Aaron Landesman: Malle's conjecture over function fields (NTWS 290)

Welcome back everyone. For today's seminar, I'll be discussing one of the fundamental questions in arithmetic statistics, Malle's conjecture. And as a precursor to Malle's conjecture, I'd like to discuss the inverse Galois problem. So this is a age-old question in number theory, which asks, if we're given a finite group, is there a Galois G extension of the rational numbers? So here's the setup. Let G Let G be a finite group. And for the purpose of this talk, I'll use Q to be either the rational numbers or FQT. So often the more classical number setting is the rational numbers, but in this talk I'll often be talking about function fields, so I take poly rational functions over a finite field FQ, Q is a prime power. And then the question is, is there a Galois G extension of Q? So classically again this is asked over the rational numbers. And Malle's conjecture is a refined quantitative version of this question. So it asks not if there's a single extension, but how many extensions there are as the discriminant grows. So let me start not with the most general version of Malle's conjecture, but just with a special case. Special case of Malle's conjecture. And so this says, here's the setup. We'll again take, as always, this Q will be either the rational numbers or FQT. And as I say, it predicts the asymptotic growth of the number of G extensions. So, I'll just focus on the symmetric group for the special case. So, it says there exists some constant C1 and C2 constants positive constants. These will depend on D that will appear. Uh so, with the property that it's about counting the number if we want to count the number of degree D extensions, degree D extensions K over Q with um with some properties. So, we want the number which have the Galois the Galois This might not be Galois, but the Galois closure K tilde over Q is SD. So, this is SD is the symmetric group on the elements. Uh K tilde is the Galois closure of K. of K over Q. And at with as stated, there could be infinitely many of these. So, we want to bound the discriminant. So, I'll take the discriminant bounded. The discriminant of K over Q is at most X. So, X is some varying parameter. So, we want to count the growth of the number of degree D extensions. And then the conjecture is this grows linearly in X. So, the first constant is bounds this below, and the second constant bounds it above. Right? So, the key point is that I Let me just write this. I mean, the number grows linearly in X. It's about the growth rate of the number of degree D extensions. The number grows linearly in X. as the discriminant grows. So, this is as I said, just a special case of Malle's conjecture, and I'd like to describe a more a more general version of this, where we're not just talking about degree D extensions, but G extensions for arbitrary groups G. So, here's a sort of more general version. You could ask even more general versions than this, but let me just state this version. So, this says I guess you want to see two on the right-hand side. C1 and C2, thank you. Thanks very much. So, this more general version says, and again, these constants will depend on the the D here. So, this more general version says there exists C1 and C2 um So, maybe I'll say, yeah. So, there given given a group G given a permutation group G So, the probably the most important case is the group acts on itself, which will be corresponding to Galois extensions, but I'm setting up this more general version that basically should think either we would be counting Galois extensions, but in this first case I wrote here, they're not exactly Galois, so somehow this permutation group allows this more general flexibility, and Q is either the rational numbers or FQT. Then, there exist these constants C1 and C2 depending on G constants, and these are unspecified and specified constants by Malle. A A of G and what I'll call VM of GQ. So, these are two constants um so that now we're going to do the analogous counting problem but for the more general type of G extensions. So if we count the number of K over Q G extensions with a discriminant at most X. So if we count these number of G extensions uh it should be growing at a predicted rate and the rate is the first constant times X to the one over the A of G. This A predicted by Malle. Um times log X to the B of G. Be the Malle predicted G BG value minus one. And so that's the lower bound and the upper bound is of the same flavor C2 X to the one over A of G. log X to the what to the BM of G Q minus one. And so again just emphasize this these A and B are specific constants but I they're kind of complicated formulas so I didn't decide not to write them down but just as an example um if you take the permutation group the symmetric group acting on the set one through D then count G extensions and the G extensions exactly correspond to these degree D extensions with Galois closure SD um so we obtain the we obtain the prior special case. The special case the prior special case. And in that case um so there with So in these constants A of G, so this SD is the group G. A of G is one and BM of G turns out to be one. And that will tell you that the asymptotic So here in general the asymptotic is X to the one over A of G log X to the BM of G Q