 Okay, so it's a pleasure to announce the basic notion seminar by Professor Baylinson from the University of Chicago who's now also a little bit associated with ICTP and So he will tell us about algebraic cycles. Thank you so much for Inviting me to give a talk and the talk will be very elementary. So if you if you Studied a bit algebraic cycles. Well, well, you would rather well go and Have a pleasant pleasant walk in this beautiful weather well so The story about algebraic cycles. Well, it's it's a pretty huge subject. So I will Try to give some ideas of a Well, considerably narrow narrow part of it of which Of which I'm acquainted with And What I would like you what I would like to assume for you to know is just some very basic notions of algebraic Apology like what homological groups of topological space are and something like this and maybe what algebraic variety is and for me It's a projective variety. Maybe you consider just as Subvariety in complex projective space which Which is given by several polynomial equations I think it's like this. Well now let me let me pass to To the subject proper so I Will consider At the moment I will consider only complex algebraic varieties. So let X over a Complex algebraic variety which I assume to be smooth projectives Well, you can consider all possible algebraic sub varieties of X and you know that well there are every algebraic subvariety is just a union of Irreducible subvariety so that cannot be Presented in such minus union of two algebraic varieties and Just by definition if we consider Let us consider algebraic cycles of dimension e and Every group generated algebraic subvariety is irreducible algebraic Subvariety of dimension e and let's consider a free a billion group with those generators And it is denoted by the eye of X and It is called the group of algebraic cycles For example, you can consider just zero cycles and the reducible zero cycles are just point of X And the zero of X will be free a billion group generated by the points of X. It's huge and Seemingly seemingly stupid to make it less as stupid Well, you should remember a little bit of algebraic topology and try to consider not just cycles But cycles up to homotopy and since we do algebraic geometry the homotopy should be algebraic and it's clear what is a Fame if you have an algebraic variety as it's clear what is an S family of algebraic cycles on X so you consider Cycle on X times as such that intersection with each fiber is cycle of given dimension. So just Clear what it what it means and what I what I want to consider is and homotopy for me will be So I consider family of algebraic varieties. So I should it over connected as and as homotopy will be Well to fibers of given family of cycles Well, I consider to be homotopic, but the very the most important one come when S is a line or projective line, which which will be say the same for us and cycles model such model Homotopes algebraic homotopes defined over Just over line a one Well, it's called the group of algebraic cycles model rational equivalent such homotopy is called rational equivalence and the corresponding quotient group Is called the Chao group algebraic variety and it is And the first assertion that it's a very nature and interesting invariant of algebraic variety and one one want to try to understand it and certainly Well, it cycles Rational equivalence means homotopy parameterized by Rational equivalence Well, so up to algebraic Parameter just the line Well in principle you could take here any will see a curve But the curve is something complicated For example usual topology of a curve is complicated and you will get different answer not much different, but But still different and that is what I would like to see so one reason for considering the things is that If you have two algebraic cycles then using such a homotopy you can bring them to a general position So that intersection of every component or one cycle with another will have Will have well, it's clear what would the minimal the minimal possible dimension and in this situation they can be Intersected the intersection of two components can be assigned. Well, multiply some multiplicities and this would make this intersection would make the direct sum of Chao groups a ring and This ring is Quite interesting and that was the first reason why people started to consider such a things and Well, and maybe one first Very extremely rough way to try to see what Chao groups are is certainly you should say that algebraic cycle is certainly a usual cycle topological and so If cycle has demand complex dimension I it has topological dimension to I And so certainly we have a map from the Chao groups of acts to just playing homology group with integral k-efficient and in a sense, maybe it's a very general vague feeling that well first Wake feeling is that Chao groups are some sort of algebraic version of the bright version of homology But at least you can do intersection theory within it But certainly the kernel the kernel of this of this map is mysterious and the image well the image also certainly not everything but I Will mostly I will mostly talk and let's talk about about the corner. So the most classical situation is was Goes back to Abel and this is a situation when X is a curve Let's consider it first and in this situation certainly the only interesting group is the Chao group of zero cycles And it can be described very Very very explicitly So first you have the map to homology and the homology of top degree is certainly the integers and and there's just a Sensation to its cycle we assign a degree Cycle is combination of points and we take the total degree Well, and that appears when X is a curve the the kernel can be described very very explicitly. So let me Let me formulate the corresponding theorem and so the first fact