 This algebraic geometry video is a continuation of the previous one, and we'll give some examples to illustrate. So recall, we proved the Hurwitz bound that if you've got a complex curve, projective and non-singular, of genus G, then the order of its automorphism group is at most 84 times G minus 1. And this is for genus G greater than 1. For genus G equals 0 or 1, we can still derive interesting information from the argument last lecture. For example, let's look at genus G equals 0. Then this is the case when the orbifold Euler characteristic we were discussing last lecture is positive. And if you look at the possible values for that, there aren't all that many. And those actually give you the orders of a finite group of automorphisms of a genus nought surface, which is more or less a sphere. For example, the smallest positive value turns out to be 2 minus a half minus two-thirds minus four-fifths, corresponding to three points, three orbifold points of orders 2, 3 and 5. And if you work this out, it's equal to 1 over 30. And since this number is equal to 2 over the automorphism group, we see that the maximum value of G for a finite group of automorphisms acting on the projective line is 60. And in fact, there is a group of orders 60 acting on the projective line. Of course, this is only the largest finite value. The full group of automorphisms of the projective line is in fact infinite. So there are plenty of infinite examples. Sorry, that's not the maximum finite value. That's one finite value there. There are lots of other finite values coming from cyclic and dihedral groups. Similarly, if G is equal to 1, the orbifold Euler characteristic should be 0. And we notice there are several ways of getting this. The numbers Pi could be, well, we could take the empty set, or we could take 2, 2, 2, 2, or we could take 3, 3, 3, or we could take 2, 4, 4, or we can take 2, 3, 6. And these actually correspond to various finite subgroups, finite cyclic groups acting on the elliptic curve and fixing a point. They correspond to points of orders 1, 2, 3, 4, and 6. And there are indeed elliptic curves with cyclic groups acting on them of this order. So that's done the case of genus 0 and 1. Now let's look at genus G equals 2, and we want to find the most symmetric curve of this genus. So first of all, we can ask, can we achieve herwitz's upper bound? So can we find a genus 2 curve with automorphism group G of order 84 G minus 1, which in this case is just 84? And the answer is no. And this is not very difficult to show. So suppose G has order 84. We can apply the Silov Theorums. So by the Silov Theorums, G has a normal subgroup of order 7. So regularly get complaints in group theory course that nobody knows of any applications of Silov Theorums outside finite group theory. Well, here's one, you can use them to eliminate herwitz surfaces of genus 2. So the quotient achieved by this group of order 7 has order 12 and has a normal subgroup of order 3 or 4, which again is easy to do by applying the Silov Theorums to it. So in the first case, sorry, second case, if it is order 4, then every element of order 2 or 7 is in the normal subgroup of order 28, which is no elements of order 3. So there are no elements of order 2 and 7 whose product is order 3. Remember, last time we said that any herwitz group actually has to have 3 elements a, b and c with a squared equals b cubed equals c to the 7 equals a, b, c equals 1. So in this case, that's not possible. This case is kind of similar. Every element of order 3 or 7 is contained in the normal subgroup of order 21. So in either case, a group of order 84 cannot be generated by 3 elements with these properties. So there are no herwitz groups of order 84. Well, you remember we said that if the herwitz bound wasn't satisfied, so either g is at most 84 g minus 1, and we pointed out that if g is not equal to 84 g minus 1, then g has order at most 48 times g minus 1. So the maximum possible order of a group of automorphisms of a genus 2 curve over the complex numbers is 48. And in fact, there is a curve of this order. We can take the double branched cover of the group p1 branched over the following 6.0 infinity 1 minus 1i minus i. So you can think of these as being something like the vertices of an octahedron if you think of p1 as being symmetric. So the point is the subgroup of psl2c, fixing this set of 6 points, is generated by the following automorphisms. We can map z to i times z, and we can map z to z plus 1 over 1 minus c. And you can easily check these generator group of order 24 acting on these points. In fact, it's isomorphic to the group s4 and 4 points. The group, the full group of automorphisms is a central extension of this because where the centric exchanges the two branches. And so altogether we get a group of order 48 acting on this hyper elliptic curve. So that's the most symmetric possible curve of genus 2. What about genus 3? In this case, there is a Hurwitz curve. So this is the Hurwitz curve of smallest possible genus. This is given by the climb quartic. And the climb quartic is x cubed y plus y cubed z plus c cubed x equals 0. So this is a curve contained in the two-dimensional project of plane over complex numbers. It's easy to check. This is nonsingular. So the genus and the genus of a degree 4 nonsingular curve is always 3. And in fact, it's the only optoisomorphism. It's the only genus 3 curve whose automorphism group achieves this bound, which I forgot to write down. So the order of the group is going to be 84 g minus 1 equals 168. Its automorphism group is in fact PSL2 of F7, which is a simple group of order 168. It's the next smallest simple group after the alternating group A5 of order 60. It's not terribly easy to see that this is the full group of automorphisms of that. There are some obvious automorphisms. We can map x to y to z to x. So this obviously gives you an automorphism of order 3. We can also map x to zeta to the 4x, y to zeta squared y, and z to zeta z, where zeta to the 7 is the seventh root of 1. And this obviously is order 7. And there seems to be no particularly easy way to find a third automorphism. So these elements of order 3 and 7 generate a subgroup of order 21, and finding a further automorphism is kind of rather tricky. So as I'm feeling lazy, I'm not going to do it. Finally, I've mentioned once or twice that we are working characteristic 0. In characteristic P, this proof doesn't work because we can't define Euler characteristics. Well, we can define Euler characteristics, but that involves etalcohomology, and they don't behave in the quite the same way as they do in characteristic 0. So in finite characteristic, the Hurwitz upper bound on the automorphism group definitely fails.