 So welcome to this basic notion seminar. It's a big pleasure for me to introduce Professor Andrea Solotar from the University of Buenos Aires. So she is a senator, as you said, here of the CPP in the math department. And she's going to talk about Hockshell Comology and Geometric Revolutionary. OK, thank you. In fact, I'm going to talk a lot about Hockshell homology today. But let me start with some intuitive notions concerning regularity. And I will use the terms regularity or symmetric regularity or smoothness, well, kind of interchangeably. So I'm not going to be very, very detailed in this part. It's just a motivation, some intuitive ideas about regularity. So for a long time, we are going to be in the commutative setting and all the algebras that we will consider will be netarian algebras. So for some moments, we are going to suppose that k is a field. And just for some minutes, we are going to suppose that it is algebrically closed. Afterwards, it will not be necessary anymore. So our algebra now is going to be the quotient of the polynomial algebra in commuting variables, let's say, finite number of variables divided by the ideal generated by a finite number of elements of the algebra. So you know that each time you have k algebra like this, you can associate to it an algebraic variety. And the idea is that we are going to say that this ring is going to be smooth or geometrically regular if, when localizing at each point, the associated variety is local. Let's say the local ring has an associated variety and this is smooth. But what does it mean? Because, well, the definition is i is regular, geometrically regular, if and only if for all maximal ideal, m in A, when localizes, one obtain this algebra. And then this algebra is this algebra that has to be regular. That means that we consider this algebra. There is a unique maximal ideal here. And then you consider the dimension as a vector space of where the field is this one. The algebra divided the maximal ideal. And the vector space that you are considering is just m, the maximal ideal divided by m square. And this dimension has to coincide with the cruel dimension of the algebra you are considering. So what is the cruel dimension? Well, the cruel dimension is the supremum of the lengths of prime ideals, of chains of prime ideals that one may have within this ring. So usually what happens is that we have that this one is greater or equal to this one. But what we are asking here is to have an equality. So intuitively, what does it mean? It means that we cannot have situations like the ones that I will draw here. So we cannot have, we are going to draw, then we are going to have two variables. So here my ring is going to be the quotient of the polynomial algebra in two variables divided by ideal generated by x3 minus 1y square. So I have something that is more or less like this. And well, here we can consider another example. I'm not going to write the equation. But for example, consider a variety like this. So if we localize the maximal ideal corresponding to the origin here, we are going to have problems. Because of course, the varieties associated to the algebras I'm considering here and here are curves, are varieties of dimension one. But this dimension of the maximal ideal divided by the maximal ideal squared corresponds to the dimension of the tangent space at each point. So here, in any point here, here, here, the tangent space has dimension one. But here, this is not the case. It doesn't have like here and like this and like this. And here also, the dimension of the tangent space at this point is bigger than one, which is the cruel dimension that we must have. So what means being regular for a ring, well, a commutative ring, it means that we don't have in the associated varieties this kind of anomaly. No multiple points, no casts like here. So the contrary of this is the situation that we would like to consider. Well, so of course, if we want to consider a non-commutative setting, this is not going to work because we have the cruel dimension, we have maximal ideals, prime ideals, et cetera. So each time one tries to generalize a concept from commutative setting to non-commutative setting, the idea is to give a definition by an homological property. So well, sir proved that under, well, some hypotheses that are true here, the fact of being regular or smooth is equivalent to having what is called global dimension that I will explain of the algebra finite global dimension. So what is the global dimension? The global dimension of A, it is the supremum of the projective dimension of A models of the A models one can consider. So here, pay attention because we are in the commutative setting up to now. And then when I say model, I don't have to specify if it's right model, left model or what because if it's right model, then the same structure can be considered for a left model. But in the non-commutative case, I will have right global dimension and left global dimension, the projective dimension. Yes, I have to define it. Thank you. What is a projective dimension? One has an algebra. And M, so for global dimension, a projective dimension, sorry. One has A and algebra. And M is an A model. No, not necessary. So