 Okay, great. Now that we are recording, it's a great pleasure to introduce Carl Lien from Humboldt University in Berlin speaking about Tevelev degrees. Thanks. Okay, thanks a lot Anders for the introduction and thanks for the invitation to speak. It's a great pleasure to be here virtually and I hope to see many of you in person at some point in the United States. Right, so the title of the talk is Tevelev degrees and I will explain hopefully in the first half for the title, where the name Tevelev degrees comes from. But this is based on work with, well, by various combinations of the people listed here. And this will come up and some of which I've also contributed to. Okay, so let me start by just introducing the basic question. So first of all, X will be a smooth projective variety I will always work on the complex numbers. And I will always call its dimension R. Okay, and I will fix general points on X which I will call X one through XN. Okay, so I'm reading here. And I will also pick the curve class on X so you can think of this as a some effective class living in the second homology. I also want to fix the data of a curve with end points on it. Okay, so the, this end here with the number of points on the curve is equal to the number of points on X. Okay, so the question I want to ask is how many ways can I map the curve to the, to the target right. Okay, and I want to impose the conditions that on the one hand the points that I've marked so the points on the curve are supposed to map to the points on the target variety. So the degree of the map is supposed to be given by the homology, homology class that shows it. Okay, so the push if I push forward the fundamental class of the curve to X I should get some curve class on X and that's the, that's supposed to be the curve class that I fixed in the beginning. Okay, so you can think of beta is like the degree of the map. Okay, so that's the question and let me just give an equivalent formulation. I've, I've made a map tau so the source of this map is the space mgn X comma beta. So this is the modular space of just maps from any curve C to X. And again I require that the, this map have degree given by beta so beta is part of the data so the push forward the fundamental class is supposed to be equal to beta, and this end denotes the number of mark points on my, on my source curve. And to that, that has a map. So on the one hand I could forget the data of the map and just remember the source curves that has a map to the usual mgn so it just remembers the source curve of the map with its mark points. And the other hand I can also just map, I can remember just the data on the target so namely I could remember the positions of the, the images of the mark points the end mark points so I have one map to mgn I've been able to map to X to the end so I can make a map to the product. And so the question of as is equivalently, what is the degree this map. Right because we're saying we'll fix a point of the target which is the data of endpoints on X, along with a fixed curve and endpoints on the curve. And we're asking how many maps are there that live over that data. So I'm going to call this map tau. And if the degree of this map is well defined, I denote it by Tev sub g beta and so G is the genus of the curve beta is the degree is the curve class and and is the number of mark points and of course this the superscript X is the target Okay, and this is what I'll refer to in this talk to the geometric Teva left degree for X. Okay, so there are questions about this, this setup as I've, as I formulated it. Okay. Okay, so of course this question is only reasonable if this degree is expected to be finite so I should do a dimension count to put myself in a situation where this is true. So if you've seen the computation of the expected dimension of mgn X beta, it's given up by some formula and you haven't seen this computation if not so important. But there's some formula for the X at least the expected dimension of this modular space of maps. And that's this line here so the expected dimension of the space mgn X beta is it's given by this formula with the purple dot and then the the dimension of the target which is which which we know in all cases it's well three G minus G plus M which is the dimension of mgn and then plus RN which is the dimension of XVM. Okay, and if I want this map to be finite then while I should, I should at least demand that the expected dimension of the, the source space equal to the target space. Okay. And all right so then you can rearrange this equation, and this gives you some numerical condition on what the number of mark points should be okay so n should be equal to one minus G plus X missing here, plus one over r times the integral of beta times the first term class the tangent bundle. Okay so remember that beta is a is a curve class it's a class living in the second homology of X. And the first turn classes of the first turn class of the tangent bundle is a class living in the second co homology of X. Right because it's a first turn class. So you compare these two things and integrate them so you get a number. Okay, so we'll see an example of this in a little bit. Okay so for this question to really make sense. I need to assume that this numerical condition otherwise I sort of can't really hope for for tau to be a finite map. And so throughout this talk I will assume this numerical condition. Right