 This talk will be about the Riemann-Roch theorem in a special case when the genus of a curve is equal to 3. So I'll very quickly review what it says. So in this case the Riemann-Roch theorem says that L of D is the degree of D plus 1 minus G which gives us minus 2 plus L of K minus D. Here as usual D is a divisor. This is the dimension of the space of functions with poles at most on D. This is the degree of the divisor and K is the canonical divisor. As usual we can find that L of K is equal to 3 and the degree of K is equal to 2G minus 2 which is 4. And now I'm going to figure out how L of D varies as a function of the degree. So let's suppose the degree of D is one of these numbers. So this is the degree of D and we're going to plot L of D. So if the degree is negative then L of D has to be 0. If the degree is 0 then L of D can be 0 or 1 as usual. If the degree is 1 then it can be 0, 1 and it can't be 2 because if it were 2 that would give us a 1 to 1 map to the projective line and the curve would have Gina 0 not 3. So the possible values if degree of D is 2 it can increase by most 1 so it might be 0, 1 or 2. And here we get to the Riemann line. So Riemann showed as his part of the Riemann-Rock theorem that L of D must be at least this number here. So we also know that the canonical divisor lives here and above this point that the L of D must actually lie on the Riemann line. So the only other possibility is that we might imagine a divisor here. So can we get a divisor with this value of degree and L of D? And the answer is you can't because if we had such a divisor then you can check using the Riemann-Rock theorem that K minus D would be here which isn't allowed so there's no divisor there. In particular we see that if we draw a line like this, this is called the Clifford line, then all points either lie on this Riemann line or they lie on the line where the degree is 0 down here or they lie inside this triangle formed by these three lines. And the Clifford line for general genus is the line passing through the the point corresponding to the divisor K and the divisor 0. So the divisor 0 has coordinate 0 and 1 and the divisor K has coordinates 2G minus 2G in general. So we see that the Clifford line has slope a half and passes through 0, 1. Next we can ask do we really get a divisor with these properties here? And the answer is we sometimes do and we sometimes don't. So we can ask is there a G22 or what does this mean? Well GMN asks if we can find a divisor with degree equal to M and L of D being equal to N or maybe I guess at least equal to N. So this is a sort of short notation for linear systems of various dimensions and degrees. Well if so we get a map from the curve C to the two-dimensional sorry one-dimensional projective line. This is usually 2 to 1 and it's one-dimensional because if we've got a divisor D with L of D equals 2 and degree of D equals 2 then we can use these coordinates these two sections of D as the coordinates for the projective lines. That gives us a map from the curve to P1 which is defined at most points and it's 2 to 1 because the degree of the divisor is 2 which means there are usually two points mapping to each point of P1. Well that means the curve is hyper elliptic because we can then write this as y squared equals x minus alpha 1 to x minus alpha N for some N and we saw earlier that the genus of this is either N minus 2 over 2 or N minus 1 over 2 and since we want genus 3 we see that N should be 7 or 8. So the theory of these curves is very similar to what we did for genus 2 hyper elliptic curves so I'll just be very brief. First of all there are eight special points these are the via strice points with y equals 0 and x equals one of these alpha i's and these points P have the property that L of 2P is equal to 2 rather than 1 as it is for most other points. We can also write down the differentials explicitly we get dx over y, x dx over y and x squared dx over y and we go all the way up to x to the g minus 1 dx over y in general so this is an explicit basis for all the holomorphic one forms on the curve. Well so the case when there is a g22 is rather similar to genus 2 and it's rather more interesting to look at what happens if there is no g22. So how do we how do we study these? Well we look at the canonical divisor and we notice that L of k is equal to 3 so it has a basis of one form say omega 0, omega 1, omega 2 and we can now have a map from c to the projective plane taking z to the values of ratios of omega 0, omega 1 and omega 2 at z. Notice that these are one forms they don't actually have values at a point z of c however their ratios are functions so their ratios are well defined so this is actually a well defined point of p2 at least provided one of these are non-zero which is the first thing we have to check so let's make sure this is well defined. Well to make sure it's well defined we've got to make sure not all the omega i vanish at some point p so how can we do this? Well this holds because L of k minus p is less than L of k and that's because L of k is equal to 3 and L of k minus p is equal to 2 and that means if we add the condition that the one form vanishes at p this reduces the number of one forms which obviously shows that there's at least one one form that doesn't vanish at p. Next thing we want to do is to show this is injective so what we do for this is we want to show that given pq in c we can find some one form omega with omega vanishing on q but not on p so we're trying to show this is injective and to do this we notice that L of k minus p minus q is less than L of k minus q that shows that adding the extra so this is one forms vanishing on q and if we add the extra condition that it vanishes on p this means there are fewer one forms so there must be a one form that's none zero on p but not on q and this follows because this is dimension 2 and this is dimension 1 well why does this have dimension 1 because there is no g2 2 because the only other possibility is that L k minus p minus q would be equal to 2 but this thing has degree equal to 2 so so if this was if this number were equal to 2 we would have a we would have a g2 2 and we're assuming we don't have one. Thirdly we should check that there's a local coordinate for c at each point so the so the