That's why I am wondering how Thomas Thiemann is going to deal with the
'LQG-string' in nontrivial backgrounds, as he has announced in his paper.

### Creighton Hogg

Mar 10, 2004, 12:27:21 PM3/10/04
to

So his paper on quantizing the bosonic string without the critical
dimension or anomalies might not be as big of a deal as it sounds?
I started trying to read it almost as soon as it hit arXiv, but it's a
little too advanced mathematically for me and I have no way to judge his
results.

Mar 12, 2004, 6:32:11 AM3/12/04
to
Urs Schreiber <Urs.Sc...@uni-essen.de> skrev i
diskussionsgruppsmeddelandet:404f064e$1...@news.sentex.net... > > suggests that LQGists should > > look for 2-cocycles of the diffeomorphism group. > > So is this exactly what you are concerned with? Yes. > > > But of course there is a big problem: So far in the LQG-like > > quantization of gravity people have simply set V=1 > > identically > > I think that there is also another big problem: Divide and conquer: solve one big problem at a time. > Just knowing the symmetry > group doesn't tell you all the physics. This may or may not be true. For 2D conformally invariant systems, the Ward identities fix all multipoint correlation functions, which more or less is all there is to physics. You can interpret the usual point moving argument along the same line: proper diff invariance is such a powerful symmetry that the Ward identities make physics trivial, no matter what the action is. Locally trivial, at least. > Claiming that all these systems can be 'quantized' by just finding some > operator rep of a quantization of the symmetry group means ignoring > essentially all physical content. Come on! You proposed a research program for other people to carry out, so you must think that there is some point to it, no? Understanding what constraints the Ward identities impose is an important part of physics. Even if this does not tell you all the physics, it tells you some of it, which is not too bad in QG. Anyway, I find it amusing that, when left to yourself, you came up with essentially the same idea as myself. :-) > > That's why I am wondering how Thomas Thiemann is going to deal with the > 'LQG-string' in nontrivial backgrounds, as he has announced in his paper. That LQG seems incapable to get any anomaly whatsoever makes me seriously concerned. An anomaly may or may not be inconsistent, depending on context, but it is genuine quantum behaviour. One may wonder how quantum LQG really is. ### Aaron Bergman unread, Mar 12, 2004, 6:32:12 AM3/12/04 to In article <Pine.LNX.4.44.04031...@dill.hep.wisc.edu>, Creighton Hogg <wch...@hep.wisc.edu> wrote: > So his paper on quantizing the bosonic string without the critical > dimension or anomalies might not be as big of a deal as it sounds? > I started trying to read it almost as soon as it hit arXiv, but it's a > little too advanced mathematically for me and I have no way to judge his > results. As best I can tell from observing Urs's and Jacques Distler's conversations about it, Thiemann claims that he can quantize any classical theory preserving all its classical symmetries. This is false. Not only is it false, its falsity is confirmed by experiment. Anomalous global symmetries have well-known physical consequences. To steal Jacques's example, classical YM is scale invariant. If you could quantize it while preserving this symmetry, you would never see the beta function. Hell, you would screw up pion decay without an anomaly. So, if the LQG quantization doesn't see anomalies, I can't help but conclude that it is, at best, something very different from any sort of quantization that has been used before, and, at worst, just outright wrong. Now, all this is rather surprising to me because I just can't believe that the people doing this don't know what an anomaly is. So, I figure there must be a response to this line of thinking [1]. I'm quite curious to know what it is. Aaron [1] I've talked to people who have said that this is (one of?) the sticking point(s) in Carlo Rovelli's infamous seminar of your, ie, you can't see the beta function in this approach to 'quantization'. ### Urs Schreiber unread, Mar 12, 2004, 7:05:18 AM3/12/04 to "Aaron Bergman" <aber...@physics.utexas.edu> schrieb im Newsbeitrag news:abergman-747D7B.21474910032004@localhost... > So, if the LQG quantization doesn't see anomalies, I > can't help but conclude that it is, at best, something very different > from any sort of quantization that has been used before, [...] > So, I figure > there must be a response to this line of thinking [1]. I'm quite curious > to know what it is. From our discussion with Thomas Thiemann I think it is clear that indeed - something very different from any sort of quantization that has been used before is proposed and that - while this new 'form of quantization' is clearly physically wrong for all systems that we have experimental access to the claim/hope is that it is after all the right 'form of quantization' for _quantum gravity_. This is at least what Thomas Thiemann said: " In any case, whether or not you should have an exact or projective rep. of a symmetry depends on the physical system under study and hence must be decided ultimately by experiment. In case of the string everything is allowed until we have quantum gravity experiments." (from http://golem.ph.utexas.edu/string/archives/000299.html#c000588) Sometimes LQG papers like gr-qc/0207106 are mentioned which supposedly show that the LQG-like 'form of quantization' reproduces ordinary quantization in some limit. But having a closer look at this paper shows that this _only_ works when the usual quantum corrections are copied to the LQG-formalism (lower half of p.14). This is however not the case in the 'LQG-string' or the LQG treatment of the spatial diffeo constraints of 3+1d gravity. ### Creighton Hogg unread, Mar 12, 2004, 10:47:40 AM3/12/04 to On 12 Mar 2004, Urs Schreiber wrote: > > "Aaron Bergman" <aber...@physics.utexas.edu> schrieb im Newsbeitrag > news:abergman-747D7B.21474910032004@localhost... > > > So, if the LQG quantization doesn't see anomalies, I > > can't help but conclude that it is, at best, something very different > > from any sort of quantization that has been used before, > > [...] > > > So, I figure > > there must be a response to this line of thinking [1]. I'm quite curious > > to know what it is. > > From our discussion with Thomas Thiemann I think it is clear that indeed > > - something very different from any sort of quantization that has been used > before > > is proposed and that > > - while this new 'form of quantization' is clearly physically wrong for all > systems that we have experimental access to the claim/hope is that it is > after all the right 'form of quantization' for _quantum gravity_. Wait, it seems like this would be a pretty strong argument *against* LQG? What's the reason for believing that this new form of quantization would work for quantum gravity? Note that I'm not trying to pick a fight or start arguments. I really just want to hear the reasoning of people who work in LQG. ### Urs Schreiber unread, Mar 12, 2004, 11:22:23 AM3/12/04 to "Creighton Hogg" <wch...@hep.wisc.edu> schrieb im Newsbeitrag news:Pine.LNX.4.44.04031...@dill.hep.wisc.edu... > Wait, it seems like this would be a pretty strong argument *against* LQG? Yes, that's what I think, too. Papers like gr-qc/0207106 have been written in order to show that this is not an argument against LQG, but, as I have tried to point out, I don't think that the 'shadow states' technique works when no quantum corrections are taken into account (which corresponds to setting the alpha^2 term to zero in equation (IV.5) of that paper). > What's the reason for believing that this new form of quantization would > work for quantum gravity? As far as I understand this is a result of the historical development: When writing down GR in terms of the connection variable and constructing the space of 'generalized Wilson loop observables' (spin networks) for this connection, one finds that the operator 'multiply by the connection' (which is the canonical coordinate) is not well defined. This means that the ordinary procedure of canonical quantization, which tells you to represent canonical coordinates and momenta as operators on some Hilbert space, fails, because half of them cannot be represented as operators at all. This led people to try a modification of the canonical quantization procedure. If the Heisenberg algebra cannot be represented, at least the Weyl algebra can. So they stick to that. But doing that looses contact with standard quantization. Furthermore, it introduces ambiguities over and above those of ordinary 1st quantization. For instance note that in equation (IV.5) of gr-qc/0207106 the quantum correction V = exp(-alpha^2/2) is introduced by hand in order to mimic the standard quantization procedure. Any other value of this correction would yield an LQG-like quantization, too, but have no limit in which the standard quantum theory were reproduced. In particular, using the factor V = 1 instead would mean to copy the classical algebra to the quantum theory. This is indeed what is done by Thomas Thiemann in the 'LQG-string' paper and what is also done for the spatial diffeomorphism constrains in the LQG quantization of 3+1d gravity. This cannot possibly capture any quantum effects, like anomalies. > Note that I'm not trying to pick a fight or start arguments. I really > just want to hear the reasoning of people who work in LQG. Unfortunately Thomas Thiemann has left our discussion at the Coffee Table at some point. http://golem.ph.utexas.edu/string/archives/000299.html#c000588 . I felt that, while we did manage to all agree on what happens at the technical, mathematical level in his paper, the disagreement about the physical viability of the 'LQG-string' couldn't be bridged and Thomas Thiemann left with the impression that we are having 'religious' objections: http://golem.ph.utexas.edu/string/archives/000299.html#c000569 . That's very unfortunate. ### Urs Schreiber unread, Mar 12, 2004, 11:23:07 AM3/12/04 to "Creighton Hogg" <wch...@hep.wisc.edu> schrieb im Newsbeitrag news:Pine.LNX.4.44.04031...@dill.hep.wisc.edu... > Wait, it seems like this would be a pretty strong argument *against* LQG? Yes, that's what I think, too. Papers like gr-qc/0207106 have been written in order to show that this is not an argument against LQG, but, as I have tried to point out, I don't think that the 'shadow states' technique works when no quantum corrections are taken into account (which corresponds to setting the alpha^2 term to zero in equation (IV.5) of that paper). > What's the reason for believing that this new form of quantization would > work for quantum gravity? As far as I understand this is a result of the historical development: When writing down GR in terms of the connection variable and constructing the space of 'generalized Wilson loop observables' (spin networks) for this connection, one finds that the operator 'multiply by the connection' (which is the canonical coordinate) is not well defined. This means that the ordinary procedure of canonical quantization, which tells you to represent canonical coordinates and momenta as operators on some Hilbert space, fails, because half of them cannot be represented as operators at all. This led people to try a modification of the canonical quantization procedure. If the Heisenberg algebra cannot be represented, at least the Weyl algebra can. So they stick to that. But doing that looses contact with standard quantization. Furthermore, it introduces ambiguities over and above those of ordinary 1st quantization. For instance note that in equation (IV.5) of gr-qc/0207106 the quantum correction V = exp(-alpha^2/2) is introduced by hand in order to mimic the standard quantization procedure. Any other value of this correction would yield an LQG-like quantization, too, but have no limit in which the standard quantum theory were reproduced. In particular, using the factor V = 1 instead would mean to copy the classical algebra to the quantum theory. This is indeed what is done by Thomas Thiemann in the 'LQG-string' paper and what is also done for the spatial diffeomorphism constrains in the LQG quantization of 3+1d gravity. This cannot possibly capture any quantum effects, like anomalies. > Note that I'm not trying to pick a fight or start arguments. I really > just want to hear the reasoning of people who work in LQG. Unfortunately Thomas Thiemann has left our discussion at the Coffee Table at ### Urs Schreiber unread, Mar 12, 2004, 12:31:40 PM3/12/04 to "Thomas Larsson" <thomas_l...@hotmail.com> schrieb im Newsbeitrag news:24a23f36.04031...@posting.google.com... > Urs Schreiber <Urs.Sc...@uni-essen.de> skrev i > diskussionsgruppsmeddelandet:404f064e$1...@news.sentex.net...
> > > suggests that LQGists should
> > > look for 2-cocycles of the diffeomorphism group.
> >
> > So is this exactly what you are concerned with?
>
> Yes.