about about complex algebraic curves is that if we can consider holomorphic forms On X, which is the same as algebraic forms and we can consider the anti-holomorphic forms and We know that every holomorphic form is closed and So we can assign to a form the corresponding class in Camology and the first assertion From well very basic watch theory is that the corresponding by the corresponding map is an isomorphism And this means that let's consider the dual the embedding from forms to Camology and let's consider the dual map. It will be the map from Nommology third gemology with complex coefficients to the onto the dual space of Glamorfic to the space of holomorphic differentials. So this will be projection and here see it the latest of Integral gemology And this is a complex vector space of dimension G and this is what G is the genus of our curve and this is a latest of rank 2g and the corresponding embedding as follows from this the composition actually identified with the latest Subletters of maximal rank so that the quotient is Is complex stores and let's call the the quotient is called the G code n of X and The first the basic the basic theorem about zero cycles on curves is that the kernel of the degree map So the cycles of degree zero Naturally identify naturally identified with the Jacobin and the identification is comes as follows So we have there is a natural map and the map comes as follows. So if I have a cycle which looks like a Linear combination of points with integral k fusions So it has degree zero and therefore I can find just the logical chain One dimensional chain whose boundary equals to the cycle. Well, and then I can can given A one chain I can consider the functional on holomorphic form just integration along this chain and this Integration certainly since this chain defined up to closed up to up to us. So we we fix its boundary. So it's defined up to Updating a cycle. So this is The functional integral over gamma which lies in the dual space to holomorphic forms is defined up to the image exactly up to the image of the slates of the corresponding element of the Of the Jacobin Depends only on the zero cycle and it's not difficult to check that Well, the cycles which are rationally equivalent to zero go to zero and therefore Everything will This is a well-defined map and a little bit more difficult, but not too difficult theorem tells that it is an isomorphism notice that one thing is Well, I told you an analytic construction Which is which is the most easy one But the fact is that this complex torus the Jacobin is actually it's an algebraic torus and one can define all this all this picture in a Pure algebraic manner that would mean that would make sense for algebraic varieties over any field and so on But the only thing is that so we can have constructed this algebraic torus By transcendental means, but certainly you can go back and say that one dimensional homology is the same as Of you Jacobin is the same as one dimensional homology of this torus So in a sense complex torus is the same as come all as a latest plus Well, essentially this type of watch the composition of its Of the corresponding complex vector space But so but you can look at it into in two ways one way is that it's pure algebraic object and another that says is a latest Plus some well some struck a linear algebra structure hodge the composition and the second point of view is Transcendental so you cannot recover from a torus as algebraic Geometric object you cannot recover so late as it requires topology, but not you cannot do it purely with an algebraic geometry But that is just a side remark which Which still will be for us important. Well, now let's try to consider what will happen If I consider the situation one dimension of x is more than one Well, and let's consider of just for simplicity again just a group of zero to our group of zero cycles and certainly there will be Almost the story will be almost the same Well, so you can at least all the constructions will work we can consider the degree map which is the corresponding This topological thing we and we can be interested in cycles of degree zero and then we can also This construction integrate so you will have same hodge the composition or one forms. It makes sense It it's also happens for algebraic varieties over any dimension And you will have the same torus the only difference is now that it's called not the Jacobin But albana's a variety so we should change the authorship and Important thing is that we should delete here Dizomorphism, but at least this map is is more or less evidently the subject of and again The remark that that I made that that the definition is transcendental But still it could be done purely also purely algebraic geometry And Well, and we have the thing and one can ask about the about the cure The kernel of this map of the subject of that was called albana's a kernel Well, and certainly the first the first guess that anyone would make is that's albana's a kernel is three Well, and people believed that it was so that it was so and even it was proved by Severe in in the beginning of 19th century chess, but then at the end of Well in the second half of 1966 momford tried to trying to understand Severe's argument and He proved the opposite theorem that I will that I will try to explain to explain now and which is And the most remarkable thing is that the proof was the same so Essentially, so it's it's probably the deep philosophical fact that that using the same Argument in different times and maybe in different countries you can come to different conclusions Well now Well, now let me formulate Momfart's Theorem so we consider the case one exit the surface and suppose that our surface carries a non-trivial global Amorphic to form and then the claim is that the Zanzalban as a kernel is