 So I wanted to talk about the veil conjectures. So this is a rather deep theorem of algebraic geometry, which was first conjectured by André Veil, and was then the proof was finally completed by Deligne. And so this theorem relates the topology of complex projective varieties to the counting of points of varieties over finite fields. And this in some sense is already quite exciting because these are two different things, quite different things. And the fact that there should be any relation is kind of surprising. And so I will here not make any attempt to explain why this should be true. I will just try to explain the statement, so tell you the ingredients of it, and then give some examples. So this is somehow erected mostly in some sense the way I set it up towards the diploma students. So I will sometimes refer to the lectures that you had. So one of the advantages of this thing is that it somehow uses several things. It shows that the mathematics is a unity and that you should kind of also see it as such. So first we talk about algebraic geometry. So let me say what is a projective variety. I'm not trying to be 100% correct, so we say we work over a field. And we have the projective space, which you know. So this is pn over k, say. So these are just the one-dimensional subspaces of k to the n plus 1. And so in other words, these are the sets of n tuples a0 to an, where the ai are in k, divided by an equivalence relation. Namely, we have such an n plus 1 tuple of numbers is equivalent to the same if we multiply everything by the same constant. The lambda is an element of k without 0. So for the moment you can think that k is the complex number, so we have this. And so this is a nice space. And now we want to look at projective varieties. So for this we take the 0 set of polynomials inside this. So if f is a polynomial in n plus 1 variables, then we, which is also homogeneous, so it contains only monomials, which have all the same degrees. Then we can look at the 0 set of f, which is a subset of this projective space. This is just a set of all, you know, such equivalence classes, so a0 to an in this projective space, such that we get 0 if we replace the variables by the ai. So this is a hyper surface in Pn. And then we can also look at the common 0 set of several hyper surfaces. So if fi are polynomials, homogeneous polynomials, then we can look at the 0 set of f1 to fL, which is just the intersection of the 0 sets of the fi. So the set where all of them are 0. And so this will be some subset in this projective space. And, you know, in the algebraic geometry course, you know that such sets are called algebraic sets. And if they are also irreducible, so if I call this thing x, so if x is irreducible, basically means I cannot write it as a union of two smaller subsets, then it's called a right, projective right. Okay, so that's this. And so you can, if you want, you can always think that we have just the 0 set of just one polynomial, z of f. And in that case, the 0 set of f is irreducible. If f is irreducible, that means not the product of two polynomials. And you also have the notion of dimension and non-singularity, the smoothness of such varieties is defined. And, you know, there's a completely algebraic definition. So the dimension, so for instance, the dimension of the projective space will be n, the dimension of the 0 set of a polynomial inside the projective space will be n minus 1. And you have non-singularity or smoothness, which means whatever that the tangent space at every point has the same dimension as x. And for, you have an algebraic criterion for that. For instance, the 0 set of f in Pn is non-singular, if and only if. There is a, if I take the 0 set of f and all its partial derivatives is empty. There is a completely algebraic criterion, this you had, I think, in your course. So now, we want to look in particular at complex projective varieties and manifolds. In the algebraic geometry course, usually when you do algebraic geometry, you use, you have these varieties with the risky topology and this is all very strange. But if you have a variety over the complex numbers, you can also view it as a complex or differentiable manifold by just taking the same set in the projective space and using it with the usual topology. So we have that if x in Pnc. Let's say that is a smooth, so non-singular, objective variety. So maybe I could put a large n here, of dimension n. So then I claim this is a differentiable manifold, like in the course that students had. So first thing is that the projective space Pnc is a two n-dimensional, dimensional, differentiable manifold. And it's kind of obvious. If I have a point, if I have, look at the set A0 to An, such that in Pn, such that, so this was a large n, such that A0 is different from 0. So then this is a chart, diphylmorphic to C to the n, which is the same, which is for this purpose the same as R to the two n, just in the obvious way. I sent this thing to A1 divided by A0 until An divided by A0. And so we, if we do this for the other AI, we get all these charts and this makes it into a differentiable manifold. And if x is, our x is supposed to be a smooth projective variety inside this thing, then x, that's the subset x, will in a natural way is a sub-manifold of dimension. When? When n, so we always talk about real dimensions here. So, okay. So let's look at some examples. If we take one-dimensional projective space over the complex numbers, you know, one can describe