 This algebraic geometry video will be about herwitz curves. These are the most symmetric curves of given genus called the genus G. So G equals 0, the only curve is the projective line, and the group is infinite. So there's nothing very interesting to say. If G equals 1, this is an elliptic curve. And again, the group of automorphisms is infinite. The curve is actually as a group structure, so you can just take translations by the group. So what we want to do is to talk about the case G greater than 1. And here, herwitz proved that the order of the automorphism group is at most 84 times G minus 1. This is if the characteristic is equal to 0. As we will see a bit later, if the characteristic is P, you can actually exceed this bound. So we're going to sketch a proof of this. And the idea of the proof is to look at the orbifold Euler characteristic. So you remember, if you've got a surface, then the Euler characteristic usually denoted by chi is equal to N2 minus N1 plus N0, where Ni is the number of cells of dimension i in some suitable cellular decomposition. So for example, the sphere has Euler characteristic 2 and the torus has Euler characteristic 0. And in general, the Euler characteristic for a compact orientable surface is equal to 2 minus 2G. So this is compact orientable. Well, orbifolds are given by a quotient of a surface by some sort of finite group. At least they are in this particular lecture. More generally orbifolds can be more complicated. And the idea is you have something called an orbifold Euler characteristic. The point is, if you've got a point fixed by the subgroup H, then the quotient of this point by the group H should be thought of as only one over H of a point. So the points divided by the subgroup H counts as one over H of a point. So this is a funny way of counting points. You sometimes count a point as being only a, it's really only a fraction of a point. So a simple example of this, suppose we've got a disk and we take an automorphism order 2 that just rotates by 180 degrees. And if we're quotient out by this, you know, imagine you get a piece of paper and cut it in half and then glue these two ends together. You can see you really get a sort of cone where this point here corresponds to this point here. And this point counts as only half a point when you're trying to work at the Euler characteristic. The key points of this is that if you've got a surface S, then the Euler characteristic of S modulo group G, where G is going to be a finite group acting nicely on it, is equal to the Euler characteristic of S divided by the order of G. It's not too difficult to check this. Now, if you've got a surface quotient out by group G, you can consider it as an orbifold. You can also just consider it as a surface again, provided the group action is reasonably nice. So this will have two Euler characteristics. It's got a sort of topological Euler characteristic. If you consider it as being a surface and it's got orbifold Euler characteristic. So the difference is the topological Euler characteristic. You count this as one point and the orbifold Euler characteristic, you only count it as half a point. So suppose the orbifold S over G, suppose you think of this as being a surface T with some conical singular points, where you've quotient out by a group of orders P1, P2, P3 and so on. So what you can do is you can think of this quotient as being something like an ordinary surface. It might be something like a torus, but then on the torus there are going to be some finite number of points that sort of look a bit like cones like that. So you might have some strange conical singularity sticking out of your torus. And then the orbifold Euler characteristic is equal to the topological Euler characteristic minus one minus one over P1 minus one minus one over P2 and so on because to get from the orbifold Euler characteristic, the topological one, we're removing honest topological points of Euler characteristic one and replacing them by these funny conical points of Euler characteristic one over P. So what we want to show is the maximum value of this is minus one over 42. So where does this funny number one over 42 comes from? Well, the topological Euler characteristic is going to be two minus two H for some H, where H is the genus of the topological surface. And each of these numbers one minus one over P will be a half, two thirds, three quarters, four fifths and so on depending on what P is. So this has to be a negative number and we want it to be as close to zero as possible. So what are the possibilities? Well, first of all, H equals zero because if H is greater than zero, then we've got a factor of two minus two H minus one minus one over P one and so on. And this is going to be either zero, which isn't allowed, or it's going to be less than or equal to minus a half because this is minus a half. So we know the orbifold must be a sphere with Euler characteristic zero with some of these funny conical points. Suppose there are most two conical points, P i, so we've got P one and P two. Then the orbifold Euler characteristic is going to be two minus one