minus one. This implies the asymptotic is this uh X to the one over one times log X to the one minus one, which is which is X. So that's why here there's this X here. Um because it's log X to the one over one times log X to the one minus one. So this is a more general setting. And the the emphasis the formula is a little complicated, but the point is we're trying to count the number of G Galois extensions, so it's a sort of souped-up version of the inverse Galois problem. So next I'll discuss the case I'll discuss what cases are known. So there's a sort of over the rational numbers there's sort of a smorgasbord of known cases. Some things are known, many things aren't known. So first um this is known if G is an abelian group. It roughly boils down to class field theory. Uh another important known case is when G is the symmetric group on three elements. So this S3 case is due to Davenport and Howegron. And then uh sure after that, so that's around 1970s, but in the 2000s Bhargava dealt with the case S4 and S5. So, that was already kind of a big breakthrough. These were two animals papers and it was part of his Fields Medal citation. Uh where he was able to do these two cases, but if you want to understand say S6 or S7 or bigger symmetric groups in the version of that special case of Malle's conjecture, we we don't know how to do that. So, there are many some things are known, but many things are open. Uh another example that we can do is we can do the dihedral group of order six. So, that's that's for me I'm using this dihedral group of order six, so that's the same as S3. We did that. Dihedral group of order eight uh we can do, but if you go to the dihedral group of order 10, that's already trying to count D10 extensions is a wide open question. D12 we can actually do as of very recently, but then once you get to higher dihedral groups, we can't. We don't know how to count the number of dihedral group extensions. And another one we don't know is the alternating group on four elements. So, there many There many cases we know, but many cases we don't know. Uh that's over the rational numbers, of course. So, me saying this is will be leading up to a result about working over function fields FQT. So, in the next uh little segment of the talk, I want to describe what's known about Malle's conjecture uh and what what the issues are and let me and then I'll tell you what the main thing we can prove over function fields is. And so, let me just Here's a restatement of Malle's conjecture. Again, given a permutation group G, there is some bounds we can predict Malle has a prediction for the growth rate of the number of G extensions. So, my favorite One of my favorite facts about Malle's conjecture is that it's One of the things that makes it very interesting is that it's not correct in general. And trying to find the correct value of this number BM has undergone a lot of work. So, there's a a theorem by Cluners, where he says that in fact, this predicted value of BM is not correct in general. So, he says if G is you if you imagine considering extensions which are Z mod 2 extensions of Q followed by Z mod 3 extensions of that, so if G and the group ends up being this for a certain permutation action, Z mod 3 squared, the semi-direct product, the wreath product, then Malle's conjecture predicts I didn't tell you the formula, but there's it there's a specific value of this B GQ uh for the rational numbers, I'll take the rational numbers here, and it predicts it to be one. So, there So, So, it predicts that there are about X to the 1/2, and it would be log X to the 1 minus 1. So, they predict there are what many. But Cluners But Cluners showed that there are X to the 1/2 log X many. But in fact, there are about X to the 1/2 log X many. So, the correct value this exponent should have been two, cuz when you take 2 minus 1, you get the first power of log X. So, so Malle did not have the correct value of BM in general. have the correct exponent of log. So, if we're trying to predict the number of extensions, it it the specific value that Malle predicted that I didn't tell you is not always correct. So, given that, it's natural to try to find the correct value. And so Turkel uh went on to try to make a prediction. So, Turkel's conjecture is basically a modification of Malo where he changes the value of this BM that I didn't tell you. Again, it's a little complicated, but um so, Turkel's conjecture but there's some specific value is the point. So, he proposes a value of BT proposes BT in place of BM. BT of G of G Q in place of BM for the exponent of log. And so, he has this this it's exactly the same conjecture, but I just wrote BT here for the number of G extensions instead of what previously BM was. So, it's a different value. And Turkel was motivated by thinking about the function field case. So, so it's natural to ask, Turkel proposed this new version. Malo's conjecture is not correct in general. By the way, the only issue with Malo's conjecture is this BM at least as far as we believe. So, everyone mostly believes AG is this value of A is correct, the power of X, but the B was under some question. And so, Turkel proposed this new value. And so, given that Turkel proposed this new value, it's natural to ask if that new proposal is correct or could it be wrong. And