IIRC these cocycles will not be scalar (i.e. proportional to the identity),
right?

following point:

Can we check how these cocylces look like in string theory? In Matrix Models
(BFSS or IKKT) the diffeomorphism group on spacetime becomes a part of the
overall U(N>>1) symmetry. I have only a very vague idea of how this works
(One way to see it is that the permutation subgroup of U(N) permutes matrix
eigenvalues and hence discrete spacetime points.) but can we read off the
quantum corrections to the diffeomorphism group from that? Is that a naive
question? Is the N->oo limit a problem?

> > Just knowing the symmetry
> > group doesn't tell you all the physics.
>
> This may or may not be true. For 2D conformally invariant
> systems, the Ward identities fix all multipoint correlation
> functions,

But if I handed you only the abstract Virasoro algebra and ask you "Compute
graviton scattering from that!", you'd rightly complain that I did not give
away all the necessary information.

> > Claiming that all these systems can be 'quantized' by just finding some
> > operator rep of a quantization of the symmetry group means ignoring
> > essentially all physical content.
>
> Come on! You proposed a research program for other people to
> carry out, so you must think that there is some point to it, no?

:-) I was trying to make constructive criticism. After the flaw was spotted
the idea was to think about how to alleviate it. At least I understand now
why it should be interesting to know the possible cocycles of the
diffeomorphism group.

But I think the problem is obvious: Assuming that you and others have
already solved this problem. This alone is like knowing that SO(d-1,1) plays
a role in special relativity. Very good. Now which role precisely? What are
the degrees of freedom of the theory? What is the dynamics? How do we find
it from knowing just the symmetry group alone?

> Anyway, I find it amusing that, when left to yourself, you came
> up with essentially the same idea as myself. :-)

Hehe. But note that I would sell this idea differently. I wouldn't claim
that knowing the possible cocycles of the diffeomorphism group means having
solved quantum gravity and I wouldn't claim that it implies that string
theory is wrong! :-)

### Josh Willis

Mar 12, 2004, 5:06:20 PM3/12/04
to
Well, since Urs keeps referring to this paper I co-authored, I feel I
ought to respond with what I see.

So I will start by replying to the remarks about gr-qc/0207106, and
then work backwards towards some more general remarks about the
relationship between group averaging and anomalies, as I see it. This
means I take the quoted article in reverse order. My apologies if this
is confusing, but I think in the long run it will be clearer.

Urs Schreiber wrote:
> Sometimes LQG papers like gr-qc/0207106 are mentioned which
supposedly show
> that the LQG-like 'form of quantization' reproduces ordinary
quantization in
> some limit. But having a closer look at this paper shows that this _only_
> works when the usual quantum corrections are copied to the LQG-formalism
> (lower half of p.14). This is however not the case in the 'LQG-string' or
> the LQG treatment of the spatial diffeo constraints of 3+1d gravity.

I do not understand your assertion that the "quantum
corrections" are copied into the model system we study, and that this
is somehow very different from what is done with, for instance, the
diffeomorphism constraint in LQG.

So, for people who haven't read page 14 of gr-qc/0207106, let me
summarize quickly what is going on.

And let me start by saying that page 14 of that paper is the wrong
place to be looking to begin with: there we are finding candidate
semiclassical states, not considering commutation relations. Instead,
look at pages 7 and 8, to start.

We are looking at the one-dimensional point particle in quantum
mechanics. We have not yet considered any particular Hamiltonian, but
are instead at this point in the paper looking only at the canonical
commutation relations.

One normally learns these as:

{q,p} = 1

and then seeks in the quantum theory for self-adjoint operators
\hat{q} and \hat{p} that satisfy the corresponding relations:

[\hat{q},\hat{p}] = i\hbar

But these commutation relations among the p's and q's of course imply
commutation relations among the exponentiated operators, which---if
\hat{q} and \hat{p} are self-adjoint---will be unitary. This is the
relation:

(*) U(l)V(m) = e^{-ilm}V(m)U(l)

where:

U(l) = exp{il\hat{q}}
V(m) = exp{im\hat{p}}

The relation (*) is that of the Weyl-Heisenberg algebra, and what we
do in our paper is look at a non-standard unitary representations of
this algebra, rather than looking at a self-adjoint representation of
the CCR. The particular representation we look at is motivated by
analogy with LQG, and the main point of the paper is to look at the
low-energy limit of this theory, using techniques that it is hoped
will be relevant for LQG. However, as that does not seem to be what
the questions are about at the moment, I will not say more about that
here.

And this is the key point: the relation (*)---including the factor of
e^{-ilm}, which is what in other places on the net Urs seems to think
is put in by hand, i.e. by "knowing" what the quantum theory should
be---is in fact dictated *classically* by the Poisson algebra of the
basic observables. Any representation by unitary operators which
purports to be a canonical quantization *must* have this relationship,
and in particular the representation we consider does (as of course
also does the standard Schroedinger representation).

How does one see this? By (as Urs has mentioned elsewhere) the
Campbell-Baker-Hausdorff theorem, which lets me calculate what (*)
should be, entirely in terms of iterated commutators of p's and q's.
For the CCR, this is in fact the easiest way to go, because the CBH
series stops after the first commutator.

But for more complicated Lie algebras, as for example the
diffeomorphism algebra in gravity, the CBH series does not terminate
and this is not a terribly enlightening way to view the problem.

But it's also not necessary, because a self-adjoint representation of
a Lie algebra exponentiates to a unitary representation of the Lie
group. So what one should check instead is that one has an honest
group representation, that is, that:

U(g_1)U(g_2) = U(g_1 * g_2)

And this *does* hold for the diffeo group in LQG. The
noncommutativity of the Poisson brackets is represented by the fact
that for two diffeos phi_1 and phi_2, in general phi_1*phi_2 does not
equal phi_2*phi_1, and so

U(phi_1)U(phi_2) != U(phi_2)U(phi_1)

So, to put it another way, we could have approached the point particle
thus: "The CCR make a Lie algebra under Poisson brackets. Rather than
looking for a self-adjoint representation of this *algebra*, let's
look for a unitary representation of the corresponding *group*."

For the one-dimensional point particle, the group corresponding to the
CCR is of course the three-dimensional Heisenberg group, so we would
be looking for unitary representations of that. There are at least a
couple of reasons wy we didn't approach it this way in the paper:

(1) Part of the point (though not the main one) of this paper was to
make some LQG constructions more accessible, by comparing them to
things people already know. And very few people think "Oh, quantizing
the point particle, that's all about finding unitary representations
of the Heisenberg group." The ones that do wouldn't have much trouble
with the existing LQG papers, I think.

(2) More importantly, we were aiming for something more specific. I
representations. If the group rep is continuous, they are there, as
for example in the normal Schroedinger rep. In the rep we consider,
there is a self-adjoint generator for the "U" operators, giving rise
to a position operator, but there is not one for the "V" operators
(which would be momentum, if it existed). The reason for this
asymmetry is to mimic what happens in LQG, where there is a
but instead only an "exponentiated" operator in the form of the
holonomy. This asymmetry is very strange if all I'm thinking about is
unitary representations of the Heisenberg group from an abstract point
of view.

You might notice that I haven't talked at all about group averaging in
any of this, which brings me back to the second main thing I wanted to
say:

> "Aaron Bergman" <aber...@physics.utexas.edu> schrieb im Newsbeitrag
> news:abergman-747D7B.21474910032004@localhost...
>
>
>>So, if the LQG quantization doesn't see anomalies, I
>>can't help but conclude that it is, at best, something very different
>>from any sort of quantization that has been used before,
>
>
> [...]
>
>
>>So, I figure
>>there must be a response to this line of thinking [1]. I'm quite curious
>>to know what it is.

[I realize that group averaging is not explicitly mentioned above, but
it has come up in this context in other posts on SPR as well as some
other internet forums]

I don't know of course what Thomas would say, but I think there's some
misunderstanding as to what group averaging is, and what it is used
for.

As I indicated above, and as I'm sure Aaron and Urs both know, when I
have a classical symmetry captured in a constraint algebra, then I
have an anomaly when the corresponding operators in the quantum theory
terms, higher order in \hbar. What this means at the group level is
that I do not have a unitary representation of the symmetry group.

So, how does group averaging allow you to construct a unitary
representation for any classical symmetry group?

Group averaging *starts* with a unitary representation, and goes from
there: if you don't have one to begin with, then it can't help you.