it may be projective or not. Projective models are nicer or better. So one wants to have projective models. But unfortunately, I don't know, it's not always the case. So we kind of approximate M by projective models, projective A models. So I have M is projective, either projective or not. But I know that there is always a projective A model, P, and an epimorphism from the projective onto M. So of course, if this one is projective, I can take the same here and this is an isomorphism. But sometimes it's not true. But what happens with this? If it's not an isomorphism, it has a kernel. And the kernel, let's write it here, A0. The kernel is inside here. The kernel may be projective or not. If it's projective, OK, I've finished. But if it's not projective, again, I can cover it with a projective A model. So I have a projective here, P1. And here, this is a projection, again, epimorphism. And this is the composition of these maps. So like this, one may continue and one may stop or not. Stop, what does it mean stop? Stop means that at some point, I can stop this procedure because the kernel that I find is projective. But maybe it never happens. So this complex is going to be exact because I constructed it like this. And if I consider only the projectives and the morphisms up to here, this is what is called a projective resolution of M as A model. So the idea is that if it's projective, OK, better. If it's not projective, then I kind of approximate it by a sequence, which may be finite or not, of projective model, exact sequence. So the projective dimension of M is the minimal length of a projective resolution. So it may be finite or not. So this global dimension being finite means that all these projective dimensions are finite. So all A models M have a finite projective resolution. This is global dimension being finite. And it's more than this because it has to be bounded. It has to be bounded by a number that will be finite. Well, so this definition is better if one wants to consider regularity in the non-commutative case, because it's the homological definition that we'll take. But for a moment, we are going to continue being in the commutative setting. So this was the intuitive part. There is Hock-Schild homology and homology in the title. They are there because one way of describing regularity is by means of Hock-Schild homology and homology also in the commutative case. And we'll see what happens in the non-commutative case. So we are still in the commutative setting. And the following part of the next part of this talk is going to be about what are called the color differential. So now we are going to have a k field as before. I'm not asking it to be algebraically closed anymore. And we are going to define a model that will enable A is k algebra, netarian. And we are going to define a model, an A model, which will be called omega 1 of A over k. k is always the background field. And what is this? So this will be the model of color differential. Recall that if m is an A model, this is commutative. An A model, so I don't say left or right. And we have a k linear map from A to m. We say that f is a derivation, k linear derivation. If and only if each time, well, it is k linear. And each time we have two elements in A. And we compute f on the product AB, we obtain, well, you all know the lameness rule, plus FAB. But since we are considering always symmetric models, action on the right is the same as on the left. This is going, we are going to write it with the action always on the left. So this is our hypothesis for the moment. So this is the derivation. So this A model that we are going to obtain, we have the following property, which is an universal property. Is that if we consider the morphisms of A models for all m A models from this one to m, this is going to be isomorphic as an A model to the derivations, the k linear derivation of A into m. And this property is going to give the definition of these objects. Because what does it says? It says that each time that you have a derivation in this sense from A to m, k linear, you can associate, in a unique way, an A linear map from this omega to m. OK? So omega 1 is defined by a universal property. And the universal property, well, first of all, we have two parts. The easy part, let's say, is that there exists a map, which is a derivation, from A, a k linear derivation, into omega 1. We are going to denote D. And the second part is that for all symmetric, I'm not writing it, A model m and for all k linear derivation from A to m. So what do we have? We have the algebra. This model that we are defining by the universal property, the derivation we already asked for. And we have m. So this is a derivation. f is a derivation, a linear. And so the universal property says that there exists a unique map here, a linear, that is going to make the diagram commute. That means that f bar composed with D is equal to f. So the good thing about objects defined by universal property is that if you have two of them verifying the same universal property, then they are isomorphic. But the thing to do always is to construct one of them. Because maybe we have none. So the thing here is that we need to construct this omega 1. So we need to show