 So, is this, yes, it's working. Okay. I'm starting at... Okay. Well, so, thank you for allowing me to give this talk here. My talk will be about counting the number of trigonal curves of TNS5 over finite fields. And first, I'll explain why we want to do this, and then I'll say a bit about how we can see these trigonal curves as plane curves. And then I'll describe how we count these plane curves using a modified sieve principle. And these methods are similar to the ones by Jonas Bergstrom, which he used when he did the case where the genus is 3 and non-hyperliptic curves. Okay. So first of all, why do we want to count these trigonal curves? Well, we are interested in the cosmology with rational coefficient of the modular spaces, and in my case of M5 bar. And there is this thing called the Hodge Euler characteristic in mixed hot structures. And I'm not going to say much about it because Ursula will speak about this tomorrow. And it is defined using compactly supported cosmology. And you can sort of think about it as the Euler characteristic, but instead of an integer, it's a polynomial. And instead of taking dimensions of cosmology, you take the dimensions of the mixed hot structure. And in the case of M5 bar, we have that it is smooth and projective. And so this mixed hot structure is actually a pure hot structure. And so we have purity, and that means that we have this equivalence between knowing this Hodge Euler characteristic and knowing the cosmology with the hot structure. And the nice thing about the Hodge Euler characteristic is that it relates to the Hodge Euler characteristic of M5 and the Hodge Euler characteristics in the boundary. But for the Hodge Euler characteristics in the boundary, we need to take as an equivalent Hodge Euler characteristics. So, well, these as an equivalent Hodge Euler characteristics in the boundary, they're all known except for two cases, which of course are the cases with the highest genus. So they are except M41 and M42. So there's still work to be done, but I'm mainly looking at M5 at the moment. And the nice thing is that if the number of points of a finite field, so the number of points of space over a finite field, if this is a polynomial, so if this is a polynomial with integer coefficients, then it gives the Hodge Euler characteristic. And, well, the question is now, of course, is this true for M5? So is M5 polynomial? And the answer is we do not know for sure, but we can also ask is M5 bar polynomial? And it turns out that this is the case if M5 bar only has state structure and the state structure that Ursula talked about today. And this is because we again here have used the fact that M5 bar is smooth and projective. And now the question, is this true? Well, this is true of two big conjectures. And the way the argument goes is there is this classification by Cheneffier-Enlon from 2014 of algebraic caspital representations of the projective general linear group. And then you can use Langlitz conjecture and Fontaine-Mazure conjecture to translate this to the Galois representations. And then this corresponds to the Hodge structures. And then what you get is that for a dimension 12, which is the dimension of M5, the only non-taked things that can appear are cusp forms. And you cannot have cusp forms on something that is unirrational and M5 and M5 bar are unirrational. So then if these conjectures hold, we have this polynomial and that means that counting the number of points of finite fields will give us this Hodge order characteristic. But this is just, this is not really a problem that this is up to conjectures because if we can count these points, then we can see if it's a polynomial or not. We don't really need the conjectures. The conjectures are just there to give us an idea that it's a good idea to actually even try this because, well, because you have this direct, then you have this nice formulation of the Hodge order characteristic and things that are polynomial are generally easier to count than things that are not polynomial. Okay, so how do we try to count then the number of points of M5? Well, we can look at the conality of the curves we have hyperliptic curves we have trigonal curves and we have to the rest. There's Q to the ninth hyperliptic curves and there's Q to the 11th plus Q to the 10th minus Q to the eighth plus one trigonal curves and the rest is still unknown. I'm currently looking at this and this is my result which is the topic of this talk. And so how do we get this result? Well, as I mentioned, we will look at plane curves. So we have this correspondence between smooth trigonal k curves of genus 5, projective k-quintics that have a unique singularity and it is a delta invariant one. So, and how do we get this correspondence? Well, if we have a trigonal curve then it has a G13. We look at its dual linear system then by Riemann-Roch this is a G25 and so we have a plane quintic then we can use the genus degree formula this is 5, the genus this is 6 so this must be 1 so we have one point with delta invariant one and the other way around if we have such a plane quintic then you get that the genus is 5 and if you take a line through the unique singularity then this will intersect the plane quintic in three other points and so the pencil of lines through the singularity will give a G13 so the curve is trigonal. So if we now denote this space by T then what we actually want to count is the number of curves in T but then we need to divide by the number of elements in the projective and tenor linear group because well, similar to what also I mentioned in the first talk when you have these projective curves you need to divide by the projective and the linear group to get the modular space so now how do we count these curves? Well, we look at the type of the singularity it can be like this a node or it can be a cusp and if it is a node we can further specify whether it's a split node or a non-split node and a split node is a node where tensions are defined over K, the ground field and non-split node is a node where they are defined over a degree-to-field extension and they form a conjugate pair well, you can count these all separately these cases and they are very similar to count but for the rest of this talk we just look at the split nodes so if we have such a split node then what we do is we apply a K linear transformation K linear change of coordinates if you will of coordinates that sends the node to this fixed point that we just choose 001 and the tensions they go to x and y where x, y, z are the coordinates of the projective plane so we want to count if we can now count all the curves that look like this then we can just multiply by the appropriate subgroup of the projective linear group that corresponds to these linear transformations so this is what we actually want to count and how do we do this well, I'll just erase this because I need a little more space we do this using the sieve principle and for this I have a little bit of notation we write curly C to be the space of all projective plane K quintics that have a singularity so basically those type of curves a singularity I'm sorry, this must be a node not just any singularity a node at 001 with tensions x and y but these places are different because these curves can also have more singularities and we only want the ones with the one singularity and so what we do is we apply a sieve principle where we take all these curves and then we add and subtract various low side of curves with singularities so we define if we have some points in P1 at 2 Pn in P2 then this is this is the space consists of that consists of curves Pn and so on Pn so now we can describe the sieving method we take all these curves and then for every point in P2 so every K point P2K but we need to be careful because we don't want to take this fixed point that we that we chose so for any such point we subtract all the curves that have a singularity at that point and now if a curve is smooth outside of the fixed point it has been counted once here and not been subtracted afterwards if a curve has one singularity outside of the fixed point it