this with the open set where the first coordinate, the zero coordinate is non-zero, which is isomorphic to C and we add just one point at infinity. And so this is really just the one-point complication of that. So it will turn out that this as a topological space or as a manifold is just S2. So it looks like this. And then we can also look at an elliptic curve. So let's say contained in the two-dimensional projective space should be an elliptic curve. So that means it's a non-singular variety which is a zero set of a homogeneous polynomial of degree three, non-singular cubic. Then one can show that it will be a TORS like this. Okay. And modernally, if you look again at TORS instead C in P2C is a curve, a non-singular curve of degree D, zero set of a homogeneous polynomial of degree D in three variables. Then it follows that C is the Riemann surface B where G is equal to D minus one times D minus two. So it looks somehow like this as such a handles. Okay. So we see that these are some nice manifolds. So now I want to talk about this because I want to talk about the topology of our complex projective variety viewed as a manifold like that. So, and what we want to look at of it about is the homology. The students had a course in algebraic topology where homology was introduced. So if X is the topological space, reasonable one, then we can associate to it the homology groups. Maybe I call them just H, I, X and maybe as coefficients, I take just the rational numbers. So this is a vector space over Q. And it's a bit complicated to say what the homology, so we can somehow associate to our topological space some groups or some vector spaces which somehow tell us something about the shape of X. So very roughly, we could say that the dimension of this vector space could somehow have something to do with the number of e-dimensional holes, so i-dimensional holes of X, whatever that might mean. So for instance, if we have something which consists of two kind of blocks like that, and this itself is just an open subset of two each of them, then we will find that it has two components and this will imply that H0 of this will be equal to Q squared, namely one factor for each kind of connected component. And if we have something like a circle so then we get H1 of the circle will be equal to Q corresponding to the fact that this thing has a hole in the middle which does not divide the thing but somehow does that and H0 in this case will be also Q because it's one piece and so on whatever that means but obviously this is not a definition so I will go over lectures on algebraic psychology you had some introduction of homology I think it was a simplicial homology so I will very briefly review how that goes obviously I cannot do it properly because it's a course if one does it properly so there are many different definitions for homology which are for nice spaces give the same groups but the simplest one even though not the simplest one to work with is the simplicial homology and it works very simple we divide our topological space X into simplices this is the first step so that means you have points edges triangles tetraedas and we somehow actually we will use oriented simplities so somehow so the point the point so we counter positive one can imagine an orientation of an edge to be like a direction and this will be the same as minus if you do it the other round or if you have a triangle you can view an orientation somehow as a way of turning it around and this will be the negative of turning of somehow the other direction in the course obviously you have a proper definition which somehow has to do with numbering the edges and the vertices but for these small dimensions it somehow seems to me more intuitive to just try this and then we can make so we have our X and we have divided it into simplices for instance we have this very simple case we have this thing we have divided into so we have here these are maybe our points so we have two triangles this and this maybe we give it an orientation so whatever give all of them an orientation whatever for instance like this and now we make a chain complex I don't know why it's so difficult chain complex so we have some vector space ck associated to our X we have divided into triangles like that which is just the direct sum over all sigma oriented case simplices and we take the vector space q times the simplex so we have a vector space which has dimension as many as the number of case simplices for instance c2 in this case is a two dimensional vector space and then we have a boundary map so we have now for each k we have this and we have a boundary map between different ones which is delta from ck to ck-1 which is a vector space homomorphism so q linear so the coefficients and on simplices itself it's given like that if we take a point this is map to zero if we have an edge with the two endpoints like this maybe I call this is number one and this is number two this would be and say it's oriented like this then this will be minus the starting point plus the end point and for a triangle we just take again the edges with the orientation given by this so this is say this one plus this one plus and so this gives us a map I don't tell you how it works for higher k because then you would again have to cannot rely on this intuition anymore with this little drawings and you can check if you apply delta twice get the zero map and so you can define the k-thomology group of x to be the kernel of the vector space