so, uh Julia Wang was thinking about this, and she found that it's also not correct in general. So, she showed that there are similar counterexamples. So, I'll just say Turkel Turkel's conjecture is not correct over Q. is not is also not correct over Q. And the only issue again is just this value of BT is is the issue. It's also not always correct over Q. And the But um she doesn't find any counter examples over function fields. It could be true over function fields. It can be true over FQT. And as I was saying, Terkeli was motivated by the function field case. And so what we the main theorem I want to talk about today is this joint work with Ishan Levy. And so we show that in fact Terkeli's conjecture is correct at least when Q over FQT when Q is large enough. So that's So let me try to state that precisely. So we say for every group G every group G So again, we need Q to be large enough for our theorem to work. So there exists some constant There is a constant C depending on CG depending on G There is a constant CG depending on G so that for for Q bigger than this constant So Q will be the size of the finite field. And if we also take it prime, we'll take Q to be have characteristic prime to the order of G. If Q is prime to the size of G, then Terkeli's conjecture is correct over FQT. Uh to spell this out, what that means is that we can calculate we can compute the number If if Terkeli's conjecture may have lost you a bit in all these constants, the point is that we can compute the number of the asymptotic number of G extensions. Compute the asymptotic number of G extensions as the discriminant grows. of G extensions for any group G. But we for any group G, but we need the size of the finite field to be big relative to the group G. And maybe I'll just mention also, you might wonder what Malles conjecture uh has to do with this. Like, what was the BM? So, it turns out that Malles prediction actually turns out in this setting to be counting the number of geometrically connected extensions, the number of extensions over FQ that remain connected over FQ bar. Um so, maybe I'll just pause here to ask, are there any questions so far? We'll we'll soon go on to the talk the part of the talk which we're planning to use some topology to understand this. So, we'll relate counting the this question to counting FQ points on certain Hurwitz spaces, certain varieties, and then we'll try to understand their cohomology to count the FQ points. A quick question. Before going to that, any questions? Yeah, so is there an analogue of a discriminant to uh to order X um or the count over uh function fields? Yeah, I mean, so you should think about Exactly. So, one way this You're asking Yeah, so discriminant makes sense over the rational numbers it sounds like you're happy, but also over function fields what that what one way you can define it is the you take the sheaf the the degree of the sheaf of relative differentials or Q to that Q to the degree It measures the ramification of the extension, basically. Just like the discriminant. Yeah, so it makes sense over any global field. the discriminant. And even more generally, but in this setting we'll Yeah, so here I'm taking Yeah. You you can make sense of it as counting the number of counting the ramification degree basically. More questions? Okay, so now let me press on and as I was saying in order to answer this question, we will pass to certain spaces over the complex numbers which will also be defined over FQ and we're and basically these spaces will be spaces of covers of curves and counting these G extensions will end up being count corresponding to counting FQ points on these spaces. So let me define these Hurwitz spaces that we'll be looking at. So this is just over the complex numbers for now. So I'll sorry. So let G be a finite group. And C and G will be a union of conjugacy classes. And now I'm going to define this Hurwitz space Hur CN. So it'll measure G extensions with monodromy line with inertia line in C. So this will be the set of extensions and since I'm working over the complex numbers, this D will just be a disk. But you should think that D is kind of like A1. This is sort of like A1. A1 over C. Uh so D will be a disk and will and X to D will be a G cover. A G We can take G Hurwitz covers with N branch points. And we'll also and the inertia lies in C. So the monodromy at these branch points lies in C. Inertia in C. And um an important thing for this setup is that we'll also have a trivialization over a boundary point of the disk. A trivialization of this cover. So we'll identify it with a group G. Trivi- trivialization. zation over a boundary point of the disk. So that's this Hurwitz space. And for the purpose of this talk, there'll be a couple variants. So one thing is that probably the most important variant is that inside here I'll have something called C Hur. So this this is the union of components. So this parametrizes connected covers. Connected such covers. So I didn't assume X is connected. But you can just pass to a union of components where they are connected. And this is not going to be very important for the talk, but there are some G actions. This group G acts on this trivialization. And so you can quotient by G. And so we can also make C Hur mod G. This will only come up at the very end of