Noncommutative topology in geometry & physics, lecture 2

0:00 All you have to do is to press the red tip. It's now recording. And nothing else. Also, as I mentioned yesterday, I put the notes up on my web page. This is not an official site yet, because I'm still making revisions, and as people point out, misprint some errors and so on, I'm trying to fix them, but they find these errors. So, there's no reason to copy these things later. Okay, let me just correct a few things from yesterday. Oh, look, we were talking about the similarity of grupoids and how that gives examples of Marie equivalences of C-star out there, and I realized after everybody started looking at the slide, the slide didn't say what I intended to say, so what I intended to say was this. So suppose you have a group G acting on a space F, so that's a classical example of groupoid, maybe the single most important case in fact. And so then I wanted to consider two different cases. So if the group acts freely, but not necessarily transitively, then the groupoid is similar to the groupoid consisting of a quotient space with no morphisms. So the only morphisms And therefore, this cross product C-star algebra, which is the C-star algebra of the groupoid, is marine equivalent to this commutative algebra of C algebra of functions on the quotient. If, on the other hand, gx transitively, but not necessarily freely,

2:30 then the groupoid is similar to the stabilizer group at the choice of base point and of course all base points are the same so it doesn't matter the point you take. So then the cross product algebra is equivalent to the C star algebra of this group. So, in other words, the groupoid C-star algebra of the transformation group G cross X, which is a cross product, C-star algebra in another language, is really equivalent in one case to an abelian algebra with no group, and in the other case to C-star algebra with a group with no abelian part. So those two special cases, the C-star algebra of the group, and just the C-star algebra of space are special cases of the C-star algebra of the group points. So in terms of trying to understand what group points do, they should be viewed as a kind of a combination of those two examples. All right, another thing I should correct is we were talking about various kinds of equivalence relations in non-commutative topology, and as I mentioned, the whole idea of looking for equivalence relations for classifying things really goes all the way back to . very first paper on topology, and he had asked the question, So in other words, he was asking the question about whether you would classify things up to homotonic equivalents by means of certain algebraic and variance in just the bedding numbers, and so that's still an important theme of topology ever since then. So that was, well, 117 years ago. So now, we want to go back and look at various equivalence relations on C-star algebras, and I mentioned,

5:00 I mentioned the problem of trying to understand what these non-commutative equivalence relations amount to when you just specialize to the commutative case, and I made a mistake. I forgot to explain what stable homotopy equivalence amounts to. So suppose you have two spaces, X and Y. So out of these, let's just take for simplicity these are finite CW complexes. here's another one. So homotopy equivalence of the C-star algebers would mean that X is homotopy equivalent to Y. But what does it mean to say that the utensils are compact, that these things are homotopy equivalent? It turns out that that gives you something that's actually very close to K-theory. So this sort of explains the comment I made that once you start looking at stable things, K-theory is almost forced on you. This is actually a fairly difficult theorem of Deterlap and McClure based on some stuff that I did, which is based on some stuff done by Graham-Siegel. So the final result is that C of X and C of Y are stably homotop equivalent, if and only if their equivariant k-homology groups are the same, as Z adjoin U-modules. So U here is the, for those who don't know what this means, which is probably most of the people in the audience, this is some kind of exotic homology theory, which is a lot like ordinary homology that it comes with a very important self-map, u, which is the bot element that shifts degree by 2. And when you invert the bot map, when you invert u on these things, then you just get complex k homology. So complex k-homology is what you get from these things by inverting U. And if these things are isomorphic as the adjoined U-modules and some other minor technical condition is satisfied, which I won't go into right now, but it's satisfied in some.

7:30 And in cases then, this is necessary, it's sufficient for these things to be a state-like homotomy equivalent. So, K-theory already starts showing up when you start looking at these equivalence professions in the non-community world. And since yesterday, I also augmented the references a little bit. So, let's see, I mentioned all those references yesterday. today. So the paper on that is this one over here. And yesterday I mentioned Karubi's Osterisk volume, but I forgot to put the references, so that's over there. Okay, so now we're ready for a new topic. So today's lecture is supposed to be about Equivariant topology, and, uh... I think that's too thick. In fact, I'm sure it's too thick. So, today I want to talk about equivariant topology, which is the topology of group actions on spaces. So, if you have a group acting on a space, you can be acting on X. A topologist just studied this situation for its own sake, after various kinds of equivalence relations again, which we'll discuss. But if you want to sort of analyze this in terms of the non-commutative world, you can pass from this to the groupoid, G cross X, and then from that to the C star algebra groupoid, which is the cross product. Let's take X compaxis, minus W compaxis. As you pass through the cross product,