homomorphism from pk to ck-1 divided by the image under delta of ck plus 1 by this statement this thing is a sub vector space of this one and this will be our vector space which is the homology so one has this definition it's not quite clear what this should have to do with holes but you can experiment a little bit with small cases you can somehow see where it comes from that somehow it has something to do if you have a hole because somehow it can have a k-cycle which goes so something in the zero which goes around it and you can only fill it up which then would mean that it is in the image here if the hole is not a hole so just to say it if you have a triangle like this but just the outer side then this delta of this will be zero but it will be delta of this delta of this will be zero but this is equal to delta of this okay and so therefore you can see that you will get a non-trivial group if this thing is empty but if you can close it up then you don't and so it has something to do with holes okay so much about that so once one has these things these groups we can look at the Betty numbers so b i of x just the dimension of the i-thomology group of x and we have the so we just take this number which is the dimension so we throw away some information and we have the Poincare polynomial which is just the polynomial which keeps these numbers so this is say p of x t sum over all i the i-th Betty number of x to the i so knowing this polynomial is the same as knowing all the Betty numbers so just in our examples maybe we will not explain it but for instance we get if we take the Poincare polynomial of s1 according to what I told you this is 1 plus t if we take the Poincare polynomial of s2 which is the same as you know obviously so I had to t as a variable here p of the projective space over c one dimensional t this is 1 plus t squared and if I have a surface of say genus g then we will get 1 plus 2g times t plus t squared so the dimension of the first homology group is 2g in particular for trolls we get 1 plus 2 plus t squared because g is equal to 1 so now so much for the complex things so I have explained one side of the theorem which I haven't told you and now I will tell you the other side which is somehow counting points over finite fields so for this to count points of so if we start with a complex variety if we want then to count points of a variety over a finite field we somehow have to make out of the complex variety over a finite field over the complex numbers 1 over a finite field and this we do by reduction modulo p p is a prime like 2 or 3 or 5 or 7 so but we do it so let x in complex projective space so maybe we do it like this be a complex projective, a non-singular complex projective variety as I explained before and we require that it is defined over q what does that mean so we know that x is the 0 set of some polynomials that's what it means to be a projective variety and we want that the coefficients of these polynomials are in q so if that is the case we say that x is defined over q now if it is defined over q it's in some sense also defined so over the rational numbers it's also defined over the integers because once we have coefficients in q we can multiply we can clear all the denominators of the fi and we get polynomials with coefficients in z and we can also do this in such a way that afterwards the coefficients are relatively prime ok so then clear denominators then we have that the fi are polynomials with coefficients in z as I said I want them to be relatively prime so once we have that we can reduce modulo p so we have the finite field fp so we have the fp which is just z modulo pz the simplest finite field where p is a prime number so this as a set is just the set from 0 1 until p minus 1 and then as you know you add things here by adding these numbers and then taking the rest by division mod p and you multiply them by multiplying these numbers and then taking the rest by division mod p so this is a nice field and so we can we have our fi with coefficients in z so we can take fi bar to be the so the polynomial obtained by reduced reduction modulo p so we have this has some coefficients ai and we just take for each coefficient we just take the rest that we get when we divide by p and replace it by that so we get the polynomial with coefficient in fp okay that's very simple so we have associated our complex variety with this assumption a variety over a finite field and actually fp now well we haven't yet but obviously the obvious thing is that x bar we take just the zero set of these polynomials which is a subset in the projective space over this field fp now note that okay so this is the reduction of x mod p it's a rather simple thing and now we want to so thus we have associated to a complex projective variety a variety over a finite field over this field fp and we want to count the points so note if we take this projective space pn over so so obviously we also so I write this maybe xfp we also have for if q is a prime power you know that the finite fields have as number of elements the powers of primes and these are all finite fields so if q is a power of p then this we also have the finite field so we have fq is a finite field was mentioned in the algebra course and so if you want so this is a field extension of fp it's actually a splitting field of a certain polynomial over fp so we have studied that and it has p to the n elements so anyway we have that fp is a subfield of fp to the n so