From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Re: Geometric Quantization Date: 12 Aug 2000 06:47:21 GMT Newsgroups: sci.physics.research Summary: [missing] John Baez wrote in large part: >Toby Bartels wrote: >>See here, John: the standard example of phase space >>is the cotangent bundle of a previously given configuration space. >>Now, do we recover configuration space as a subspace of this? >>No! we recover configuration space as a *quotient* space of this. >The projection p: T*M -> M exhibits the configuration space M >as a quotient space of the phase space T*M. And the zero section >s: M -> T*M exhibits the configuration space as a subspace of the >phase space. O, sure, I forgot that the 0 vector is special. >I won't deny that the projection p is somehow more "fundamental" >than the zero section s, but they are both perfectly natural in >the category-theoretic sense, and they are both interesting. And sp: M -> M is the identity map, which shows that the natural transformations s and p are split monic and split epic, respectively. Whatever good that does. >>So, I would define a coLagrangian quotient space of a Poisson manifold >>to be a maximal quotient space satisfying the property that, >>if f and g are functions on the alleged Lagrangian quotient space, >>then {f,g} = 0 when f and g are interpreted as functions on phase space. >>(A not necessarily maximal quotient space with this property >>would be a coisotropic quotient space.) >But I never got around to finishing the definition of a real polarization! >Here it is: we want Lagrangian subspaces L_x of the tangent spaces T_x(X), >these subspaces should vary smoothly with x, and they should fit together >in a nice way: they should be "integrable". That means they are the tangent >spaces of submanifolds that foliate X. (And yes, it follows that these are >Lagrangian submanifolds... whoops, so they *are* somewhat relevant here.) >Let's do an example, quick! Let X be the cotangent bundle T*M - the >most traditional example of a phase space. Then at each point x of X >we let L_x be the space of "vertical" tangent vectors: those that >project down to zero when we apply the projection pi: T*M -> M. It's >easy to check that L_x really is a real polarization, and the submanifolds >foliating X are just the cotangent spaces of M. >Toby will enjoy pondering this and seeing how it fits in with the >yoga of quotient spaces versus subspaces... I won't deprive him of >this pleasure by doing it for him here. Well, like I said, you were really dealing with the dual space all along. See how you're treating M as a quotient space of X, not a subspace? Now, first, is M a coLagrangian quotient manifold of X as I defined it? Well, if f and g are functions on M, then they must pull back to commuting functions on X, since they're functions only of q, not p. Therefore, M is coisotropic; any extension to M introduces p dependance, so M is maximal among coistoropic quotient spaces. Thus, M is coLagrangian. What does this have to do with the Lagrangian subspaces L_x of T_x X? Well, these subspaces are what we're sending to 0 in M. (IOW, they tell us the directions where the covariant derivatives of the line bundle sections which we want to keep *vanish*.) So, if we want to recover M and learn the identity of the space on which our line bundle sections will actually appear to live, we must look at the quotient spaces (L_x)+ := (T_x X)/L_x. (L_x)+ will be coLagrangian because L_x was Lagrangian. Thus, if all you want to do is specify the polarisation, sure, go ahead, just mention the Lagrangian subspaces. But, if you want to calculate the correct analogue for M, you're going to have to deal with coLagrangian quotient spaces. Exercise: If M is an arbitrary coLagrangian quotient space, as I defined it, of the arbitrary symplectic manifold X, then M arises in this fashion from a valid real polarisation of X. I guess the lesson here is that the quotient space corresponding to an isotropic subspace need not be coisotropic, nor vice versa. But the maximality conditions assure that this does follow in the case of Lagrangian and coLagrangian. This lets us look at everything from either side. >Suppose you start with a symplectic vector space V. We >get a map V -> V* given by v -> w(v,.), and this is an >isomorphism. Using this isomorphism we can transfer the >symplectic structure on V to one on V*. And now we can >play the game all over again! We get a map V* -> V**, >which turns out to be an isomorphism. >Composing these isomorphisms V -> V* and V* -> V**, we get >an isomorphism V -> V**. Now there happens to already be >a famous isomorphism from V to V**, which we use to identify V with V** (or, without finite dimensions, with a subspace of V**). I anticipate the punchline already (because I've thought of this): This isomorphism, under this identification, is additive inverse. This makes me want ww (that's omega omega) to always be -1. This comes up when deciding on a notation for the isomorphism w induces from V to V*. First, suppose w has been nicely defined on V, so that w(v,v') makes sense when v,v' in V. Then we'd like wv in V* to be the functional such that = w(v,v'). Fine, done. Now, if