From: baez@galaxy.ucr.edu (John Baez) Subject: Re: C*-algebras Date: 1 Feb 2000 06:37:53 GMT Newsgroups: sci.physics.research In article <872s78$9d9@gap.cco.caltech.edu>, Toby Bartels wrote: >John Baez wrote something like: >>for any operator a there is some isometry v for which >>a = v |a|, >>which is the polar decomposition of a. >Actually, we don't want this for our isometry I. >This is because I*I = 1, so |I| = 1 and v = I. >What we want is to replace the above by >|a| := sqrt(aa*), a =: |a| v. >The theory, of course, is identical. >(Alternatively, we *do* want the original version, >only applied to I*, not I.) Right, that might be more informative. However, I'm getting the feeling that this is not the easiest way for you to solve this puzzle. >>I guess this means that the polar decomposition is not something >>we can do in a C*-algebra. However, we can do it in a W*-algebra >>(aka von Neumann algebra) - which is why I was able to do it above >>in the algebra of all bounded operators on a Hilbert space. And >>every C*-algebra has a universal enveloping W*-algebra, so this is >>good enough. >O, is it? Maybe I should learn about W*algebras. You don't need to know about them for this puzzle, but they're nice to know about. The basic idea is simple: just as C*-algebras generalize the algebra of bounded continuous functions on a topological space, W*-algebras generalize the algebra of bounded measurable functions on a measurable space. In particular, just as Banach algebras come equipped with a "holomorphic functional calculus" and C*-algebras come equipped with a "continuous functional calculus", W*-algebras come equipped with a "measurable functional calculus". In other words: Given any element x of a Banach algebra and any bounded holomorphic function f on the spectrum of x, we can define f(x). Given any normal element x of a C*-algebra and any bounded continuous function f on the spectrum of x, we can define f(x). And given any normal element x of a W*-algebra and any bounded measurable function f on the spectrum of x, we can define f(x). In each case this "functional calculus" has similar nice properties. Now, the most important functions that are measurable but not continuous are the characteristic functions: functions that are 1 on some measurable set and 0 off it. So a very fun thing to do in a W*-algebra is to take a normal element x, take a measurable subset of the spectrum of x, let f be the characteristic function of that subset, and define f(x) using the functional calculus. For hopefully obvious reasons having to do with the "nice properties" alluded to above, f(x) is a projection, i.e. a self-adjoint element whose square is itself. Since projections represent "yes-or-no questions" in quantum mechanics, W*-algebras are very handy in physics. I hope you see why C*-algebras don't cut it here: we can't usually get a yes-or-no question out of an observable in a C*-algebra, since saying "yes" or "no" depending on the value of an observable is a discontinuous thing to do, and C*-algebras only have a *continuous* functional calculus. But don't get the wrong impression that W*-algebras are drastically different from C*-algebras; they're not. They go hand in hand just like measure theory goes hand in hand with topology. >>U is unitary iff UU* is the identity and U*U is the identity. >>I is an isometry iff II* is a projection and I*I is the identity. >>P is a partial isometry iff PP* is a projection and P*P is a projection. >To round this out, what do you call I* if I is an isometry? I *knew* you were going to ask that. As I was writing that little chart, I thought: "He won't be able to resist asking about the missing fourth entry." The thing you're asking about is usually called "the adjoint of an isometry", but if you insist on a snappier name, we can call it a "coisometry". I'm not sure anyone else calls it that, though! >>>Now let's look at the most vanilla flavoured isometry. >>>A fairly ordinary isometry is the shift map S on l^2. >>YES!!! This is, in fact, exactly what I was hoping you would guess. >>If you can figure out what this has to do with the unit disk, you will >>be almost done. >That's pretty much what I've been trying to figure out from the beginning. >Bringing l^2 into it doesn't give me any clues. No? Aha. That's what I feared, which is why I said you might have to reinvent some well-known mathematics. So here's the real puzzle behind the puzzle: what does l^2 have to do with the unit disk? Since I don't mean to torture you unduly, I'll give you a big clue. You've already answered my puzzle in one way: you've noticed that the free C*-algebra on an isometry is the same as the C*-algebra of bounded operators on l^2 generated by the "left shift operator". But I want a more geometrical way of thinking about this, which is related to the unit disk somehow. So you gotta figure out how l^2 is related to the unit disk.... You may know that any decent function on the unit circle T has a "Fourier expansion" in terms of powers z^n, where z is the identity function on the unit circle. (We'd been calling this function t, but I'm trying to nudge your thoughts in the direction of the complex plane, the unit disk, and stuff like that). This trick lets us see, for example, that L^2(T) is secretly the same thing as L^2(Z) - i.e., square-summable sequences that go off infinitely in *both* directions. But now instead you've got l^2 - i.e., square-summable sequences that go off infinitely in *one* direction - and I'm claiming this has something to do with the unit disc. ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: path integral - sum over all paths? Date: 3 Jun 2000 00:57:44 GMT Newsgroups: sci.physics.research In article <8gmi09$jsv$1@nnrp1.deja.com>, wrote: >John Baez wrote: >> Let the *-algebra of bounded cylinder functions on V be denoted >> Cyl(V). Cyl(V) is a normed *-algebra with the sup norm. We can >> complete it to a commutative C*-algebra Fun(V). >This brings up an interesting thought: commutative C*-algebras are >known to correspond to locally compact Hausdorff space via the >Naimark-Segal (I hope I don't confuse the names!) construction. It's Gelfand-Naimark. Gelfand and Naimark were the Russian duo; Segal was the guy who did some similar things here in America. >On the >other hand, V is infinite-dimensional and thus not locally compact, so >this space has to be something else. What may it be? Well, it's tough to describe this space *explicitly*, since its points are the maximal ideals in Fun(V), and it's not easy to describe all of these. But it's a compact Hausdorff space in which V is dense subset: points in this bigger space are certain "limits" of points in V, in the topology such that x_i -> x iff f(x_i) -> f(x) for every cylinder function f. There are some very general facts at work here: Suppose X is a topological space and A is some algebra of bounded continuous complex-valued functions on X that's closed under pointwise complex conjugation. Then there's an obvious way to complete A and obtain a commutative C*-algebra. By the Gelfand-Naimark theorem this C*-algebra is isomorphic to C(Y), the algebra of all continuous complex-valued functions on Y, for some compact Hausdorff space Y which is unique up to canonical isomorphism. Okay, now how does X relate to Y? Well, there's a continuous map i: X -> Y since any point of X determines a maximal ideal in C(Y). So in some very rough sense, Y is a "compactification" (and "Hausdorffication") of X. The image of i will always be dense in Y. But i won't be one-to-one if A doesn't contain enough functions. That's bad. So let's say i "separates points" if for any two distinct points x and x' in X there is a function f in A such that f(x) is not equal to f(x'). The map i: X -> Y will be one-to-one iff A separates points. If i is one-to-one, we have more of a right to say that Y is a "compactification" of X, since we can think of it as including X as a subset. However, the topology induced on X from being a subset of Y may be weaker than the original topology of X, so Y need not be a compactification in the usual sense, which demands that the induced topology on X match its original topology. This stuff is pretty important for understanding integration over infinite-dimensional spaces - not only vector spaces, but spaces of connections mod gauge transformations and other such spaces that are important in quantum field theory. Typically you have to go from some space X of "smooth fields" to a larger space Y of "singular fields" to get ordinary measure theory to work nicely. And the way to do this is to pick some algebra A of bounded continuous functions on X, and do the above trick. Typically A should contain functions whose integrals you really want to be well-defined. And typically A should separate points. In ordinary quantum field theory, folks often take A to consist of "cylinder functions" when X is a vector space. But in loop quantum gravity, we take A to be something very different, because X is not a vector space - it's the space of SU(2) connections mod gauge transformations. ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: path integral - sum over all paths? Date: 5 Jun 2000 00:18:46 -0700 Newsgroups: sci.physics.research In article <8hc0t9$kgs@gap.cco.caltech.edu>, Toby Bartels wrote: >John Baez wrote: >>squark wrote: >>>This brings up an interesting thought: commutative C*-algebras are >>>known to correspond to locally compact Hausdorff space via the >>>Naimark-Segal (I hope I don't confuse the names!) construction. >>It's Gelfand-Naimark. Gelfand and Naimark were the Russian duo; >>Segal was the guy who did some similar things here in America. >I have seen books that call it "Gelfand-Naimark-Segal"; >I even have one that goes so far as to abbreviate it "GNS". I think you may be mixing up the Gelfand-Naimark theorem and the Gelfand-Naimark-Segal construction. The former is a classification of commutative C*-algebras; the latter leads to a classification of all C*-algebras. The Gelfand-Naimark theorem says that every commutative C*-algebra is isomorphic to the algebra of continuous functions on a compact Hausdorff space. (If we allow commutative C*-algebras without multiplicative identity, as squark seems to prefer, we must use continuous functions vanishing at infinity on a locally compact Hausdorff space.) The Gelfand-Naimark-Segal theorem is a way of getting a representation of a C*-algebra from a state on it. With a little trickery this implies that every C*-algebra is isomorphic to a closed *-subalgebra of the bounded linear operators on a Hilbert space. [construction of space Y from C*-algebra A deleted....] >If A is all of C(X), then isn't Y the Stone Cech compactification of X? Yeah, but don't people only talk about the Stone-Cech compactification of a locally compact space? ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: path integral - sum over all paths? Date: Tue, 6 Jun 2000 23:03:15 GMT Newsgroups: sci.physics.research In article <8hc0t9$kgs@gap.cco.caltech.edu>, Toby Bartels wrote: >John Baez wrote: >>Suppose X is a topological space and A is some algebra of >>bounded continuous complex-valued functions on X that's closed >>under pointwise complex conjugation. Then there's an obvious >>way to complete A and obtain a commutative C*-algebra. By the >>Gelfand-Naimark theorem this C*-algebra is isomorphic to C(Y), >>the algebra of all continuous complex-valued functions on Y, for >>some compact Hausdorff space Y which is unique up to canonical >>isomorphism. There's a continuous map i: X -> Y since any point of X >>determines a maximal ideal in C(Y). So in some very rough sense, >>Y is a "compactification" (and "Hausdorffification") of X. >If A is all of C(X), then isn't Y the Stone Cech compactification of X? Some extra comments here... First of all, you really meant to ask: "If A is all of the *bounded* continuous functions on X, then isn't Y the Stone Cech compactification of X?" - unless you are secretly using C(X) to stand for bounded continuous functions on X. We need boundedness for A to be a C*-algebra. And then the answer is: "Morally speaking yes, but most people use the term `compactification' only if the original topology on X matches the induced topology it gets by thinking of it as a subspace of Y using the map i: X -> Y. And this will be true only if some conditions hold. I think the usual condition people consider is that X be locally compact, but your favorite functional analysis textbook (by Conway) seems to favor the condition that X be completely regular, which seems to be necessary and sufficient."