See, the point of group averaging is to construct states that are
invariant under the symmetry. You begin with a Hilbert space of
states on which you have some operators, among them operators
corresponding to your constraints, or at least exponentiated versions
of them. The states in this Hilbert space are not annihilated by your
constraints, and thus not gauge invariant, or, if you have only
unitary group operators, they are not left invariant by these.
However, *if* you have a unitary group rep, group averaging will try to
construct for you another Hilbert space of gauge invariant states.
The basic idea is simple; if |psi> is any state in the non-gauge
invariant Hilbert space, then

\int_G U(g)|psi> dg

should be a gauge invariant state, where G is the whole gauge group
and dg is a bi-invariant measure on G. Obviously, even if one has a
unitary group rep, one can see lots of places where this simple idea
can run into trouble, starting with finding an invariant measure, and
then worrying about the convergence of the integral. The devil is in
the details, and various tricks must often be employed, and naturally
these won't always work, thought they do for the diffeo group in LQG.
I'd recommend the paper:

Abhay Ashtekar, Jerzy Lewandowski, Donald Marolf, Jose Mourao,
Thomas Thiemann, "Quantization of diffeomorphism invariant theories
of connections with local degrees of freedom." gr-qc/9504018.

This is the paper where the construction of the diff-invariant Hilbert
space of LQG is first carried out, but it also has several simpler
examples of group averaging first, as well as some useful references.

But my main point is the following: group averaging has essentially
nothing to do with whether or not you have an anomalous symmetry
representation, since that is equivalent to whether or not you have a
unitary group rep. Group averaging starts from such a rep, it does
not give you one. In the case of the diffeo group in LQG, one can
explicitly check that the Hilbert space used, which carries an
appropriated rep of the Weyl-Heisenberg algebra and other observables,
like the area operator, does in fact have a unitary rep of the diffeo
group as well. But there's no claim that for any classical system
with a Lie algebra of symmetries, there must exist some quantum rep in
which the corresponding group action is unitarily implemented. Even if
there is such a rep, there can of course be other pathologies of the
quantization.

--Josh

### DickT

Mar 12, 2004, 5:07:36 PM3/12/04
to
Aaron Bergman <aber...@physics.utexas.edu> wrote in message news:<abergman-747D7B.21474910032004@localhost>...

>
> As best I can tell from observing Urs's and Jacques Distler's
> conversations about it, Thiemann claims that he can quantize any
> classical theory preserving all its classical symmetries.
>
I think this is misunderstanding Urs and Jacques. The "quantize
everything" comes from their examples of what they think Thiemann's
claim would entail. He himself only claims to quantize the Nambu-Goto
action on the closed bosonic string.

### Urs Schreiber

Mar 13, 2004, 9:31:06 AM3/13/04
to
"Josh Willis" <jwi...@gravity.psu.edu> schrieb im Newsbeitrag
news:c2slkk$g2k$1...@f04n12.cac.psu.edu...

> Well, since Urs keeps referring to this paper I co-authored, I feel I
> ought to respond with what I see.

> I do not understand your assertion that the "quantum
> corrections" are copied into the model system we study, and that this
> is somehow very different from what is done with, for instance, the
> diffeomorphism constraint in LQG.

I'll try tomake my concern more precise below.

> (*) U(l)V(m) = e^{-ilm}V(m)U(l)

[...]

> The relation (*) is that of the Weyl-Heisenberg algebra, and what we
> do in our paper is look at a non-standard unitary representations of
> this algebra, rather than looking at a self-adjoint representation of
> the CCR.

Ok.

> And this is the key point: the relation (*)---including the factor of
> e^{-ilm}, which is what in other places on the net Urs seems to think
> is put in by hand,

No, sorry, that's a misunderstanding of what I was saying. I am grateful
that you took the time to answer so that I can try to clarify this.

I do understand why the Weyl algebra (equation III.1) in your paper looks
the way it does, including that factor. I understand that this factor is not
put in by hand. My point is that another factor is.

> How does one see this? By (as Urs has mentioned elsewhere) the
> Campbell-Baker-Hausdorff theorem, which lets me calculate what (*)
> should be, entirely in terms of iterated commutators of p's and q's.
> For the CCR, this is in fact the easiest way to go, because the CBH
> series stops after the first commutator.

I understand this and totally agree. But I think for those following this
one should emphasize that the BCH theorem applied to exp(p)exp(q) can only
serve to motivate certain steps in the LQG construction, because the p's and
q's do not exists (not both at least) in your framework.

> But for more complicated Lie algebras, as for example the
> diffeomorphism algebra in gravity, the CBH series does not terminate
> and this is not a terribly enlightening way to view the problem.
>
> But it's also not necessary, because a self-adjoint representation of
> a Lie algebra exponentiates to a unitary representation of the Lie
> group. So what one should check instead is that one has an honest
> group representation, that is, that:
>
> U(g_1)U(g_2) = U(g_1 * g_2)
>
> And this *does* hold for the diffeo group in LQG.

Yes, as it does for Diff(S^1) x Diff(S^1) in Thomas Thiemann's 'LQG-sting'.
By construction. I.e. operators U can be found which do satisfy this
relation. I do understand and agree that these operators can be constructed.

The problem is, that these operators don't capture the usual quantum
effects. They do not come from what ordinarily is called 'canonical
quantization'. That's because the ordinary prescription of 'canonical
quantization' at some point says that we are to promote the classical
canonical coordinates and momenta to self-adjoint operators on some Hilbert
space. This step is explicitly violated in LQG, since half of these objects
are not represented as operators at all.

So with this step violated, something else has to be found to replace it.
Among other things, I claim that this replacement introduces a large
ambiguity.

Namely whenever there is now an object in the classical theory in the
enveloping algebra of p and q, you don't know precisely how to quantize it
(even modulo operator ordering issues) since one of p and q is not
represented on the Hilbert space.

I think that this is a problem both in gr-qc/0207106 as well as in the
'LQG-string'. Here is why:

In the text leading to equation (IV.5) of gr-qc/0207106 you are looking for
a way to express the notion of 'coherent state' in the LQG-like language,
where you characterize a coherent state by its property of being an
eigenstate of the annihilation operator

a ~ x + ip

(which is one of several possibilities of characterizing coherent states).

This is an object in the algebra generated by p and q. It happens to have no
operator ordering problems so the usual quantization of this guy is obvious.

But now in your framework p is not representable, so something else has to
be done. Some object in the algebra of operators that you do have must be
identified as being the 'quantization' of the annihilation operator.

Due to the Weyl-algebra context, what you really want is the exponentiation
of the adjoint of the annihilation operator, i.e. an analog of

exp(a^dag) .

This has to be expressed somehow using the operators x and V that do exist
on your Hilbert space. You do this by looking at the usual quantization of
this object and read off the expression above equation (IV.5) from this,
e.g.

exp(a^dag) -> exp(x)V(-1/2)e^(-1/4) .

But why don't you use for instance

exp(a^dag) -> exp(x)V(-1/2)

or

exp(a^dag) -> exp(x)V(-1/2)e^(42) ?

This is the factor which I am talking about and which is put in by hand. It
is _not_ put in by hand in the _usual_ quantization, because there the
e^(-1/4) comes from the BCH theorem. But in your rep there is no p and q
which could enter the exponent in the BCH theorem. You are just copying this
plausible looking result to your representation.

And from reading the paper I got the impression that you would agree that
here you are fixing an ambiguity by hand. Because on p.14 it says:

"Collecting these ideas motivated by results in the Schroedinger
representation, we are now led to seek the analog...".

I do agree that this is indeed the most natural choice for this simple
system. My problem is that in more difficult cases, where there is mayber no
working Schroedinger representation from which ideas can be motivated, how
do you fix such ambiguities then?

I noted that in the 'LQG-string' as well as for the diffeo-constraints of
3+1d LQG gravity the analog of this ambiguity is fixed by using ideas
motivated by results not of the Schroedinger rep but of the classical
Poisson algabra.

Here is why:

When quantizing the Nambu-Goto action by LQG-like methods we are faced with
a problem very similar to the representation problem of the annihilation
operator "a" in the above example. Namely, there we have the classical
Virasoro constraints L_m, which are objects bilinear in the canonical p and
q.

In the standard quantization these p and q are promoted to operators and no
matter how you deal with the operator ordering ambiguity the resulting
quantum versions of the L_m feature the anomaly, which is a commutator
effect very much like the factor discussed above.

But now in Thomas Thiemann's 'LQG-string' quantization, as in your paper
above, the p and q are not represented as operators. Hence he has to look
for an _analog_ of the usual quantization, much like you construct an analog
of exp(a^dag), as discussed above.

So Thomas Thiemann considers the classically exponentiated Virasoro
generators and models his quantum operators U(phi) on these. He _could_ have
proceeded along the lines of your paper and "collected ideas motivated by
the Schroedinger representation to seek an analog of" the exponentiated
Virasoro generators in the usual quantization instead. This way he would
have had the anomaly (in terms of the group instead of the algebra, but
still).

But he chooses not to. He instead collects ideas motivated by the classical
action of the Virasoro generators. And that's how the anomaly disappears.

I'd very much enjoy hearing your opinion on these assessments. If I am
wrong, I'd very much want to be corrected. Thanks.

### Urs Schreiber

Mar 16, 2004, 1:43:01 PM3/16/04
to
Josh Willis <jwi...@gravity.psu.edu> wrote in message news:<c2slkk$g2k$1...@f04n12.cac.psu.edu>...

> In the case of the diffeo group in LQG, one can
> explicitly check that the Hilbert space used, which carries an
> appropriated rep of the Weyl-Heisenberg algebra and other observables,
> like the area operator, does in fact have a unitary rep of the diffeo
> group as well.

This is apparenly a crucial point for the communication between LQG
people and
others. In LQG there is a lot of emphasis on finding such unitary
reps. For instance Thomas Thiemann shows how such a rep exists for the
string. But I believe that it is important to emphasize that the mere
existence of such reps is not the issue. They do exist, and it is not
hard to see why and how.

