From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Re: This Week's Finds in Mathematical Physics (Week 151) Date: Sun, 2 Jul 2000 17:11:16 GMT Newsgroups: sci.physics.research,sci.physics,sci.math Summary: Classifying spaces of groups and the Right categories for topology John Baez wrote: >If G is any topological group, there is a topological space BG with a >basepoint such that the space of loops in BG starting and ending at >this point is homotopy equivalent to G. This space BG is unique up >to homotopy equivalence. IOW, LBG = G, where "=" means homotopy equivalence. But this reminds me, how do I define the topology on LX? And what's the difference between LX and pi_1(X)? >Even better, >two G-bundles that we get this way are isomorphic if and only if the maps >they come from are homotopic! Is there a correspondence between isomorphisms of bundles and homotopies of maps? >H^n(X,A) = [X, K(A,n)] >where the right-hand side is the set of homotopy classes of maps from X >to K(A,n). In short, K(A,n) knows everything there is to know about the >nth cohomology with coefficients in A. So, we can define cohomology with coefficients in a *topological* Abelian group by H^n(X,A) := [X, K(A,n)]. >I can't resist explaining why the unit sphere in an infinite-dimensional >Hilbert space is contractible. It seems very odd that a sphere could be >contractible, but this is one of those funny things about infinite >dimensions. Well, if it weren't contractible, what homotopy group would be nontrivial? pi_1(S^oo) is trivial, since only pi_1(S^1) is nontrivial. pi_2(S^oo) is trivial, since only pi_2(S^2) is nontrivial. pi_3(S^oo) is trivial, since only pi_3(S^3) is nontrivial. And so on. >We can visualize CP^infinity quite easily this way. A rational function >of a single complex variable has a bunch of zeros and poles - think of >them as points on the Riemann sphere. We should really stick an integer >at each of these points: a positive integer at each zero, and a negative >integer at each pole, to tell us the order of that zero or pole. This >gives enough information to completely specify the rational function up >to a constant factor. So a point in CP^infinity is the same as a finite >set of points on the sphere labelled by integers - which must add up to >zero. For the algebraic geometers, then, we just have a divisor on the Riemann sphere of degree 0. >The space >of Hilbert-Schmidt operators is a Hilbert space in its own right, >with this inner product: = tr(AB*). Speaking like a mathematician, eh? Here on sci.physics.research, we say = tr(A*B). >Now, so far I've been restraining myself from talking about "gerbs", >7) Lawrence Breen, On the Classification of 2-Gerbes and 2-Stacks, >Asterisque 225, 1994. Throughout, you say "gerbs", but all your references say "gerbes". What's going on here? -- Toby toby@ugcs.caltech.edu ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: This Week's Finds in Mathematical Physics (Week 151) Date: Thu, 6 Jul 2000 16:51:46 GMT Newsgroups: sci.physics.research,sci.physics,sci.math toby@ugcs.caltech.edu (Toby Bartels) wrote: >John Baez wrote: >>If G is any topological group, there is a topological space BG with a >>basepoint such that the space of loops in BG starting and ending at >>this point is homotopy equivalent to G. This space BG is unique up >>to homotopy equivalence. >IOW, LBG = G, where "=" means homotopy equivalence. Right! >But this reminds me, how do I define the topology on LX? Jim Dolan came up to me the other day and told me you asked this question. My newsserver has dead since last Tuesday, so I'm having trouble reading articles here, and I didn't see this one of yours. He then proceeded to give me a half-hour lecture on how I should reply to this question. Writing up what he said would take at least least two hours, so I'm afraid I won't do that. I'll just start with a straightforward answer your question and then mention some of the deeper issues it leads into. Straightforward answer - the compact-open topology. Given topological spaces X and Y, let hom(X,Y) be the set of continuous maps from X to Y. In the "compact-open topology", the open sets in hom(X,Y) are generated from sets of this form: {f in hom(X,Y) : f(K) is contained in U} where K is compact and U is open. Deeper issues - There's a general question lurking here: what's the right topology to put on the space of continuous maps from a space X to a space Y? So we should think about this. Wait a minute! Actually you are interested in the space of *pointed* continuous maps from a *pointed* space X to a *pointed* space Y, since I was talking about loops that start and end at a special point. Luckily, this does not affect the basic issues: once we know the right