From: baez@galaxy.ucr.edu (John Baez) Subject: Geometric quantization / Noncommutative unit disc Date: 14 May 2000 03:26:54 GMT Newsgroups: sci.physics.research Summary: Bergman, Bargmann, and quantum field theory (QFT) In article <8f4ggf$k8j$1@pravda.ucr.edu>, wrote: >Actually, you're mixing up Bergman and Bargmann - both of them >invented nice Hilbert spaces of holomorphic functions! > >Bergman: take the holomorphic functions on the unit disc and >define an inner product by > > = integral_T f*(z) g(z) > >where we integrate only over the boundary of the disc, that is, >the unit circle T. We use the usual rotation-invariant measure >on the unit circle to do the integral. We get a Hilbert space >with an orthonormal basis given by the functions z^n. > >Bargmann: take the holomorphic functions on the complex plane >and define an inner product by > > = integral_C f*(z) g(z) > >where we integrate over the whole complex plane C. We use a >Gaussian measure on the complex plane to do the integral. If >we pick this measure correctly, and only consider square-integrable >holomorphic functions, we get a Hilbert space with an orthonormal >basis given by the functions z^n/sqrt(n!). Actually I was getting my attributions a bit mixed up myself here. The space I attributed to Bergman is actually due to Hardy; Bergman came up with *another* nice Hilbert space of holomorphic functions on the open unit disk. So it really goes like this: Hardy: take the holomorphic functions on the unit disc and define an inner product by = integral_T f*(z) g(z) where we integrate only over the boundary of the disc, that is, the unit circle T. We use the usual rotation-invariant measure on the unit circle to do the integral. We get a Hilbert space with an orthonormal basis given by the functions z^n. Bergman: take the holomorphic functions on the unit disc and define an inner product by = integral_D f*(z) g(z) where we integrate over the whole disc D. We use Lebesgue measure on the disc to do this integral. If we normalize this measure correctly, we get a Hilbert space with an orthonormal basis given by the functions z^n/(n+1). Bargmann: take the holomorphic functions on the complex plane and define an inner product by = integral_C f*(z) g(z) where we integrate over the whole complex plane C. We use a Gaussian measure on the complex plane to do the integral. If we pick this measure correctly, and only consider square-integrable holomorphic functions, we get a Hilbert space with an orthonormal basis given by the functions z^n/sqrt(n!). These Hilbert spaces have a lot in common and you can read about all of them in Brian Hall's paper "Holomorphic methods in analysis and mathematical physics". Now let me sketch my answer to that old "noncommutative unit disk" problem that Toby and I were discussing. I'm sure everyone except possibly Toby has forgotten that problem, so let me state it again. What's the free C*-algebra on an isometry? I.e., find a nice description of the C*-algebra freely generated by one element u such that u*u = 1. Toby came up with a nice algebraic answer: it's the C*-algebra of operators on l^2 generated by the operator u with u(a_0,a_1,a_2,a_3,... ) = (0,a_1,a_2,a_3,... ) This operator is called the "right shift operator" and its adjoint is called the "left shift operator" since u*(a_0,a_1,a_2,a_3,...) = (a_1,a_2,a_3,... ) But here's a more analytic way to think about this answer. The Hilbert space l^2 is isomorphic to the Hardy space via (a_0,a_1,a_2,... ) |-> sum_n a_n z^n and via this isomorphism u turns into an operator on the Hardy space. What is this operator? Just multiplication by z! And what is the C*-algebra it generates? The C*-algebra of "TOEPLITZ OPERATORS". A Toeplitz operator is an operator on the Hardy space where we first multiply by a continuous function on the unit disc and then project back down to the Hardy space. Eh? Well, if we multiply a holomorphic function in the Hardy space by a continuous function, the result need not be holomorphic, but its restriction to the unit circle is a well-defined element of L^2(T), which we may then project back down to the Hardy space. For example, the operation of multiplying by zbar and then projecting down to the Hardy space is just the operator u* in disguise. Now, it's probably not clear why the C*-algebra of Toeplitz operators really deserves the name "noncommutative unit disk", and I seem to be forgetting some of the reasons I once knew... but there's a paper by Lesniewski and Klimcyk [I'm probably mangling these names] which explains this stuff better, and really makes a good case for this "noncommutative unit disk" idea. ============================================================================== From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Geometric quantization Date: 14 May 2000 03:28:05 GMT Newsgroups: sci.physics.research John Baez wrote: >Toby Bartels wrote: >>John Baez wrote: >>>Someone hands you a symplectic manifold M and says "quantize this >>>classical phase space!" The symplectic structure gives a volume >>>form which allows you to define L^2(M). Classical observables >>>easily become operators on this Hilbert space, and their Poisson >>>brackets go over to commutators. This is called "prequantization". >>>Unfortunately, this Hilbert space L^2(M) is too big. You have to >>>chop it down to size. >>Right, I understand this. >Okay, good! Unfortunately, I lied slightly when I said we start >with L^2(M). In general we should start, not with L^2 *functions* >on M, but L^2 *sections* of some complex