so in particular and as always determined by the other characters in this story so it's determined by the genus. Okay, and well. So I've been using the term expected dimension for the space mg and X beta so there's some notion of the expected dimension there's a naive dimension cut that you would. You hope that the space has. But it can happen that. Well, this the space, this has the space has to mention higher than expected. So this is a story I'm not really going to get into. But I do want to note that well the space mg and X beta. Because I'm asking for the degree of tau I only care about the demand the components of this modular space that that actually dominate the target. Right so mg and X beta can be a very complicated space with many irreducible components. So I'm not going to say that this is the correct dimension but some of the and some of them don't. But if I'm only interested in the degree of tau, right, if I'm only interested in the degree of a map coming out of the space that I only care the components that actually dominate. Right so have components that do not surject onto the target then they're not going to contribute to kind of the number of points in a fiber. Okay. So there are dominating components that have too high a dimension, right so if this space, if this, if this map tau in fact is a some kind of vibration where the fibers positive dimensional, then I really cannot ask about the degree in which case this type of degree is not, it's not defined. Okay, so in order to make this dimension I'm this definition I'm assuming two things. I'm assuming one that I have this numerical condition here right in this box. And this numerical condition comes from some calculation of the expected dimension of this space. And so the second thing I'm assuming is that the space actually has the expected dimension, at least when you respect to components that that dominate the target. Okay, so I'll give a more concrete example in a bit but are there other questions but what is that here. Okay. So, let's do concrete example in some sense this is the main example of this talk. And probably the simplest example you can ask for this example of projective space. Okay, so for projective space, I need to fix the curve class and well, the second homology of projective space is has ranked one. So the only the only invariant of a second homology of a curve class in projectives is it's degree. Okay, so the data of beta is simply the data of a degree of a curve living project space. And, well, and the space MGN PRD is simply the space you can think of as some kind of Hilbert scheme of curves of degree D inside projective space. And, okay, and if you haven't seen this, the stable map spaces before well you can you can still think about this the dimension of the space of maps to PR of degree D. You can still think of this in terms of the Riemann rock formula. So if you fix a curve for instance, then you can ask, well what's a what's a map of degree D to PR. Well it's like a line bundle of degree D plus r plus one sections. And Riemann rock at least give you gives you some naive calm for how many of those you'd expect there to be, right at least if this if the line bundle you chose is non special. Okay, so if you haven't seen this before this is actually a good exercise to work out. So you can pretend if you want that D is very large. And if you fix a curve there's some, there's some space of maps to a projective space and you can count the dimensions of that space. And then you can count the dimensions of the space of curves, and then then you'll end up with this formula. Okay, but in any case expected to mention the space is given by this number three g minus three plus and plus D times r plus one minus r times g minus one. Well, I said well and again in order for this. In order for this, this this tower to be kind of reasonable I need. I need my find my my fibers to be finite so I should require that this number this dimension is equal to the dimension of the target. And moreover, the classical results from Bill knows the theory tells us that in fact, the space of maps to projective space actually does have the correct dimension, at least if you respect the components that dominate MGN. Right so another way to say this is that if you take a general curve, the space of maps to a projective space a general fixed curves the space of maps to a projective space of some degree has expected dimension. So what this tells you is that when you compare this formula to the, as we did here to the dimension of the target so in this case you get just MGN cross PR to the end. And you kind of take this numerical condition which I've written down for a general x. Whenever and is sort of the the expected number of points where you would hope you were you'd hope that Tao is finite just based on the nine dimension counts. In fact, there is a there is a well defined degree so this map Tao is, in fact, generically finite, and it's degree is some well defined number that you can ask for. Okay, so in the end if we asked this question for projective space we always at least if the degree is positive we always get can hope for a reasonable answer. You can, if you fix a curve and you fix a projective space and you fix a bunch of points, you can ask for the number of maps from the curve to the projective space of some degree. And the naive dimension counts kind of imply that there's a finite number