From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Geometric quantization Date: 17 May 2000 19:23:37 GMT Newsgroups: sci.physics.research Summary: [missing] In article <8ft3lq$q6f@gap.cco.caltech.edu>, Toby Bartels wrote: >John Baez wrote: >>Given a smooth manifold M and a symplectic structure w on M, we >>want a recipe to pick a complex line bundle L over M. We will not >>quite succeed in choosing L in a *functorial* way, but we will be >>able to choose L in a way that's *unique up to isomorphism*. As >>you know, these are very different things, and the latter is but a >>poor substitute for the former - but hey, I never promised you that >>quantization was straightforward. >Like the algebraic completion of a field is unique but not functorial? Hmm, I haven't thought about that much, but I do recall endless arguments on sci.math about whether the algebraic completion of the rationals is a subset of the complex numbers or not. As so often the case in such endless arguments, the crucial thing is your definition of words "is" and "the". If there's an embedding of the algebraic completion of the rationals into the complex numbers, but no canonical one, one is abusing language to argue about whether it "IS" a subset - one should say it "can be made" into a subset. And if there's no functorial way to algebraically complete the rationals, our very use of the term "THE" algebraic completion is somewhat sloppy! We can use "the" for a uniquely defined element of a set. We can also use it for an object in a category that is defined up to canonical isomorphism - as when we speak of "the" tensor product of vector spaces. And so on up the n-categorical ladder - James Dolan speaks of the "generalized the" in this case. But if we know an object in a category up to isomorphism, but not canonical isomorphism, we are courting danger when we speak of "the" object. We just don't have a firm enough grip on it in this case: we have it in our hands but it's wiggling around in an uncontrollable way, like a greased pig. The problems with quantization arise from issues like these! Operator ordering ambiguities, phase ambiguities, projective representations... all are problems of not having a tight enough grip on things. You might enjoy thinking about just what it would take for us to get ahold of "the" line bundle L up to canonical isomorphism. So far I've said we can get ahold of L up to isomorphism using the cohomology class [w] defined by the symplectic structure w. I haven't actually said how we do it - I've actually just shown that starting from L we can get [w]. But once you see how the whole story goes, you could think about these things some more. For example, one might hope that if we used all the information in w, we could actually get ahold of L up to *canonical* isomorphism. I don't think this is true. But there are fallback positions: one can try to figure out exactly what, in addition to w, one needs to canonically get ahold of L. If you did this, you might make some progress on the deepest mysteries of quantization. Okay... I'll skip most of the stuff where you went through what I said and checked it and fixed it up here and there... I'll just comment on some of the interesting bits... you can assume anything I don't comment on is correct. >>If we trivialize the bundle locally, we can think of A as a >>u(1)-valued 1-form, and we can think of its curvature F as a >>u(1)-valued 2-form. In fact, since U(1) is abelian, how we >>think of F as a u(1)-valued 2-form is independent of our choice >>of trivialization! >In fact, I defined F above directly in terms of the connection A >rather than in terms of the u(1)valued 1form A. >Therefore, it had to be independent of trivialisation; >I only used a trivialisation to prove that it *was* a u(1)valued 2form >(rather than having some other value or not being a tensor). >This makes me wonder why U(1) must be Abelian for this to work; >as I recall, all an Abelian group gets you >is that the formula is F = dA instead of F = dA + [A ^ A]. >(Of course, the formula is still F = dA + [A ^ A]; >it's just that [,] = 0 when the group is Abelian.) For a general gauge group G with Lie algebra L, we can think of the curvature F as an L-valued 2-form after we trivialize our G-bundle. Suppose we do this (e.g. locally) and then change our trivialization via a G-valued function g on the base manifold of our bundle. Then F changes like this: F' = g F g^{-1} When G is abelian this reduces to F' = F So when G is abelian, we can really identify the curvature with an L-valued 2-form in a trivialization-independent way. In this case there is no harm, even globally, in treating the curvature as an L-valued 2-form on our base manifold. But when G is nonabelian, we cannot globally treat the curvature as an L-valued 2-form on our base manifold. Instead, we can treat it as a 2-form on our base manifold which takes values *in a vector bundle* - a vector bundle whose fibers are isomorphic to L, but not canonically. I partially explained this in "Gauge Fields, Knots and Gravity", so you might try that - but a deeper explanation would require more about principal bundles than I had time for in that book. Anyway, here we are in the lucky case where the curvature F really is a u(1)-valued 2-form on the whole manifold M. >>So we can *globally* identify F with >>a u(1)-valued 2-form. Since there's a god-given best isomorphism between >>u(1) and the real numbers, >Really? Why is i |-> 1 better than -i |-> 1? >For that matter, why not use 7i |-> 1 instead? I didn't say that there was a *unique* god-given best isomorphism. Remember, by "god-given best" I just mean "canonical" or sometimes "functorial". In this case, there are infinitely many canonical isomorphisms! (Yes, I am weaseling out of what I said, but very elegantly!) >In fact, I charge that this construction really uses 2ipi |-> 1; >if you did that in the beginning, you wouldn't divide by 2pi later. >And 2ipi |-> 1 is really more natural when you think of u(1) as ln U(1). You are perfectly right! Some people working on Chern classes, like Quillen, have made this remark before. We should really