10:00 and what I want to discuss today is how you can get information over here, in the usual sense, from the analysis of the non-commutative topology of that non-commutative space over there. The point I'm trying to make in these lectures is not just that non-commutative topology is, shall we say, an abstraction of ordinary topology, but it actually can can be used going backwards to get additional information about concrete topological problems, which the formulation of which doesn't have anything non-commuted in it. So let me say a little bit more about this cross-product algebra since this thing is going to keep coming up over and over again. So how do you construct it? Maybe I should start with a case where g is discrete. So if G is discrete, then it's very easy to explain what this algebra is. It's just, you take all products, G times A, where A is an element of the Abelian algebra, G is an element of the group, and you take the finite linear combinations of those things, and you build in a relation that when you move the G across the A, that just amounts to the action of G on A. Since g is acting on x, it acts on the space of the algebra functions on the space also. So g dot a is the action of g on a, and this is just multiplication in the algebra. So that's the definition of the algebraic cross product. And then you take all finite linear combinations, so it simply gives you an algebra. So that gives you the definition of the algebraic cross product of a group acting on an algebra. And that definition, with a slight change of terminology perhaps, goes all the way back to Betterburn about a hundred years ago. Now when G is not discrete, so that initially, the algebraic cross product doesn't have any topology, but if the algebra that you're acting on is a C-star algebra, then it's easy to

12:30 see that it becomes a star algebra and you just complete in the greatest C-star norm, which is easy to see as well. Now, if G is locally compact and not compact, then you you can take a convolution algebra of L1 functions on g cross h for suitable, g cross x for suitable convolution. That's basically the point of view that was taken in Jean-Louis' talk on Wednesday. Or alternatively, this cross product algebra can be viewed as living inside its multiplier algebra is consisting again of products B times A, but now the B has to be an element of C star G, not just an element of the group ring. And A is in the Abelian algebra, and again you have this. Well now the commutation relation is a little more complicated because, you see, this is the commutation relation for group elements. The group elements don't actually live in the C star algebra if G discrete, because if G isn't discrete, they're just sort of formal multipliers on the C-star algebra. So the C-star algebra sort of smeared out averages of group elements, but once you know this commutation in relation to group elements, that determines what the relation is here. It's harder to write down if you can think of it, but that determines the multiplication, Okay, so now one of the very first tools that comes up in studying equivariant topology is equivariant k-theory. And the first paper on this was by Graham Siegel back in the 60s. So you start with the notion of a g-vector bundle on a space. So suppose x is a space, it's a compact space, and g is a compact group for now. So an equivariant vector bundle, g-vector bundle over x, is a vector bundle in the usual sense,

15:00 together with the g-action on the bundle that's compatible with the g-action on the base. And the only thing is that the action on the bundle has to send drivers linearly with drivers. I mean, so in other words, it has to be an action in the category of vector bundles, not just in the category of spaces. So then you just take the group of the isomorphism classes of such bundles, and that's called the equatorian k-group. Now, let's look at one special case. suppose the space involved is just a point, all right, so an aquivariant, what's an aquivariant vector bundle over a point? A vector bundle over a point is just a vector space, and it comes now with a linear g action, so that's just a representation of a group. So, and then we're taking a group, so what you get is the set of formal differences of finite dimensional representations of g, which which is called a representation ring. And also, you have both the internal and the external tensor products. The external tensor product is sort of more elementary. So if you have a vector bundle on x and a vector bundle on y, you naturally get a vector bundle on x cross y. And then when you restrict to the diagonal inside here, the cup product on kg of x, which makes this thing into a ring. And one thing you see from this is that, for example, if you take x equal to a point or y equal to a point, then you just get x back here, so that means that kg of a point is naturally acting on kg of x. So that means that equivariant cake theory is always a module over the representation range. So, an important thing to keep in mind is that when you want to understand the nature of equivariant k-theory, you should think of it not just as an Abelian group, but you should think of it as an R of G module, and it's the homological algebra of this ring, R of G, that becomes important. Okay, so now a few more things about equivariant k-theory.