of such maps. So there are questions with this example. Okay, great. So this is sort of the classical way to formulate the question again so I'm just asking a very. So I'm just asking you sort of what I would consider to be a very basic question if you fix a curve and you fix a target variety how many maps can you make between them. And of course you can formulate that that basic question in terms of these modular spaces. Now if one allows themselves to work in the language of global wouldn't theory there's actually a very natural alternative formulation of this question. So before on the previous slide I had the spaces mg and x beta and mg and without bars. But in groom of wind theory, there's a very natural compactification of the space mg and x beta bar, and this is the modular space of stable maps. And the advantage of this somehow is that now I have a proper modular space right so in general mg and x beta is typically not proper but there there is some compactification called the modular space of stable maps. And one feature of this modular space of stable maps is that if I forget the source curve, I don't just get a curve mg and but I rather get a curve mg and bar. Okay so I get some kind of stable curve. And all right so I'm not going to get into the definition of stable maps it doesn't really matter for this talk but it's just some compactification space of maps in the usual sense, where the target is some kind of singular nodal curve instead of just a smooth curve. Okay so in particular if you if you work with this modular space of stable maps you can compactify the morphism tau, which I defined earlier. And you get this picture so I have a, I have a map from the space of stable maps to this to the space of stable curves cross points on x. Okay so if you just do this. Okay so so far so good and if you, you could you could do the same dimension counts as before, and you could kind of think about when you want this map to be generically finite and that's for its degree, but in fact you're not. At least if you kind of at least if you take the closure of the, the, the space of maps with smooth target you're not going to get any different answers here, right because somehow you're just, you're asking for a degree so if you if you just compactify everything they're not going to the answer will not change. But on the other hand what this modular space of stable maps also gives you is because you access to is a virtual fundamental class. I have alluded to before the, the space of maps to x even without the bar can have the wrong dimension in which case the dimension case counts that I wrote down will be incorrect. And the table of degrees that I've defined will be undefined. But when you pass to the space of the modular space of stable maps this still might have the wrong dimension compactifying can in fact make things worse. But what you do have is a virtual fundamental class, which lives in the expected dimension of this modular space. This is somehow a completely canonical object. And this virtual fundamental class this is a long story which I also will make it on to get into, but this is supposed to be some kind of correction, or some kind of replacement of the usual fundamental class, when the, the actual dimension of the space not agree with the expected dimension. But because this virtual class lives in the expected dimension. Well the expected dimension is something I always have control over right so the the actual dimension might be something I have. I have no control over it I might not know what that is it might just be too large, but the expected dimension is simply given by the formula that we had on this second slide. Which means that well, if you assume that the the virtual dimension, or the expected dimension of mg and x beta is equal to the dimension of the target. Then if I push forward this virtual class which lives in the expected dimension, I always get something downstairs that's just that losing that the top homology. Right because I've put myself in the situation where the virtual dimension agrees with of the top agrees with the actual dimension of the bottom. I don't know what the demand what the actual dimension of the space of stable maps is. I can push forward this virtual class. I know I'm going to get a multiple the fundamental class, and that multiple is, is, is, you can think of as some kind of virtual degree. Right so if mg and x beta bar is kind of smooth of the expected dimension at least on dominating components, this will agree with the actual degree. So this definition for wouldn't vary is that well, I don't need to care, at least in principle about what the actual dimension of the space is because this virtual fundamental cost we defined, nonetheless. So we define the virtual fundamental the virtual level of degree to be to be. Well, the rational number such that when I push for the virtual class. It's possible of the fundamental class downstairs. Okay, and the upshot again as I've said is that, no matter what the actual geometry of this much like stable maps is this number will be well defined. And okay so this is sort of a diversion this will not sort of play a role in the actual Schubert calculus that will come up in this talk at least right so so not play a role in the Schubert calculus results I want to mention. But I do want to mention this because this is sort of a parallel story to be the one I do want to talk