topology on the space of all continuous maps, we can restrict it to get the topology on the space of pointed continuous maps. So I will work in the unpointed context. Okay: let hom(X,Y) be the set of all continuous maps f: X -> Y. We want to know the right topology on hom(X,Y). But what does "right" mean? What properties do we want? Here's one: a continuous map from X to hom(Y,Z) should be the same thing as a continuous map from X x Y to Z. Why is this so good? Well, it says the functor hom(Y,-) is right adjoint to the functor - x X. And that's a very nice thing. Technically, it says our category of spaces is "Cartesian closed". What it really means, though, is that we're going along with the Tao of mathematics. Okay, now for the problem: there's no way to pick a topology on hom(X,Y) that gives this nice property. At least, not any way that I've ever heard about! If there is a way, there must be something awful about it, because nobody ever talks about it. I bet there isn't one. What people do instead is replace the category of topological spaces by a slightly nicer category. There are lots of ways to do this but one popular approach involves "compactly generated Hausdorff spaces". You know what a "Hausdorff" space is: it's one where you can find disjoint open neighborhoods for any pair of distinct points. A space is "compactly generated" if a subset C is closed iff its intersection with every compact set is closed. It's easy to check that any locally compact space is compactly generated, and so is any space whose topology comes from a metric - even ones that aren't locally compact, like an infinite-dimensional Hilbert space. So the only spaces we're excluding here are ones that are too pathological to be interesting in homotopy theory anyway. Remember, homotopy theory is NOT about arbitrary topological spaces! It's only about spaces that are locally pretty nice. In fact, the category of compactly generated Hausdorff space leans on the side of being too generous - it includes things like the space of rational numbers, which we do not want to think about here. Other categories are better suited to eliminating such junk, but they are less like the category of topological spaces. Anyway, there's a category CGHaus whose objects are compactly generated Hausdorff spaces and whose morphisms are continuous maps. There's a nice functor from Top to CGHaus which is left adjoint to the inclusion from CGHaus into Top. And there's a way to make CGHaus into a Cartesian closed category. It's a bit sneaky: given X and Y in CGHaus, we first take their product in Top and then map that object back into CGHaus to get their product X x Y in CGHaus. And similarly, we first make the set of continuous maps from X to Y into an object of Top using the compact-open topology, and then map that object back into CGHaus to get the object hom(X,Y) in CGHaus. With this rigamarole we get a Cartesian closed category with all the nice properties you could want. Typically you learn about this stuff near the beginning of a heavy-duty algebraic topology course and then you forget about it, because the machinery runs so smoothly that you don't need to think about it. But anyway, you should not think this is the only solution to the problem! You also need to learn about some other "convenient categories" that can serve as substitutes for Top, like the category of CW complexes or simplicial sets or Kan complexes. And when you get to wondering what exactly makes a category sufficiently similar to Top to count as a "substitute" for it, and how we ignore spaces with local pathology when using Top or CGHaus, you'll need to learn about model category theory. Quillen invented "model categories" to answer this sort of question. Jim's answer to this question would pick up right around where I just left off. Posets, distributive lattices, inverse limits, Lawvere, Grothendieck... those are some of the words he said. >And what's the difference between LX and pi_1(X)? That's easy: pi_1(X) is the set of path components of LX. In other words: pi_1(X) = pi_0(LX) And I'm sure you can guess what I'll say next: pi_{n+1}(X) = pi_n(LX) So you see, L is just a way of pushing the level up by one. In the lingo of n-categories, it turns n-morphisms into (n+1)-morphisms, and turns an omega-groupoid into a groupal omega-groupoid. (Remember, an omega-groupoid captures the homotopy-theoretic essence of a space, while a groupal omega-groupoid captures the essence of a topological group.) Similarly, B is just a way of pushing the level down by one: pi_n(G) = pi_{n+1}(BG) >>Even better, >>two G-bundles that we get this way are isomorphic if and only if the maps >>they come from are homotopic! >Is there a correspondence between >isomorphisms of bundles and homotopies of maps? Umm, I don't think so, but you'll have to learn the proof and check it out for yourself someday. >>H^n(X,A) = [X, K(A,n)] >>where the right-hand side is the set of homotopy classes of maps from X >>to K(A,n). In short, K(A,n) knows everything there is to know about the >>nth cohomology with coefficients in A. >So, we can define >cohomology with coefficients in a *topological* Abelian group >by H^n(X,A) := [X, K(A,n)]. Right! People usually say this with more jargon - they say: a topological abelian group is a special case of an infinite loop space, and an infinite loop space determines a spectrum (a connective one, in fact), and we can define cohomology with coefficients in any spectrum. In a previous issue of This Week's Finds I ran you through all that stuff, but it's a lot easier to understand in this special case, so I thought I'd do it from scratch this time. >>I can't resist explaining why the unit sphere in an infinite-dimensional >>Hilbert space is contractible. It seems very odd that a sphere could be >>contractible, but this is one of those funny things about infinite >>dimensions. >Well, if it weren't contractible, what homotopy group would be nontrivial? >pi_1(S^oo) is trivial, since only pi_1(S^1) is nontrivial. >pi_2(S^oo) is trivial, since only pi_2(S^2) is nontrivial. >pi_3(S^oo) is trivial, since only pi_3(S^3) is nontrivial. >And so on. Right, exactly! An infinite-dimensional hole ain't no hole at all. Of course, you are using a theorem here, that noncontractible CW complexes have nontrivial homotopy groups. If you didn't know this, you might think that even though all maps from S^n to S^infinity are homotopic, this ain't true for all maps from S^infinity to itself. But in fact, all maps from S^infinity to itself *are* homotopic. >>We can visualize CP^infinity quite easily this way. A rational function >>of a single complex variable has a bunch of zeros and poles - think of >>them as points on the Riemann sphere. We should really stick an integer >>at each of these points: a positive integer at each zero, and a negative >>integer at each pole, to tell us the order of that zero or pole. This >>gives enough information to completely specify the rational function up >>to a constant factor. So a point in CP^infinity is the same as a finite >>set of points on the sphere labelled by integers - which must add up to >>zero. >For the algebraic geometers, then, >we just have a divisor on the Riemann sphere of degree 0. Yes, those nasty algebraic geometers would probably say it like that. I didn't want to give them the pleasure of saying it that way. >>The space >>of Hilbert-Schmidt operators is a Hilbert space in its own right, >>with this inner product: = tr(AB*). >Speaking like a mathematician, eh? >Here on sci.physics.research, we say = tr(A*B). Ulp! >Throughout, you say "gerbs", but all your references say "gerbes". >What's going on here? Gerbe is a French word, and lots of English-speaking mathematicians have adopted that word unchanged. For a while I thought people were gonna start changing it to "gerb" in English, but now I'm getting the feeling that I'm the only one who writes it that way, so maybe I'll switch and do what everyone else does. By the way, I keep meaning to restart the geometric quantization thread, but the fact that my newsserver is down somehow makes it harder for me to get up the energy to do this. I'm gonna mail this post to you as well as to the s.p.r. moderator, to speed things up a bit. Please reply by posting - though it's okay if you email me *in addition*. ============================================================================== From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Re: This Week's Finds in Mathematical Physics (Week 151) Date: 8 Jul 2000 18:14:41 GMT Newsgroups: sci.physics.research,sci.physics,sci.math John Baez wrote: >Toby Bartels wrote: >>But this reminds me, how do I define the topology on LX? >Jim Dolan came up to me the other day and told me you asked this >question. My newsserver has dead since last Tuesday, so I'm having >trouble reading articles here, and I didn't see this one of yours. >He then proceeded to give me a half-hour lecture on how I should >reply to this question. Writing up what he said would take at >least least two hours, so I'm afraid I won't do that. I'll just >start with a straightforward answer your question and then mention >some of the deeper issues it leads into. >Straightforward answer - the compact-open topology. Given topological >spaces X and Y, let hom(X,Y) be the set of continuous maps from X to >Y. In the "compact-open topology", the open sets in hom(X,Y) are >generated from sets of this form: {f in