 So I want to talk about work in progress with, unless I'm going to count in terms of an arbitrary field. So the plan is for the first two thirds of the talk, I want to talk about counting what I mean by this, and then the last third, I hope to get to talk about that. So let's start with counting. What I want to count today is just I'll restrict the case of saying rational first. And for this I want and be the number of plain. Rational degree D curves through given 3D minus one given. It's a very easy example that's not very easy. When he's equal to one, then 2D minus one is, it's true so we want degree one curve so lines with two different points and there's only one. And one. And we want to count on x to five points. And for between this before that the conic is that type. You can do this. But there's only one. But in degree three, not one. So this is actually. So, I want you to view these counts now and I want you to use these counts to be the counts over an other right to both fields and then you can always find the kind of the I started with, but let's go to another field. And what was not right to go to the real numbers. Already real know that well known. So the question is, when all points are real I'll restrict the case today. How many of the curves are real. Different color. But in this case. This number is so long. The point where you live on your line with numbers to one. But this number is not, it depends on the choice of points. It could be over R this is. It depends on where view curve. And so, and this depends on configuration of points. But I want to tell you today how to still get an invariant. So this doesn't depend on the configuration of points. And this works in the following way this was observed by a nation week. And this version G observed that one can count your curves with a plus or minus one and this depends on the, on the, on the types of notes so we see curves will have one note. And this could either look like this, or it could be something like this and you can find over the company, let me write this down. And so there are two types of nodes to find over R. And so, you know, what I want to call it is what I want to call split note. That's the this one here. So locally looks like this and given by the equation x squared minus y squared to zero. And this is given by x squared plus y squared is equal to zero. So it has one real point at zero with you to know, but the branches are over the public numbers and that and this observation was that, well, instead of counting this we occur we don't then we don't get anything invariant but if we assign a plus or minus one and some of all these thoughts or minus one we get something in there and me in the fourth column weight. And I wanted to find the type of a note, you know, it's either plus one or minus one plus one in this case. And minus one in the solitary week. And I wanted to find the very sign the following. So, take a curve C, take the product over all the real notes. The ones that actually have really few, the real numbers. Over the types. No, won't contribute anything to this product, which is minus one to the number of. We showed that all the WT. We take the sum over all of them. They show that she's fine. And we see a real perfect kind of reveal numbers rational plane. We agree. Through given and three minus one. But this is actually independent of the choice. Pick one generic point consideration and the here and said that this is independent of the point. So, we go back here. Here it is. One is W one. And now that we're very interesting question here is what it's going to be three. And this turns out to be eight. There could either be just a real courage. They all have a good note. There could be 10 real curves, nine of which have a split note and one have a solitary note, or they're 12 real curves. And then they have a good note and the other one has the other two have a solution. And now the pictures are for later. Before I get to the pictures. I now want to ask the question. What to do over other questions. And I want to start with a very, very nice approach and I'll tell you later why. So, the patient. And now you don't know what, but this is really random. If I take the units of the conference number to see without zero, and I take it as a set, and I mod out by unit spared. That means the set of units where where I have an equivalent relation I say that to two public numbers are equivalent if they differ by a spare. Well, it doesn't really make much sense here because any two complex numbers. They provide spread. So this is just one element and let me call it. Here it's a little more interesting. So I want to look at this, this set of the equivalent pattern. And here, minus one doesn't have a square root. So, for example, one at minus one are not in the same. However, there are only two. Otherwise I have so they're two elements. Plus and minus one. So, the observation is, well in the first case, where I just didn't care about the field or just counted over and I did both field I counted everything and when this is all the right to focus always one element. So, I counted everything with the one. But in the real case I counted the curve with either possible way. So the naive idea is found. There's with the weight and the weight should live in units of the field. What units where and what should this wait. Well, let's let's do this very naive. Here in the real case I looked at the types of notes. So let's look at what kind of what types of note over K they are. So, I claim that we still have like a flip note. And I want to give the picture. I want to say that the equation is, I can choose of the form x squared minus the alpha y squared. But now alpha lives. I can change this equation to this equation by just replacing y by alpha. But I want to write it like this, actually matches with a different class. Or we could have a solitary