therefore we can also look at the zero set of the same polynomial in fp to the n so we can also also have x bar fp to the n which is just the same zero set of the same polynomials in the projective space over fp to the n and now we want to count the points of these things so we want to count the number of points of x over fp to the n was it x of x bar over fp to the n so we should remember that the projective space of any dimension over a finite field is finite because it's a quotient of just the n plus 1 dimensional vector space over fp to the n elements and this is points of this variety are subsets so it's certainly finite and we want to count it and so we write down a generating function the generating function is a formal power series which contains all these numbers this is called the zeta function of x this would be so let's say x z our prime number of x t is a variable so what we do is we write down for all n bigger equal 1 the number of points of x over fp to the n some nice number then to make it more complicated we make a generating function like this to somehow confuse the issue even more we take the exponential so exponential just means the usual thing so the exponential of a formal power series of a variable x is just the formal power series sum n bigger equal to 0 x to the n over n factorial and as you see the thing inside here is a power series starting with t so certainly we can do this to this power series to get a formal power series this is this generating function for instance to see what this exponential does if I take I can call this exercise although it's trivial if I take the exponential of say sum n bigger equal 1 alpha to the n t to the n over n this will turn out to be 1 over 1 minus alpha times t so it's the power series which is the development of this rational function you know this just comes from the fact that this thing inside basically is the logarithm of this and so then you take the exponential so we can therefore we can use this to write down one example so what is the zeta function of p1 so I claim so where do I have it so we know that p1 over failed k can be written as k union the one point infinity so in our case over fq we have here fq and one more point so we see that the number of points of p1 over fq is equal to q plus 1 okay so if we write this zeta function here we get therefore this is the exponential of some n bigger equal to 1 so 1 plus p to the n t to the n over n now the exponential is multiplicative so it's the exponential of this first thing with the 1 and the one with the p to the n and to both of these this applies so according to the exercise this is equal to 1 over 1 minus t times 1 minus p times t that's actually relatively simple in particular we see it's a nice rational function and so now I want to say that this somehow always happens and in a very interesting way which is related to the topology of the variety we are talking about so this is the theorem very conjectures so let's say x smooth projective variety of dimension n over the complex number and so we assume that the reduction it should be defined over c defined also works anyway we will not come to that complication so defined over q and we assume that the reduction mod the prime p is also non-singular no mod p so the reduction mod p is non-singular then we have the following result first this thing this zeta function is a rational function so zp t can be written as a rational function like that so we have some polynomials with odd indices q q1 of t q of t and so on until q 2n minus 1 of t this is a polynomial and then below we have even indices q2 q0 of t times q4 of t and so on until q 2n of t where this qi are polynomials with complex coefficients in t yeah that's q2 yes they are just the even numbers there is no further complication so these are this and one can actually say one of them two of them are particular zero particular simple q0 of t is just 1 minus 1 is just 1 minus t and q 2n of t is just 1 minus q to the n you can see that for instance the case of p1 we are in this instance where we have only these two so we have only q0 and q2 and it fulfills this thing and what else can we say about these polynomials that is then more the more interesting part namely we can we are over the complex numbers we can somehow factorize these polynomials so we can write qk of t we can write this as a product over some i I will soon say more about this i 1 minus some complex number alpha k i times t so there okay where the alpha k i are some complex numbers and they have one interesting properties these are complex numbers so I can take the complex absolute value and this will be equal to p so here here I have worked over zp so everything is over p so this q becomes a p okay so this will be p to the k halves and what else can we say now comes this actually is very deep but the thing which is more that we actually can see what the index is the 0 the index goes from i equal 1 to the kth petty number of x over the complex numbers so this is very amazing so that means the degree of this polynomial qk of t is equal to the kth petty number of xc and this is really quite amazing because it doesn't look like there's very much relation between counting these points and the topology of this thing okay and the last thing which I'm not sure we will use is that there's actually some relation between these different polynomials namely if I take q2n minus k it goes on to n of t this can be written as the product again i