L is a linear functional in V*, what should wL be? Is it v where wv = L? No! for then we'd have wwv = v, which is wrong. Rather, wL is v where wv = -L, so wwv = -v. Then define w(L,L') to be , and we're done. Note that w(wv,wv') = = - = -w(v,v'), so again ww becomes -1 here. Also, w(wL,wL') = -w(L,L'). It all works out! >And eventually, pondering this, one sees that it's all part >of the analogy: >bosons: symplectic vector spaces :: fermions : real inner product spaces >The difference between bosons and fermions is just a sign..... And anyons ... -- Toby toby@ugcs.caltech.edu ============================================================================== From: baez@galaxy.ucr.edu Subject: Re: Geometric Quantization Date: 12 Aug 2000 21:50:57 GMT Newsgroups: sci.physics.research In article <8n2rtp$2be@gap.cco.caltech.edu>, Toby Bartels was talking about the situation where we start with a configuration space M, form the phase space T*M = X, and then do geometric quantization to get a Hilbert space of functions on X that are constant along the fibers of p: X -> M. These functions are really just functions on M.... >See how you're treating M as a quotient space of X, not a subspace? Yup. >Now, first, is M a coLagrangian quotient manifold of X as I defined it? Yes! >Exercise: If M is an arbitrary coLagrangian quotient space, >as I defined it, of the arbitrary symplectic manifold X, >then M arises in this fashion from a valid real polarisation of X. Yes, this is pleasant. But it's also worth noting a less pleasant fact: if we start with a real polarization of a symplectic manifold X, we get a foliation of it by Lagrangian leaves - but the space of leaves, a certain quotient space of X, need not be a manifold! Exercise: Let X be the phase space of a particle on the line, with its usual symplectic structure. Let H be the harmonic oscillator Hamiltonian: H(q,p) = (p^2 + q^2)/2 Using this data cook up an obvious real polarization of X, and work out the space of Lagrangian leaves. Guess how geometric quantization will work with this polarization. >I guess the lesson here is that the quotient space corresponding to >an isotropic subspace need not be coisotropic, nor vice versa. By the way, I'm afraid you're gonna run into trouble with this "coisotropic quotient space" terminology - the word "coisotropic" is already in widespread use to mean a particular sort of *subspace* of a symplectic vector space. Let me describe this, since it's sort of fun. Given a symplectic vector space (V,w) and a subspace L of V, there's a subspace called the "perp" of L, which consists of all vectors v such that w(v,x) = 0 for all x in L. Note: this is defined just like the "orthogonal complement" of a subspace of a space with an inner product, but it acts quite differently, because in a symplectic vector space, all vectors are perpendicular to themself! I'll denote denote the perp of L by L+, since my keyboard doesn't do a real "perp" symbol: _|_ Now, folks say a subspace L is "isotropic" if L is contained in L+, and they say it's "coisotropic" if L+ is contained in L, and they say it's "Lagrangian" if it's both isotropic and coisotropic. This is equivalent to 4 other definitions of a "Lagrangian subspace": 1) a maximal isotropic subspace, 2) a minimal coisotropic subspace, 3) an isotropic subspace whose dimension is half that of V, 4) a coisotropic subspace whose dimension is half that of V. As you've emphasized, it's fun and worthwhile to think about what happens when we mod out by any one of these sorts of subspaces. I guess you can define V/L to be a coisotropic quotient space iff L is an isotropic subspace, and an isotropic subspace iff L is a coisotropic subspace - or maybe the other way around, if that seems better.... >>Suppose you start with a symplectic vector space V. We >>get a map V -> V* given by v -> w(v,.), and this is an >>isomorphism. Using this isomorphism we can transfer the >>symplectic structure on V to one on V*. And now we can >>play the game all over again! We get a map V* -> V**, >>which turns out to be an isomorphism. >>Composing these isomorphisms V -> V* and V* -> V**, we get >>an isomorphism V -> V**. Now there happens to already be >>a famous isomorphism from V to V**, >which we use to identify V with V** >(or, without finite dimensions, with a subspace of V**). >I anticipate the punchline already (because I've thought of this): >This isomorphism, under this identification, is additive inverse. Hey! You really thought of that already??? Cool! >This makes me want ww (that's omega omega) to always be -1. >This comes up when deciding on a notation for >the isomorphism w induces from V to V*. >First, suppose w has been nicely defined on V, >so that w(v,v') makes sense when v,v' in V. >Then we'd like wv in V* to be the functional >such that = w(v,v'). Fine, done. >Now, if L is a linear functional in V*, >what should wL be? Is it v where wv = L? >No! for then we'd have wwv = v, which is wrong. >Rather, wL is v where wv = -L, so wwv = -v. >Then define w(L,L') to be , and we're done. >Note that w(wv,wv') = = - = -w(v,v'), >so again ww becomes -1 here. Also, w(wL,wL') = -w(L,L'). >It all works out! Gadzooks!