that there exists at least one. So we can prove existence here in two ways, at least, in two different ways. So the first one is that we need to, well, it's kind of not so intuitive construction. So we can take omega 1 as the quotient of the free A model generated by a set of symbols, let's say, that we are going to call D of A, or A in A, divided by, well, the submodel generated by some relations. And the relations here are those making the linear derivation. So the relations here are, well, first of all, it is k linear. So D lambda is 0 for all lambda in k. Second thing, it will behave with respect to sums for all A, B, and A. And third, it is a derivation in that sense. So is that this plus this? Yes, yes, OK, yes, I haven't said it, but yes. OK, yes, during all the talk. This is OK, and it provides the existence of omega 1, but it's not so, I don't know, I don't like this presentation. So let's look for another one, so existence 2. So for existence, other way of constructing this object, nicer, in my opinion, is that one can consider, we are considering A, which is the k algebra. We are taking the tensor product with A over k, and we consider the multiplication, which is surjective, of course. This is a map of A models, symmetric A models, AB models, no, it's not symmetric. It's a map of A by models, OK? And it has a kernel. The kernel is going to be called I. This is the equation, OK, right. So how can we describe this I, OK? Look at that I can be generated as left A model by elements of type 1 tensor A minus A tensor 1 for A in A, for A in A. Why? Because suppose that you have a finite sum of elementary tensors that belongs to the kernel, OK? So since it belongs to the kernel, this is the same as writing, of course, same thing, minus this, that is 0, because it's in the kernel, tensor 1. Now you write the sum over I, and you use the left A model structure here, and you take out the AI, and then you get this, OK? So it is generated as left A model by these elements. Of course, there may be too much here, but it's enough, OK? So suppose that now, yeah, up to now, yes. So now suppose that I want, I will say, well, is this my object? Well, I will try to define, let's say, kind of D tilde here, saying that, well, D tilde of A is going to be exactly this, that it's inside, OK? So is this a derivation, OK? So in order to verify if it is a derivation where I have to do a computation, so I compute D tilde of A times B, then this is, well, minus what is supposed to be to verify for a derivation, minus B D tilde of A, OK? So I compute all these. It's 1 tensor AB minus AB tensor 1 minus A tensor B plus AB tensor 1, and then minus B tensor A plus BA tensor 1. This is eliminated, and then I get this. And what is this? It is exactly the product of two elements in the kernel. So this is not, D tilde is not a derivation, but the difference, whether the defect, let's say, to have a derivation, is these elements that belong to I squared, OK? So instead of considering I, I go to the quotient upstairs. I divided by I squared with a projection. And then my map is going to be now from here up to here. I D, well, I cannot write it, OK? And so now D of A is going to be not this, but the class in I divided by I squared, OK? And now, since we are in I divided by I squared, it will be a derivation, a linear derivation. And so we have a kind of map like this, OK? So afterwards, each time that you have, well, I should have written in the other side, because I cannot write upstairs. But if you have, let's say, M here, OK? And you have a derivation S from A to M, OK? And then you can always define an A linear map F bar here, making the diagram commutative. And it is very easy because you define F bar of phi of 1 tensor A minus A tensor 1, exactly as F evaluated in A. And this works. So the second proof of existence is that verifying that this object I divided by I squared and this map D have the universal property. So since it is a universal property, we know that we can consider we have this isomorphism, OK? Great. So if here, I am still in a commutative case. But in the non-commutative case, I would not be allowed to make this commutation in the definition of derivation. And so what will happen here is that the derivation is going to be something such that F of a product, A times B, is going to be A F of B plus F of a times B, OK? And then I will need a bi-module because I will act on the left and on the right, OK? And in this case, what will happen, I will not write it, is that I will still consider the product, but I will do the show, OK, without necessarily dividing by I squared, OK? So I have this A-module. And since, well, this is an A-module and I am in the commutative case, I can construct the exterior algebra on this A-module. And this is going to be omega star. So this is the exterior algebra. That's another universal property. And, well, I'm going to consider this option, OK? And we are going to consider this object. And this is a very nice object. Well, there will be a lot of these here. But for each n, I consider the omega, the n component of the exterior algebra. I can consider the n plus 1 component. And there will be a map, still called d, sending. What do I have here? Something