 This lecture will be about the etalcohomology of a curve. And I'm going to take a rather unusual approach to this. So I'm not actually going to tell you what etalcohomology is before calculating the etalcohomology of a curve. So the problem is that most introductions to etalcohomology, you sort of spend large amounts of time doing abstract category theory and talking about growth and diptypologies and so on. And this is all very important, but it's also kind of a bit mind-numbing. So I'm going to skip over this and just go straight to the interesting part of the calculation. The disadvantage is that you won't actually know what etalcohomology means. So I guess this is just a sort of overview as motivation for finding the precise definition. So let me give a very quick summary of what an etal sheaf is. So an ordinary sheaf in the Zariski topology is a contrarian functor from open sets of X to sets. So here U is open and X is whatever space we are working over. And furthermore, there is some sort of covering condition that I'm not going to worry about because this is a very informal introduction. The difference between that and an etal sheaf is that an etal sheaf is a map from the etal open sets of X where this map is now an etal map and need not be injective. So obvious question is what is an etal map? And I'm not going to tell you. Probably the next lecture I'll give a sort of introduction to roughly what etal maps are. But there is sort of generalization of inclusion of open sets. So an etal sheaf is some sort of contrarian functor from these mysterious etal maps to sets. And instead of defining an etal sheaf, I'll just give you some examples. So these are the examples we're going to use. First of all, we have the constant sheaf. So here, if I'm given an abelian group A and I'm given an etal open set of X, this is just going to take the etal open set to set A, at least if U is connected. If U isn't connected, you have to fiddle around a bit more. Secondly, we have the sheaf of regular functions. So this just takes an etal open set of X to the space of regular functions on U. So this is very similar to the sheaf of regular functions in the Zariski topology, where we just map any open set U to the regular functions on U. One that we're going to use is the sheaf of invertible functions, which is, I'm used to denoting this by GM, which stands for the multiplicative group. And this takes an etal map to the units of the regular functions on it. Fourthly, we could have the sheaf mu N of Nth roots of unity. And this just takes the map etal set to the Nth roots of one in O of U. So what we want to do is to calculate the etal cohomology groups of a curve and try and show that they're very similar to the usual singular cohomology groups. So let's recall the singular cohomology groups of a curve. So this is with integer coefficients. And we're going to work over the complex numbers. Otherwise, singular cohomology is kind of stupid. And these groups are as follows. They're Z for Q equals nought, Z to the 2G for Q equals 1, and Z for Q equals 2, and nought otherwise. Here, G is just the genus. And X is a non-singular projective curve over the complex numbers. Now, the etal cohomology groups denoted like this of a curve. Here, we're going to take coefficients mod n. And we want to show that these are Z over nZ for Q equals nought, Z over nZ for Q equals 1, so that should be to the 2G, and Z over nZ for Q equals 2. Here, n is invertible in whatever field K we're working over. And this is very important. In particular, you can't take n equals 0. You can, but you don't get this answer. And you can't take n to be divisible by the field. Otherwise, again, you get something very weird. So here, X is going to be non-singular projective curve over some field K, which will take to be algebraically closed. And the first step in calculating these groups is to find HQ of X with coefficients in GM star. So you remember this was invertible, the chief of invertible functions on X. It turns out that although this looks more complicated than the constant chief C over n, in fact, these groups are easier to calculate. And by reducing them, we'll see they get reduced to calculating a Picard variety. And what we do is we write down an exact sequence of sheaves. And the exact sequence looks like this. One goes to GM, goes to, let's call this invertible rational functions sum over X of Z. Let's explain what these are. So this stands for invertible regular functions. So in other words, if you've got any et al map, you go to X, its value on this is going to be the invertible regular functions on U. This stands for the invertible rational functions. And this stands for a copy of Z for each closed point, X of the curve X. And let's explain what's going on here. So this map here just takes a rational function for the orders of the zero or pole at each point. And what this exact sequence is saying is just saying, if you've got a rational function in here and the order at every zero and pole is zero, in other words, if it's got no zeros and poles, then it's an invertible regular function. So that's a kind of obvious statement. And this is just a sort of complicated sheath theoretic way of saying this. Now, for any short exact sequence of sheaves like this, we get a long exact sequence of cohomology groups that I will write out in a moment. And what we want to do is to calculate the cohomology groups of this from the long exact sequence knowing the cohomology groups of this and this. So we need to know what are the cohomology groups of these two objects. Well, first of