the talk, but I'll just include it here for now. So you can quotient by this G action. Or C mod G. So these these correspond these are connected covers still. And this quotient by G just corresponds to forgetting the trivialization. So these parametrize over the boundary. So these parametrize just G covers with N branch points with inertia in C, and they don't have a trivialization. So these spaces exist over the complex numbers, but it turns out you can also define scheme structures on them and and you can make sense of them over FQ. So, I want to just let me try to give you a bit of a picture of how I think about these spaces. So, here are a few examples. First, let's study the case that C is the group G is is Z mod 2. So, G is Z mod 2 in the first example and C is just the element one, which indicates in this case that corresponds to degree two covers with and the branching is if it if it's branched, it's non-trivially branched. And so, this is my picture of a hyperelliptic curve, in other words, a degree two cover of the disk with four branch points. And so, um that's in her C4. So, the next example is a little more complicated than here G G is this group S3 in the second example and I'm going to take transpositions monodromy. So, that means that the cover is simply branched. And and here here in the picture I have three branch points, but only two come together at a time in the first the first two sheets one two come together, so I'm labeling it by one two, then one three and two three. And so, this would be in her C3 because there's three branch points. And the trivialization corresponds to labeling the covers one two three on the left. So, here's a the third in the third example, here somehow there would be this is not in her C3 and the reason is here there's somehow one two three monodromy. So, here I'm still taking C to be transpositions. C is still that same class of transpositions and this is not in her C3 because here there's a triple branch point and I only wanted simply branched covers. Um and uh the So, that's that's a non-example. And now for the last example, I'll have these covers, which is which only has monitor me 2 3 at these three points. And so it's a disconnected cover because the first sheet is never getting mixed up with the other sheets. And so that would be in her, but not in the C her, which is the connected covers. Um so, those are these Hurwitz spaces. And now I want to explain how these Hurwitz spaces are related to Malle's conjecture. In other words, to our proof of Malle's conjecture in this function field case for Q large enough. So, how do you connect it? Well, remember our theorem. Let me just go back. Right here. Remember the theorem is that for every group G, there's some for large enough to tell you this conjecture is correct, so we can count the number of G extensions. So going back, our goal is to count the number of extensions K over FQT. Uh we want to count this number where with some condition on the discriminant. The discriminant of K over FQ is sub X is at most X. Let's say let's just let's just sort of simplify this and count the number that are Q to the N. It In the function field case, X always has to be a power of Q to the N. So you can kind of try to count this number. This is roughly what Malle's conjecture is asking about. And uh by passing to some more global geometry, instead of counting the number of field extensions, you can count the number of rings of integers in the or specs of the rings of integers. In other words, we're trying to count the number of curves over A1 over FQ. of discriminant N of discriminant q to the n. And the correspondence between these is just if you have this cover x to a1, this cover x to a1, you can take the function fields. So here you send this to the function field k of x um over k of a1 fq. But the function field The point being that the function field of a1 fq is fqt. So counting these covers of curves is the same as counting these extensions of fields. And then the key point why this is related to Hurwitz spaces, this is not exactly true, but roughly this is counting the number of fq points on the space c Hur. So this c Hur is some variety, and if you allow arbitrary monodromy, then the fq points of this variety roughly count the number of these extensions of discriminant q to the n. So going back, uh if you go back to the definition of c Hur right here, it's it covers x to a1 of the complex numbers with which are g covers with n branch points. Uh but if you kind of put in fq there, it'll be covers of fq, and so these extensions over a1 fq are roughly counting the fq points of this Hurwitz space. So we've sort of rephrased this question, this arithmetic question about counting extensions, in terms of terms of a more geometric question about counting fq points on Hurwitz spaces. And now let me just Recall there's this relation between the fq points of any variety or and the and its cohomology, which is called the Grothendieck-Lefschetz trace formula. So what it says is that if x is smooth over fq um of dimension n, then it relates the fq points of x uh to the cohomology. So it's equal to the sum from I equals 0 to twice the dimension of -1 to the I times the trace of Frobenius arithmetic Frobenius on the cohomology. This is a top cohomology. Um so, the point