non-trivial Well, in fact, it is well one can It's a Reutman theorem that this map is an isomorphism on the torsion subgroup So the kernel is actually a q-vector space. So I will I will explain just give Sketch of the argument, but it's a different argument since it's a little bit more on the style of But I will be talking about later and it's due to Spencer blog, but if you want to To study the thing I am very much suggest to read to read the original Momfart's work for the reason one reason is that this beautiful and second that it's philosophically deep as I explained well blocks a profit is based on two liammas and So the first lemma is this that the claim is that if we have this that this condition implies that or every the risk you open on certain an attempt to The map let us take age for I will consider Comology of the rational coefficients and I will skip the coefficients for future So consider the maps restriction to an open subset now inside the fourth Comology, there is the class of the diagonal. So consider the diagonal inside of x times x Well, and well, you can consider its class in fourth homology, but by one credit I can consider this Well, and the claim is that this restriction map sends data to some non-zero class Well, so so here is Here is a reason for this You can first you can write the class of diagonal using one career duality namely you can If you fix if you take a base well The Camology of x times x can be decomposed by q-net formula and if you fix a base in Camology, then the Chris is a class of the diagonal will be in q-net terms It will be just the sum of the tensor product of element of the base and elements of the dual base So if you wish it is Well, Kizimir element for the For the intersection pairing So that is a class of the dead now Let's see what will go to the Camology of you times you well first In order to see that the thing will Go to non-zero. It's enough to see that it's only That one of its q-net components will go to non-zero and I consider the q-net component Which is h2 tensor h2 so consider the map from H2 tensor h2 So let's just decompose this group by q-net and consider them up and let's show that just the q-net q-net component of Delta Here will go to non-zero and that is very easy to see if we this condition actually it can be Rephrased well, that's a standard result of hodge theory that if we consider inside of Camology second Camology of s the classes of Devisors the classes of all possible curves consider the vector subspace Generated by them then it will be not at the whole h2 and that's orthogonal Well, actually the precise well so it will be not the whole thing and let's consider its Artisanal complements this will be something which is called transcendental cycles So it's enough to check the distortion when you is small enough and If you is small enough then the projection from h2 of x to h2 of u has Kernel exactly the all the the whole group of algebraic cycles and its image will be the group of Precisely the subgroup of transcendental cycles and Well, and it's easy to see that the Cosmere element here will go exactly to the Cosmere element of h2 transcendental Then there h2 transcendental which is non-zero. So it is elementary topological lemma Corollary of punker adults now here is As a second lemma and the combination is Mumford theorem well, I Recall that what we want to check that the albana's a kernel is non-zero, but inside of certainly We can consider for every maybe given a cure inside of x We can consider the group from the child group of our curve And we know perfectly well what they are because of I will say them and albana's and Certainly, this is well, this is a complex torus There's a complex torus and we know that this is controlled by the third gemology of the curve and this is controlled by the third gemology of algebraic variety and No, no, no, just just any curve Consider any curve at the moment Show zero of a curve. Yes child zero of a curve. Yes Yes cycles of degree zero child zero of C Yes, okay well, you know Well, it's again a usual result of pump topology of algebraic varieties that if C is Is a hyperplane section then it's first homology groups maps on to the first gemology group of eggs and therefore since you know how Jacobian of C and albana's a how they connected with this homology we see that the corresponding composition will be surjective Well, no So therefore to establish that the albana's a kernel is not trivial. It is enough To show that this map is never is now a surjective. So for any curve is not surject So let me sketch Let me sketch a proof. Sorry. So, so this is this would imply sorry maybe So this will be reformulation of this is just plain reformulation of The formulation of Mumford theorem, but what I will prove what I will prove is the following that Is any cure of touch that This assertion such as the corresponding map is Surjective and the claim is that then for as efficiently small the risky open you inside of of X the image of the class of data in Well by the map the first component will be actually X minus C times you is Zero the so as there is also the combination of the first and the second lemma implies This assertion and this implies And this implies Mumford theorem So so it remains to prove to proof lemma 2 and lemma 2 is follows by The idea of the proof is this it comes from something called spreading So let me let me explain the idea. So here comes algebraic geometry So we know that every algebraic variety. Well, so we have X which sits inside of the projective space and so we can we can We know when a subvariety is defined over some Subfield of complex numbers. So just if it can be given by equations with coefficients in the subfield and certainly any algebraic Variety of projective space Is defined over some subfield which is Finitely generated