note. There we could have different types. Now I also want to write this equation, the same equation, but now alpha is not squared. And here we have a different type for each class in here which is not the square. Because I can't do a border change. There might be more types. And this this alpha this really detects where these conjugate these double conjugates. So earlier I defined the type of a note to be plus or minus one. And this is the observed that this actually plus or minus one is the alpha. First case alpha one and here we have minus one so naive idea. Now, would be to define type of a note. To be the class of alpha. So this matches actually this matches the real case. Remember the class of alpha. So what should the, the key version of the session J. Wait, the veteran G sign B. So it should somehow be the product over all the notes of these types. Well, there might not be any notes assigned over a day so what we actually want to do here is, I want to take the norm, the huge norm of the residue fields. Over K. And then I take into account all the notes that it's not decided okay we just need to go to the residue field and then we take the norm and we land and take. But this really generalizes. What I did here. So that will assume that for a real curve, there could be no defined over the context numbers we would have a complex conjugate here, and then the norm of this would be something positive. So, we just forget about it. So this naturally generalized this and I claim that this actually works. So. So this actually works and the input comes from a one homogeneity theory and I will say a little bit about this, but maybe I should stop here and not. You have some time to think about it. That was the motivation and now it gets a little technical, and then maybe later when I get to public stuff will get a bit more familiar. And now I want to argue that this naive approach actually works in the following way. Like the nor an orbit of nodes. This will be one. One K point, you can either view this as one K point and then notice has that we will say and then the orbit. And you could either view this as the product of the type of this and you do the correct product and the color conjugates, or you just take the normal big point, whatever you. And so I want to argue that this naive idea actually works and the input comes from a one homogeneity theory. So we're much about this. This is what it was for the idea. So what should this be. This is a topic theory on smooth. Variety favorite. Okay, so some feel that way. And we want to do a homotopic theory on smooth. And this actually works that we developed by my violin. But I will spare you the. We will need one particular thing from here so this tells me that everything I know from all of the theory I can do also with smooth. And now I want to tell you what I want to use some from classical. Let's go back to classical. Let's assume you have a map. X FX to why. And I want this to be very nice. Close contact without boundary. And that's important. And then what I can do is, you can look at the floor stock from the end. Homology. And because I, I'm in the oriented case. This is I can work big to see, and this is I can work deep. And here. Then all that. I've arrived at the quality class that one here is sent to what is called the degree of it. But all one more thing, which I won't go into in detail. This actually can be written as the sum of notes. Where you sum up over all x in the pre image of some point why. Take a point here. We look at all the points of the pre image. And there are kind of many. And you take something with the local degree, what do you do to take a pre image with a very small ball around it. We only have one point in there. You get SN. The same you do in the image, so you're getting that from SN to SN. We take and take the degree again. Maybe community. Otherwise, you can write it into some of local. So, and now, let me just tell you that there's a version for this in this A1 homotopy. In A1 homotopy theory, and we have something I want to call the A1 degree. So for that, now f x to y, now f to y should be proper and still say n dimensional k variety. And we also need some orientation assumption. And I also brought connected here so let's also assume that they are something similar to connected or the A1 version. And then what we can do is we can also assign an A1 degree. And this should really just be the analog of this, but I'm not telling you how it works, but I tell you where it lives. And this one here is not an integer. But it lives in something called the growth and bituring of this new thing. And it turns out what I will tell you next that the elements in here look very similar to this. Let me tell you why. What is this growth and bituring of k, of gw of k, not from a particular growth and bituring of that. You can define it as the new confusion of the set of isometric methods of non-degenerate quadratic form. For this I might want to which we will see later. With respect to taking the dark. So you take the set of isometric classes of non-degenerate and we have quadratic forms and you can take two quadratic forms of k and you can take the direct sum and you get a new quadratic form. So you have a monoid structure on this and then you just make this into the group. This is actually ring. It becomes a ring. But this is maybe not so easy to compute with. So let me tell you a different representation of this. Namely I claim that the w of k is generated by class of bracket a where a is a unit or unit squared. This is something familiar. Sticking brackets around a is the member of the class. Why is that? This should be the class of the quadratic form a and here it also makes sense. It is the same as the class of a b squared and then the class of a b into the coordinate. So these are the generators. That's pretty great. And we have a couple of