17:30 So it can be made into a cosmology theory. extend it to locally compact spaces, not just compact spaces. Simply, you add a point at infinity and then you take it out by taking the internal of the natural. And in a case where x is already compact, when you take a one-point compactification, you're just taking a disjoint union with a point. And then you're subtracting this point off, so started with. So this doesn't change what we have already. It now makes it possible to define this for locally compact spaces. And then you get a cohomology theory by taking products and completing spaces. And see, the proof of excision and all the other properties is the same as for ordinary K-theory. So it's easy to see if you get a cohomology theory. And you also get a product, as I mentioned. It makes KG of X, the varying index here, into graded commutative RNG algebra, and you have bot periodicity, too. The bot periodicity theorem is slightly harder to prove in the equivariant case than in the non-equivariant case. Well, it depends on which proof you use. There are about 25 different proofs of bot periodicity. Roughly half of them don't generalize in half of them do, so you got that in? No, this one is an inventory. This one? Well, what's not an inventory is when the stuff is more present. This one? Yeah. It's just . Right. The original bot proof, for example, doesn't generalize so easily. Depends, but yeah. This one's not so hard, yeah. There are various ways of doing it. OK, now the same thing also works for Bonn algebers. If you have a, let's say you have complex Bonn algebers with a continuous G action on them, then the equivarian K group now is defined to be the Grotin D group of isomorphism classes of finitely generated projected A modules equipped with, again, with continuous G actions

20:00 that are compatible with the G action on A itself. So if you think about what that means, that means that your module is supposed to be a G equivariant direct sumand in A tensor B, where B is a finite dimensional representation of the group. I mean, an ordinary finitely generated projective module is just a direct sum in an A tensor vector space, a finite dimensional vector space, and now you want a G action, but the G action on this vector space can be non-trivial, so the V now can be any finite dimensional representation. And then you extend the theory to non-unital algorithms in the same way, and equivariate the same proof as in the commutative case so all the theories are saying again kg is always a module over r of g the one thing that you don't have for general bonnick algebers that are not commutative is a cut product so you have the external product but not the internal product because there's no diagonal map And you have an equivariant version of Swan's Theorem that if you had a compact g-space and a compact group and a g-vector bundle, then the set of sections of this vector bundle, of course, set of sections is clearly a module over the continuous functions on x. And this turns out to be a finitely generated projective C of X module, but because of the G action, you also have a natural G action on here, and that G action is compatible with the G action on X, so you get an isomorphism between the equivariant k-theory of X and the equivariant k-theory of the algebra functions on X. So the bonnet algebra situation, in In other words, really just gives you back what you had in the situation for spaces. So, let me just sketch the proof of that. So this theorem's in Siegel's original paper. And I should mention, where is it that we're using the assumption that g is compact here? So it's used in a crucial way for application of the following thing.

22:30 So I assume most people are familiar with the Peter Weill theorem that says that basically that in the case of a compact group, well, there are lots of different formulations in the theorem, but one way of thinking about it is that it says that if you have a compact group that acts continuously on a bonnet space, then the set of elements that generate finite dimensional G subspaces is dense. Okay, so that's what we're going to use here. So there's two steps to prove. The first is that if you have a surjective G vector bundle map from X cross V onto your given vector bundle, Then it's very easy to see that the sections in the vector bundle is a g equivariant direct sum in here. Okay, so that gives you the fact that this thing is a projected module with compatible g action. So how do you do that? Well, you take a g invariant commission metric on the bundle just by taking any metric you want and just averaging it out of the group, which induces the group's compact. And then you take the orthogonal complement of that bundle map, of the kernel of the bundle map, and then that will be a splitting per phi, so that shows that you get a direct summing. So that's one part of the proof. And then the other part of the proof is to construct that bundle map in the first place. That's the harder part. That's the part that uses the theory of out there. So, see, because of that statement that I mentioned, that in a, you have a g-action on a bomb-off space, are dense. Given a point in S, there's a finite set of sections that can be taken to be G finite, so they generate a finite dimensional G subspace of the set of all sections. And they have the property that they span the fiber for every point in the neighborhood of x.