The crucial problem is rather that many people don't want to call the
procedure of finding such a unitary rep a 'quantization' of the system
at hand. Rather, they would want to see a canonical quantization in
the ordinary strict sense where the symmetry generators are
represented on some Hilbert space.

> But there's no claim that for any classical system
> with a Lie algebra of symmetries, there must exist some quantum rep in
> which the corresponding group action is unitarily implemented.

On the other hand, it seems that when you allow non-seperable Hilbert
spaces then constructions analogous to those used in the 'LQG-string'
do give you unitary reps for virtually everything. Can you give an
example of a symmetry group which, by the methods used in LQG and in
the 'LQG-string' could _not_ be represented unitarily?

> Even if
> there is such a rep, there can of course be other pathologies of the
> quantization.

Could you please explain what kind of pathologies you have in mind
here?

For instance, do you think that the method of 'shadow states' could be
applied to Thomas Thiemann's 'LQG-string', thus showing that his
approach does reproduce the usual quantization in some limit?

I believe that this won't be possible, because all the information
about the usual quantum effects have been eliminated in the
'LQG-string' and they won't reappear in any limit. Is this the kind of
pathology that you are referring to above?

Mar 16, 2004, 7:15:09 PM3/16/04
to
Aaron Bergman <aber...@physics.utexas.edu> wrote in message news:<abergman-747D7B.21474910032004@localhost>...
>
> As best I can tell from observing Urs's and Jacques Distler's
> conversations about it, Thiemann claims that he can quantize any
> classical theory preserving all its classical symmetries.
>
> This is false. Not only is it false, its falsity is confirmed by
> experiment. Anomalous global symmetries have well-known physical
> consequences. To steal Jacques's example, classical YM is scale
> invariant. If you could quantize it while preserving this symmetry, you
> would never see the beta function. Hell, you would screw up pion decay
> without an anomaly.

It would be interesting to see how LQG can handle decay of
neutral pions into two photons. Not to get the exact decay
rate, but rather to show that this process really can occur.
But it would actually be equally interesting to see how string
theory handles the same problem. This is an old question raised
by Roman Jackiw in hep-th/9911071.

Classically, the standard model has a chiral symmetry which
forbids that a neutral pion decays into two photons. However,
this process does occur in nature, because the chiral symmetry
is broken by an anomaly. This anomaly can be traced back to QFT
infinities. To extract information from the standard model, one
must regularize the theory and carry out renormalization.
However, it is impossible to find a regularization which
preserves chiral symmetry, and it turns out that chiral
symmetry remains broken when the regulator is finally removed.

What Jackiw sees as a problem for string theory is that the
anomaly is an indirect effect of QFT infinities. In a theory
which is perturbatively finite, there is no need to introduce a
regularization in the first place, so chiral symmetry should
remain manifest. AFAIU, string theory is a finite theory which
has no need for regularization. So how can chiral symmetry be
broken to fit experiments?