relations. We won't do it again. We have a plus b the same as a plus b and a times b the same as a b and this is for a b should be 0 and the first base a plus b should be 0. Maybe one thing I will need two is then whenever we have a finite system we will do the extension. I can define a map from bw of l to bw of k which takes the quadratic form I will use the first definition again. So this is the quadratic form and I can just send to what should be the quadratic form of the l and then propose it with the p trace. So we note this map by the way and I will use this in a second. And then we want to take a theorem about the invariance but maybe let's use some examples first. We have already seen that this set here is just one element it has one generator it is an integer we have already seen that the set of generators here consists of two elements one is the polynomial ring and the other is the cyclical loop of order. That's also the finite field then at least as group it brings it to the more complicated this is z plus unit not unit squared or this I could also identify and now let's define this weight to really be whatever was here but to stick it pretty much the same thing remember everything else now this is a stick graph and now we want some kind of invariance statement and we want to use this to write down curves with this weight through different configuration of points then this should be independent of the choice of points that's what I write next and the reason for this will be that this count is actually equal to the degree but let me write this down this is the degree so this is just because we need the column on and here is the diagram which came out at least you can find it on this home page you can see the assumption that this is a case and then let's define it be the following I take some of all curves or these weights but now I actually want to take into account all the curves so I take these traces I define here so I trace down to think about this again like this and here to sum this is in WK and here to sum over all plane rational degree D so this is through different configuration of 3B minus 1 point defined in this case point defined over K you can also make it more complicated but today it's all point defined over K and the theorem is that this one is independent of the choice of points so I give you an idea why this is true this NDA1 is actually this A1 degree which I have of the following map evaluation map of the modernized space 3B minus 1 marked stable map 3B and this one they show that it's well defined by 1 degree and just like the A1 degree it splits up as the sum of local degrees local A1 degree so the left hand side doesn't depend on your point configuration but the sum of local A1 degrees again it's taking configuration of points and here and you look at the pre-image and for each of the points of the pre-images this one here turns out to be different so this is all I wanted to say about this NDA1 for this counter of the arbitrary field and now the big question I want to answer in the last 12 minutes is how to count this and for this I want to use what people do on the tree of correspondence theorem for example yeah yeah yeah that's it turns out that when you do this here and you have a complex this is the complex of the real number then this will be this will give for each of these it will give me minus one and then when you do it over the real number you forget the bracket I have I can also take and then this will still be invariant yeah it's the same yeah it's the same thing but today I only want to look at this thing I have to figure out the other thing so I've seen earlier in the talk there was also some some tropical geometry I looked at the abstract and I assume that you have at least seen tropical curve before because I don't have time to define them but I drew some examples so we want to count we want to compute this NDA1 now and one way to do this is actually translate the question to the tropical word for each of these plain rational curve words for each curve I can associate the top of the curve this will be graphed like this and then now and we can identify this with the count of tropical so let's gamma simple tropical curve so as I said it will be the graph like this so this is the weighted graphs so each edge has a weight and the weight once I don't write the weights of all these unbounded edges here is one as in all of this example and also it means that we only have two valent and four valent vertices here we have four valent vertices and all of these other vertices and for these one can define complex and dual multiplicity namely as follows over all the let's say triangles in dual subdivision and you take the double equilibrium area I never know how to do this so for this curve I have drawn the dual subdivision here I have a a lattice polygon so this is a subset of R2 in R2 we have a lattice I drew the lattice point in here so it's 0.00 this is the point 0.30 this is the point 0.3 and I I subdivide this in the following way to each of the edges I draw another polygonal edge to this I can draw this polygonal edge to this I draw this polygonal edge so the edges point to the edges here and the vertices correspond to this vertex here corresponds to this triangle and this four valent vertex corresponds to this one and I want to take the product over all the triangles so we have a lot of triangles in here of the double subdivision area here so let's do this example so here this multiplicity is of all the areas here so the area to half to the double area is one here here here these triangles have all area a half these two have area one so double area two so the multiplicity is and here the multiplicity is these have all area one and this has area and there goes the real multiplicity of it and this one we just ignore them we don't care about them this is just we really don't care