all, let's look at the cohomology groups of this. We'll hear the cohomology groups. So hq of sum over z is equal to zero for q greater than naught. The reason for this is that this is just a sum, a direct sum of sheaves with support, a point, and furthermore, this point is the spectrum of an algebraically closed field. And in etel cohomology, the spectrum of an algebraically closed field, all the higher etel cohomology groups vanish. I think that should be a direct sum sign rather than a summation symbol, but never mind. So these cohomology groups are easy to figure out. Now we want to know the cohomology groups of these, and these again all vanish for q greater than naught, but this takes considerably more effort to see. So I'll sketch why the cohomology groups of this all vanish. So what we want to know is we want to work out the cohomology of x with respect to the sheaf of non-zero rational functions. And we can reduce this to working out the cohomology of the generic point of x with coefficients in the rational functions. So it's really spectrum of k where k is the field of rational functions on x. And etel cohomology of the spectrum of a field turns out to be the same as Galois cohomology. So Galois cohomology means you take the cohomology groups with respect to the Galois group of the algebraic closure of k with respect to whatever you're taking cohomology of. It turns out that etel sheaves over the spectrum of k turn out to be more or less the same as modules of the Galois group. So working out this turns out to be a problem in Galois cohomology. And we do this in several steps. So I guess step naught is to say that etel cohomology of the spectrum of field turns out to be Galois cohomology. Incidentally, we used a special case of this last time because if the k's algebraically closed, then all these Galois cohomology groups vanish for i greater than zero. So in order to work out these groups, we first used Sen's theorem, which says that suppose k is field of rational functions on a curve over k where k is algebraically closed. Then k is C1. Well, what on earth does C1 mean? Well, C1 is this funny technical term, meaning any homogenous polynomial in n variables of degree less than n has a non-zero solution or non-zero root. So for finite fields, this is the Chevalet warning theorem and Sen proved it for fields of transcendence degree one over an algebraically closed field. The next step is that a C1 if a field is C1, then it implies the Brouwer group is trivial. So what's the Brouwer group? Well, the Brouwer group is a group formed out of the central division algebras over the field. So we need the theorem of the Brouwer group, which is another of these things I'm going to skip. So in Galois cohomology, the Brouwer group turns out to be isomorphic to the Galois cohomology of k bar over k with coefficients in the non-zero elements of k. So the Brouwer group is the second cohomology. If you're wondering why we don't start with the first cohomology, the reason is very simple. Hilbert's theorem 90 says that the first cohomology is always trivial. So Sen's theorem says that the second cohomology is sometimes trivial. Thirdly, there's a theorem from Galois cohomology which says that if the Brouwer group equals one for all finite extensions, this implies that hq of k bar over k with coefficients in k bar star is equal to zero for all q greater than or equal to one. And using the correspondence between Galois cohomology and etal cohomology, this implies the etal cohomology of hq of x with coefficients in the non-zero rational functions is equal to zero for q greater than or equal to one, which is what we wanted to do. Okay, so we've proved some vanishing themes from cohomology. Now we get on to the long exact sequence of cohomology. So let's write out what the long exact sequence corresponds to the short exact sequence of sheaves. So it starts off with the zeroth cohomology of the groups in the short exact sequence. So this is going to be a sum over copies of z. And then this map here isn't necessarily onto and instead it maps onto the first cohomology. And then we get h1 of the rational functions and then we get h1 of whatever this thing was and then that maps onto the second cohomology of gm and so on. And now we use the vanishing theorems we've proved. So all this junk vanishes using to sends them and brow groups and all that other stuff. And all this vanishes for rather trivial reasons. And so we see as a consequence that all these groups vanish. So it just leaves h0 and h1 to work out and that's very easy to work out. So h0 is just the rational functions with no zeros or poles. So this is just isomorphic to k star. And here we notice that h0 of this just means we assign an integer to every point. So these are just the divisors on the curve. That's linear combinations of points. And these are just rational functions. So the image of this is just the principal divisors. So this group here is just the principal divisors. Sorry, it's just all divisors. Modular, the principal divisors which is the divisor class group. And now we have to work out the divisor class group. And first of all, every divisor has a degree which is an integer, so that gives you a factor of z. And that's the product of z and the divisors of degree zero which are called the Picard group or Picard variety or whatever. And that has a fairly complicated structure that we'll talk about a little bit later. So the cohomology of the multiplicative group of the field can be worked out explicitly in terms of the Picard group which is the group of line bundles or the