equals 1 to pk of p so 1 minus p to the n times alpha k i so this is something which is related to probability for any faults I don't think you know that but anyway there is a further property that this thing has but you can see this in particular as you see implies that pk and p2n minus k are equal and this is quite a equality which one knows from algebraic topology but maybe not in a beginning course like that okay so what one can easily check that one can reformulate this in a simpler way so this formula so this formula 1 and 2 we can reformulate in a different way so 1 and 2 are equivalent to the following if we just take the number of x over x bar over fp to the n then we can write this as the sum k from 0 to 2n sum i equals 1 to the i th petty number of the our complex variety minus 1 k times alpha k i to the n where the alpha k i are the same as we had here so this is just if you think of it it's just again applying this exercise it's really an exercise that these two statements are equivalent but what this tells us however is that if we count the number of points over these fp to the n for all of them then we can determine these numbers I mean you can determine the terms in this formula and we can read off the petty numbers we just have to see when we know all these numbers we know what the alpha i are and we have to count how many alpha i have absolute value p p to the power k half this is the case petty number of our variety okay so this can read off the petty numbers or if you want the pancrepolynomial of our complex variety from the number of points over fp to the n if we know it for sufficiently many n and in particular in particular a simple case so if we are in the case that this is very very simple that the alpha i are just powers of p so if number of x over say f so q is p to the n is a polynomial say g of q then it follows that the pancrepolynomial of xc you know we just we have to count the things where we have the power p to the k half so we just replace q by t squared but in general you know we don't need that so even if it's not like that we can do things okay so i want to give some examples so i should say so for one thing it's exciting that this action is there between complex geometry and counting points but it's also actually useful it has been applied so the field of algebraic geometry i'm working in is modular spaces so modular spaces are varieties which parametrize something we are interested in so the points of the thing corresponds to certain objects say vector bundles curves whatever so then usually you can prove that such a modular space exists and maybe you can prove it's non-singular and projective but maybe you cannot say very much about its shape but you know what its points are so therefore you can count the points so without knowing anything more than the existence and the kind of smoothness you can compute the homology of the space we have no idea how it looks like but it is based by just counting points so and so let's look at some examples of such modular spaces in some sense for instance if we take the n-dimensional projective space so if we take this over fp fq so this is just parametrizes the one-dimensional subspaces of fq to the n plus 1 and so we use a basis vector so this is fq to the n plus 1 we take a non-zero vector and we divide by multiplying by constant by so we have the so we have equivalence classes no n tuples non-zero divided by by this and so this means if we take the number of points this will be n plus 1 minus 1 points and this has q minus 1 so we get q to the n plus q to the n minus 1 plus and so on plus 1 so we get this so this is a polynomial in q so according to the story we get that if we take the Poincare polynomial of pn this is just say whatever t to the 2n plus t to the 2n minus 2 plus and so on plus 1 and in fact it's kind of well known that the homology of p2 has one is z in every even degree and so this is over the complex numbers and so we get this let's look at this somewhat more complicated but related example which is the Grasminian so this is a projective variety which parameterizes so g I call g ln so this will so over any field this will parameterize the l-dimensional subspaces of k to the n so the projective space is just g1 n plus 1 how can we count points so we want to compute the number over a finite field fq so how does it go so if we want such a subvector space so it's a subvector space we first can choose a basis for the subvector space d of dimension l so we choose just l linearly independent vectors how does it go so the first vector we can just choose any nonzero vector so we get q to the n minus 1 so this is v1 is an element in k to the n 0 then the second vector we can take any vector which does not lie in the span of this so this would be q to the n minus q so the v2 is an element in k to the n minus the span of the first vector and obviously you can see how it goes on so we get in the end q to the n q to the l minus 1 because our last vector vl is an element in k to the n so k was fq without the span of the first l minus 1 which is in l minus 1 dimension of vector space and then we have to forget the basis so we just want to remember the vector space so if we have two bases of the same vector space v1 to the l depends the same space as w1 to wl they are both bases of a l dimensional sub vector space then this is like this if and only if there exists a matrix which is also then unique and l by l matrix which kind of transforms