 Okay, great. So great. So just to recap where we were, the theorem I stated was at this count, which is at I hope at the top of the page of branch covers of degree D from a fixed genus G curve to P1, satisfying an appropriate number of incidence conditions, so conditions that one point maps to another point is equal to this number in the theorem. Okay, and what I'd like to explain now is the proof that of this, so there are two formulas and these formulas are of course formulas for the same number, but the proofs are very different. You know, and I'd like to explain how the, how this Schuber's cycle formula comes up. Okay, but I actually also want to explain where this two to the G comes from in the second formula. So I'll try to do both of these things at the same time. Okay, great. So this goes through something called the, the theory of limit linear series. So this is a beautiful theory that was written down by Eisenbud and Harris in the 80s. And in fact, this is in some sense a very old idea going back to the 19th century. So the idea as well, I fixed a genus G curve and I want to compute some maps to P1 out of that curve. And well, because anytime you're computing something out of a fixed object, you can always degenerate that object and hope that, and hope that there's some, there's some limit objects on that degenerate thing that can count. And hopefully this is some kind of combinatorially tracular problem. Okay, so specifically I want to take a curve, my curve of genus G and I want to degenerate it to this curve on the right, which is singular. And so it has a vertical component, which is isomorphic to P1. And it has, and in order for it to have genus G, will I attach G elliptic tails to it? Okay, so this is a nodal curve of arithmetic genus G, it has a rational component and has G elliptic components. And remember that C also had N mark points on it. And I, in this degeneration, I put the mark points on the rational component. So what do I want to do here? Well, so what you can show using the theory of limit linear series is that well, or if you like also the Harris Mumford theory of admissible covers, is that, so maps from C to P1 degenerate to some kind of objects on the singular curve. So those objects can be regarded as what's called limit linear series, or they can be regarded as admissible covers. And so in that degeneration, what you find is that you reduce to the problem of actually counting maps out of a P1, right? So now somehow in a way I'm not really good at explaining, the elliptic tails are kind of don't really contribute anything. And in the degeneration, somehow all of the interesting, all the interesting geometry boils down to studying not maps out of this genus G curve C, but rather maps out of this P1 that lives on the, which is a component of the degenerate curve. So what happens, so the count I'm interested in ends up being equal to the number of maps now out of P1 also of degree D. And I still have the condition that these N mark points are supposed to map to the N mark points downstairs, which are still fixed. But then I pick up the additional condition now that these G points Q1 through QG, which are the nodes of my degenerate curve, or ramification points of this cover. Okay. So this is kind of a, I'm taking a bit of a leap of faith here, but what I'm claiming is that this geometric problem that I start with, when I degenerate it, I reduce to a problem about maps out of a curve of genus zero. Right. So, and the hope is that when you reduce the genus of your curve, you get something simpler. Okay. And well, why is the simpler? Well, maps of degree D from a P1 to P1, I mean, this is, this is something very simple, at least in principle. So what's a map of degree D from P1 to P1? Well, it's simply two polynomials of degree D. So homogenous polynomial is a degree taken up to simultaneous scaling. So this is mod scaling. Okay. And these polynomials are kind of parameterized by their D plus one coefficients. And if I have two of them on scaling, this is simply a projective space is demented to D plus one. Okay. So I've, I've essentially reduced to the question of, to this, this same question, genus zero, but not quite because I have these additional G ramification points, which have entered the picture. Those are kind of the attachment points of these elliptic tails. So now what you say is okay, now you have a projective space of two D plus one, and I want to impose two kinds of conditions. So on the one hand, I, I want to impose the condition that well, when I take these two polynomials and plug in the point P1, or is there a PI, I get out the point XI. And if you're going to write down what that's saying, that's, that's, that's simply a linear common, linear condition on PD 2D plus one that defines a hyperplane in 2D plus P2D plus one. And then a slightly less obvious, but also not such a hard thing to check is that the condition that FB ramified, right, this is some condition about the derivative of F. So the condition that FB ramified at QJ, well, that's, that's a quadric condition of the coefficients of these, of these two polynomials. So what you find is that, well, my, my space of maps is a projective space. And the conditions that F sends the points of the source of the points of the target that I want are a bunch of hyperplanes. And then these new conditions coming from the degeneration that F is ramified at a bunch at G points Q1 through QG, are ramified, well, that's G quadric hyper surfaces. And then the maps that I want, well, there's just the, there's the points that lie in all of those conditions that satisfy all those conditions. So by Bezos theorem, I get simply two to the G break safe, I have hyperplane and hyperplanes of degree one, and then G quadrics of degree two. Whoops, sorry. So I did that a little bit fast, but other questions about what I've just said here. Okay, so of course, this cannot be quite correct, because if we already seen as we've already seen two to the G is the wrong answer. Right. So two to the G is this virtual answer. But what I'd really like to get at is this geometric answer. And we've seen that the geometric answer, at least I've claimed to you from the theorem of the formula of Chellep-Hundermann-Pundesch-Schmidt that