about in fact, in some senses. At least one sense it's actually an easier story than the, the one I want to discuss, due to the results on the next page. So I haven't told you anything about so somehow these geometric heavily left degrees are kind of the geometric question that I think is much is most natural. You can reformulate the problem some kind of fancy way with these virtual classes whatever that means a group of wouldn't theory. And the funny thing that happens is that actually the group of written numbers are much easier to compute. And I haven't told you anything about how to compute the geometric numbers that will come later, but I do want to mention this result of Anders and Rob under a pond a which to my knowledge is soon forthcoming. So there's just some formula for the virtual template degree that it's just completely universal. Okay, so I'm not going to explain really what this formula means. So, it's, it's given in terms of some classes that are defined in the quantum homology of x which is some kind of deformation. In German put in theory of the usual homology of x, but I do want to write this formula just to just to show you that there is one. If you have a complete understanding of the quantum homology of x, which we do in many examples. Then computing these virtual type of left degrees becomes complete is really a completely combinatorial problem. Right so the proof goes through some kind of some kind of standard yoga and groom of wouldn't theory that I won't get into. But the upshot then is that if you understand this if you kind of do the do a black box the kind of hard work of defining and and understand the quantum homology of x. Then when you get out of this package is some is some very concrete combinatorial problem. When one asked for these virtual degrees. Okay, so as an example. So as examples of this applications of this formula. So if you ask the virtual table of the question for projective space. So this is again the virtual number. So this is the, this is sort of the virtual number of maps from see which is genius G to PR with some number of incident incidents conditions. You get the very simple answer R plus one to the G. R plus one is in fact the other character PR. Okay, and this is actually a pretty easy calculation because the quantum cosmology projective space is something easy. Okay, and it's someone considerably harder computation is that for hyper surfaces and PR plus one so they also show that if you take a hyper surface of agree not too large so I think degree about the dimension of the project is of the dimension. Then again there's just some formula for this virtual time tell have a lot of degrees so this is the virtual number of maps of some curve of genius G to hyper surface of the degree. Okay, and then for some technical reasons I have to say that the degree is at least three here but there's some other formula for hyper surfaces of degree to. Okay, so another example that comes up in their papers that a flag varieties and those are also spaces X for which we know something the quantum cosmology. But there I think the formulas were not so not as easy funny. I'm not mistaken. Okay, so I think this paper will be out soon if I'm my information is not correct, not incorrect so you can read more about this that. Okay, but I want to mention this was all. Okay, so in the end this this is this virtual table of questions some sense is. And it kind of theoretical sense is completely answered by this, this very nice formula. But then there's a there's a kind of very rich combinatorial question which is well for particular examples of X can you actually compute this. And this paper of Anders and Rahul is sort of the first series attempt to do this. Okay, so where do I want to go into stock. I want to go back to the the geometric have left degrees which I've claimed are on the one hand, and more geometrically natural but on the other hand, they're at least at the moment we don't have a formula that's nearly this nice, or any formula. So at least for the perhaps simplest example project is basis can we compute the geometric numbers. So we have some kind of mysterious virtual count of these of these maps. Can we get, can we get the actual count. Right so that's direction one and that's the direction I want to spend the most time on. In direction to as well we have these very nice virtual formulas at least in some cases. And so well when does the virtual, when does the virtual table of degrees so when does the virtual account agree with the actual account. Right so what is this kind of mysterious. So a virtual fundamental class is actually producing the, the number that has kind of the most obvious geometric meaning. Okay, now there was one or the virtual content number. Okay, I should say in the kind of usual Gorma Witten theory setup the answer is usually is often kind of never. In fact, one nice feature about these table of degrees is that there's somehow so constrained that, in fact the virtual table of degrees do tend to do seem to be equal to the actual table of degrees the geometric table of degrees. At least in some quantifiable sense, more often than the Gorma Witten, the usual Gorma Witten variance. Okay so what I'd like to do I think is state a result for the geometric table of degrees of P1. And kind of defer the proof to after the break so I'll state the result and then maybe we'll take the five minute break there. Okay, so