 This talk will be about plane curves. So what we want to do is to talk about curves in the project of plane P2. Before doing this, I'll first show that all curves can be more or less made isomorphic to plane curves. So suppose you've got a curve in n-dimensional projective space you can project to P to the n minus one from some point P. And we will usually get a curve inside P to the n minus one, except this curve might not be the same as C. For instance, if the curve looks like that and you project from a point P that lies on a line passing through two points of the curve they will both end up as the same point in P to the n minus one. So we want P not in any secant of C. That's a line passing through two points. Similarly, we want P not in any tangent line to the curve C because if we've got a tangent line and we project to it then this point of the curve will often become a singular point in P to the n minus one. Well, if we sort of count up we find that the secant lines sweep out a three-dimensional space. It's two dimensions because we can choose these two points and then we can choose any point on the line through them giving a third point. Similarly, tangent lines form a two-dimensional space. So if n is greater than or equal to four we can find a P not in any of these spaces here. So we can project C isomorphically into P to the n minus one. And obviously we can keep going like this until n becomes three. So any curve C is isomorphic to a curve in three-dimensional projective space. And the question is can we get into two-dimensional space? And the answer is in general you can't. Most curves are not isomorphic to curves in two-dimensional space. However, we can if we're willing to allow mild singularities. So the secant lines sweep out of three-dimensional space so we can't avoid those. We can avoid tangent lines and we can also avoid lines where P is in a line with three points of C. So these sweep out a two-dimensional space. And so what we find is that any curve C is isomorphic to a curve in P two with only normal crossing singularities. These are double points that look like two lines crossing each other. By the way, saying it's isomorphic to this curve in P two isn't quite accurate. I should say it's sort of birational to a curve in P two because to get to see you have to resolve the singularities of this curve which means you sort of pull them apart like that somehow. So we can get any non-singular curve by taking a curve in P two possibly with double points and then just resolving the singularities of these double points. So the next question is, how do we find the genus of these curves? So we want to find the genus of a curve C in P two which we're first going to take it to be non-singular and then we will allow it to have a few double points. And for this, we use the Riemann-Harewitz theorem. Riemann-Harewitz theorem, I can't spell. What this does is suppose you've got a map from C one to C two where these are say compact Riemann surfaces and we want to relate the genus of C one to the genus of C two. Well, it's a bit easier to relate the Euler characteristics. So we find the Euler characteristic of C one is N times the Euler characteristic of C two where this is an unrammified N fold cover. So that means every point in C two has exactly N points in this inverse image. And this is pretty obvious because chi is equal to the number of zero cells of a triangulation minus the number of one cells plus the number of two cells. And if you triangulate C two, then you can lift that up to a triangulation of C one, which will obviously have N times as many of everything. And from this, you can see how the genus is related because you recall the Euler characteristic is equal to two minus two times the genus. Well, in general, a map from C one to C two won't be unrammified. It will have ramification points. So ramification points are where C two does something like this. So C two might look like that. And you see it might look like something like X equals Y squared, where there should be two points mapping to a point of C one, but suddenly there's only one point. Similarly, you could have a three-fold ramification point of C one looks like this. Then C two might be a sort of cubic X equals Y cubed. And there's three branches because we're thinking of these as being complex curves. And again, you see this would be a three-fold ramification point, so we say E equals three. This is two-fold, so we put E equals two, where E is the ramification index. And what you see is that every time we have a ramification point, there are fewer points in C two than there ought to be. So the number of points of C two is reduced by E minus one at each ramification point E. So we find the Euler characteristic of C one is equal to N times the Euler characteristic of C two, where this is the most points of N inverse images, minus sum of the EI minus one, where you sum over all the ramification points. So this is the Riemann-Harewitz formula for working out the Euler characteristic of a curve C one. For example, if we've got a hyper elliptic curve, this is a double cover of P one, so N is equal to two, and there are two K points. And at each of these points, EI is equal to two. So the Euler characteristic of the hyper elliptic curve is two times two minus two times K, where this is the Euler characteristic of P one. So the genus is equal to K minus one, which is what we found earlier. Well, we can now use this to work out the genus of a non-singular plane curve. So we take C one equals non-singular plane curve. And let's take it to be a degree D. And then what we do is we just map C one to C two, which we're going to take to be the projective line. So we're just mapping projective plane to the projective line. So this might map X, Y to X, if we ignore the points at infinity. And this is a default cover. So we take, so the Euler characteristic of C one is equal to D times the Euler characteristic of C two minus sum of EI minus one, where these are the ramification points. And the Euler characteristic of C two is easy, that's the projective line, so that's just two. And now we need to look at the ramification points of C one. Well, we're going to take C one to be in sort of general position, whatever that means. And if you kind of look at C one, we're mapping it onto the projective line. And you can see it is going to have a ramification point whenever the tangent line becomes vertical. So here we will have E equals two at all these points here. So the ramification points, we're going to have EI equals two as long as it's in general position. We might occasionally get points where EI equals three, but then we can just projection in a slightly different direction and find EI equals two and so on. And the number of these, well, this occurs whenever the tangent line is vertical. So if the curve is given by fxy equals naught in the xy plane, what we want is partial derivative of f with respect to y is equal to zero. So we want to solve these two equations. This has degree D, this has degree D minus one rather obviously. So the number of intersection points is D times D minus one by Bazou's theorem. So we find the Euler characteristic of C one is D times two minus D times D minus one, which is equal to two. Two