25:00 So just take sections, find many sections that generate the fiber at x, and then look at the g-space they generate, and then take a sort of dense g-finite set inside there and then you can construct this. And then you use compactness and partition of unity to patch these things together, and therefore you get a finite subset of the sections that generate a finite-dimensional g-subspace and that generate all the sections of the C of x module. And then when you think about that, then that set of sections gives you this map over here. proof. I mean, you can read it in Siegel's original paper. Okay, so now, since I promised to show you what non-commutative topology has to do with commutative topology, we want to make a relationship between equivariant k-theory, which is something that basically lives in the commutative world, over here, the equivariant k-theory of x, and the k-theory of the cross-product, which is something genuinely non-commutative. So there's a theorem which is proved independently by Green and Jules. Green did this in the 70s and never wrote it out, and there's a nice paper, Joule, that explains this. So if you have a C-star algebra equipped with a continuous G action, where G is a compact group, then you get an isomorphism between the equivariant K-theory of the algebra and the K-theory of the cross-product. By the way, the C-star condition, I think, is not so essential. I think you can improve this permonic algebra results. So, alright, because of suspensions and so on, you can come down just to considering the case of K0.

27:30 And so the idea is that you have to, well, let's construct the map going in this direction. So given a finitely generated projective A module that has a compatible G action, you want a finitely generated projective module over the cross-product. And you want to see that everything over the cross-product comes from that construction. So there's one technical point here, which is a kind of technical nuisance. to stop and think about this. Again, the Peter-Bott theorem is what comes to your rescue here. But for a compact group, of course, if it's not finite, then the C-star algebra is not unital. It's an infinite direct sum of matrix algebras associated to the various irreducible representations. But for a general compact group, there are infinitely many irreducible representation, so the C star algebra doesn't have a unit. But fortunately that doesn't really matter much because we just started working at one little sum and in the C star algebra at a time. So we construct this map phi and the map phi will have that sort of obvious property that it's compatible with direct sums. So So if something is a direct sum and something else, then the P of the individual sum ands goes to things that add together to the P of the product. So for that reason, it's just enough to really consider this case here. So the basic case is a case of A tensor B. This is sort of a basic case of a projective of A module with compatible G action. The important thing here, though, is that the V can have a G action. The V is a finite arbitrary monodimensional representation of G. Yes, there is an equivalence of categories and textures. But it's a little complicated algebra, this algebra is not unital, so, you know, the notion of finally generated projected motto has to be properly interpreted. But anyway, if you take something like this, you

30:00 just have to see what that goes to over here. So, if you think about it, there's one and and you just have to check if that has the right properties. So, C star of G is a big direct sum of matrix algebras, and if you cut down by projection in one of those matrix algebras, if there's a minimal projection in one of the matrix algebras, then your representation space of E is just isomorphic to C star of G times P as a left C star of G marker. So, on the other hand, the cross product, let's write the cross product, it doesn't, because of the commutation relation, it actually doesn't matter whether you put the g on the right or the g on the left. So the cross product is actually, can be written in the form A dot C star of G. In other words, everything in the cross product is a product of something in here and something in here, but the product is taken inside the multiplier algebra, since this thing doesn't actually directly live in here. It doesn't matter. So, all right, we had A tensor V, and V was C star of G times P, and you wanted to send it to a left module for this algebra, so the sort of obvious thing to do is to send it it. Send it to that. I mean, there's one and only one obvious map. All you're doing is just replacing the tensor product by your multiplication in the algebra. So you just have to check that this thing has the right properties. So that's what I've done here. So this thing is obviously compatible with direct sums. And this maps to a module. Now, Now, as I said, this algebra over here is not unital, but since p is a projection, it still defines a class in k0.

32:30 And furthermore, you can check that k0, the algebra, is generated by p's of this type, and therefore that gives you the isomorphism. So that's the idea of proof. So, as Max mentioned, actually there's a little more that's true. Not only is this, you get an isomorphism in the papers, there's an equivalence of categories and also the map is compatible with the RNG action in the sense that there's a way of of letting R of G act on this, well, just by the obvious thing, so the map is an intertwining map of that. All right, now we're getting to something a little less elementary. equivariant topology, you need to use one more result, which is the localization theorem of Siegel. So, localization theorem says, suppose you have a compact lead group and a prime ideal in the representation ring. So remember, the representation ring is going going to be a finitely generated commutative ring, a theory in commutative ring. And so the idea is now to apply ideas from commutative algebra to bear on the problem of understanding So in commutative algebra, look at 4 by P commutative algebra, look at chapter one, you see the very first thing you want to start studying is you want to understand the prime ideals and localization of the prime ideals. So suppose you take a prime ideal in a representation range, so Siegel showed that there's a subgroup H which is minimal among subgroups for which P is induced from R of H.