over Q just take All coefficients of all equations Generators of the subfield of X and certainly and so so we can find a Finitely generated field such that both X and C are defined over Over this field now Now the claims that so we can find actually you also Well, it will be also defined over Well, so now so so now No one proceeds as follows choose a point On X. So let well, let me just give a sketch of the argument So choose a point of X which does not lie on any subvariety defined over the subfield so again subfield is countable and there are Countably many subvariety is defined over as a subfield. So so we just exclude a countable number of subvarieties and So there is you can choose a point So this point will be as a result this point will be the field of Definition of this point will be the field of rational functions on whole X. This is generic geometric point of X over this field K now if we have if Our condition means that this point can be there is an algebraic Homotopy from this point to a zero cycle which is supported on C So the space of this algebraic homotopy is also well they form Sort of an algebraic variety and so you can find one which is defined over the finite extension over The field of definition of the of the point X which is a field of rational functions of X over the field Well suppose for simplicity that the thing is actually defined over this field of rational functions now You can spread this family from the generic point Which is the field of definition to some the risky open you And use and as a result you we will we will get an algebraic homotopy Which will push the diagonal cycle to a cycle which is On X times you which is supported on C times you and this is This will be exactly well and therefore the corresponding class of the diagonal passing from to base C will in passing to the Topological class of the diagonal we will see that it lies That it's restriction is zero well, so So therefore we come to this strange So that is the end of the sketch of the proof and therefore we come to a strange situation that that there is seemingly Well, it's not it's not clear how to describe What does the meaning of the albana the cure how to describe so the albana the variety is clearly something Geometric and which is defined by the first homology, but the albana the kernel is not and that's not clear what it means now one shortly after probably Mumford wrote this work Spencer block asked well probably in connection with this proof acts as himself a question if Indeed the assertion of Mumford theorem is not only sufficient for Non-trivality of the kernel of albana Not but If if the opposite theorem is true that if There is no homomorphic to forms on the surface then actually the kernel of albana this triple This assertion is called blocks conjecture it goes back to to the beginning of 70s, and it's still and it's still not proven in any generality now What I will try to explain in the remaining 20 minutes is some Well, since then blocks conjecture was somehow put in a general General framework which was in the beginning was was rather philosophical, but then Then it became well pretty concrete since the advent of triangulated category of motives, but I will try to explain somehow at least some sort of ideas behind precise assertions which at least Hopefully tell the language how you can describe how you can describe Chao groups. Well, maybe First first sort of a weak question that one can ask so in both in Abel theorem and in Mumford theorem the main objects were homology groups of algebraic It's a plus Plus what structure there? Well, and well, maybe not well At least something like this and you can ask the following question if So first first I will make it pretty vaguely that if there is a linear structure On homology groups of an algebraic variety natural linear structure which determines the Chao groups well So again Abel's theorem and Mumford theorem and or rather blocks conjecture they tell that something like this might hold so here is a test question and What means could mean the word linear structure suppose that you have a finite group G X on X. Well, then it acts on and then it acts on Homology of X and it acts on the Chao groups Suppose so we want to we want to see to define actually linear structure on homology now suppose that in a reducible representation irreducible representation of G does not the homology of X and If such a sort of a Such sort of a question has positive answer then this would imply that it also does not occur in the Chao group Well dancing this on this test question is not known though. So you see just a degree of Lack of understanding of the subject Well, so if you so that is if you wish it's a version of Bloch's conjecture certainly and a typical typical example, which is not well suppose that you are that Not many shing of the forms means that I mean that you can as a transcendental part of Camology equals second Camology equals to zero well You can you can drop algebraic For some reason algebraic part of Camology Well, if you wish I could put here also transcendental part of Transcendental part of Camology and this would be equally true and then blocks conjecture will be sort of a part of this Assertion for the trivial group. Well, I don't I don't say that it's exactly the same but it is But it is a but it is of the same of the same spirit maybe Now let me try to to give some feeling what what the word linear structure means and Well, so in order well Camology comes when we we look at algebraic variety and then we see it only as topological space We forget about algebraic structure, but we see it as a topological space now Well now there is let us see Well, let us formulate a sort of a dream And to some to some extent this dream is true due to the words of wewotski and others on modus So a dreamy