map is actually an isomorphism from c to its image and that's quite easy all we have to do is to check that L of k minus 2 p is less than L of k minus p and that follows in exactly the same way that we did here except we just take q equal to p rather than distinct from p so so this means that there's a local coordinate at each point in other words the image of c the map from c to its image is actually an isomorphism and doesn't have and its image doesn't have some sort of funny singularity so what we see is there are two cases either there is a g2 2 and this means the curve is hyper elliptic or there is no g2 2 and this means the curve embeds into p to the g minus 2 so this is called the canonical embedding so we're looking at the case g equals um so that should be g minus 1 we're looking at the case g equals 3 so this would be an embedding into p2 so let's just have a quick look at what happens for small g so if g is not 1 2 3 or 4 well if g is equal to 0 we get a map from c to p minus 1 no doesn't make sense um what actually happens here is the canonical embedding is not defined for g equals 1 we get a map from c to p 0 which is just a point so this is the case in an elliptic curve and the canonical map from elliptic curve just maps the elliptic curve to a point which doesn't really give us much useful information about the curve for genus 2 we did last lecture this maps the curve to p1 and is generically a 2 to 1 map and is the representation of the hyper elliptic curve as a double cover of p1 for genus 3 which is the case we've just done we get a map to p2 and it's either hyper elliptic and its image is a double conic in p2 which is i similar to a line so we've got a map from the curve to a line or it's a quartic in p2 so this is the hyper elliptic case and this case is everything else and for genus 4 we get a map from the curve to p3 and it's either we get a double copy of the twisted cubic or we get a degree 6 curve in p3 which happens to be the intersection of a quadric and the cubic if you care about that sort of thing and similarly for as g gets bigger and bigger this gets more and more and more complicated anyway um we're just going to do the case of genus 3 so let's look at a couple of examples of it so then there are two famous examples the first is the Fermat curve so called because of its relation to Fermat's last theorem another famous example is the Klein quadric and this actually is the unique curve with the largest possible symmetry for genus 3 curves its symmetry group is in fact a simple group of order 168 um in general the automorphism group of a curve of genus greater than one is is finite and this is an upper bound for genus 3 another thing we can do is we can write out the one forms explicitly so we've done this for hyper elliptic curves so now we're going to do it for degree 4 curves um let's take them in the affine plane for the moment um and we know that l of k is equal to 3 so there should be a three-dimensional space of one forms and the degree of k is equal to 4 which means that each of these one forms should have four zeros somewhere on the curve so let's try the most obvious thing let's just try dx and where are its zeros well if we've got a um a quartet curve dx is zero whenever the tangent line is vertical and this happens whenever the derivative of f with respect to y is equal to zero so we want to solve the equation f equals zero f y equals zero and this is degree four and this is degree three so by bazoo's theorem we expect three times four equals 12 solutions well we seem to have a bit of a problem here because one forms according to the reamon rock theorem especially of four zeros and here we found a one form that apparently has 12 zeros well it does indeed have 12 zeros the reason it doesn't contradict the reamon rock theorem is that this also has eight poles or rather it has four double poles at the points at infinity so there are four points at infinity in general because this is a degree four curve and it turns out that dx actually has a has a double pole there so so we seem to have failed rather badly because we actually want to form with four zeros and no poles and we've got a form with 12 zeros and eight poles well this is actually quite easy to fix all we can do is just take dx over f y and the zeros of this will cancel out with the zeros of that so this is no zeros on a two intersection c well we should just have a quick look at what happens at infinite points so if we choose projective coordinates we can map this to say one one over y one over x which would be y over x so let's make that equal to y and this equal to z so we see that dx is then just minus one over z squared dz and you see this has pole of order two at z equals naught so dx has a pole of order two at the infinite points and f y has a pole of order three at infinity because it's a degree three polynomial so dx over f y has four zeros at infinity as it ought to have and no poles or zeros anywhere else so we found a one form that is holomorphic and now we want to find some more well an obvious choice is dy over f of x except that doesn't give us a new form because this is equal to minus dx over f of y as you can easily check from the fact that f of x dx plus f of y dy equals zero so this is just df so how can we find a three-dimensional space of one forms well since this is a zero at infinity you can still multiply it by a linear function and it will still be holomorphic at infinity so we can take a plus bx plus cy times dx over f y and this will be a holomorphic one form so we've now got a three-dimensional space of holomorphic one forms on our curve so we've got the holomorphic one forms explicitly um we can also see what the canonical divisors are so these are just the zeros of one form well the one form is a plus b x plus cy times dx over f y so you can see zeros are the zeros of a line a plus bx plus cy equals zero so this gives us a picture of the canonical divisors on our quarter curve all we do is we take any line intersect it with the curve and these four points will form a canonical divisor um we can also do things like find the odd theta characteristics what we want to do for these is to find the divisor d such that 2d is equivalent to the canonical divisor so how do we do this well that's