minus one over P one minus one minus one over P two and this will be greater than zero. So this case isn't allowed. Suppose there are at least four conical points, then the Euler characteristic is going to be two minus one minus one over P one up to minus one minus one over P four and so on. And each of these numbers is a half or two thirds or three quarters and so on. So you can see that this since it's negative must be at most, must be less than or equal to minus a sixth. So minus a sixth comes when you take these numbers a half, a half, a half and two thirds. If they're all a half and this would be zero, which isn't possible. And minus a sixth is less than minus one over 42. So we don't allow this case either. So we see that there are exactly three conical points P one, P two and P three. So now we examine the possibilities of what P one, P two and P three can be. Suppose there are two conical points. Sorry, suppose there are no conical points of order two. Then what are the possibilities? We could have three, three, three. Well, that's no good because that gives Euler characteristic two minus two thirds minus two thirds minus two thirds which is equal to zero. The next case would be three, three, four which gives us two minus two thirds minus two thirds minus three quarters, which is equal to minus a 12th. And you can see that this is the if we take three, three, anything else so any numbers other than three then the number is going to be even smaller than minus a 12th. The 12th is less than minus one over 42. So that case is okay. So we can't have zero conical points of order two. So there's at least one point of order two. So our points of orders two, P two and P three. If there are two points of order two then what do we get? So it's going to be two minus a half minus a half minus one minus one over P three and this is greater than zero. So that's not possible. So there is exactly one point of order two. So our numbers are two, P two, P three where these are greater than or equal to three and that's greater than or equal to three. If neither is order three. So let's look at the case where P two, P three are greater than or equal to four. Well then we've got one possibility is the points of orders two, four, four. Well this gives an order characteristic of two minus a half minus three quarters minus three quarters equals naught which isn't possible. The next case is two, four, five which gives an order characteristic of two minus a half minus three quarters minus four fifths which is equal to minus one over 20 and this is less than minus one over 42 and two, anything bigger than four and five will be even worse than this. So we can't have both points of an order greater than or equal to four. So we can assume that P one is of order two P two is of order three and the P three is at least three. Now we look at the possibilities for P three. So suppose P three is equal to three, four, five, six, seven and so on and what is the order characteristic two minus one minus one over P one minus one minus one over P two minus one minus one over P three. So you remember P one is equal to a half and P two, sorry P one is equal to two and P two is three. Well the possibilities for P three equals three, four, five, six, seven you can work these out there one sixth, one 12th, one 30th, zero minus one over 42, minus one over 24 and so on so that comes with eight and these cases are not allowed because the order characteristic has to be negative so we see that the largest possible value is minus one over 42. So in other words the order characteristic if the orbit fold order characteristic is less than naught it's less than or equal to minus one over 42 and we've also seen the next best case is minus one over 24 so if the order characteristic is not equal to minus one over 42 the order characteristic is less than or equal to minus one over 24 so we'll use that a little bit later. So the order characteristic which is two minus two G over the order of G is less than or equal to minus one over 42 if it's less than zero. So there are lots of cases when it can be zero or positive which correspond to groups acting on surfaces of genus one or zero but if it acts on the surface of genus at least two then we get this bound here and this just says the order of G is less than or equal to 84 times G minus one which is the how it's bound. Moreover if this bound is satisfied then G must be a quotient of the orbit fold fundamental group so this implies G is generated by elements alpha, beta, gamma with alpha squared equals one, beta cubed equals one gamma to seven equals one, alpha, beta, gamma equals one. So a finite group generated by three elements with these properties is called the how it's group. So it's a, you can also show that if you've got a finite group generated by three elements with these properties then it is in fact comes from a how it's surface. So this gives you the problem of finding all finite groups which are generated by elements of order two and three whose product has order seven. So in the next video we will give some examples of how it's curves and how it's groups.