hom(X,Y) : f(K) is contained in U} >where K is compact and U is open. OK, I've heard of this. >Deeper issues - There's a general question lurking here: what's the >right topology to put on the space of continuous maps from a space X >to a space Y? So we should think about this. >A continuous map from X to hom(Y,Z) should be the same >thing as a continuous map from X x Y to Z. >Technically, it says our category of spaces is "Cartesian closed". >What it really means, though, is that we're going along with the Tao >of mathematics. >Okay, now for the problem: there's no way to pick a topology >on hom(X,Y) that gives this nice property. At least, not any >way that I've ever heard about! If there is a way, there must >be something awful about it, because nobody ever talks about it. >I bet there isn't one. I'm pretty sure Saunders MacLane (... for the Working Mathematician) said there wasn't. He said we should use Kelley spaces instead. >What people do instead is replace the category of topological spaces >by a slightly nicer category. There are lots of ways to do this but >one popular approach involves "compactly generated Hausdorff spaces". This sounds familiar. Perhaps this is what Kelley spaces are. >You know what a "Hausdorff" space is: it's one where you can find disjoint >open neighborhoods for any pair of distinct points. A space is "compactly >generated" if a subset C is closed iff its intersection with every >compact set is closed. Given Hausdorff, iff its intersection with every compact set is compact. >It's easy to check that any locally compact space is compactly generated, >and so is any space whose topology comes from a metric - even ones that >aren't locally compact, like an infinite-dimensional Hilbert space. So >the only spaces we're excluding here are ones that are too pathological >to be interesting in homotopy theory anyway. Remember, homotopy theory >is NOT about arbitrary topological spaces! It's only about spaces that >are locally pretty nice. So, how much homotopy theory can you do over something warped like the space of prime ideals of a polynomial ring over a finite field? >Anyway, there's a category CGHaus whose objects are compactly generated >Hausdorff spaces and whose morphisms are continuous maps. There's a >nice functor from Top to CGHaus which is left adjoint to the inclusion >from CGHaus into Top. Any simple description of this? >And when you get to wondering >what exactly makes a category sufficiently similar to Top to count as >a "substitute" for it, and how we ignore spaces with local pathology >when using Top or CGHaus, you'll need to learn about model category >theory. Quillen invented "model categories" to answer this sort of >question. >Jim's answer to this question would pick up right around where I just >left off. Posets, distributive lattices, inverse limits, Lawvere, >Grothendieck... those are some of the words he said. I can handle those words. Are you listening, jim? Post something! >>And what's the difference between LX and pi_1(X)? >That's easy: pi_1(X) is the set of path components of LX. In other >words: pi_1(X) = pi_0(LX). Since LX is a topological group (up to homotopy), pi_0 (LX) is a group. >And I'm sure you can guess what I'll say next: >pi_{n+1}(X) = pi_n(LX). As soon as you said pi_1 (X) = pi_0 (LX), I remembered this. >>John Baez wrote: >>>I can't resist explaining why the unit sphere in an infinite-dimensional >>>Hilbert space is contractible. It seems very odd that a sphere could be >>>contractible, but this is one of those funny things about infinite >>>dimensions. >>Well, if it weren't contractible, what homotopy group would be nontrivial? >>pi_1(S^oo) is trivial, since only pi_1(S^1) is nontrivial. >>pi_2(S^oo) is trivial, since only pi_2(S^2) is nontrivial. >>pi_3(S^oo) is trivial, since only pi_3(S^3) is nontrivial. >>And so on. I guess the above is not quite right, since there are nontrivial things like pi_3 (S^2). But pi_n (S^m) is trivial whenever n < m. >Of course, you are using a theorem here, that noncontractible CW >complexes have nontrivial homotopy groups. If you didn't know this, >you might think that even though all maps from S^n to S^infinity are >homotopic, this ain't true for all maps from S^infinity to itself. >But in fact, all maps from S^infinity to itself *are* homotopic. So, there are noncontractible spaces with only trivial homotopy groups? I guess that's OK, if they're nasty. >By the way, I keep meaning to restart the geometric quantization >thread, but the fact that my newsserver is down somehow makes it >harder for me to get up the energy to do this. That's OK, you have by now. >I'm gonna mail this post to you as well as to the s.p.r. moderator, >to speed things up a bit. Please