Mar 16, 2004, 7:16:36 PM3/16/04
to
"Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message news:<4051f3fc$1...@news.sentex.net>... > IIRC these cocycles will not be scalar (i.e. proportional to the identity), > right? This is correct. vect(N) has no *central* extension when the number of dimensions N > 1. The Virasoro-like cocycles are proportional to closed (N-1)-forms, which transform non-trivially except when N = 1. Dzhumadildaev's cocycles are proportional to tensor densities, but two of them are closely related to the Virasoro algebra in several dimensions. > > BTW, when talking about this with somebody else, I began to wonder about the > following point: > > Can we check how these cocylces look like in string theory? In Matrix Models > (BFSS or IKKT) the diffeomorphism group on spacetime becomes a part of the > overall U(N>>1) symmetry. I have only a very vague idea of how this works > (One way to see it is that the permutation subgroup of U(N) permutes matrix > eigenvalues and hence discrete spacetime points.) but can we read off the > quantum corrections to the diffeomorphism group from that? Is that a naive > question? Is the N->oo limit a problem? I really haven't the faintest idea what you are talking about, but please give it a try. What I do know is that cancellation of gravitational anomalies is a big deal in string theory - Green-Schwartz mechanism. However, that kind of diff anomaly is of a different kind, which only exists in 4k+2 dimensions. In particular, Weinberg II asserts in chapter 22 that there are no pure gravitational anomalies in 4D. This statement does not apply to the 4D Virasoro algebra because it cannot be constructed using spacetime fields alone. > But if I handed you only the abstract Virasoro algebra and ask you "Compute > graviton scattering from that!", you'd rightly complain that I did not give > away all the necessary information. This is true. However, it might be possible if you also handed me the relevant representation (not for me, but for someone who knows how to compute scattering amplitudes). One may regard the space of functionals of the metric, modulo Einstein's equation, as a diffeomorphism group module. Classically, this module is very reducible, but it encodes information about the physics. > What are > the degrees of freedom of the theory? What is the dynamics? How do we find > it from knowing just the symmetry group alone? This sounds like somebody complaining about M theory - except that there one does not know the symmetry group. > > > Anyway, I find it amusing that, when left to yourself, you came > > up with essentially the same idea as myself. :-) > > Hehe. But note that I would sell this idea differently. I wouldn't claim > that knowing the possible cocycles of the diffeomorphism group means having > solved quantum gravity and I wouldn't claim that it implies that string > theory is wrong! :-) Alas, you are not me. In particular, I am older and have witnessed firsthand two decades of extreme overselling - Witten's assertion that string theory is "more predictive than conventional quantum field theory" ( http://www.arxiv.org/abs/hep-th/0212247 ) is only one recent example. Maybe I am overreacting. If I were a different person, I might have tried to sell the multi-dimensional Virasoro algebra as the symmetry of the long-sought background free formulation of M theory. After all, this algebra is closely related to stringy math (Virasoro was a string theorist), diffeomorphisms are closely related to background independence, and multi-dimensional even starts with an M. ### Thomas Larsson unread, Mar 16, 2004, 7:20:55 PM3/16/04 to Josh Willis <jwi...@gravity.psu.edu> wrote in message news:<c2slkk$g2k$1...@f04n12.cac.psu.edu>... > As I indicated above, and as I'm sure Aaron and Urs both know, when I > have a classical symmetry captured in a constraint algebra, then I > have an anomaly when the corresponding operators in the quantum theory > do not reproduce the Poisson brackets but instead have additional > terms, higher order in \hbar. What this means at the group level is > that I do not have a unitary representation of the symmetry group. While I technically don't disagree with the truth of this statement, I think that it is somewhat misleading. In the presence of an anomaly you don't have a proper rep of the group at all. Rather, you have a rep of a group extension, which in some cases may be regarded as a projective rep of the original group. But unitarity is not really the issue from a purely group theorectical point of view. In some cases the anomalous group does not have any unitary rep at all, at least no unitary faithful rep on a separable Hilbert space. This happens e.g. for the Mickelsson-Faddeev algebra, which describes the chiral anomaly, and probably also for the superstring diffeo anomaly that is cancelled by the Green-Schwartz mechanism. In that case, the extension does imply lack of unitarity. However, in the case of the Virasoro group, or at least the Virasoro algebra, there are certainly unitary irreps with c > 0. These reps may be unsuitable for other reasons, but they are unitary. ### Arnold Neumaier unread, Mar 16, 2004, 7:21:16 PM3/16/04 to Urs Schreiber wrote: > Josh Willis <jwi...@gravity.psu.edu> wrote in message news:<c2slkk$g2k$1...@f04n12.cac.psu.edu>... > > >>In the case of the diffeo group in LQG, one can >>explicitly check that the Hilbert space used, which carries an >>appropriated rep of the Weyl-Heisenberg algebra and other observables, >>like the area operator, does in fact have a unitary rep of the diffeo >>group as well. > > This is apparenly a crucial point for the communication between LQG > people and > others. In LQG there is a lot of emphasis on finding such unitary > reps. For instance Thomas Thiemann shows how such a rep exists for the > string. But I believe that it is important to emphasize that the mere > existence of such reps is not the issue. They do exist, and it is not > hard to see why and how. > > The crucial problem is rather that many people don't want to call the > procedure of finding such a unitary rep a 'quantization' of the system > at hand. Rather, they would want to see a canonical quantization in > the ordinary strict sense where the symmetry generators are > represented on some Hilbert space. Once you have a unitary representation, you have a Hilbert space, namely that spanned by any orbit of the group. The generators of continuous symmetries then automatically act on a dense subspace of that Hilbert space. But the real problem is to construct a representation that is 1. unitary, and 2. in which an energy-momentum tensor T exists which satisfies the minimal causality property that u^j T_jk u^k >=0 for some vector field u that is nonzero everywhere. These properties are satisfied both in standard quantum field theory and in conformal field theory. Unless 1. and 2. holds, I think any claim to have a quantization in any physically relevant sense is unfounded. Arnold Neumaier ### Urs Schreiber unread, Mar 17, 2004, 9:09:16 AM3/17/04 to "Arnold Neumaier" <Arnold....@univie.ac.at> schrieb im Newsbeitrag news:40575338...@univie.ac.at... > Once you have a unitary representation, you have a Hilbert space, > namely that spanned by any orbit of the group. The generators of > continuous symmetries then automatically act on a dense subspace of > that Hilbert space. If the unitary rep is weakly continuous, at least. But this is not the case in LQG and LQG-like quantizations. There the generators are not operators an a Hilbert space. For instance in LQG of 1+3d gavity the generators of spatial diffeomorphisms are not represented as operators on the Hilbert space. In Thiemann's quantization of the string the Virasoro generators are not represented as operators. ### Aaron Bergman unread, Mar 18, 2004, 11:40:24 AM3/18/04 to In article <24a23f36.04031...@posting.google.com>, thomas_l...@hotmail.com (Thomas Larsson) wrote: > Classically, the standard model has a chiral symmetry which > forbids that a neutral pion decays into two photons. However, > this process does occur in nature, because the chiral symmetry > is broken by an anomaly. This anomaly can be traced back to QFT > infinities. To extract information from the standard model, one > must regularize the theory and carry out renormalization. > However, it is impossible to find a regularization which > preserves chiral symmetry, and it turns out that chiral > symmetry remains broken when the regulator is finally removed. This is one way of seeing the anomaly. There are other ways, however. The real point is that the measure cannot be made invariant. This shows up perturbatively when regulating the triangle diagram, but one can see it in the Fujikawa approach just by examining the measure directly. Anomalies are really an IR effect, not a UV effect. In that I don't think it's correct to say that anomalies are a consequence of the infinities, I don't see a problem with finding them in string theory. The literature on anomlies in string theory is quite extensive. Aaron ### Josh Willis unread, Mar 20, 2004, 4:04:30 PM3/20/04 to "Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message news:<4053...@news.sentex.net>... > "Josh Willis" <jwi...@gravity.psu.edu> schrieb im Newsbeitrag > news:c2slkk$g2k$1...@f04n12.cac.psu.edu... > [...] > > But it's also not necessary, because a self-adjoint representation of > > a Lie algebra exponentiates to a unitary representation of the Lie > > group. So what one should check instead is that one has an honest > > group representation, that is, that: > > > > U(g_1)U(g_2) = U(g_1 * g_2) > > > > And this *does* hold for the diffeo group in LQG. > > Yes, as it does for Diff(S^1) x Diff(S^1) in Thomas Thiemann's 'LQG-sting'. > By construction. I.e. operators U can be found which do satisfy this > relation. I do understand and agree that these operators can be constructed. > > The problem is, that these operators don't capture the usual quantum > effects. I completely disagree. The point is that one could do *Schroedinger* quantum mechanics this way, if one wanted to; i.e., for the kinds of things we consider in the paper, one need never consider the self-adjoint generators. And you would get exactly the same physical effects as if you did consider them, because you're looking at exactly the same representation of exactly the same algebra. We are therefore just taking the unitary group rep, rather than the self-adjoint algebra rep, as the "jumping off" point. > They do not come from what ordinarily is called 'canonical > quantization'. That's because the ordinary prescription of 'canonical > quantization' at some point says that we are to promote the classical > canonical coordinates and momenta to self-adjoint operators on some Hilbert > space. This step is explicitly violated in LQG, since half of these objects > are not represented as operators at all. > > So with this step violated, something else has to be found to replace it. > Among other things, I claim that this replacement introduces a large > ambiguity. Well, if you don't like ambiguity, you're going to find quantization in general very frustrating :) > Namely whenever there is now an object in the classical theory in the > enveloping algebra of p and q, you don't know precisely how to quantize it > (even modulo operator ordering issues) since one of p and q is not > represented on the Hilbert space. Here is a good example of what I just said about ambiguity. You're used to constructing the space of observables from the enveloping algebra of the quantization of p and q. But of course this is hardly ideal, from the point of view of "quantization," since the enveloping algebra will not agree with the Poisson algebra of polynomials in p and q, in general. So there is already ambiguity, in choosing what subalgebra of the Poisson algebra one represents. People who study the general mathematical problem of quantization worry a lot about this sort of thing. You are misunderstanding several things here. First, as I indicated in my original post, this is not the point in the paper where we define the physics. That has already been done, in choosing the algebra and representation; the rest of the paper (more or less) is about finding out what the physics of this rep is---not specifiying it. Even though there is no p operator (there is in fact a q operator) we still know what the commutation relations should be, and therefore how to apply the CBH theorem, even if we had never heard of the Schroedinger rep. That's because we know the what the commutation relations should be from the *classical* theory, the Poisson algebra. This alone tells us that exp(a^dag) -> exp(x)V(-1/2)e^(42) is wrong, provided we know that a^dag = q +ip classically. But how do we know that? In other words, what does a state being an eigenstate of this particular operator have to do with that same state being a semiclassical state? But we could just as well ask the same question in Schroedinger quantum mechanics! The point is that as far as semiclassicality is concerned (because of course as you know coherent states have many other interesting and important properties), it's just a fact that the states which satisfy this eigenvalue equation also have minimal uncertainty and saturate the Heisenberg bound. But it is the latter that makes them semiclassical, not the fact that they are eigenstates of the annihilation operator. Likewise, in our paper the states are semiclassical because they meet the criteria on pp. 20-21. It just turns out that states which meet this criteria also satisfy the eigenvalue equation (IV.5). > And from reading the paper I got the impression that you would agree that > here you are fixing an ambiguity by hand. Because on p.14 it says: > > "Collecting these ideas motivated by results in the Schroedinger > representation, we are now led to seek the analog...". You're misunderstanding our point, I think. We've introduced in this paper a non-standard rep of the Weyl-Heisenberg algebra, which we claim should be thought of as a viable (at least for a non-relativistic theory) quantization of the particle on the line. If we wanted to do the analagous thing to quantum gravity, all we would be looking for is the semiclassical limit: that is, to show that Newton's laws are recovered in an appropriate sector of the theory. Of course, for quantum mechanics (as opposed to quantum gravity, at present) this is not nearly enough; we must also show that the quantum effects that are confirmed by experiment are also reproduced. But in fact we aim to do slightly more even than that. We are trying to show that a great deal of the structure of the Schroedinger representation can be carried over to this polymer particle representation. A priori that might seem impossible, since the two Hilbert spaces aren't even unitarily equivalent, much less the representations. The "conventional wisdom" would therefore dictate that these must be physically different theories. And indeed they are: there is some regime in which their physical predictions will disagree, as there must be. But not only is there a large regime in which their predictions are physically indistiguishable, but there is much more similarity in structure than that superficial observation of unitary inequivalence might have suggested. And in order to compare the structure of the polymer particle rep to the Schrodinger rep, we obviously have to say what the latter is. But if we didn't know or care about the Schrodinger rep, there would be no need to rely on it. But we do, and that is why the emphasis on constructing an analogue of the coherent state eigenvalue equation; if all we cared about was semiclassicality we could have gone directly to positing the state and verifying its properties. This might have been inelegant, but there's nothing per se that requires an eigenvalue equation to select semiclassical states. And indeed most semiclassical states will not satisfy any particular such equation, because the class of semiclassical states is much larger than the class of coherent states. But in fact these states, which are exactly the states one might have "guessed," also satisfy the same criterion as coherent states in the Schrodinger representation. > I do agree that this is indeed the most natural choice for this simple > system. My problem is that in more difficult cases, where there is mayber no > working Schroedinger representation from which ideas can be motivated, how > do you fix such ambiguities then? Two things: again, I don't really think there's an "ambiguity," as I explained above. Second and more importantly, there are guidelines to what one should be looking for in other quantum theories, including gravity. Specifically, one would be looking for the "best approximation" to Fock space, among other things. If you're really interested in this then you should read the papers that led up to the one we wrote, namely: gr-qc/0001050, gr-qc/0104051, and gr-qc/0107043 Some other stuff has been put on the web since then as well. > I noted that in the 'LQG-string' as well as for the diffeo-constraints of > 3+1d LQG gravity the analog of this ambiguity is fixed by using ideas > motivated by results not of the Schroedinger rep but of the classical > Poisson algabra. As I said ealier, you're comparing apples to oranges if you compare the selection of coherent states in the polymer particle paper to the representation of the diff-algebra in LQG. The former isn't representing any algebra at all; the choice of algebra has already been made, and it was made based on the classical Poisson algebra. In this respect the polymer particle is not a complete model because we didn't look at constraints, but if we had they too would be dictated by the Poisson algebra; you would fix that before you ever looked for semiclassical states, coherent or otherwise. --Josh ### Josh Willis unread, Mar 22, 2004, 3:43:03 PM3/22/04 to Urs Schreiber wrote: > Josh Willis <jwi...