35:00 sitting inside G, there's a sort of obvious map from R of H to R of G. So if you have a representation of G, you can restrict it to H. You have the map here. And by being induced from R of H, I mean the fact that we want prime ideals that live over a prime ideal over here. So look at the smallest minimal subgroups such that this prime ideal lives over a prime ideal over here. But when you say induced, you're usually in the induction now. in the other direction. That's not an inclusion, but there is a map going the other way. Well, there's a kind of adjunction formula between you, so these things are related. I'm going to read Siegel's paper to see, but the map going the other way comes from inducing representations. Let's not worry about that right now. So then there's a minimal H with this property, and the H turns out to be topological cyclic automatically, which means it has a dense cyclic subgroup inside, and it's unique up to conjugacy. So since it's unique up to conjugacy and conjugate subgroups do the same thing, And it makes sense to talk about the support of P, although the H is not as unique as you need it to be. And then, for any G space X, the inclusion of X upper H into X induces an isomorphism on the equivariant K theory after you localize it P. So I have to explain what this thing is. First of all, X upper H, without the parentheses, is just a fixed set for H. Okay, so if you look at the elements of X that are fixed under H,

37:30 that's a closed subset. But if G is not abelian, then that set is not necessarily G invariant. So you take the G invariant set that it generates, and that's called X upper H with parentheses, Don't worry about the parentheses, because that makes things a lot easier. And the statement is that this inclusion, so this is a G-space and this is a G-space, and that inclusion induces an isomorphism on equivariant k theory localized at p. Now, there's another result also, which goes along the same lines, and that basically shows that after you throw away a certain amount of torsion, loss of generality in assuming that everything is abelian, and when you combine those facts together, then that makes it possible actually to come down to fixed sets for cyclic subgroups, and then that makes things very nice. Because there it's much easier to do calculations. Well, so remember that K star G of X is an R of G module. This is a commutative Ethereum ring, it's a modulo of the rings. I just mean localized in the sense of localization and commutative algebra. So effectively what we're doing is we have this prime ideal P, and basically you formally invert all elements of the ring that don't lie in the prime ideal. When you do that, you end up with a local rate, it's only one maximal ideal, and then you get a module over this local rate, and that's what I'm talking about. Is that the same as taking X of H, taking the normal bundle to that, and looking at the action on that normal bundle to X of H? There's no normal bundle. This is just a space. There are no manifolds here. No, I mean, we can use the topological rule. No, no, no, no, no. I mean, because... We'll get to normal bundles later if we haven't gotten there yet. That's not the same thing. I mean exactly what I wrote there.

40:00 So, it's just a big set. That's something very allegory. The X doesn't have to be a manifold. This could be an arbitrary complex set. Okay, so now let's look at equivalence relations. So in equivariant topology, just as in ordinary topology, you want to classify things up to certain natural equivalence relations, so let me explain what some of these equivalence relations are. So people who do equivariant topology deal with these things. So one, the most obvious equivalence relation is equivariant homomorphism. I mean, when two spaces are equivariantly homeomorphic, then for all practical purposes as G spaces, they are identical. But this is such a strong equivalence relation, it's not really very useful. Because it's sort of hopeless, in most cases, to do classifications after that. Something that's more manageable, in which there's a huge literature in the topology literature, is classification of equivariant homotopy equivalence. So an equivariant homotopy is just a homotopy, which is invariant under the group, in other It's a continuous path of maps in the category of G-spaces. And spaces are equivariantly on the top equivalent if they're G-maps in both directions, such that the composites are G-hematopping to the identity of the two spaces. And the natural factors in equivariant topology, like equivariant k-theory, or all equivariant homology is another example, those things are all preserved under this But now there's some other equivalence relations, these two, which are less well-known. If you don't already work in equivariant topology, you may not have heard of these two. The ones on the previous page are sort of well-known to everybody. So something that's not so well known is something called pseudo-equivalence. suppose you have a G-map from X to Y.