reply by posting - though it's >okay if you email me *in addition*. I'm emailing this to you as well as posting, just in case, but it looks like you're doing OK now. -- Toby toby@ugcs.caltech.edu ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: This Week's Finds in Mathematical Physics (Week 151) Date: 9 Jul 2000 01:07:57 GMT Newsgroups: sci.physics.research,sci.physics,sci.math In article <8k6va5$47f@gap.cco.caltech.edu>, Toby Bartels wrote: >John Baez wrote: >>A continuous map from X to hom(Y,Z) should be the same >>thing as a continuous map from X x Y to Z. >>Okay, now for the problem: there's no way to pick a topology >>on hom(X,Y) that gives this nice property. At least, not any >>way that I've ever heard about! >I'm pretty sure Saunders MacLane (... for the Working Mathematician) >said there wasn't. He said we should use Kelley spaces instead. Yeah, I asked Jim and he said there wasn't. To prove this, we just need to find some horrible space Y such that Cartesian product with Y doesn't preserve all (small) colimits. That'll imply the functor - x Y can't be a left adjoint. Or in other words, it can't have a right adjoint hom(Y, -). >>What people do instead is replace the category of topological spaces >>by a slightly nicer category. There are lots of ways to do this but >>one popular approach involves "compactly generated Hausdorff spaces". >This sounds familiar. Perhaps this is what Kelley spaces are. Yes! Well, Kelley spaces are either compactly generated spaces or compactly generated Hausdorff spaces - I've seen people try to use both of these as a substitute for Top, and I forget which one is called "Kelley spaces". But G. W. Whitehead uses compactly generated Hausdorff spaces in his book "Elements of Homotopy Theory", so that's what I use. No point in trying to shave off an extra assumption just to make life tough on yourself. >So, how much homotopy theory can you do over something warped like >the space of prime ideals of a polynomial ring over a finite field? A lot - *if* you do everything in a completely different way that you normally otherwise would in topology!!! This is why Grothendieck and Deligne invented stuff like "etale cohomology" when proving the Weil conjectures. I don't understand this stuff, but the idea is basically: you can define cohomology groups of complex algebraic varieties using their topology as manifolds, but suppose you want to do something similar for a variety over a finite field. If you just use the Zariski topology and turn the crank you get junk - i.e., stuff that's not really very similar to the complex case. (In fact, even in the complex case you'd get weird junk if you used the Zariski topology.) But if you're Grothendieck, you can cook up some really sneaky tricks involving topoi that get the job done! More recently, Voevodsky has developed an analogue of homotopy theory for schemes. He is blowing everyone away because he's solving important problems in algebraic geometry using ideas from homotopy theory, and practically nobody else understands both subjects deeply enough to keep up with him. >>Anyway, there's a category CGHaus whose objects are compactly generated >>Hausdorff spaces and whose morphisms are continuous maps. There's a >>nice functor from Top to CGHaus which is left adjoint to the inclusion >>from CGHaus into Top. >Any simple description of this? Well, let's chop the problem in two. We have forgetful functors U U' CGHaus ---> Haus ---> Top with left adjoints L' L Top ---> Haus ---> CGHaus so the left adjoint of the forgetful functor UU': CGHaus -> Top will be the composite L'L: Top -> CGHaus. (Beware: I am using the sensible notation for composite maps here, not the usual notation.) What's L? How do you turn a Hausdorff space into a compactly generated one? Well, you just define a new topology on it, where the new closed sets are the sets whose intersection with every compact set is closed in the old topology. (Here "compact" is also defined using the old topology, of course.) And what's L'? How do you turn a topological space into a Hausdorff space? I leave this one to you.... >So, there are noncontractible spaces with only trivial homotopy groups? Yeah. Take the curve y = sin(1/x) for 0 < x < 1 together with the point (0,0), and then draw a nice big curve connecting the point (0,0) to the point (1,sin(1)) to get a curve that looks vaguely like a circle with a seriously wiggly bit thrown in. This space is not contractible but all its homotopy groups are trivial. We need Cech cohomology or something like that to detect the fact that it's vaguely like a circle. Its first Cech cohomology is nonvanishing! This is an example of the kind of space homotopy theorists don't want to worry