divisor class group or whatever you want to call it. Now we use that knowledge of the cohomology of the multiplicative group to work out the cohomology of z modulo nz coefficients. So first of all, we notice that the chief C over nz is isomorphic in a non-chronical way to the group of roots of unity provided n is invertible in our field K. This is the key point where n has to be invertible. And now we have the following comma exact sequence. One goes to nth roots of unity, goes to gm, goes to gm. So this is raising things to the power of n, goes to one. So this is a comma sequence. And there's one very important point here which is this map surjective. And this is surjective in the etal topology but not the Zariski topology. So in some sense, the surjectivity of this map is the main reason why we're using the etal topology instead of the Zariski topology. Much of what we've done so far works pretty much the same way in the Zariski topology. For instance, we could calculate the cohomology of x of coefficients in gm and we get the same answer. It just gives us something involving the Picard group. In fact, calculating this is much easier in the Zariski topology because you remember calculating h1 of this group here was really quite difficult in the etal topology. We had to use to send theorem and Brau groups and whatever. In the Zariski topology, this group here is really easy to calculate and it's very easy to see at zero. The problem is that the same very easy argument that shows that zero also shows that the cohomology of this group vanishes in the Zariski topology which is not what we want at all. So the surjectivity of this map is essentially related to the fact that the map taking Z to Z to the n is an etal map from K star to K star. So the surjectivity of this is a sort of key point that you need to pay attention to. Now what we do is we look at the long exact sequence of this. So we know the cohomology of these two groups so the long exact sequence is going to give us information about the cohomology of this which is what we want. So let's take a look. We get nought goes to h1, so h0 of mu n, goes to h0 of gm. I'm missing out x here because I'm getting bored of writing it out. Goes to h0 of gm and this is multiplication by n and then here we have h1 of mu n goes to h1 of gm goes to times n h1 of gm goes to h2 of mu n and now this goes to h2 of gm and everything from this point onwards is just zero. So what can we see from this? Well, first of all, we notice that all these higher groups are equal to zero because they just fit into this exact sequence of zeroes in front of them and behind them. Now we can recall what these groups are. So this group here was just the multiplicative group of k and this group here was the Picard group z times pic zero, z times pic zero. So if we write this out a little bit more neatly in one line so we can see what's going on, we get the following. We get the following, we have nought goes to h0 of mu n goes to k star goes to n k star goes to h1 of mu n goes to Picard group goes to n, I guess that should be z times pic zero goes to h2 of mu n goes to zero. And now let's think about this a bit. This map here is surjective. So this map here is the zero map and this gives us the rather uninteresting information that this group here is just a group of nth roots of unity which we already knew. Now since this map is zero, this group here is the kernel of the map from z times pic zero to itself. So it's the kernel of the of the map saying x to n times x. Now the kernel of the map taking z to n z is zero so we can forget about that and we need to know the group of elements of order n in the in the Picard variety and this takes a fair amount of calculation. This is in fact an abelian variety of dimension equal to g as shown by v in his work on the v conjectures and the elements of order n provided n is invertible is isomorphic to z over nz to the 2g. This takes a fair amount of work as I'm just going to quote. Now finally this group here well the map from pic zero to pic zero is onto surjective again you need to think about abelian varieties in order to prove this so we're left with the map from z to z which is multiplication by n so all we left with here is the group z modulo nz so finally we see that h zero of mu n is equal to z isomorphic to mu n h one of mu n is isomorphic to z modulo nz 2g and h2 of mu n is isomorphic to z over nz actually that probably shouldn't be a z over nz but never mind if yeah we since we're just assuming mu n is isomorphic to z modulo nz this doesn't really quite matter how we write this and I guess it shows that hq of mu n is equal to naught for q greater than 2 so this is the result we want remembering that mu n is isomorphic to z over nz so that gives you the etalc homology of a curve this allows you to get hold of etalc homology of high dimensional varieties because what you can do is you can map an n dimensional variety and the fibers will be curves and you can get hold of the co homology of the fibers by using this result they might be singular so you need to do a bit of extra work and then you can apply spectral sequences to relate the co homology of your variety to co homology of the base of the fibers and so using this basic result about curves you can sort of climb up to etalc homology of higher dimensional varieties okay well one thing that was really rather seriously missing in this lecture was I never actually explained what an etal map or an etal sheaf was so next lecture I plan to say a little bit about what an etal morphism of varieties or schemes looks like