easy um all we have to do is we want to have the canonical divisor having two double zeros well that's easy all we do is we take a by tangent to our curve um and um in fact if you study theta characteristics you find there are 28 odd theta characteristics so you find that there are in fact 28 by tangents to a degree four quartic and they're just the odd theta characteristics so um we can also look at via strice points p so you remember via strice points are where l of np is unusually large sum n so we better find out how usually how big usually large is so if we've got this picture minus two minus one zero one two three four five let's remember what n of p can be what l of n of p can be well its graph has to go like this and it jumps up to zero here and then it might go all the way up to the Riemann line and then go up like this so this is this is the minimum we would expect but it can also get a bit bigger um for example for a hyper elliptic curve um it can go like this so here we have one of the eight via strice points on a hyper elliptic curve um so um that's what via strice points on a hyper elliptic curve look like um we can now ask what do they look like on a degree four quartic and in this case they sometimes look like this so these actually correspond to inflection points of a quartic um so they're the points with um l three p equals two rather than one which is the usual number that we expect um so let's see why this is true well what we do is we draw a picture of our curve c something like this and suppose we've got an inflection point um so this might be an inflection point and um if this inflection point is called p and here's f equals zero and let's take this line to be g equals zero and now you notice that um one over g cubed has a pole of order three at p which is what we want however uh since this curve has degree four the line intersects c in a fourth point where it also has a pole and we don't really want this however we can get rid of this as follows suppose we just draw a line through here and have this as h equals zero and now we can take h divided by g cubed and this will have a pole of order three at p and three zeros on h intersection c so it has three poles and three zeros and and and has the proper and and shows that l three p is now equal to two because we found an explicit non-constant function there um we can ask how many inflection points um well an inflection point is given by the vanishing of the Hessian of the curve which is this huge three by three matrix delta squared f over delta x squared delta squared f over delta x delta y and so on all the way down to delta squared f over delta z squared um and so you want the Hessian to be zero and if you look at the Hessian f is degree four so the second rivers have have degree two so this is degree six so the number of inflection points ought to be four times six equals 24 um you have to be a bit careful with these sorts of arguments because sometimes you find things like some of the points turn out to be double points so you might get less than you expect but in the case of inflection points on a non-singular cubic we these problems don't turn off and we do indeed get exactly 24 points so there are 24 via strice points um and if we go back to this picture here um we see there are now 24 of these and you notice the sum of the weights of the via strice points is always 24 well what do I mean by the weight of the via strice point well I mean how far does it fail to live on the screen line so these blue points um you only have to sort of move one point upwards by one on the pink line we have to move three point there are three places where where things are one bigger than expected so we say these points of weight three so here we have eight points of weight three which gives us 24 and here we have 24 points of weight one which is 24 um in general the sum of the weights of the via strice points for a curve of genus g is g squared minus one by g which in this case is three squared minus one times three and um so curves of genus two this number would be six which is what we found and for genus zero and one this predicts correctly that there are no via strice points um so um finally we can have a look at the moduli space for genus three curves and we're just going to be very informal and just guess what its dimension is so there are two sorts of genus three curves they can be hyper elliptic and let's try and estimate how many of these there are well they're given by choosing eight points on p one which gives us an eight dimensional space except we have to divide by the automorphism group of p one which is p gl two of c and this is dimension three so the dimension of the space of hyper elliptic curves of genus three is eight minus three equals five now let's look at the quartics well the quartic is given by you have to take the coefficient of x to the four x cube y x squared y squared all the way up to y to the four um x cubed z all the way up to z to the four and so on so we we have um all together five plus four plus three plus two plus one which is 15 coefficients so the dimension of the space of quartics is 14 because we have to subtract one because if you multiply this by constant you get the same same um um quartic however we should then quotient it out by the dimension of the automorphism group of p two because if you just act on p two then we get an isomorphic curve and the automorphism group of p two is p gl three of c which is dimension equal to eight so all together the dimension ought to be 14 minus eight which is equal to six so um what's going on is we seem to have two different components of the moduli space one of dimension six and one of dimension five however that's misleading they're not really different components at all um really the hyper elliptic curves are all limits of quantics so what you should think of the moduli space as is some sort of six dimensional space and inside that six dimensional space is a sort of divisor which is corresponds to this five dimensional space of hyper elliptic curves so that's a very crude picture of the moduli space it's something six dimensional with a five dimensional divisor of hyper elliptic um so hyper elliptic curves okay next lecture we'll be looking at higher genus curves a bit