about. It's not locally well-behaved. Btw, my newsserver seems to be working fine again. Also btw, you should stop asking me about this stuff if you want me to have the energy to talk more about geometric quantization! ============================================================================== From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Re: This Week's Finds in Mathematical Physics (Week 151) Date: 11 Jul 2000 01:21:40 GMT Newsgroups: sci.physics.research,sci.physics,sci.math John Baez wrote: >Yeah, I asked Jim and he said there wasn't. To prove this, we >just need to find some horrible space Y such that Cartesian product >with Y doesn't preserve all (small) colimits. That'll imply the >functor - x Y can't be a left adjoint. Or in other words, it can't >have a right adjoint hom(Y, -). Do you think this is the sort of thing that would be in that book Counterexamples in Topology that I've heard about? (Do you think I should get that book?) >Well, Kelley spaces are either compactly generated spaces or >compactly generated Hausdorff spaces - I've seen people try to use >both of these as a substitute for Top, and I forget which one is >called "Kelley spaces". I'll bet it's the Hausdorff version, because, while the compactly generated stuff rests fuzzy in my mind, I'm certain the spaces MacLane recommended were Hausdorff. >Toby Bartels wrote: >>So, how much homotopy theory can you do over something warped like >>the space of prime ideals of a polynomial ring over a finite field? >A lot - *if* you do everything in a completely different way that you >normally otherwise would in topology!!! ... >... If you just use the Zariski topology >and turn the crank you get junk - i.e., stuff that's not really very >similar to the complex case. ... But if you're Grothendieck, >you can cook up some really sneaky tricks involving topoi that get the >job done! More recently, Voevodsky has developed an analogue of homotopy >theory for schemes. ... OK, I guess I'll leave that for later in my education. >>John Baez wrote: >>>Anyway, there's a category CGHaus whose objects are compactly generated >>>Hausdorff spaces and whose morphisms are continuous maps. There's a >>>nice functor from Top to CGHaus which is left adjoint to the inclusion >>>from CGHaus into Top. >>Any simple description of this? >Well, let's chop the problem in two. We have forgetful functors > U U' L' L >CGHaus ---> Haus ---> Top with left adjoints Top ---> Haus ---> CGHaus >so the left adjoint of the forgetful functor UU': CGHaus -> Top >will be the composite L'L: Top -> CGHaus. (Beware: I am using the >sensible notation for composite maps here, not the usual notation.) (I'm beginning to wonder if we should start pointing arrows to the left just so the usual notation becomes the sensible one.) >What's L? How do you turn a Hausdorff space into a compactly generated >one? Well, you just define a new topology on it, where the new closed >sets are the sets whose intersection with every compact set is closed >in the old topology. (Here "compact" is also defined using the old >topology, of course.) OK, that's pretty straightforward. >And what's L'? How do you turn a topological space into a Hausdorff >space? I leave this one to you.... Well, every topological space has an binary relation on it which measures the space's deviation from Hausdorffness: p ~ q iff p and q have no mutually disjoint neighbourhoods. This may not be transitive and so may not be an equivalence relation, but of course it generates an equivalence relation # (p # q iff p ~ r_0 ~ r_1 ~ ... ~ r_(n-1) ~ q for some sequence r). So, mod out by # to get a Hausdorff space. >>So, there are noncontractible spaces with only trivial homotopy groups? >Yeah. Take the curve y = sin(1/x) for 0 < x < 1 together with the point >(0,0), and then draw a nice big curve connecting the point (0,0) >to the point (1,sin(1)) to get a curve that looks vaguely like a >circle with a seriously wiggly bit thrown in. This space is not contractible >but all its homotopy groups are trivial. O, sure, the topologists' sine curve! I guess I never really thought of this as pathological, because it's always an example in algebraic topology. But, of course, it's really presented as a *counterexample*. >This is an example of the >kind of space homotopy theorists don't want to worry about. It's not >locally well-behaved. And it's not compactly generated. >Also btw, you should stop asking me about this stuff if you want me >to have the energy to talk more about geometric quantization! O, it's all good stuff. Just don't *forget* about geometric quantisation. (I keep a few files sitting in my home directory of responses I need to finish writing.) -- Toby toby@ugcs.caltech.edu