next, exterior algebra, OK? So it's an A0, dA1, dAn. And the map is going to send this to d of A0, dA1, dAn, OK? And of course, you can already see that this square is 0, OK? So we get the complex, which is the RAM complex, OK? So the good thing is that in the regular case, in the smooth case, this complex is very, very closely related to oxylogmology. And this will not be the case in the non-regular case, OK? So now I need to talk about oxylogmology, of course, because you don't know. I suppose, what is that? Why? Because this exists. And I'm computing this in the commutative setting. But I want to replace by another thing in the non-commutative setting, OK? No, no, no, no, no, no, no. I can define the color differential. I can define the color. Yes, I can replace it by another universal property, OK? Because change in the definition of derivation in the non-commutative setting, I'm going to have i, the current of the multiplication, in that case, OK? No, I cannot. I shall have the omega 1, OK? But it, well, yeah, OK. So now, oxylogmology, OK? So, oxylogmology. So for some minutes, k is going to be a commutative ring with unit. A is going to be an associative k algebra with unit. And we are going to consider this map, sending lambda here to lambda times the unit of a. And so let's think for a moment. Ah, yeah, a opposite. A opposite in the commutative case is the same of a, because it's the opposite algebra. The only thing I do is to change the sense of the product, OK? So this is the opposite algebra. I have not said yet that a is commutative, OK? So I consider the opposite algebra. And I have what is called the enveloping algebra in this setting that is just a tensor over k a op. And it is a k algebra. And the product of elementary tensor is like this. On the left side, one multiplies in a. So here, I have ac. But on the right side, one multiplies this by this in a op. So what I get is the product in the other sense. That's db, OK? db product in a. This is the same db product in a, which is the same as b times in a op d, OK? Well, so this is my algebra, generated by this elementary tensor. And, well, now, since a is not necessarily commutative, I need to consider not just a models, but a by models. What is an a by model? It is a model, and a model, let's call it, and on the left, an a model on the right. And both structures should be compatible in the sense that if I have an element in m, and I use the structure on the left, and then this is in m, and on the right, it's the same as the other way around, OK? But the point is that working with by models is not so nice, because one knows that, just a sentence, the category of a models of models over an algebra is a abelian. But for a by models, in principle, we don't know it. So we want to look at this as a category of models over some algebra, OK? And, in fact, what happens is the same having a by models. This is in one-to-one correspondence with a with left models over the developing algebra and the same on the right. This is model. And the object, the m, remains the same. But what changes is the structure. So here, I have an element in m, and I can multiply on the right, on the left, and on the right. It's the same in the parenthesis. I associate here this element acting on m, and here this other element acting on the right, OK? So it's useful, OK? So now, each kind of algebras, this is, I'm going to say it as a philosophical statement, but it's more precise than that. Each family of algebras, let's say, associative algebras, all commutative algebras, all real algebras, all lameness algebras, has an associated homology theory, a natural homology theory. Hochschild's homology is not the homological theory adapted, let's say, to commutative algebras, but to associative algebras. It's more general. So example of A by models, first example of A by model is A itself is an A by model. Taking as the action on the left and on the right the multiplication in A. And of course, this condition is true because A is associative, OK? But more generally, if I consider A tensor over K always, A n times for any n, this is an A by model with the action that A tensor B multiplying, extending this action, something here. This one acts on the right and the left. This one acts on the right. So it's this. So this is a kind of natural by model. So with this, the thing, the fact that A is an A by model and that these are by models, I can construct a complex. Because I have A. A, in general, most of the cases, is not a projective A by model. It's not projective over this ring. So I have to construct a resolution. So my resolution starts with A tensor A and the multiplication here. This is an epimorphism and this is in case. Now I have to ask for some hypothesis because up to here, K is a commutative ring, OK? So let's continue a little more. So here I have this map, the multiplication. And then, well, here there are twice, two times A. Now here there are three times, three copies of A. And here I'm going to put a map that's going to be called B prime for some historical reasons. And here I'm going to have four and the same map, but defined here. And this map B prime, when I compute it in an elementary tensor like this, is going to be a sum