well, two to the G is the correct answer when D is large. But when D is small, there are all these correction terms coming from binomial coefficients, right? So something has to be going wrong here. And what's going wrong here is that well, I've told you a little bit of a lie. So this projective space is not quite parametrizing maps from P one to P one, because if, for example, if not an F one, there's nothing preventing them from being linearly dependent. Right. So in this projective space, I could certainly have, for example, or F not or F one could just be zero. Right. In which case, I have a perfectly good point of this projective space. But I'm not going to get a map of degree D in any kind of sensible way. Right. Because if you just have a, if F not and F one are multiples of the same polynomial, you're not going to get a map. Right. You're just going to get constant maps. And in fact, when D is small, what you can check is that this intersection, this intersection that I've set up actually produces points that you don't want on this locus, on this degenerate locus. And what's worse, those that you get some kind of excess intersection. Right. And you can actually see why this is happening, because well, when F not and F one are linearly dependent, and well, so what does that mean? It means that if not an F one, well, so, so, right. So how would you impose the condition that f of PI is equal to XI? Well, you would plug in some point and hope that you get the point XI in the end. But if you plug in the point PI and PI is a common root of F not and F one, which there are many, right. In other words, if this is a base point, then you're just going to get zero comma zero. Right. Which is, and when you get zero comma zero and you write down the equation defining the condition F of PI equals XI, this will just always be satisfied. Right. So when F not and F one are kind of divisible by say all of the if not an F one or both vanish say on all of the points, then all of these linear conditions are going to be satisfied automatically. Right. Even though you have some kind of constant map without clear geometric meaning. And this ramification condition will have some condition on the derivative. And well, for instance, if F not is zero, well, then the derivative of F is just zero. So F will always, F will look like it's kind of ramified everywhere. So what will happen, and in fact, this only happens when D is small, is that these conditions will kind of be satisfied for dumb reasons. Right. They will this sort of bad locus of constant maps will lie in the intersection of all of these, of all of these hyperplane sections. But you don't want to count this. Right. So this two to the G is not giving you any kind of geometric number. So how do you get around this? So how do you get around this? You can you can try to use the excess intersection problem. You can say, well, I can control where my, where the extra points are. But this will lead you to kind of difficult territory because the excess locus are not smooth and they intersect each other. And they're very high dimension. But what you can do instead is you can do some kind of controlled blow up of this bad locus. So you can say, well, I have some I have some locus of constant maps and locus of points in this P2D plus one, parameterizing pairs of polynomials that are actually just given multiples of each other. And I can blow that up. And the, and somehow the point is somehow that you can blow it up in a modular way. So you can actually say what the points are in this blow up. So namely, what I want to do is I want to take, okay, so here's the construction, I want to take the product of my P2D plus one with which is the space I had before with the gross monion of two place in a, in the D plus one dimensional vector space of, so this, this D plus one is really the D plus one dimensional vector space of, of polynomials of degree D homogenous polynomials of degree D. And well, my kind of nice locus, my open locus on P2D plus one where nothing is going wrong is the locus where my two polynomials are linearly independent, right, where they actually define some kind of non-constant map. And so when you have a non-constant map, and when, when F0 and F1 are linearly independent, well, then they, then their span is a two dimensional vector space, right? And that gives you a, that gives you a point in this gross monion. And when they're linearly dependent, well, then there are many, there are many two dimensional vector spaces that contain both polynomials. So what I want to do is I want to take the incidence correspondence where I impose the condition that the point of my gross monion, so my two dimensional vector space, contains both of my polynomials of degree D. Okay, and again, on the locus where the two polynomials are linearly independent, V is uniquely determined, it's simply the span of F0 and F1. But when F0 and F1 are linearly dependent, well, then I have some choice of, for what I can take for V, right? So I get some, I get some kind of projective space for the choices. Okay, and this is a space I call call2d plus one. This is not quite standard notation, but this is a special case of what's called the space of complete collineations. Can I ask a quick question, Carl? Yes, please. Just the, can you clarify again, what's the, this way rocket F0, F1 notation? Right, so F0, F1, so I'm now counting maps from P1 to P1, right? And so a map from P1 to P1 is given by two polynomials of degree D, because it's like a rational function. It's like a rational function. Right, right, sorry, thanks so much. Yeah. And so that numerator and denominator, if I take the coefficients that forms this projective space, yeah. But I have some locus where that numerator and denominator are actually the same polynomial up to scaling, so I want to blow up that locus. And that blow up I'm claiming is the space called 2d plus one, because on the open locus where F0 and F1 are linearly independent, I have a nice morphism, right? Because my V is uniquely determined. And then on the locus where F0 and F1 are scattered in multiples of each other, then I have this choice of, somehow the, the normal bundle term is this choice of two-dimensional vector space containing both polynomials. Okay, so