42:30 X and Y are both G-spaces. This is a G-map. And suppose that after you forget the G-action, that's a homotop equivalence. Then you say that X and Y are pseudo-equivalent. Now, that's not a symmetric relation because there's not necessarily a G-map back in the other direction. But you just take the formal equivalence relation that it generates, and that's called pseudoequivalence. So let me give you a sort of example of that. So for example, suppose for any group, there's always a g-space called eg, on which g acts freely and properly. And the quotient of EG by G is called the classifying space of EG. And now, for compact groups, this EG is actually huge. It's not finite-dimensional, but never mind. So the EG is, in this case, actually not what we're compact. Just forget about that for the moment. They're just sort of working, let's say, in a category of CW complexes or something. So then you obviously have an equivariant map from eg to a point, gx over here, gx over here. That map is certainly equivariant. And for getting the g action, that map is certainly a homotop equivalence because this space is by definition contractible and so is that. So that means that these spaces, eg and a point, are pseudo-equivalent. But they're definitely not on the top equivalent because there's no g-map going back in the other direction. Here the g-action is free and here the action is increased. There's no way you can map this into that. So this is, introducing pseudoequivalences instead of equivalences gives you a lot more, much bigger equivalence classes. So the equivalence classes are bigger, and therefore there are fewer of them. And so it might be easier to do classification up to this. and then finally there's something else which also shows up in the equivariant apology literature and it's called isovariant chromatophic equivalence

45:00 what does the word isovariant mean as opposed to equivariant a map is isovariant if it preserves isotropy so it preserves stabilizers so an equivariant map doesn't have to preserve stabilizers I mean once again over here the stabilizers are trivial and here the stabilizers are big. So this is an equivariant map which is not isovariant. So an isovariant homotopy equivalence is stronger than an equivariant homotopy equivalence. Then you require that these, the homotopies actually be isovariantly homotopic and that's a stronger condition that the homotopy groups preserve. So that's just introducing some terminology. Okay, so now I'm going to start getting to some really interesting things. What I was doing up until now was fairly elementary, and now from this point on the material becomes a little bit more advanced. So, the question you can ask is, is equivariant cake theory a pseudoequivalence invariant? Well, I mean, a space like this has to be disallowed because it's not locally compact. Let's say you have proper pseudoequivalence. Proper pseudoequivalences. of locally compact spaces, or pseudo-tumacism on compact spaces. And this actually comes down to a question in non-commutative topology, and there are a lot of other related questions which are actually quite interesting. So I'll mention these. Let's look at these questions over here. And that will give you an idea of what some of the sort of interesting questions are. So first one is, suppose you have a G c-star algebra. In the literature, something like that is also sometimes called a C-star dynamical system. I don't like that term so much. It just means a C-star algebra with a G action. I mean, you can think about it from the point of view of dynamical systems, really, but it's a matter of point of view.

47:30 If G is compact, then a compact group acting on a space doesn't give you a very interesting dynamical system. point of view that's important, but on the other hand, if A is not compact and the action isn't proper, then the dynamical system's point of view might, in fact, be fruitful. So that point of view is useful sometimes, but not all the time. So these questions, by the way, make sense even if G isn't compact. In order to understand the sort of general framework of these questions, it's important to understand the locally compact groups in general. Now, I'm temporarily dropping the assumption that G is compacted on the back of the layer. So I suppose A is a G C star algebra, and A is contractible. In other words, A is on the top equivalent to zero. Now, I should mention, by the way, there's one thing that sometimes confuses people right from the start, and that is that the word contractible, as used by the C-star algebra people, is not exactly the same as the word contractible, as used by topologists, for the following simple reason, that it depends on the difference between working with reduced and unreduced homology and homology theories. So for ordinary topologists, they usually work in a category of spaces with base point, and then for them the sort of basic object is reduced homology, which is the relative homology of a space relative to a base point. And so therefore you say that a space is contractible, not if it's hamatope equivalent to nothing, but if it's hamatope equivalent to the base point. So there's a certain essential difference between a point and nothing. So that means a reduced symbology is trivial. So this space is contractible in the topologist sense because this is in the category of the base point. Now, when we pass to the non-commutative world, it's more natural to deal with non-unital C-star algebras and allow locally compact spaces that are not compact, and that basically means that we're dealing with a category of spaces with base points and we've taken the base point out. So now, you no longer work with reduced homology, you work with the unreduced theory.