that I need one component less. So I will do all the possible multiplications. And there is an alternating sum A plus 1 from 1 to A minus 1. And then I multiply at each place, OK? So this complex is exact. And I can, it's possible to prove it in different ways. One way is to do it by hand, but it's too long. And then you find a simple structure, et cetera, et cetera. And another way is to construct a contracting homotopy, OK? So the contracting homotopy that is going to be called S here. For example, I can choose different ones, but we can choose an homotopy that is just putting, adding a one at the beginning, for example. So it is a map not of by models, OK, but of A right models, OK? So well, it is just a computation to prove that it's a contracting homotopy. So we have that the complex is exact. But it is called then a resolution, but a what, which kind of resolution? Well, now I need to add some hypotheses. So if A is k, either flat that I'm not going to define, or projective, or free, then this is going to be, respectively, a flat resolution of A as A by model, a projective resolution, a free resolution. Well, so now the definition of Hock-Shield homology take M and A by model. So the Hock-Shield homology of A with coefficients in M. So the resolution is up to here. We replace the by model by the resolution, OK? This part is what is called the bar complex of A, OK? So by definition, this is the homology of the bar complex of A, which is a complex of A by models, tensor over the developing algebra by M. And the map, we are going to compute an homology. We need a map. It's the identity of M tensor, the B prime that we have. That we are going to call now B. And if A is k flat, this is going to be isomorphic, because now it is going to be a flat project, either projective or a free resolution. This will be isomorphic to the tau with respect to k of A, or AE of A with coefficients in M. And we need this hypothesis. So if we are going to have a field, it's going to be free, because as I said yesterday, computing the Hock-Schiller homology using the bar resolution is almost impossible. So we need to replace by a smaller resolution. And we need to have this description as a homological factor, OK? Fine, OK? So just one particular case, if I may say, then we just denote it by HH of A, OK? Just a notation, OK? Fine, OK. I'm not going to compute examples, because I don't have enough time. But now, why I have been speaking about scalar differentials and now Hock-Schiller homology? Because there is a map going from the, now in the commutative case, there is a map going from the omega 1 to the Hock-Schiller homology in degree 1, I think I'm going to write. So now, we are again with k field and A commutative. So we have a map from omega 1 into HH 1. And the map sets sense and element, let's say, A1, D, A2 here. 2, the class, this is a homology. So it's a kernel divided by an image, OK? So it's the class in the quotient of A1 tensor A2. OK? Well, it is important. The thing to do here is to check that this map is a morphism of A models, and it is well-defined, OK? So I can define it just like this in 1 tensor A2. And it is well-defined because here we are taking the class in the homology, OK? So this is in size A tensor A divided by the image of the corresponding B, OK? If I do not divide by the image, it is not well-defined, OK? And in the commutative case, the whole Hock-Schiller homology has a multiplicative structure given by a product which is called the Schaffel product, OK? So if I commutative HH with the Schaffel product, but I will not detail, is a anticommutative graded algebra, OK? This is not the case when I is not commutative. And so because it's an anticommutative graded algebra, OK? And because of the universal property of the exterior algebra, this can be extended to the map from the exterior algebra of this one, which is omega star, into HH, OK? So by universal property, the exterior algebra, this map that is called gamma exists, fine. So in degree one, this map is not difficult to see that it is an isomorphism of a model, OK? It's easy to construct the map on the other way, and the only thing to do again is to prove that it is well-defined, OK? And the problem and the question, let's say, is this map an isomorphism? So is, so question, is gamma an isomorphism? Well, it is not always the case, OK? So a very nice result about this what is called the Hochschild-Gostand-Rosenberg theorem, Hochschild-Gostand-Rosenberg from 1961, 1961. That says that if k is a perfect field and a is what is called an essentially finite, commutative algebra of essentially finite type, what does it mean? It means that it is either a quotient as I have been considering before at the beginning of the talk or maybe a localization of this, OK? This is essentially a finite type, OK? So if this is the case, so if a is regular, then gamma is an isomorphism. So the interesting, well, it's a very nice theorem. It's an easy proof, very interesting. And the problem is, well, does this property of being gamma an isomorphism characterizes a smooth commutative algebra? And well, there are some results saying that, yes. But look at one thing before continuing. It's that if this map is an isomorphism, since