has been counted once and subtracted once which is fine because then we are not counting it if a curve has two singularities outside the fixed point and they are defined over K then it has been counted once subtracted twice once for each singularity so we need to add it again we get a pair of such points so well the same thing here it's equal and then so we then subtract all the curves that have singularities here and then if we have three K points we need to add them so we need to subtract them again if we have three and then if we have four we need to add them again and we can go on and if we have for example a conjugate pair of points we need to add those cases and you can also have for example a conjugate triple and a conjugate pair and the K point and you have all these different combinations of points and you need to consider all of them and then when you go on and on you are the ideas that you are left with the curves that are completely smooth outside of the fixed point but the problem is that this does not terminate because there are curves that have infinitely many singularities for example this curve you can have and then one big double line and so this has this is a quintic you can take one of these three to prefix node and then every point on this double line will be a singularity so you need to stop at some point and then for all the curves that you have not considered yet you need to actually compute them by hand or at least that's what I did and then so you compute by hand how much of those curves there are and then you just do the combinatorics and then add or subtract them to make sure that in total they have been counted zero times and then when you do this for all these types you get in the end only the curves that are completely smooth outside of the fixed point so now the question becomes where do you stop? well a natural point to stop would be at ten points in total so the fixed point and nine others because you can have at most ten points when you don't have a double component you have at least where you have five lines but the problem is that these spaces these CPs, CPQ, etc they become increasingly difficult to compute as the number of points grows so in fact I stopped at five points so the fixed point and five more points go up to the fixed point and five more points and then I did the rest by hand so that's a lot of cases basically what I did and then you get this nice formula which I said at the start plus Q to the tenth, minus Q to the eighth plus one and well because there's so many cases it's nice to have some kind of check that you're doing the right thing and so I wrote a computer program and that was able to check if this is correct for Q is two and Q is three so that's a nice check and of course when you can if you can compute everything the other parts of M5 and M41 and M42 then you can combine all this information together and then you would have M5 bar which and there you have point coraguality so that would also be a check because then you would need to have symmetry in the homology so if you do not have that then you know you are wrong so now I can try to give an example of one of these cases and show how you count them so well we just look at I'll just take a case we have we take a conic we take two lines tensioned to the conic and we take a third line that is not tensioned to the conic we have in total we have one, two, three, four, five, six, seven points well we just take a case where everything is over K because that is the simplest seven K points so and one of these is the node and you need to consider that these two tensioned points and then you can count okay how many of these curves are there in P2 well first you take the conic so take the conic well I'll just write number of conics because I am lazy but that is it's very simple you just take all the conics which is a P5 and then you subtract all the singular ones so those are the pairs of lines the double lines and the pairs of lines and then you get this formula and then you take two points on the conic so you can take this point and this point for example and then you take the tensions at those points so in how many ways can you do this there are the conic in P2 is a genus zero curve so there is a P1 of points and then we just choose two and then we need to choose the third line but we need to take it in such a way that the intersections with the conic are k points because if you just take any line then this might be a conjugate pair so we just choose two k points on the conic and we take the line through them so that gives P1k of course minus two because we don't want to choose the points that we already chose before and then we choose two more points but now we need to be careful because if the characteristic is not two then we can have this that when we take these two points that the line we take through them will go through this point this intersection of the two tensions and this is clearly a very different case so we do not want to count this so we need to subtract this case so how do we do that well we just take when we have this we just take any k point on here and then for every such k point that intersects here in another k point so we need to subtract again the number of k points but now we do not need to choose two and again we do not want to choose the one that we had before I should erase this but now if we choose this point or this point we end up with the same line so we get every line twice so we divide by two here so we have this but when the characteristic is two so characteristic is two then basically this cannot occur oh, I am sorry when you have this then any line through this point will be tensioned to the conic in characteristic two so this case cannot occur so then it is just enough to just take this part basically and do not take this part so that is basically this case and okay now I can if I have some time left I can maybe say a bit about why these spaces become harder to compute as you go on okay so the thing is we first look at what this space is okay so the total space of plane quintics is a p20 because there is 21 monomials of degree 5 then when you when you take look at that space, well what was the definition you have this point 001 so that needs to be on the curve which is the condition so set to the fifth is zero and here I mean that the coefficient of set to the fifth is zero and then it needs to be a singularity so we have this condition and this condition and you have the conduction that is a node and the tensions are x and y so you have more conditions and this is a non-zero condition because that tells you that the tensions are x and y and if this would be zero then we would not have a node but something of higher multiplicity so you see well you have these five conditions that the coefficients are zero so you get to pay p15 and then you have this non-zero condition and you get an a15 and then every point you add here will be another singularity so it will be three more conditions because one for the point being on the curve and two more for it being a singularity and so you can add five more points and then you end up at a zero and then if you want to add more points well you would get negative dimension so this does not really work and this corresponds to the fact that up to five points you can always find a way where these points are in general position but if you take more than five points this is not so you just it doesn't work so when so these guys are very nice to compute because up to five points you can find general position but then you might ask what if the points are not in general position and if they are not in general position then it turns out that there's some combinatorics at play and they all cancel and this has something to do with has a while set of formula because all the sums here if you take all the sums of the coefficient of combinatorics and you look at the weights the things you get for a certain weight then this somehow generates the one is generated by one divided by the has a while set of function okay so I think that's that's it. Are there any questions?