line bundle over M. Of >course, when this line bundle is trivial, we can think of these >sections as functions. For example, this is what happens when M >is just R^2 - the phase space of a point particle on the line. Great, so now we have another possible nonfunctorial aspect -- a choice of which line bundle to use! How do you make this choice? >>>Luckily, you know how Bargmann and Segal dealt with this problem >>>in the special case when M was R^2 - they thought of it as >>>complex plane and threw out everything except the *holomorphic* >>>functions. >>Do you have to complete this to form a Hilbert space? >>It doesn't really matter, but it doesn't hurt to know. >>Wait, maybe this is that Bergman space we talked about before ... [*] >>yeah, it is; holomorphic functions which are square integrable >>form a complete Hilbert space all by themselves. >Actually, you're mixing up Bergman and Bargmann - both of them >invented nice Hilbert spaces of holomorphic functions! Actually, I figured they were different people, but who am I to complain if Bargmann wants to use Bergman's space? [skip definitions of Bergman and Bargmann spaces, which turn out to really be Hardy and Bargmann spaces] Neither of these is the Bergman space I was thinking of. In this space, if U is an open subset of the complex plane C, L^2_a(U) consists of all analytic functions f: U -> C such that ||f||^2 := \int\int_U |f(x+iy)| dx dy < \infty. I realise we never hashed out what the Bergman space is, because the first time you mentioned the name "Bergman" was the last post that I never responded to. A half finished response is sitting on my hard drive, and it's a good thing too, because it's hard to search for the Bergman space in Deja with Aaron around here! [and, of course, John got that Bergman space too in a later post] >[Erudite note: You may wonder why we use a Gaussian measure to define >the Bargmann space instead of Lebesgue measure. After all, Lebesgue >measure is the measure naturally associated to the symplectic >2-form! The problem is that the only holomorphic function on the >the complex plane that's square-integrable with respect to Lebesgue >measure is ZERO. Right. If we take Bergman space and set U = C, L^2_a(U) = 0. >So we need to tweak the prescription in a somewhat >ad hoc way - destroying (or at least hiding!) the translation-invariance >of the set-up, while maintaining the rotation-invariance. We don't >need to pull this trick when our symplectic manifold M is compact. >However, in that case we really need to work with a nontrivial line >bundle, since the only holomorphic functions on a compact connected >Kaehler manifold are the constants.] So, you've made it clear why the nontrivial line bundle is needed. But I still don't know how to choose it! >By the way, it's fun to think about the underlying algebra here. >Suppose you have a real vector space V with >1) a complex structure J: V -> V >2) a symplectic structure w: V x V -> R >3) a real inner product g: V x V -> R >Let's say items 1) - 3) are "compatible" if V becomes a complex >Hilbert space with ix = Jx for all x in V >and = g(x,y) + iw(x,y) for all x,y in V. >Show that if items 1) - 3) are compatible, any two of them >determines the third one. Figure out the conditions on any >two items for there to *exist* an item of the third sort that >makes V into a complex Hilbert space. Note that |x|^2 = = g(x,x) + iw(x,x) = g(x,x). That means knowing only g and J makes V a Banach space. But a Banach space which satisfies the parallelogram law has a unique Hilbert space structure, defining w. Specifically, w(x,y) = Im = i/2 - i/2 = /2 + /2 = Re = g(ix,y) = g(Jx,y). Similarly, g(x,y) = Re = /2 + /2 = -i/2 + i/2 = -Im = -w(ix,y) = -w(Jx,y). Finding J from g and w is easy because g and w are nondegenerate. Just reverse either of the above formulae. If we write "gx" for the covector \mu such that \mu(y) = g(x,y), "wx" for the covector \mu such that \mu(y) = w(x,y), "g\mu" for the vector x such that gx = \mu, and "w\mu" for the vector x such that wx = -\mu, then the above formulae give wx(y) = gJx(y), or wx = gJx, so Jx = gwx, and gx(y) = -wJx(y), or gx = -wJx, so Jx = wgx. Now, the compatibility condition for g and w is simply that gwx and wgx be equal. Alternatively, this may be written x = -gwgwx or x = -wgwgx. The compatibility condition for w and J may be found in section 2 of part II of Schmotons; it is that w(Jx,Jy) = w(x,y) and -w(Jx,x) >= 0. The compatibility condition for g and J is simpler; it is only that g(Jx,Jy) = g(x,y). >>Thus, the Fock space on M approach is a short cut to the power series, >>but nonlinear M requires that holomorphic functions be specified directly. >Righto! The only big lie left to undo is when I said we start >with L^2 functions on M. I gotta explain sometime why we need >L^2 sections of a line bundle. Well, I see *why*: otherwise, we'd just get 0 or C for the Hilbert space! I need to know *how*: which line bundle do we use? >>This works only if M is a vector space -- noninteracting QFT. >>So, interacting QFT is left without a rigorous treatment. >Yes, for that we need to work a bit harder. OK, so here is the question: Were you lying to me when you said canonical quantisation could be manifestly covariant in quantum field theory (secretly meaning only *noninteracting* quantum field theory), in which case I shall tell squark he was right after all, or is there still a way -- perhaps not very well defined -- to go from a manifestly covariant view of the classical phase space (as simply the space of