being that it relates the FQ points to the cohomology. And so, hence, if you want to understand >> understand the number of FQ points on this C her, uh it suffices to understand the cohomology. So, so the conclusion we want to understand the cohomology of C her. Understand the cohomology of C her of G minus of the C her G minus identity M. And if we can understand those cohomology groups, it'll be enough to compute the FQ points here. And then that will let us compute the number of these extensions that we were interested in originally. So, now what this will let us do, this maneuver, is that we related what we wanted to understand to understanding certain cohomology groups. And hence, we can now the cohomology groups over FQ will be the same as those over the complex numbers. And so, it's going to be enough to understand a statement about these Hurwitz spaces over the complex numbers. So, now we've sort of transitioned to topology. And here's the main theorem that will let us understand these cohomology groups. So, um now we're this is just over the complex numbers. This is over the complex numbers. So, we'll let C in G be a conjugacy class. There's many more general versions, but let me just try to state the simplest version. So, C is a conjugacy class generating G, let's say. conjugacy class. Guess you don't need to generate in Q case. And then the statement is that there exists constants depending on uh I and J depending on C. So that the homology stabilizes. So so that if for N bigger than N big enough if you look at the homology of C her N even with integral coefficients is fine. This is the same as the homology of C her N plus 1. So this um Let me try to draw a picture to indicate what this means. So if we Here is my picture. I have N growing here and I growing here. And there's this line. This is the I I plus J line, so it's slope I. And the intercept is J. This is the I I I plus J line. I I plus J. N equals I I plus J. Then what it's saying is that if you're under this line, so here I'll put H I of C her N. And in the next slide it'll be H I of C her N plus 1. And what it's saying is that if you're under this line, the adjacent groups stabilize. So they're they're isomorphic. So there's to the right underneath this line, there's sort of isomorphisms going to the right. The homology stabilizes once you pass this line. So And it turns out that just this sort of modest control of the cohomology is enough and the point counts well enough using that trace formula in order to deduce this version of Malle's conjecture we wrote. I guess technically it's slope 1 over I, the way you've drawn >> Oh, yeah. Sure, sure. Okay, anyway. Yeah. Yeah, yeah. So, this yeah. Exactly. Thank you. Yeah, any more questions? Um let me So, let me just mention that there was some uh really amazing precursor work related to this, which I really one of my favorite my favorite paper as a grad student was this paper by Ellenberg, Venkatesh, Westerland. And they basically proved a slight weakening of this when the group is uh proved a version of this when the group is the dihedral group. Uh and they have a slight general slightly more general than the dihedral groups and C inside G is the conjugacy class of reflections. Is reflections in the dihedral group. And L was an odd prime. So, in that sort of special case they previously proved this this uh stabilization of homology and that was enough to deduce some consequences for a different set of conjectures in arithmetic statistics called the Cohen-Lenstra heuristics which is closely related to counting dihedral group extensions. So, this this is sort of enough to prove Malle's conjecture, but there is if we're here, there's a very natural next question, which is well, we've shown these homology groups stabilize, but you can ask what are the dimensions of these homology groups? That could give you maybe that could give you some finer information. And so we also can answer that question. So here's the here's the the sort of refined version where we compute the stable value. So it says just as before there exist these constants I and J depending on G, depending on C. The same ones as before. Um and it says that uh we compute the stable value of the homology. So we compute what these numbers are. So if you take Z is a component, there could be many components, but if you just choose one a component, a connected component we take a component, then we can compute the homology of each component. So the homology, let's say the rational homology in fact, you only need to invert the size of the group is it's very simple. It's Q if I is zero or one and zero otherwise. Uh if n is bigger than I plus J. So let me draw kind of draw the picture again. What does it look like if we if here n is this and I is in this direction, what we're computing is the numbers that appear here, and so here's that line of slope I or one over I and if we kind of put here the homology of Z, which is one of these components, we know how many components there are, but you have to so here we put HI of Z um so it'll the HI of Z looks like it will be Q in this range. So Z will be yeah, Q it'll be kind of Q's here. In the first two. And zero past it. So maybe yeah, maybe Maybe I'll let me call it this. So if there's yeah. So there'll be zero past here. There'll be