the, so the claim here is that the projection to P2, 2d plus one is the blow up of this constant map locus. And well, if I consider the other map to the gross monion, this is a projective bundle, right? Because if I fix a two-dimensional vector space, then the space of, so then what's the fiber over that? Well, it's just the space of choices of F0 and F1, both of which are in V, right? And that's sort of, that's, that's, that's nothing more than the, well, that's the projective bundle associated to V plus V, right? Because I'm taking two, I'm taking two polynomials in V. And now the claim is that the, that this, that this, this blow up somehow results in the intersection on the previous slide. So what happens now is that, so what you see now is that the ramification conditions and the incidence conditions, well, the ramification can just somehow live more naturally on this gross monion. And so the condition that you're, you're, now your linear system is ramified at some point that corresponds to some, the Schubert-Sekel sigma one. Okay, so that condition when you pull it back to the, the, is, is, is on, on the gross monion sigma one to the G, and then I pull that, pull that back to my collineation space. And then the end incidence conditions, those are again linear conditions, but now I have a projective bundle as opposed to a projective space. So here now I get the, the nth power of the relative hyperplane class. Okay, so in the end, when you multiply these things, well, you can think of these, you can think of this intersection as being on the collineation space, but then when I push forward, I can apply the projection formula and get an intersection number just on the gross monion. So I have sigma one to the G coming from the ramification conditions, and then I have to push forward of some power of the, the relative hyperplane class of my projective bundle. And that's precisely the segregate class of the, well, it's, it's up to some shift in degree. It's the segregate class of that bundle, which in this case is V plus V. So what I get in the end is I get the n minus third segregate class of the bundle V plus V because it's a P three bundle. And that's the sum of sigma i sigma j. And okay, so somehow the content of this is that this intersection really now does is somehow proper in that it all of the points in the intersection are actually points of geometric meaning. So this is some non-trivial step that you have to do, but this is where this formula comes from. Okay, so the questions. All right, so let me just say a couple words about some further directions. So, so again, this is I've spent most of this time talking about covers of P one. And so there's a variant of this problem where you start imposing arbitrary ramification profiles for this map. So this is a, this is something that I've been working out with Alessio cello and they should hopefully be posted soon. And of course, the the next step is also to try to push this to projective space to higher dimensional projected spaces. So what we can show at the moment is that in fact, the geometric Tevelev degree equals the virtual Tevelev degree, again, when the degree is sufficiently large. And I've put this number in red because it's not clear to me if this bar is sharp. But so I've used this the notation call to D plus one and somehow some kind of project in some kind of suggestive way. And in some sense, you can reduce in the same way to some kind of question about intersection theory on some space of modular space, complete collineations of higher rank, whatever that means. But there's some kind of new ideas that remain to be worked out. Okay, so I did promise I would say a little bit about the question of intermortivity. So let me just just say, I think I think I only have one slide on this. Okay, several sides, but maybe I only I only I won't even get same thing with the last one. But the observation here in this, there's a result that I've mentioned is again, that the the virtual Tevelev and geometric Tevelev degrees are equal to each other. In some in some range where D is large, but not when D is small. Okay, and then you can interpret this in terms of the modular space of stable maps, but kind of not say much about this. So the expectation is that this is actually a general phenomenon. Right, so the expectation is that when you when you take any target variety, that this, we've already seen that the virtual Tevelev and geometric Tevelev might not agree, right, they already disagree for in some range for for P1. But the conjecture is that well, if you take some sufficiently positive variety, or if you're in some sufficiently positive situation, and it's not so clear what hypothesis we want, but for example, you might ask for X to be funnel. And if the degree of the map you're considering is sufficiently large. So in the most general situation, what you want to say is that the integral of beta against the cannot the anti canonical is sufficiently large. Then the virtual Tevelev numbers are but not necessarily when the degree is small. Okay, so the most recent result, which will hopefully hit the archive in a couple days, is that we have managed to check this with so Rahul and I have managed to check this for some examples, which do come up in the paper with with Anders and Rahul. So it is true for all flag varieties. And it is true for all hypersurfaces of in a very small degree. So so degree kind of no more than the square root of the dimension for X. Of course, the expectation is that this is true for all fun varieties, which would be less than less than R, but somehow they the methods do not achieve this this bound. Okay, so again, we already know that these virtual Tevelev numbers are not always enumerative, sometimes that sometimes they give you sometimes they give you numbers for which you kind of have no geometric interpretation. But the but the theorem is that at least in these cases and in conjecturally more cases, that this is only something that happens in low degree. So if your curve class is sufficiently positive, then actually, this virtual answer that you can compute through through German wooden theory is actually giving the correct geometric count. Okay, so I think I'm already over time, so I'll stop there. Thanks.