solutions to the equations of motion) to the interacting quantum field theory? >>So, what *can* be said about interacting QFT? >Ah, that's a short question with a very long answer! Too long >for this post, anyway. Well, it gets to the heart of what squark and I were arguing about. -- Toby toby@ugcs.caltech.edu ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Geometric quantization Date: 16 May 2000 18:09:48 GMT Newsgroups: sci.physics.research In article <8fkm5a$osh@gap.cco.caltech.edu>, Toby Bartels wrote: >Neither of these is the Bergman space I was thinking of. >In this space, if U is an open subset of the complex plane C, >L^2_a(U) consists of all analytic functions f: U -> C >such that ||f||^2 := \int\int_U |f(x+iy)| dx dy < \infty. Just to be 200% clear: in the case where U is the open unit disc, this space is indeed called the "Bergman space". I wouldn't be surprised if people had generalized this terminology to other open sets in C, or even C^n, but I'm not sure they have. (They have generalized related concepts, such as the "Bergman kernel" and "Bergman metric", both of which first appeared in the special case of the open unit disc.) >>>This works only if M is a vector space -- noninteracting QFT. >>>So, interacting QFT is left without a rigorous treatment. >>Yes, for that we need to work a bit harder. >OK, so here is the question: >Were you lying to me when you said canonical quantisation >could be manifestly covariant in quantum field theory >(secretly meaning only *noninteracting* quantum field theory), >in which case I shall tell squark he was right after all, >or is there still a way -- perhaps not very well defined -- >to go from a manifestly covariant view of the classical phase space >(as simply the space of solutions to the equations of motion) >to the interacting quantum field theory? So you're really interested in *interacting* quantum field theories, eh? Just to be 200% clear, I wasn't lying when I said canonical quantization could be manifestly covariant in quantum field theory: I was just talking about the only case we really understand very well, namely the noninteracting case. And just to repeat.... In this case you can take the space of classical solutions of a linear wave equation as your "classical phase space", improve it from a symplectic vector space to a Hilbert space H in the only Poincare-invariant way you can that makes energy nonnegative, and then take the Bargmann-Segal space of H as the Hilbert space for your quantum theory. Manifestly invariant at every stage! >>>So, what *can* be said about interacting QFT? >>Ah, that's a short question with a very long answer! Too long >>for this post, anyway. >Well, it gets to the heart of what squark and I were arguing about. Okay, okay, so you really want to know about the interacting case. In short, you're asking: Can canonical quantisation be manifestly covariant in an interacting quantum field theory? If I have to give you a completely rigorous answer where I can prove that everything I'm saying is correct, it'll be nice and short: "I don't know." But you seem to be willing to cut me some slack and let my procedure be "not very well defined". This gives me more scope for creativity! In particular, I can say this: Take the space M of classical solutions of a nonlinear wave equation as our "classical phase space". This will typically have a god-given symplectic structure that's invariant under all the symmetries of the problem. Then pick a Kaehler structure extending it. This will typically *not* be invariant under all the symmetries, but if we're lucky, it will be "close to invariant". This concept can actually be made precise... but I won't do so here. Then cook up some Hilbert space of holomorphic functions on M and use this as the Hilbert space of our quantum theory. We have to say how our symmetries act on this Hilbert space. It's trickier, now that they don't exactly preserve the Kaehler structure. But people have thought about this stuff a lot and have good ideas. Has this program ever been completely worked out in examples? Not that I know of! But people have come close. Usually people break down and use the crutch of a *noncovariant* description of the phase space M in terms of initial data at a given time. They use this to identify M with the vector space of solutions of some *linear* equation and then proceed from there. Do you consider this cheating? Probably. Basically, constructive quantum field theory is so hard that people don't try too hard to do it *elegantly* - they are happy if they can do it at all! Ideally one would find a Kaehler structure invariant under all the symmetries of the problem and use that. That's why I was happy when I found a Poincare-invariant Kaehler structure on the space of solutions of the equation (BOX + m^2) phi + c phi^3 = 0 in 3+1 dimensions. I thought I might use this to rigorously construct the massive phi^4 theory. However, then I noticed there were infinitely many different Poincare-invariant Kaehler structures on this space, all with the same symplectic structure... and then I got a bit disgusted and quit. If I wanted to carry out a program of "manifestly invariant geometric quantization" for field theories, I would now probably start with examples in 1+1 dimensions - like conformal field theories. People know so much about these examples that one might be able to do it quite easily. First read "Loop Groups" by Pressley and Segal, then take the phase space of the WZW model and start looking for nice Kaehler structures on it....