a is finitely generated, this omega n start to be 0 when we have more than the number of variables. The index is more than the number of variables. So if this is an isomorphism, the Hochschild homology also has to be 0 starting from some degree. So if it's not 0, it will never be an isomorphism. So a kind of, OK, just kind of converse that we proved. So there are two proofs. One that we prove in Buenos Aires is called the group of this Buenos Aires commutative homology group that we had at that time and by Abramov and Beguet also. Is that this is for characteristic 0 and this is general characteristic, arbitrary characteristic, saying that if you have an algebra of this kind over a perfect field, then if there exists a prime ideal in A, such that the localized ring is not regular, then you will have that the Hochschild homology of A, let me write in some way, will not be 0 for all i, such that i is congruent modulo 2, to the maximum of the shape such that the omega shape of the localized ring is not 0. So in particular, you will have that if it's not regular, then all the odd Hochschild homology spaces or all the even will not be 0. So this will be an infinity number of non-zero homology spaces, and so gamma cannot be an isomorphism. So the fact that in commutative case, Hochschild homology stops or not shows you, reflects the fact that the algebra is or not regular. So afterwards, Abramov, I'm just finishing in three minutes. And Iyengar proved a similar result is for homology, Hochschild homology. Hochschild homology is defined over a field as an X factor. So now if we go to the non-commutative case, I'm going to write it here, we go to the non-commutative case. We can say that A, which is a k-algebra and aetherian k-algebra, k is always a field, and A is no longer commutative. So A is regular if and only if this is definition. The global dimension is finite. So we take the homological description. And so the problem, there was a question which we called Hapel's question. Hapel was more interested in homology than in homology. So we know that if the global dimension is finite, Hochschild homology stops. And then the question is, if Hochschild homology stops, does this imply that a global dimension of A is finite? So this is from 1989. And there has been a counter-example in 2005. And a counter-example, I don't know why it took such a long time, because it's very easy. The counter-example is very easy. It's an algebra in two non-commuting variables now. And then you divide by the ideal, to say the ideal generated by x squared, y squared. This guarantees that you have infinite global dimension. And then kind of commuting relation, which is yx minus qxy, where q is an element in the field and is not a root of unity. So in the counter-example, the thing is that the dimension over k of the whole Hochschild homology of this algebra is 5. It's very small. But the algebra is infinite global dimension. So of course, people turn their attention to homology again, because it's more natural. And there is a conjecture, which is called Hans' conjecture. And it's not solved up to now. It's just verified for a lot of families of algebra. So Hans' conjecture is that, well, it's the same, but for homology, if HH stops, so let's say, 0 for in n big, does it imply that global dimension of A is finite? And well, it has been proved, as I said, for different families of algebras. He states a conjecture only for finite dimensional algebra. But I think that it should be considered for all k algebras. So it's known, for example, if the algebra is Cosul, if the algebra is n Cosul, it's known for monomial algebras. It's true in all that cases. Another algebra that are called cellular algebras is for lots of families of algebras. But there is no general proof of this. It's still a conjecture. My impression is that one should consider, construct kind of a model of an algebras in the commutative case. But well, this is another subject. So that's all. Yeah. The other direction is true. The other direction is true. And is it ground to come back? No, no, no. Well, of course, it's a hand in the paper where he states the conjecture, the first thing he does is to compute Huxley's homology for this algebra. And the Huxley's homology is infinite. No, no, because if a is finite dimensional, you have a kind of duality, which is that if you consider the Huxley, let's say, homology in degree n of a with coefficients in a, this is isomorphic to the dual with respect to the field, let's say, om k of hn of a with coefficients in the kind of doing, well, this is om k k k as a by model. But there is a dualization here. But this is only for a finite dimensional case. So no, it's not. And the problem is that, well, a homology is not functorial. Homology is functorial. Homology is, in a sense, much better than comology. So maybe this will be true for homology. I suppose that it will be true, but I can't prove it. Non-associative algebra, but there you need a homology theory. And then, well, for Leo-Chordan algebra, I don't know if there is a similar statement. I don't know. Yeah, in that case, one can ask the question, I suppose. But for non-associative, well, you have to have an homology theory. OK, thank you.