picture what Linear structure means Well, so first we are looking for We are looking for a topological space. Maybe topological space should be this word should be understood with in some not Precisely defined sense, but Well, let's call it M such that to every Every X every algebraic variety X there corresponds a Fibration let me call it curvy X over X Over M. Sorry So that's the third thing and second that there is this thing should have somehow purely Geometric origin, but what we also has we also want a point in a base point Well, let's call it. Well mu say inside inside of M and and the identification of The fiber over this point topological space that corresponds to X So that's sort of a well, so if you wish then the homology of X Will be a local system over M. So they will carry the representation of the fundamental group of M and this is this representation is this sort of a linear structure that That I'm talking about now Okay, so so that is what sort of a linear structure means now I would like to describe how say what we want from this a linear structure in order to recover Recover the child groups. Okay, so given such a situation. We have a map from If we have a vibration over M we have a map from the Camology of X to the Camology of the fiber and Roughly speaking what we want to be what one wants to be that the Camology of X Will be the child groups and the corresponding maps the restriction to the Camology of the fiber will be our old map from The child groups to the topological homology or Camology of X. Well, so this this is about About the reconstruction No M depend on the point in case when X is a point this will be M itself now What we want to against that this picture would be people have purely geometric origin It will be defined for if you wish for if we consider a variety of various fields and M will be Something that would be assigned to spectrum of a field and that will be assigned to spectrum of algebraic variety and there are also some other hopes and for example, one hope is that if We consider a field that correspond to if a base field We consider has transcendental degree n over q then the dimension of this topological space Will be M plus 1 So here is well, there are reasons for this prediction. So Maybe let me say one corollary of such a prediction which is which is In the sense it is an opposite to An opposite to Manfred's theory. No, this this picture this picture is very general So I would like to to make a sort of sort of a hypothetical framework. How one can understand Chow groups Well, I can well I can say I can formulate Okay, so maybe before before continually let let me say it like this What is what is known due to due to works of Leivocki this that this picture is In some sense it is known namely what is known is not not the topological space and again the work topological space as I told you should be understood with sort of a grain of salt but But it's algebraic topology the category of shifts on this topological space with all and the corresponding Picture like this Knowing category of shifts is enough to compute to compute Camology and so on and this is what exactly what mixed What? Motivic theory is doing the only grain of Remaining grain of salt in Leivocki work is that all his constructions are They produce triangulated category of shifts the derived category of shifts and there is Currently there is no way to To define a billion category of shifts themselves on the lectures of the main lectures of this school There was some basics about derived categories and what is What lacks in order to make the theory Well applicable to make Leivocki theory applicable to the study of Chao groups is Existence of some very canonical t-structure on the category of motives. That's uniquely characterized by many properties But its existence is Is still is still a mystery now Well, I'm at the end of my of my hour, but let me let me formulate just a very concrete Assertion that follows from from existence of such a of such a picture and this is in a sense it is a Assertion that Mumford's theorem actually requires Complex numbers or something like this for the base that over field of algebraic numbers The opposite to Mumford's theorem will see various theorem always hold So let me formulate again, it's it's it's an old conjecture, but it comes from this General philosophical picture and nobody knows the answer and namely As assertion is this that if you have an algebraic variety X over K Where K is a number field then The group of zero-cycle cycles on X K of degree zero maps isomorphic to Albanese Well, so here is a very a very concrete example most concrete example of Such a situation So suppose that we have our X is an abelian variety. No, this is conjecture Supposite X is an abelian variety well, so consider a couple of points a and b in X and Then consider the following zero-cycle. So consider a plus b minus a minus b Plus zero well, this is a zero-cycle of degree zero and certainly it goes into It lies in the Albanese kernel it goes to zero if you just add this Expressions inside of a billion variety. You will get zero Well, and it's easy to see that such things actually generate Generate the Albanese kernel, which means the following that if we consider so therefore if My billion variety has dimension also say it's an abelian surface then we know that for generic If it's a billion surface over complex numbers then for generic a b in X Such a cycle will not be rationally equivalent to zero But if it is defined over a number field and we consider point over a number field then the thing is always equivalent to zero rationally equivalent to zero to my knowledge nobody Nobody proved this assertion even in case one X is a product of two elliptic curves of K So it's number field is rationally equivalent to zero well So that's it