50:00 From the point of view of the non-commutative world, that's actually somewhat more natural. So that's the reason for the conflict in terminology. So when I say A is contractible, I don't mean that A is equivalent to a point. I mean that A is equivalent literally to nothing. In other words, that A is homotopy equivalent to zero. But that corresponds when you formally adjoin a unit to A, by saying that A with unit adjoined is homotopy equivalent to the scalars. That's right. So it means the k-theory of A, for example, then is G the cross-product of G acting on A also contractible? That's a sort of natural question. Or maybe contractible is too strong. Maybe you want to just K-contractible, in other words, contractible from the point of view of K-theory. So in other words, is the equivariating K-theory zero, or is the K-theory the cross-product zero? If G is compact, by the way, we know these two things are equivalent by the K-Juhl theory, If the g non-compact, they're not exactly the same, although they're related to each other, so there needs to be kind of separate questions. So you have all sorts of questions like this, and questions like this are actually rather non-trivial, as you'll see in the moment. Second thing, suppose you have a homotopy of g actions. So you fix the algebra, and now you vary the action on the same space. This is something that topologists don't do very often, that you can think about this situation. Then, does the equimariant K-theory remain constant in such a family, or can it change? In other words, is equimariant K-theory continuous in the action, not just continuous in the space? Again, you might think, well, of course that ought to be true. But we'll see, actually, in most elementary cases, it's not true. It's actually a very subtle thing. Well, it'll turn out that there are some groups for which this is true and some groups for which it's false. And same thing for p-theory cross-project. And we'll also see that question one and question two are actually closely related to each other.

52:30 I mean, this is not contractability in the sense that there is some contraction in the space because, I mean, even vector space... No, no, no. A is not a vector space. A is a C-star algebra. A is a C-star algebra. So, it means that, obviously, you have the homomorphism that sends A to zero. Okay? and you have the inclusion from 0 back into A, and the composite this way, of course, is the identity, so there's nothing to prove there, but you want to know that the map that sends A to 0, followed by the inclusion of 0 back into A, is homotopic to the identity. In other words, that there's a family of self-maps of A, which are G equivariant, in the category of C star algebers that go from the identity map to the zero map. So, for example, an example of something which is equivariant contractual, maybe you can't visualize this, would be the following. For topologists, let's just take G to be the circle group, and just take the X to be the closed disk. Now this thing is contractible in the topologist sets, equivariantly, because you can just shrink everything to the identity. But there, there was a base point, so that was maybe a bad example because if you look at the c of x, so c of x is g homotopy equivalent, not to zero, but to the scalars. So that homotopy is a g homotopy equivalence between this algebra and that algebra But when I say contractible I mean homotopy equivalence to 0 not homotopy equivalence to the scalars So you just have to take the scalars out So in other words, take out a base point So there's a g fixed base point at the origin here

55:00 And when you take that out When you take the functions that vanish at infinity on that locally compact space, the disk with point removed, then that is equivariantly contractible. First question, do you mean A is contractible as a GC-star algebra or as a C-star algebra? Well, if it's contractible as a GC-star algebra, then there's nothing. Then this is obvious. So this is really a question about pseudo-equivalence invariance. You see, since K-theory is a G-homotov equivalent, equivariant K-theory is invariant under G-homotov equivalence, so there wouldn't be anything to prove if A was equivariantly contractible. Suppose it's contractible but not equivariantly contractible, that is the script. That was just off-scene because you were not told it was an equivariant. of an equivariant one, and then a non-equivariant one is a little harder to start. In cases where the isocrophy changes, then you can get an example of something like this. This doesn't quite work for G. Compact, because then the space isn't locally compact. There are things like that. Alright, so those are some basic questions you can ask. They're fairly obvious questions. I don't know, maybe we should take a short break now. Let's spend about an hour. So we'll We'll come back and start talking about the answers to the questions in a moment. Why don't we take a short break so people can go out and go to the photos that are comfortable. So we'll continue again, maybe at, uh, it's two minutes before 11, so we'll continue maybe at 8 a.m. or within a 10 a.m. break. Thank you.

57:30 Thank you.