minus two G. So from this, you find G is equal to D minus one, D minus two over two. So this gives us the genus of a degree D plane curve. So we've already done the cases where D is zero, one, two or three, for instance, if D is three, the genus is one, which is an elliptic curve. And if D is four, the genus is three, which is what we looked at before. And by the way, I should put a bit of warning here. We looked at the cases where D was at most four and we found in each case that this accounted for all curves of genus G, which is zero, one, three or six. However, curves of, none single plane curves of degree five and above do not account for most curves of the corresponding genus. So for genus five, this would correspond to plane curves of degree six. And most curves of genus six are not plane curves of degree five. Anyway, that's done the case when C is non-singular. And now look at the case when C has degree D, but has M double points. So as we saw, only a few curves are isomorphic and non-singular points of degree D. So the curve might have some double points looking like this. And we now see this account, each double point accounts for two points with delta F over delta Y equals zero because there's sort of one point on this branch and one point on this branch. And it gives no ramification points because if you pull these two branches apart, you can see they're both un-ramified. So the order characteristic increases by two because we've got two more points than we might expect. So there's increases by two for each double point. So the genus G decreases by one for each double point. So we find the genus of C is equal to D minus one, D minus two over two. And then we've got to subtract off the number of double points. So this gives the formula for the genus of a plane curve with N ordinary double points. As you can imagine, you can get more complicated versions of this formula when you have more complicated sorts of singularities, but we're not going to worry about that because double points are enough to account for all curves. So that's worked out the genus of plane curves with most double points of singularities. Now let's look at the canonical divisors on these curves. So let's first of all take C to be non-singular of degree D. And what we want to do is to find the canonical divisors and find out what their zeros are and so on. And we saw earlier that if you take a form DX on C, this has the following zeros and poles. It has poles of order two at the D points at infinity of C and it has zeros of order one at zeros of C. Delta F over delta Y. We're assuming our curve C's in general position so all the points at infinity are single points and so on. On the other hand, the derivative delta F over delta Y has a pole of order D minus one at the D points at infinity because this is a polynomial of degree D minus one and it has zeros of order one at zeros of itself, of course, completely trivial. And now you see the zeros of this cancel out with the zeros of this at the finite plane. So DX over FY, let's call this FY for short, has zeros of order one at the D points at infinity of order zero order one. And so we've got a pole of order two and we've got to divide that by the pole of order D minus one. So we end up with zeros of order D minus three at the D points at infinity. So the total number of zeros is D minus three times D so this represents the canonical divisor K so we see the degree of the canonical divisor is equal to D times D minus three and which is in fact, two G minus two. And well, that gives us one example of a canonical divisor we can find all the other canonical divisors if we take a polynomial PXY DX over FY. So this has zeros of order D minus three at the points at infinity and this is a pole of order degree D, sorry, degree of P at most points at infinity. So we can see that this is holomorphic at all points if the degree of P is at most D minus three and now the dimension of the space of polynomials of degree D minus three is D minus one, D minus two over two. So this gives us the dimension of the space of holomorphic one forms and you can see that this is just equal to the genus that we calculated earlier. So this shows that for non-singular plane curves the topological genus given by the number of handles is equal to the geometric genus which is the dimension of the space of one forms. We can also describe exactly what the canonical divisors are. So the canonical divisor, so a typical canonical divisor is just the zeros of a holomorphic one form PXY DX over FY which is usually just the zeros of P if we don't worry too much about points at infinity where we recall this is a degree D minus three polynomial. So if we put the points at infinity back in we find that if C is the zeros of some polynomial FXYZ which is homogenous of degree D then a typical canonical divisor effective canonical divisor will just be the intersection of C with a curve A where A is the zeros of a polynomial PXYZ which is homogenous of degree D minus three. So this gives a very simple explicit description of the effective canonical divisors on degree D curves. For example, when D is equal to four we find D minus three equals one so A is just a line. So the canonical divisors on a degree four non-singular plane curve are just given by the intersection of the curve with the points on a line. So we have a completely explicit description of the canonical divisors from non-singular curves. We can do the same for curves with double points. So now suppose the curve has singularities which are all double points like this. Then we see that DX over FY has a pole of order one at these points. This is quite easy to see for instance if this looks locally like X squared equals Y squared we find that DX over FY is going to look locally like DX over two Y. And you can see that DX over two Y on both of these has a simple pole. So PXY DX over FY is holomorphic if P vanishes at all double points. So this describes the canonical divisors of the curve. They just correspond to the common zeroes of our curve and a polynomial PXY where PXY vanishes at all double points. You have to be a little bit careful about the order of the zeroes at double points. What you find is that P will have a zero of some order along this curve and a zero of a possibly different order along this curve and this differential will have zeroes of order one less than that on the two components of these curve. Anyway, this county informally suggests the dimension of the space of these one forms should be D minus one, D minus two over two. And then we should subtract N because we seem to have N different conditions about this differential vanishing at the N double points. However, you have to be a bit careful because it's not entirely clear that these conditions are independent. It's conceivable that if we say force it to vanish at four of these points then that would force it to vanish at another one. So strictly speaking, we've only proved the dimension of the space of one forms is at least equal to this number. In fact, these numbers aren't equal but proving this takes a little bit more effort. In fact, I'm not going to prove it now because it'll be much easier to prove once you've proved most of the rest of the Riemann-Roch theorem.