about the triangle also for the real place here we take either the product of minus again the product over all the triangle of minus one to the number of interior left this point let me write in of delta the triangle the interior left this point in case all edges have odd weight for all edges of the curve here alternatively all of these links here are odd so you see two points one to the edge here which has links and all of these are odd and otherwise it's zero for the example so here all the edges have weight one so we're in the first case so we look at all the triangles and we look for interior points but there are none so it's minus one to the zero for one here the real multiplicity turns out we're in the second case because we have this weight two edge so we need the multiplicity zero here the multiplicity that's why this example here we have one interior point that point in one of the triangles with minus and the contents for quantum theorem and now by now there are many more correspondence theorems which generalize this but I don't want to start mentioning names because I will get half of them but we're talking about the first so the following namely that the Nd from the beginning is the same as NdTrop where NdTrop is the sum over all tropics of curves who are given configuration of points and they're counted with these complex multiplicity and WdTrop is the same as WdTrop where we take the same sum to feel multiplicity and we're done with the sum sum over rational we can also define the rational the topical curve are rational topical curves through americ consideration of 3d-1 points in R2 in R2 so you take the sum from americ enough in R2 and you count all of these kinds of graphs which are not completely arbitrary they have to follow some rules to these not good points americ will tell you that it's actually simple and we can identify the counts we wanted to compute with these complicated counts but the upshot here is that finding all of these curves can be done with simple combinatorics so this is really nice and the last two minutes I want to say that I'm going to talk about my broken progress where we show a correspondence theorem but with it so first I need to define a one multiplicity the topical curve and I want this to be part of the group to be I take the real multiplicity and stick it into bracket plus the context multiplicity divided by two times h and h is one plus minus one and this is in the first case that we only have on weight edges and otherwise I take the complex multiplicity I divide it by two and I'll think about h this is really surprising this is either one or one minus one this is only one minus one this is really not obvious because here there could be maybe not anything but there could be other classes when we're not in R over R and then our quadratic form says that this one here I can compute as a practical count namely as envy a one trap and here I just do the equal this and this is defined to be the same sum as here over now all the a one multiplicity the same sum of the multiplicity and give me one more minute I already said that you can see that these multiplicities are completely determined by the real and the complex multiplicity so as a corollary I don't need to find a new combinatorial formula for this I can just express it in terms of the real and the complex count and you get the following envy a one equals envy divided by two and H so we need to times one so this is to do what you you start it and then you figure out that formula for it and here I know so let's do the first four cases where D is 1, 2, 3, 4 and here we have envy 3WD and here envy a one then this was one this was one and this was this bracket one this was also one this was one bracket one here was more interesting here remember we had the 12 we had the 8 and here we have 8 times 1 for 2 times H and the last one I got 624 and it's 240 and 1 plus the difference divided by 2 with this in 19 and I think there are also recursions for the real count they should also be I just didn't I know I was happy when I realized but yeah there should be there are recursions also I mean then you could just interpolate it like this yeah yeah it gets yeah actually when you have we looked at the next correct we looked at the simple case where you have like 2 and then 200 points there and then you do not only get what we might as well get like you get the piece in here yeah then yeah this also we looked a bit a a bit so I don't know this one here this one this one or this is like really hard this is like this is the into the correspondence here it's actually not this but it's to come up with what should this be and this is really you have several curves that tropicalize to the same topic recurve the correspondence theorem works like this you have five in a couple of curves and you construct all the other bar curves that tropicalize this and then you compute then I compute this for this because this thing of it is really hard and then magically this comes out but this is like many page factor you have more questions so there are no restrictions on the field k now that oh I forgot so yeah thank you I want the characteristic of k to be bigger than the degree for example we only have several field extensions because otherwise all the traces not even make sense and also like with the degree if the characteristic is bigger than the degree the equations are nice so in the computation like a lot of things but with these constraints or characteristic but okay so but with this constraint you're saying the nd formula actually doesn't have any of the interesting classes from the grid and grid string and if you have k-point if you have like points that come in the gala over it then you have more interesting but then the combinatorics get out of very very complicated seems very complicated we're working on the simple where we have to quickly