@gravity.psu.edu> wrote in message news:<c2slkk$g2k$1...@f04n12.cac.psu.edu>... > > >>In the case of the diffeo group in LQG, one can >>explicitly check that the Hilbert space used, which carries an >>appropriated rep of the Weyl-Heisenberg algebra and other observables, >>like the area operator, does in fact have a unitary rep of the diffeo >>group as well. > > > This is apparenly a crucial point for the communication between LQG > people and > others. In LQG there is a lot of emphasis on finding such unitary > reps. For instance Thomas Thiemann shows how such a rep exists for the > string. But I believe that it is important to emphasize that the mere > existence of such reps is not the issue. They do exist, and it is not > hard to see why and how. I don't think this is nearly as easy as you claim. The key point is finding all of the things I mention in the quoted para. in one rep. To back up a little, it's best to outline some of the criteria normally needed for a "quantization" of a given classical system. Normally, for a prequantization one needs: (1) Poisson brackets go over to commutators, {f,g} --> (i/\hbar)[\hat{f},\hat{g}] (2) The constant function on phase space is mapped to the identity operator. (3) If the Hamiltonian vector field of f on phase space is complete, then \hat{f} is essentially self-adjoint. If we have only these three postulates, then there are "quantizations" such that the entire Poisson algebra can be promoted to an operator representation preserving the classical Poisson brackets. However, even for finite dimensional systems this is not enough, and so for a prequantization to be a true quantization, one demands another criteria, usually either that the representation of the "basic observables" (i.e., the p's and q's) be represented irreducibly, or that for instance a certain class of observables (functions only of q, for instance) be multiplicatively represented. This severely constrains allowed quantizations, and in general one can then only implement (1) for a certain subalgebra of the classical symplectic algebra. Now consider the case where one is quantizing a constrained system by the Dirac quantization "program." Then it is essential that the constraints be among the classical observables that can be promoted to operators respecting the Poisson algebra, and in general this is nontrivial. So far, everything I have written is for what I think you would call "canonical quantization." What we are doing is now demanding instead of (1) and (3) that a subgroup of the symplectic group on the manifold---corresponding to the subalgebra that is represented by e.s.a. operators---be unitarily represented on the Hilbert space. Appropriate irreducibility criteria must still hold. When the unitary representation is continuous, one can obtain a representation of the self-adjoint operators from the given unitary representation; when it is not, no such rep exists, and we are now allowing for this possiblity Now, it seems that you think that this relaxation makes it easy to satisfy the criteria I list above. Why do you think this? Can you give a general construction? I'll tell you why I don't think it's so easy, again using LQG as an example. The way that these criteria are met in LQG is as follows. One takes as the basic algebra of observables the "flux-holonomy algebra"; this is supposed to be analagous to the Heisenberg algebra in ordinary QM. One then demands that the holonomy observables be implemented multiplicatively. This is done by constructing a certain space Abar of "generalized connections" on which the holonomy algebra is the algebra of all bounded, continuous functions. Then the Hilbert space you have is an L_2 space on Abar with respect to some measure, and the choice of representation is dictated by the choice of this measure. *Any* regular, borel measure by construction will represent the holonomy algebra faithfully. But that's not enough. We must represent the entire flux-holonmy algebra, and to do Dirac quantization we must also represent the diffeomorphism group unitarily. Take the latter restriction first. This means we can no longer consider any measure on Abar, but just diffeomorphism invariant measures. That already throws out a lot of possibilities. However, there are still loads of diff-invariant measures on Abar. But there is only *one* for which it is true that one gets a representation of the full flux-holonomy algebra. This is the measure usually referred to as the Ashtekar-Isham-Lewandowski measure (or sometimes just Ashtekar-Lewandowski). That this is the unique measure Abar allowing one to represent both the flux holonomy algebra and the diffeomorphism algebra is shown in recent work of Sahlmann, Thiemann, Lewandowski, and Okolow. So, given this I am highly dubious that one can make a general construction for any subalgebra of the Poisson algebra; it is not, for instance, even clear that one can enlarge the flux-holonomy + diffeomorphism algebra in LQG without breaking the construction. > The crucial problem is rather that many people don't want to call the > procedure of finding such a unitary rep a 'quantization' of the system > at hand. Rather, they would want to see a canonical quantization in > the ordinary strict sense where the symmetry generators are > represented on some Hilbert space. I suppose some people may think that, but it's worth pointing out that people other than the LQG community (for instance many working in various rigorous approaches to QFT) often focus on unitary reps, for several reasons. (1) Unitary operators, being bounded, are much nicer to deal with than unbounded self-adjoint operators. (2) The unitary criteria is easier to justify physically, since it is what corresponds to preservation of expectation values by the action of a symmetry. When the rep is continuous and the phase space finite-dimensional, the Stone-von Neumann theorem guarantees that the only continuous unitary rep is unitarily equivalent to the Schroedinger rep. But in infinte dimensions the Stone-von Neumann theorem fails, and it is for this reason that in QFT one often wants to look at the unitary algebra. Haag mentions this early on in _Local_Quantum_Physics_, IIRC. (3) When the symmetry group is infinite dimensional, as for instance Dif(S^1), any neighborhood of the identity contains group elements that cannot be obtained by exponentiating the Lie algebra. Thus, the algebra doesn't carry the full information about the group. >>But there's no claim that for any classical system >>with a Lie algebra of symmetries, there must exist some quantum rep in >>which the corresponding group action is unitarily implemented. > > > On the other hand, it seems that when you allow non-seperable Hilbert > spaces then constructions analogous to those used in the 'LQG-string' > do give you unitary reps for virtually everything. Can you give an > example of a symmetry group which, by the methods used in LQG and in > the 'LQG-string' could _not_ be represented unitarily? Again, I would instead challenge you to give examples of how you can do this for "virtually everything." >>Even if >>there is such a rep, there can of course be other pathologies of the >>quantization. > > > Could you please explain what kind of pathologies you have in mind > here? Failing one or more of the criteria I mentioned above, for starters. > For instance, do you think that the method of 'shadow states' could be > applied to Thomas Thiemann's 'LQG-string', thus showing that his > approach does reproduce the usual quantization in some limit? I don't know---we haven't even tackled full GR yet in this framework. And I haven't looked at Thomas's paper. > I believe that this won't be possible, because all the information > about the usual quantum effects have been eliminated in the > 'LQG-string' and they won't reappear in any limit. I don't understand this statement at all. See my other response (that I hope to write :) ) to your other reply to my post. But at various points you seem to me to have implied that the absence of anomalies means that one won't get the "usual quantum effects." But there are lots of quantum effects besides the existence of anomalies! After all, plenty of systems don't show anomalies in standard quantum prescriptions. Surely you wouldn't say that these show no quantum effects, that the classical and quantum systems are physically indistinguishable? For instance, one of the first quantum effects one learns about is the inability to measure momentum and position simultaneously, because they are not commuting observables. This in turn arises because one is quantizing using the Poisson brackets, rather than the usual commutative algebra structure of real-valued functions on phase space. And this effect is certainly preserved in these "unitary group" quantizations, precisely because (for a particle on the line, for instance) one chooses to represent the Heisenberg group, rather than the commutative group R^2. > Is this the kind of pathology that you are referring to above? No. --Josh ### Urs Schreiber unread, Mar 22, 2004, 5:18:11 PM3/22/04 to "Josh Willis" <jwi...@gravity.psu.edu> schrieb im Newsbeitrag news:41284420.04032...@posting.google.com... > I completely disagree. The point is that one could do *Schroedinger* > quantum mechanics this way, if one wanted to; i.e., for the kinds of > things we consider in the paper, one need never consider the > self-adjoint generators. And you would get exactly the same physical > effects as if you did consider them, because you're looking at exactly > the same representation of exactly the same algebra. We are therefore > just taking the unitary group rep, rather than the self-adjoint > algebra rep, as the "jumping off" point. I don't doubt that you can choose the U in such a way that the usual quantum effects are respected. But they are not always chosen that way. The U(phi) in Thomas Thiemann's paper are chosen in such a way that the usual quantum effects (the anomaly) are absent. He could have chosen U(phi) which reproduce the anomaly. But apparently in the LQG-framwork there is the freedom to make different choices. This is the ambiguity that I mentioned. I stated that I think that obtaining or not obtaining the anomaly in Thomas Thiemann's 'LQG-string' is completely analogous to including or not including that factor exp(-alpha^2/2) in your paper. Since you disagree it would be helpful if we could reverse the burden of proof: Do you think, and if you do could you please show, that the technique of 'shadow states' which you have tested on the 1d nonrelativistic particle, can be applied to Thomas Thiemann's 'LQG-string' such that the results of the usual quantization are reproduced in an appropriate limit? This would be very helpful. > Well, if you don't like ambiguity, you're going to find quantization > in general very frustrating :) Right. Because I knew that I was emphasizing the ambiguities of first quantization from the very beginning of the discussion (http://golem.ph.utexas.edu/string/archives/000299.html). But the usual operator ambiguities are something different from the ambiguities that are introduced in the LQG framework. And I was trying to stress that in my last post. Neither the equation a |psi> = alpha|psi> characterizing coherent states nor the Virasoro anomaly [L_m,L_n] = (m-n)L_m+n + (D/12)m^3 + O(m) are sensitive to operator ordering issues. So there is no ambiguity here in the usual canonical quantization prescription. But as soon as you are proposing a quantization where neither the operator "a" nor the operators "L_m" even exist the above are no longer universal facts. Thomas Thiemann claims that he may use operators U(phi) analogous to (but not related to) exponentiated generators L_m such that there is no anomaly. In the same vein one could choose the usual exp(L_m) such that the anomaly is there. So in such a context there is apparently a much larger ambiguity. > Even though there is no p operator (there is in fact a q operator) we > still know what the commutation relations should be, and therefore how > to apply the CBH theorem, even if we had never heard of the > Schroedinger rep. If this were true, then why don't you apply this reasoning to the spatial diffeo generators D_i(x) in LQG, or why didn't Thomas Thiemann apply this reasoning to the 'LQG-string', thereby rediscovering the anomaly? If you can guess the analogue of the BCH theorem even in the absense of a Schroedinger rep (namely no usual consistent quantization of the ADM constraints D_i(x) is known) then why isn't that discussed and applied in LQG? What is the CBH analogous correction to exp({int f(x) D_i(x) , .})exp({int f(y) D_j(y) , .})? > As I said ealier, you're comparing apples to oranges if you compare > the selection of coherent states in the polymer particle paper to the > representation of the diff-algebra in LQG. The former isn't > representing any algebra at all; the choice of algebra has already > been made, and it was made based on the classical Poisson algebra. The analogy that I am seeing is that in both cases you have an operator equation which characterizes certain states, namely coherent states in one case and physical states in the other case. Now the operators entering these operator equations are not available in your framework and so something else is sought for. But if you don't agree that the crucial non-standard step of the 'LQG-string' and of LGQ of 3+1d gravity is also present in the 'shadow states' paper (although dealt with differently in both cases) then I'd consider it more productive for our discussion if we could reverse the way of proof: Instead of me arguing why the 'shadow states' technique won't work for Thiemann's 'LQG-string' you could demonstrate that and how it does. For instance it would be great if you could show that, by taking some limit if necessary, you can find the tachyon 'shadow state' of the 'LQG-string'. But having said that, I recall that probably the absence of this state will be considered an advantage. This then would leave me completely puzzled: Why is it ok for the 1d nonrelativistic particle that the LQG-like quantization approximates the usual quantization, while for the Nambu-Goto action it is not? ### Thomas Larsson unread, Mar 24, 2004, 4:03:45 AM3/24/04 to Josh Willis <jwi...@gravity.psu.edu> skrev i diskussionsgruppsmeddelandet:c3goq8$18le$1...@f04n12.cac.psu.edu... > (3) When the symmetry group is infinite dimensional, as for instance > Dif(S^1), any neighborhood of the identity contains group elements that > cannot be obtained by exponentiating the Lie algebra. Thus, the algebra > doesn't carry the full information about the group. > There are also exponented vector fields that are not diffeomorphisms. My impression is that this distinction is a mathematical subtlety without real physical significance. The important thing is that the diffeomorphism group and the algebra of vector fields share essentially the same representations; the classical irreps are tensor densities and exact forms. Most interesting Lie algebra cocycles can be integrated to group cocycles, e.g., the Virasoro algebra gives the Schwarzian derivative. Yuly Billig has recently written down the analogous group cocycles in higher dimensions. > > I believe that this won't be possible, because all the information > > about the usual quantum effects have been eliminated in the > > 'LQG-string' and they won't reappear in any limit. > > I don't understand this statement at all. See my other response (that I > hope to write :) ) to your other reply to my post. But at various > points you seem to me to have implied that the absence of anomalies > means that one won't get the "usual quantum effects." But there are > lots of quantum effects besides the existence of anomalies! Urs Schreiber's arguments have some interesting logical consequences. If canonical quantization of the string leads to conformal anomalies (only as an intermediate step before introducing ghosts, but anyway), canonical quantization of gravity should lead to similar intermediate diffeo anomalies. However, pure gravitational anomalies in field and string theory can only arise in 4k+2 spacetime dimensions. Thus string theory is also incapable of producing such diffeo anomalies in 3D and 4D. Note the difference compared to anomaly cancellation: here no anomalies can be written down at all. The reason is that the relevant anomalies are functionals of the observer's trajectory, which is simply not available unless you introduce it. It thus seems that Urs Schreiber's argument, if correct, proves that string theory is wrong. ### Urs Schreiber unread, Mar 24, 2004, 5:54:38 PM3/24/04 to "Josh Willis" <jwi...@gravity.psu.edu> schrieb im Newsbeitrag news:c3goq8$18le$1...