Q's in the first two rows. So what this is saying So again, Z is in her end. So Z will be kind of in one of these rows, but as you're varying end the Z is varying. So maybe I should call this Z sub n. So the homology of each component Z H I of Z n goes in the nth slot. And it's Q in the first two degrees, but after that it becomes zero. So what it's saying is that actually the homology of these spaces are very simple. It's not some huge mystery. It's just some simple things in degree uh And maybe I should say here C again, C and G is the conjugacy class. The conjugacy class generating G. So it's very simple. Um and maybe let me just say a little a word or two about the proof. The proof uses some difficult homotopy theory ideas, but the the concept of where these Q's are coming from is pretty simple. So if we have this this Hurwitz space, which is a space of these branched covers with n branch points with monodromy in C. So here I was picturing that C was transpositions in S3. Then there's this branch locus map. And it just sends the It sends the Hurwitz space to comp n. This is the configuration space of n points in the complex numbers. And unordered And unordered points in the complex numbers. And it just sends the cover to its branch locus. So, in this picture there are three branch points, and then we take those three point unordered points in the complex numbers. And the the key thing that we prove, so we show that we show, what we really show is that HI of Z, this component is isomorphic to HI of Conf N. So, HI of Z with rational coefficients is HI of Conf N. Um and then this, so in other words, we have this comparison map taking the branch locus, and so the cohomology of Conf N will map to the cohomology of Hur N, and that creates some source of cohomology, and the statement is actually that's everything. So, all this at least stably. So, if N is bigger than I I plus J, stably all the cohomology comes from this simple map, and that's and then um the point is that it's sort of well known that the cohomology of this Conf N is what I said on the left. So, well known it is well known that the cohomology of Conf N that the stable homology of Conf N is this Q in degree 0 1. Conf N Q is Q if I is 0 or 1 and 0 otherwise, at least when let's say N is bigger than 2. So, that So, the idea is all the cohomology comes from Conf N, and of course the difficult part is to show why that's true. So, it's a little the I won't have time to go into this in the last 8 minutes of the 10 minutes or so of the talk, but instead uh I'd like to explain lastly, how this sort of as we saw the understanding of the showing that the homology stabilized gave us the smallest conjecture and so you could ask what does understanding the stable value give us. Okay, but before going on maybe I'll just pause for a few moments to ask if there any questions on this topology part of the talk or anything. All right, so now let me move to this concluding application that I wanted to talk about which is called Bargava's conjecture and it tells us what is the benefit of what do we get by having computed the stable value? It gives us what application of that. So here I just wrote to remind you the main theorems I had been discussing the last 10 minutes or so. So the first one said we have these Hurwitz spaces these C Hur which are spaces of covers of the disk or covers of A1 with and n and C and n branch points and the first theorem said that the homology stabilized. And then the sex next the next theorem computed the stable value of the homology. It was just very simple Q in degree zero and one and zero otherwise for each component. And so um so to so to summarize this so far um we only used this first theorem. So maybe I'll call it theorem one. Theorem one knowing the stable value, theorem one which is knowing the stable knowing that the homology stabilized knowing the homology stabilized this gave us Mallus' conjecture. Gave us this version of Mallus' conjecture for Q large enough. For Q large enough and we you didn't need to know the stable value, it turns out. Um but so the question is what, as I was saying, what does knowing the stable value give us? Question, what does theorem two give us uh in in terms of arithmetic applications? And so um if you remember uh it it's Remember, Malle only predicted these constants C1 and C2. And basically it's it's going to be telling us about what these constants are. So just recall this original statement of Malle's conjecture in the simple case, we had G um G is the symmetric group acting on the set 1 through D. And so this G extension is K over Q, this is corresponding to just degree D extensions with Galois closure uh SD. Um and so the related question is that that is the subject of Bhargava's conjecture. What what are C1 and C2? What are these constants? What are C1 and C2? And so now let me tell you our result, which which sort of answers how many extensions there are in this sort of large Q regime. So here's the statement. Uh there exists some constant KD, a constant depending on D. depending on D so that if Q is bigger than this constant and we still have this hypothesis that the GCD of Q it will be prime to the order of the group, the group will be the symmetric group, so maybe I'll just write the order of SD, that's D factorial. Um then I'll