think of our u(1)-valued 2-form F as taking values in the imaginary numbers - since that's what elements of u(1) really are - and then we should divide by 2 pi i to get a real-valued 2-form. This makes formulas much nicer. I tend to forget that. In short, I commend you for spotting this defect in my exposition and fixing it! >>Since [the cohomology class of F/2 pi i] doesn't change when we apply >>an isomorphism to L, it really depends only on the isomorphism class of M. >You mean of L. Right, thanks. >>For reasons soon to become clear, we get something called the >>FIRST CHERN CLASS of the line bundle L and denoted c_1(L). >The *1st* Chern class belongs to the *2nd* cohomology group :-). Well, this is not so bad, since the *nth* Chern class belongs to the *2nth* cohomology group. To briefly sketch the theory: the nth Chern class of a complex vector bundle is formed by picking a connection on it, forming the curvature F, and doing something with F ^ F ^ ... ^ F ^ F |------n times-----| to get a real-valued closed 2n-form which defines an integral element of the 2nth cohomology group of the base manifold, independent of our choice of connection. Line bundles are classified by their 1st Chern class and all the higher Chern classes are trivial. Higher-dimensional vector bundles have interesting higher Chern classes but, alas, are not classified by these Chern classes. For more see "Gauge Fields, Knots and Gravity" - you'll see your remarks about 2 pi i are borne out. For even more try Milnor and Stasheff's book "Characteristic Classes", starting with the appendix. This stuff is incredibly cool - your mind will be blown several times before you master all of it. >>Considerably deeper thought leads to the realization that c_1(L) is >>not just a member of the deRham cohomology H^2(M): it is actually a >>member of the integral cohomology H^2(M,Z). Eh? Well, you may recall >>that H^2(M,Z) can be defined for any space M, not just any manifold - >>but when M is a manifold we can tensor H^2(M,Z) with the real numbers and >>get a vector space canonically isomorphic to the deRham cohomology H^2(M). >Happily, I already knew this. Whew - otherwise the story would take a long long time. >>This allows us to think of H^2(M,Z) as a lattice sitting inside the deRham >>cohomology H^2(M). And it turns out that c_1(L) sits inside this lattice! >I don't think I'm going to prove this now. Yeah, this is where it starts getting a bit deeper - but like I said: >>In fact, the best way to see this is to give an alternate definition >>of c_1(L) which makes it clear from the start that it's an element of >>the integral cohomology H^2(M,Z). This alternate definition doesn't >>require that M be a manifold; it works for any space M. And when we >>proceed this alternate way, it also becomes evident that: >>1) c_1(L) determines L up to isomorphism and >>2) any element of H^2(M,Z) is c_1(L) for some line bundle L. >>In short, isomorphism classes of line bundles on a space M are really >>just the same thing as elements of H^2(M,Z)! This is the beginning >>of a long and beautiful story with endless ramifications... which, sadly, >>I haven't the time to tell right now. >If you had, we could have skipped all of the above >*and* benefitted by the understanding of why c_1(L) must be integral, >so clearly the story must be quite a bit longer than what we just did. Well, the purely topological approach to the first Chern class is incredibly important, but so is the approach involving connections! When we do geometric quantization, we ultimately need a line bundle L *together with a connection* whose curvature is i w, so the connection stuff is not just a clunky way of doing something we could have done more elegantly. >>Note that we will only *succeed* in finding our line bundle L if [w] >>is 2pi times an integral cohomology class - otherwise we are just out >>of luck! This is an updated version of the BOHR-SOMMERFELD QUANTIZATION >>CONDITION. >Sure, that also involved dividing dq ^ dp by 2pi and getting an integer. Right - all this cohomological stuff is really just an explanation of why this integrality condition arises. Basically, when you take the phase of something and start changing it, it's back to where it was after you add any *integer* multiple of 2 pi i - that's what it all boils down to. >>Once you get good at geometric quantization, you can see >>how this condition implies all sorts of familiar things like the >>quantization of angular momentum. >Whatever "angular momentum" means in such an abstract context. Oh, it's not so abstract: start with the phase space of a "classical spin-j particle", which is a 2-sphere of radius j. This describes all the ways we can have an angular momentum vector of length j. Now, there's a nice way to take this sphere and equip it with a Kaehler structure that's just a multiple of the usual Kaehler structure for the Riemann sphere. When you do geometric quantization to this sphere, you then get the space of states of a *quantum* spin-j particle! But here's the cool part: only when j is of the form 0, 1/2, 1, 3/2,... can you actually DO the geometric quantization, since only in this case does the symplectic structure w, after dividing by 2 pi, give an *integral* cohomology class. >>Now, the big question is: WHY? Why use such an elaborate procedure >>to pick the line bundle L? Well, once you learn about line bundles >>this procedure doesn't seem so elaborate, but it's still important to >>see why we use it! However, I am running out of energy to write this >>post, and you probably already ran out of energy to read it, so I'll >>make sure you understand what I'm saying and want to know more before >>telling you more. >Run out of energy to read it? Are you kidding? >I hope I've shown that I understand it; I want more! Hmm, I'm worn out just replying to your previous comments, so I can't say more now. Please don't reply to this post, or if you do, only say very dull things. Then I can go ahead and give you a clue about why the first Chern class of a line bundle is *integral* - and how you get the line bundle (up to isomorphism) from its first Chern class.