one basis into the other so such that wi is equal to sum over all j m aij times vj for all i from 1 to l that's how it goes so we can change the basis so we have to count how many such matrices there are but if you think of it the first in l matrix of rank l obviously as opposed to it should be matrix of full rank if you take the first column of the matrix it's just a non-zero vector from k to the l the second one is a vector from k to the l from the compliment in k to the l of the one dimensional sub vector space the first one and so on so it's the same story as here so the number here matrices is q to the l minus 1 times q to the l minus q times and so on q to the l minus q to the l minus and so we find that the number of points of g q is what we obtain if we divide this thing by that and we can kind of factor out the extra factors q that we have here so we will get that this is that this is just anyway it seems obvious q to the n minus 1 times q to the n minus 1 minus q to the n minus l plus 1 minus 1 divided by q to the l minus 1 q to the l minus 1 minus until q minus 1 and so thus it means if we want to see what is the Poincare polynomials what are the Betty numbers the Poincare polynomial of which we don't know anything else than that it exists and that parameterizes these things and I claim it's also non-singular and so on the Poincare polynomial of this thing over the complex numbers will be this thing where we replace the q by t squared so this is t to the 2n minus 1 and then it goes on until 2n minus l plus 1 1 divided by 2l minus 1 and it goes on in the same ways here until t squared minus 1 and if you are familiar with some combinatorics and so it's maybe not completely obvious that this is a polynomial but you can easily check that it's one factor by the other and we know that by this thing the odd Betty numbers of this thing are 0 and the even ones are given by this formula so if I take the 2k Betty number of this thing you can read off with some effort from this formula I will not even exercise that this will be equal to the number of partitions of our number k into if I'm not mistaken n at most n minus l parts and all of which are smaller each of which is smaller equal to l so you partition means just you write the number k as a sum of smaller numbers and I require that I should have at most n minus l numbers and each number should be at most l and for instance the first non-trivial case is the g2 4 so when it's not a projective space in this case we get the primary polynomial of this thing will be t8 plus t6 plus 2t to the 4 t so I used up my time I had some I wanted to say also something else but maybe I will not so ok so maybe that was all I wanted to tell you for now thank you very much I hear you no I mean what no you see when I do this thing I do the reduction modulo p inside this finite field but there's nothing to do the real parts the points over the finite field do not lie in this other one because you reduce modulo p so it's not you cannot see it in this way ok I said it in the beginning so the conjecture was maybe made in the 40s and it was proven I think 74 or something by the proof was finished by the linear 74 it got the fields medal for that there had been work on that for many years by Gotendieg and others to set the basis but it has been proven for some time somehow it's always called the veil conjectures because it's because it's not the veil theorem because he didn't prove it he proved it for curves and then conjectured it in general no no I mean I was I understand that I was trying to say it in a simpler way so the correct word is good reduction mod p I just said pretended that meant it's just smooth but I somehow yeah okay but I don't know it's maybe a bit complicated to say what good reduction precisely means but it roughly means that it's non-singular certainly it must be non-singular of the correct dimension and maybe it should be irreducible and whatever but anyway so it's I require that the model but you know if you have a natural so so this is just for simplicity so the whole thing works also when it's not defined over q over the rationales it's a bit more complicated so if your variety is not defined over q but just like that over c you have some polynomials then you can look at the field which is generated by the coefficients of the polynomial so this is a finitely generated extension of the rationales and then you can you can look there you can find some residue field which is a finite field of characteristic p and then you work with that so to do it just that I can work with fp like this and just reduce mod p for this I need to that is defined over q but I don't actually practically need it but obviously when you have a naturally defined generalized space most of the time it will be defined over a rather small field because it's naturally defined for instance like if you take the Hilbert scheme of points it will be defined more or less over the same field over which the variety was defined and so then obviously you cannot make such assumptions and you have to do this thing you take the field generated by the coefficients and then you reduce you find a finite field which is a residue field of that well if you do it in a suitable sense quasi-projective is enough maybe one has to know what one means by the Betty numbers then I think you might be getting the Borel Moore homology or something like that but it certainly makes also sense then in fact well maybe I will not say that