as a reminder so what so what is the geometric table of problem for P1 and sometimes this is the most concrete question I will put forth in this talk. So again the, the players are a general pointed curve of genus G so it genus G curves with endpoints on it. And now I'm going to take my target to be P1. So this is just the data. So my P1 is x, and I'm just going to take endpoints on that. Okay, and the question is how many maps now do I have that have degree D so this this this model to class beta is just going to be D times the fundamental class so how many covers of P1 are there of degree D that send. Well the points I started with upstairs to the point system downstairs. Okay and then the numerical condition I need here is that. And is equal to 2D minus two plus one. Okay, so are there other questions about just the formulation of the geometric problem for P1. Okay great so let me state state the theorem. So let me start with the first part. So, so the first complete computation of this of this degree was given by Alessio cello Rahul Pundarapande and Johannes Schmidt. Yeah, this year so I think this was March. And, well, they gave this formula. So, okay so so this is for now just a formula but I want to just point out a few features. So first is that so basically what I want you to get get it from this is that this formula saying to G and then minus a bunch of correction terms. So this two to the G, we saw has so has already come up in one place in this talk so this two to the G is the virtual depth of degree. Right because the virtual to have left degrees of PR we saw our r plus one to the G were G's the genus of the source curve. And so here this is P1 so the R is one, so I get two to the G is the virtual degree. And so, and what are all these binomial coefficients well if you. So here, I'm denoting the number D minus G minus one is L. So these binomial coefficients, notice that the number on the bottom is always kind of. Well here it's negative L, and here it's negative L minus one, and then in the first term it's ranging from I equals zero to minus L minus two. Okay so in particular these binomial coefficients only make sense of L is negative. Right so what this is saying is that so when I was positive actually all these binomial coefficients are just interpreted to vanish because you're taking G choose some negative number. So what is this really saying this is saying that when L, which is so when when L is positive which is to say that D is bigger than G plus one. Then all these corrections vanish and the vert and the table of degree of P to the of P1 for genus G and degree D is just two to the G. Okay and then when L is negative there's some correction terms. So in other words when D is large when I'm counting curves of when I'm counting covers of large degree. So the number to have left degrees equals to the virtual table of degree, which is, which has already been computed to be two to the G. Or rather, I mean, yeah. And well when the degree is small then this is not quite correct and there's some correction terms given by these binomial coefficients. And so then inspired by this result. A couple of months later so this is March this I think this is maybe May or June. And for our case and I gave a different proof of this result that goes to Schubert calculus and we gave this kind of, and we gave the shoe with calculus formula. Okay so I wanted to at least say this in the first half the top because this is the Schubert seminar. So the theorem is that using different methods we should we compute this degree we compute this count in terms of intersection number on the grass money. Okay so this is the grass money and of two planes in a D plus one mental vector space. So the Schubert class Sigma I is the, the ith segregate class of the universal bundle. And the claim is that this, this intersection number is also equal to the table of degree. Okay so in particular is also equal to the number, the formula above, which is not completely office. Okay, and I should say all of this was somehow inspired. So the paper of telepond or punishment was inspired by a result so this is December of 2020. And this explains the name. So table of among many other things, compute he didn't call them table of degrees but he computed the number of maps of degree g plus one out of a genius G curve with with sort of the correct number of mark points to get a final answer. And you found the answer simply to the G. So in this case D minus G minus one is is zero right because he is G plus one. And so, all these correction terms vanish. Okay and his proof is also very different it goes through some kind of the generation that goes through real logic by geometry. Okay, and so I'm not going to say much about the proof of telepond or punishment but I will say it. This is really a question about, you know, counting branch covers of P one. And those are parameterized by a nice modular space namely the hermit space of branch covers. And we've made some advances in the intersection theory and such modular spaces in recent years and the proof of the last year Rahul and Johannes uses some of this machinery. Okay so what I want to talk about in the second half. After the break is is the proof of Gaby and myself. Right so maybe I will stop there for the break and I don't know if you do questions right now or, I suppose if there are questions I should answer any. Yeah, thanks very much.