@f04n12.cac.psu.edu... > (1) Poisson brackets go over to commutators, > {f,g} --> (i/\hbar)[\hat{f},\hat{g}] > (2) The constant function on phase space is mapped to the identity operator. > (3) If the Hamiltonian vector field of f on phase space is complete, > then \hat{f} is essentially self-adjoint. [...] > So far, everything I have written is for what I think you would call > "canonical quantization." Yes. > What we are doing is now demanding instead of > (1) and (3) Just for the record: You are now saying that what is done in LQG ('we') is not canonical quantization as sketched above but a modification ('relaxation') thereof. > that a subgroup of the symplectic group on the > manifold---corresponding to the subalgebra that is represented by e.s.a. > operators---be unitarily represented on the Hilbert space. Appropriate > irreducibility criteria must still hold. When the unitary > representation is continuous, one can obtain a representation of the > self-adjoint operators from the given unitary representation; when it is > not, no such rep exists, and we are now allowing for this possiblity Agreed and understood. The point is that it is not clear that allowing for this possibility is physically viable. > Now, it seems that you think that this relaxation makes it easy to > satisfy the criteria I list above. Why do you think this? I think that if you could satisfy the above criteria without the relaxation for 1+3d gravity then you would do that instead of proposing a speculative relaxation of these criteria. Not? Isn't the problem with doing a ("what I would call") strict canonical quantization of gravity, along the unrelaxed lines sketched above, precisely the reaon why these relaxations are considered in the first place? > Can you give a general construction? See below. > > The crucial problem is rather that many people don't want to call the > > procedure of finding such a unitary rep a 'quantization' of the system > > at hand. Rather, they would want to see a canonical quantization in > > the ordinary strict sense where the symmetry generators are > > represented on some Hilbert space. > > I suppose some people may think that, but it's worth pointing out that > people other than the LQG community (for instance many working in > various rigorous approaches to QFT) often focus on unitary reps, for > several reasons. Ok, yes. It appears that this is the main result of all the discussion initiated by the 'LQG-string': LQG uses methods that many people would not call quantization while some people (most notably in the fields of LQG and AQFT) would. For me this is already a very useful information. Before this whole discussion I was told and did believe that LQG is a very conservative approach towards quantizing gravity. I had always thought "Well, just doing canonical quantization of something cannot be that wrong." Now I learned that LQG is 'canonical' only in a generalized sense which is not generally accepted and which furthermore produces strange results when applied to systems that we actually do understand (as opposed to full nonperturbative quantum gravity). > Again, I would instead challenge you to give examples of how you can do > this for "virtually everything." What I have in mind is that the way exact reps of the classical groups are constructed for the 'LQG-string' as well as for LQG of 3+1d gravity follows the following presciption: - Given any (infinite dimensional) group G with elements phi. - Pick any set of objects psi(s) such that the group acts faithfully on these like, say, phi[psi(s)] = psi(phi(s)) . - Construct a Hilbert space such that the phi(s) are orthonormal elements <phi(s),phi(s')> = 1 iff s=s' and = 0 otherwise . - On this Hilbert space the classical group is represented _exactly_ (not projectively or something) by the operators U(phi) defined by U(phi)|psi(s)> := |psi(phi(s))> . Wether or not the U(phi) are unitary is a different question. The important point is that this way the classical symmetry group can always be transferred to the 'quantum theory' exactly, by construction. This is the essence of what Thomas Thiemann does in the 'LQG-string' paper. And I think it is also what is done for the spatial diffeomorphisms in LQG of 3+1d gravity. > And I haven't looked at Thomas's paper. Oh, too bad! :-) Thomas Thiemann's paper has the virtue that it applies LQG methods to a system which is not quite as trivial as the free 1d nonrel particle, but still much better understood than full quantum gravity. It seems to be an ideal testing ground for LQG methods. > > I believe that this won't be possible, because all the information > > about the usual quantum effects have been eliminated in the > > 'LQG-string' and they won't reappear in any limit. > > I don't understand this statement at all. See my other response (that I > hope to write :) ) to your other reply to my post. But at various > points you seem to me to have implied that the absence of anomalies > means that one won't get the "usual quantum effects." This stratement is a little vague, true. But I think from the context it ishould be quite clear what I mean. Both the corrections that come from applying the BCH theorem to exp(p)exp(q) as well as the anomaly in the Virasoro algebra come from the noncommutativity of the quantum algebra which is absent in the classical algebra. These effects cannot be found if the respective algebra is not present. So in the 'LQG-string' they are simply absent while in your 'shadow states' paper you argue that they must be taken into account. > But there are > lots of quantum effects besides the existence of anomalies! After all, > plenty of systems don't show anomalies in standard quantum > prescriptions. Surely you wouldn't say that these show no quantum > effects, that the classical and quantum systems are physically > indistinguishable? No, of course not. It just happens that for the perhaps most intersting testing ground example of LQG methods, namely the 'LQG string', the anomaly is at the center of attention. The whole point for Hermann Nicolai to propose that LQG methods should be applied to the string was to see if the anomaly can be found this way. He told me that had considered it a "breakthrough for the field of LQG" if it would have been found. ### Urs Schreiber unread, Mar 26, 2004, 2:36:57 AM3/26/04 to "Thomas Larsson" <thomas_l...@hotmail.com> schrieb im Newsbeitrag news:24a23f36.04032...@posting.google.com... > Urs Schreiber's arguments have some interesting logical > consequences. If canonical quantization of the string leads to > conformal anomalies (only as an intermediate step before > introducing ghosts, but anyway), canonical quantization of > gravity should lead to similar intermediate diffeo anomalies. That's not what I am saying. I don't even know what canonical quantization of gravity even is until I am shown a way how to consistently quantize the ADM constraints. What I am trying to say is that when the 'relaxed canonical quantization' (http://groups.google.de/groups?selm=c3goq8%2418le%241%40f04n12.cac.psu.edu) used in LQG is applied to systems that we know how to strictly canonically quantize, like the NG action in 2d, it misses the well-known results. > However, pure gravitational anomalies in field and string theory > can only arise in 4k+2 spacetime dimensions. If this is a theorem which also applies to the setup used in LQG then maybe this is the answer why they may quantize 3+1d gravity the way they do? But I have never seen this mentioned in the LQG literature. On the contrary, there are LQG treatments of 1+1 and also of 1+9d gravity. The general techniques they use are pretty insensitive to the number of dimensions, as far as I can see. But I don't claim to understand this issue with respect to string theory. For one, I have unfortunately only a vague understanding of section 12.2 of Polchinski. It's only fermions considered there. What happens in a purely bosonic theory? As I have said, I believe that Hermann Nicolai told me ( http://golem.ph.utexas.edu/string/archives/000330.html#c000799 ) that you'd need closed string field theory to properly address this question. (There is of course the possibility that I misinteroreted him.) > Note the difference compared to anomaly cancellation: here no > anomalies can be written down at all. The reason is that the > relevant anomalies are functionals of the observer's trajectory, > which is simply not available unless you introduce it. > > It thus seems that Urs Schreiber's argument, if correct, proves > that string theory is wrong. Sorry, I don't understand these two paragraphs. :-) ### Creighton Hogg unread, Mar 26, 2004, 2:37:36 AM3/26/04 to On 24 Mar 2004, Thomas Larsson wrote: > > > > Josh Willis <jwi...@gravity.psu.edu> skrev i > diskussionsgruppsmeddelandet:c3goq8$18le$1...@f04n12.cac.psu.edu... > > > (3) When the symmetry group is infinite dimensional, as for instance > > Dif(S^1), any neighborhood of the identity contains group elements that > > cannot be obtained by exponentiating the Lie algebra. Thus, the algebra > > doesn't carry the full information about the group. > > > > There are also exponented vector fields that are not > diffeomorphisms. My impression is that this distinction is a > mathematical subtlety without real physical significance. The > important thing is that the diffeomorphism group and the algebra > of vector fields share essentially the same representations; the > classical irreps are tensor densities and exact forms. > > Most interesting Lie algebra cocycles can be integrated to group > cocycles, e.g., the Virasoro algebra gives the Schwarzian > derivative. Yuly Billig has recently written down the > analogous group cocycles in higher dimensions. > > > > I believe that this won't be possible, because all the information > > > about the usual quantum effects have been eliminated in the > > > 'LQG-string' and they won't reappear in any limit. > > > > I don't understand this statement at all. See my other response (that I > > hope to write :) ) to your other reply to my post. But at various > > points you seem to me to have implied that the absence of anomalies > > means that one won't get the "usual quantum effects." But there are > > lots of quantum effects besides the existence of anomalies! > > Urs Schreiber's arguments have some interesting logical > consequences. If canonical quantization of the string leads to > conformal anomalies (only as an intermediate step before > introducing ghosts, but anyway), canonical quantization of > gravity should lead to similar intermediate diffeo anomalies. > However, pure gravitational anomalies in field and string theory > can only arise in 4k+2 spacetime dimensions. Thus string theory > is also incapable of producing such diffeo anomalies in 3D and > 4D. I'm confused as to what you mean here. Are you saying that a string theory, formulated in 3D or 4D spacetime, can not give rise to such anomalies or are you saying that the usual 10D theory cannot give rise to such anomalies after 6 dimensions are compactified. > Note the difference compared to anomaly cancellation: here no > anomalies can be written down at all. The reason is that the > relevant anomalies are functionals of the observer's trajectory, > which is simply not available unless you introduce it. > > It thus seems that Urs Schreiber's argument, if correct, proves > that string theory is wrong. Again, I don't quite understand why this last statement holds. Can you explain a little further? ### Thomas Larsson unread, Mar 26, 2004, 5:16:52 PM3/26/04 to Urs Schreiber <Urs.Sc...@uni-essen.de> skrev i diskussionsgruppsmeddelandet:c3rp2g$2aqqdb$1...@ID-168578.news.uni-berlin.de... > > However, pure gravitational anomalies in field and string theory > > can only arise in 4k+2 spacetime dimensions. > > If this is a theorem which also applies to the setup used in LQG then maybe > this is the answer why they may quantize 3+1d gravity the way they do? Well, yes and no. If you look at the original proof, in L Bonora, P Pasti and M Tonin, The anomaly structures of theories with external gravity, J Math Phys 27 (1986) 2259 - 2270 you'll see that they restrict attention to the space F of local functionals, which are integrated polynomials of the connection, gauge fields, matter fields, vielbeins and inverse vielbeins. If you relax this condition, anomalies can and do arise. Actually, here is a very simple example. Consider a phase space with 2N coordinates q^i and p_i and CCR [p_i, q^j] = delta^j_i. The vector field X = X^i(x) d/dx^i acts on the space of functions of q, as the operator L_X = X^i(q) p_i. A slight modification of this constructions does give you an extension. Consider the infinite-dimensional phase space of 2N functions q^i(t) and p_i(t) of a single variable t in S^1. The CCR are [p_i(s), q^j(t)] = delta^j_i delta(s-t) and L_X = \int dt :X^i(q(t)) p_i(t): satisfies an N-dimensional generalization of the Virasoro algebra. The double dots indicate normal ordering wrt frequency: expand q^i(t) and p_i(t) in a Fourier series in t and move positive modes to the left and negative modes to the right. OK, there is no Hamiltonian and thus no dynamics in this example, but there is a phase space and an anomalous action of the diffeo algebra, which shows that you must be careful not to apply the Bonora et al theorem where it does not apply. But it evidently applies to string theory: I don't have GSW with me right now, but a few days ago I looked up anomaly, gravitational in the index, and found the relevant info on page 336 in part II. They state that pure gravitational anomalies (implicitly, in string theory) only exist in D = 4k+2 dimensions because SO(D) only has complex spinor reps for such D. Apparently pure gravity anomalies in string theory can only come from chiral fermions, which gives this condition. But there is certainly no reference to chiral fermions, or any fermions at all, in my construction above. I thought that whole point with the LQG string is that canonical quantization of 3+1D gravity should be analogous to quantization of the string. Otherwise, why should one expect that LQG quantization applies to the string? So if an anomaly arises in the latter case, a pure gravity anomaly should also arise in the former. But GSW explicitly state that such anomalies only exist when D = 2, 6, 10, ... . One may think of several explanations. 1. Perhaps not all extensions of the diffeomorphism algebra count as gravitational anomalies. If so, I would like to know which ones do count. And why is the term diff anomaly used synonymously? 2. Canonical quantization of 3+1D gravity should see the anomaly but string quantization should not. But this seems weird if anomalies are real physical effects. 3. Maybe there shouldn't be a anomaly in 3+1D gravity after all. But what is then the point with the LQG string? I thought that the purpose with this toy problem was that one should learn something about 3+1D gravity from it. > > Note the difference compared to anomaly cancellation: here no > > anomalies can be written down at all. The reason is that the > > relevant anomalies are functionals of the observer's trajectory, > > which is simply not available unless you introduce it. > > > > It thus seems that Urs Schreiber's argument, if correct, proves > > that string theory is wrong. > > Sorry, I don't understand these two paragraphs. :-) > The standard treatment of the bosonic string involves anomaly cancellation; this leads to the prediction of 26D. The LQG string is manifestly anomaly free, so there is no constraints on the number of dimensions. Thomas Thiemann explicitly states this. If there should be a pure gravity anomaly in 3+1D, which I believe (and so did you, at least until I started to agree with you), and string theory cannot give us such an anomaly, how can it be right??? ### Aaron Bergman unread, Mar 30, 2004, 11:45:28 AM3/30/04 to In article <24a23f36.04032...@posting.google.com>, thomas_l...@hotmail.com (Thomas Larsson) wrote: I have no idea which anomaly you're referring to in this post. Nonetheless: > 3. Maybe there shouldn't be a anomaly in 3+1D gravity after all. > But what is then the point with the LQG string? I thought that > the purpose with this toy problem was that one should learn > something about 3+1D gravity from it. The point of the LQG string is to learn about 1+1D gravity. Aaron ### Thomas Larsson unread, Mar 30, 2004, 12:35:01 PM3/30/04 to Creighton Hogg <wch...@hep.wisc.edu> wrote in message news:<Pine.LNX.4.44.040324...@dill.hep.wisc.edu>... > > > > It thus seems that Urs Schreiber's argument, if correct, proves > > that string theory is wrong. > > Again, I don't quite understand why this last statement holds. Can you > explain a little further? In the beginning of this thread, Urs argued that correct quantization of 3+1D gravity should give rise to anomalies, eventually to be cancelled by ghosts. GSW state that no gravitational anomalies exists except in 4k+2D in string theory, although diff anomalies do exist in any dimension in a different framework, as I showed in another post. Thus, if both GSW and Urs' argument are right, string theory cannot be the correct quantization of 3+1D gravity, for essentially the same reason why LQG probably is wrong; it cannot exhibit the expected anomalies. But lack of anomalies is only a symptom. Urs real argument is that LQG is not what usually is called canonical quantization. However, canonical quantization typically gives rise to representations of lowest-energy type, so the problem is that diffeos are not given by LE reps in LQG. However, all interesting LE reps of the diffeo algebra give rise to anomalies. This is well known in 1D, and it follows in higher dimension by restriction to the many Virasoro subalgebras living on 1D lines. In particular, it follows from the GSW statement that LE reps of the 3+1D diffeo algebra cannot arise from string theory. ### John Baez unread, Apr 5, 2004, 2:59:39 PM4/5/04 to In article <24a23f36.04032...@posting.google.com>, Thomas Larsson <thomas_l...@hotmail.com> wrote: >Josh Willis <jwi...@gravity.psu.edu> skrev i >diskussionsgruppsmeddelandet:c3goq8$18le\$1...@f04n12.cac.psu.edu...