state There's a more general version, but let Let just sort of state one that's that's a little simpler to state. So, let's look at these simply branched covers. Simply branched covers. Um uh extensions K over FQ. So, that this corresponds to the case that This is just to mention what simply branched means that the branching is as in these pictures I was drawing before. So, this corresponds to the case that G the group G is SD the symmetric group and C is transpositions. When we are studying these her C, it's got G cover SD covers of degree D with transposition monodromy. Um so, these number of these simply branched extensions of degree D And I'll also assume uh that we'll have simple branching at infinity. And then we have to compute those with bounded discriminant. So, discriminant of K over FQT is Q to the N. Uh if we compute this number So, this number ends up being 1 minus 1 over Q times times Q to the N plus little O of Q to the N. Right? So, maybe I should say here Q to the N Q to the N is X in what I was saying before. So, this is 1 minus 1 over Q times X plus O of X. So, what this is saying is that you should think of the constant This is saying that the constants should should be thought of as this 1 minus 1 over Q. This gives the constants in this version of Malle's conjecture. Um C1 equals C2 roughly is this constant. You should think about that. About it that way. And by the way, we have various generalizations, but I just wanted this turns out to be a particularly simple one to state. So, let me Let me try to indicate the proof just in the last couple minutes. The proof idea is just that um, basically the question is where is this 1 minus 1 over Q coming from? And the idea is that it's coming from that those two stable homology groups. So, let me just write this. So, we want to count the number of FQ points on the space parameterizing these extensions. So, the number of FQ points Z of FQ, where Z is a certain component of a Hurwitz space. So, where Z um, is a component of this this C Hurwitz space, C Hurwitz CN mod SD. This was this where you forget the trivialization, which is the is the unique component of this Hurwitz space. It turns out there's a unique component um, which parameterizes these component um, parameterizing with simple branching at infinity. And simple branching at infinity. And here the C is the transpositions in the symmetric group. In the SD. The symmetric group of order D acting on the elements. And now, uh, um, basically if you go back to that So, where is this 1 minus 1 over Q coming from? So, we saw that the homologies using what we explained before, that H1 H1 of Z H0 of Z is Q with rational coefficients. H1 of these is Q. is Q and HI of Z is zero for I bigger than um or is zero for I bigger than one. And so this corresponds to the If you kind of There's this Grothendieck trace Lefschetz trace formula that relates these homology groups to the point counts, and this kind of gives the one, and this gives the minus one over Q, and this gives There's no more stuff from this, and so the total number of FQ points Z of FQ ends up being roughly the one minus one over Q from this from these two contributions. Um and then it's roughly this times Q to the N, which was what the statement was. So the point is, if you want if by understanding these stable values, the H0 and the H1, that was enough to get a more refined count, not just the asymptotic count, but also understand the constants in the count. Um so then maybe I'll just conclude the talk by summarizing with a table of how these results fit into the bigger picture of various statements in arithmetic statistics. So there is Today we were talking about Malle's conjecture. That was sort of the main result today, and our theorems we computed the the stable value, we showed that we we showed that the homology stabilizes and we computed the stable value. And I also mentioned earlier this result of Ellenberg, Venkatesh, and Westerland as this case in the dihedral group. And so that was related to this other set of conjectures, Cohen-Lenstra, and sort of we've similarly computed the large Q version. We've We've Sorry, we've similarly uh computed the stable value of homology there, and there is previous work by Jeff After where he computed the the number of components, which was this uh Q which turns out to have a Q to infinity version. Enter Kelly, as I mentioned, uh computed the number of components. He can He He had some work relating to computing the H0, as I mentioned earlier, related his conjecture. And then there's sort of one more set of conjectures in arithmetic statistics, which follow similar lines, and there the sort of first case was has some work by Park Park and Wang myself, and work with Tony Feng and Eric Raines. In uh we understood the stable homology related to that in work with Jordan Ellenberg, and then most recently we computed the stable value of the homology as well. It with Rupesh Shankar. So, just to summarize, we were interested in understanding this conjecture by Malle, which is related to counting the number of G extensions. We translated that to a question in topology about counting covers of a disk and related um via counting FQ points on Hurwitz spaces, and finally we counted those FQ points by understanding the homology. Thanks very much.