>> (3) When the symmetry group is infinite dimensional, as for instance

>> Diff(S^1), any neighborhood of the identity contains group elements that

>> cannot be obtained by exponentiating the Lie algebra. Thus, the algebra
>> doesn't carry the full information about the group.

examples where drastically different infinite-dimensional
Lie groups had the same Lie algebra.

Of course, even in the finite-dimensional case the Lie algebra
doesn't carry the full information about the Lie group!
You can take the universal cover of a Lie group, or take its
identity component, without changing its Lie algebra.

But presumably here you're not talking about *that*. You're
suggesting that infinite-dimensional Lie groups are "worse" in
some way. Do you know two infinite-dimensional Lie groups whose
Lie algebras are isomorphic, but where their identity components
don't have isomorphic universal covers? I don't.

need to distinguish between various flavors of "infinite-dimensional
Lie groups". Some are better than others.

>There are also exponentiated vector fields that are not
>diffeomorphisms.

If the vector field is smooth and the manifold is S^1, or more
generally compact, its exponential will exist and be a diffeomorphism.

If the manifold is noncompact there will be smooth vector fields
that don't exponentiate to give well-defined maps, like the vector
field x^3 d/dx on the real line. But, this just means you should
put some growth conditions on your vector fields if you want to
go around exponentiating them.

>My impression is that this distinction is a
>mathematical subtlety without real physical significance.

I tend to agree, though it's *always* dangerous to make general
vague claims that some bit of math has no "physical significance".

Here's one reason I tend to agree:

The problem Josh Willis mentions about Diff(S^1) is really just a
spinoff of the fact that it's a Frechet Lie group - i.e. it's covered
with charts that are Frechet spaces. Frechet Lie groups are tricky
in certain ways.

So, instead of working with diffeomorphisms, work with something
a little less smooth. Use maps

f: S^1 -> S^1

having inverses

g: S^1 -> S^1

such that f and g have 1st, 2nd,..., nth derivatives in L^2.
The maps form a Hilbert Lie group - i.e., a group covered with charts
that are Hilbert spaces. For Lie groups like this, elements near
the identity are ALWAYS exponentials of guys in the Lie algebra!

People who want to get serious about the *analysis* aspects of the
Virasoro group, loop groups, and other infinite-dimensional groups
tend to use Hilbert Lie groups.

I'm completely sick of arguing about loops versus strings, so I will
confine my remarks to the above relatively uncontroversial topics. :-)