 Great, thanks very much. And thanks for the invitation to speak here again. I have given variants of this talk a couple of times before. So if you are one of the people that have suffered through something like this before, I hereby swear this is the last time that I speak about this material. The paper associated to it is now on the archive. So if there's anything in here that sort of piques your interest, you can go and take a look. I want to begin, so this is going to be joint work with Julian Leitschek, a former postdoc here who's now in Bath, Roman Sarapin, who was an intern at IST for a while and Arne Schmetz in Lebanon. And I want to go back and talk about where this problem comes from before I pause and speak a little bit about some basic SIV problems. And then I'm going to return to the problem and motivate it a bit more and ultimately end up explaining some of the new conjectures that we've been developing. Sorry. Yeah, so this problem really goes back to this very influential AIM meeting in 2002. I think it's the only AIM meeting that took place over two weeks. It was about rational points. It had lots of motivating open problems. And in particular, there was this survey written by the late great Peter Sunuddin Dyer. He had a Hilbertian list of 23 open problems that he presented and this was problem 21 and sort of motivated by the Manin's conjecture, which is about counting rational points of bounded height. He phrased this question. So you suppose that you have a variety which has a map down to P1 whose fibers are all conics and you just try and count the number of points in P1 of bounded height, number of rational points in P1 of bounded height such that the fiber has rational points. And he asks, what can you say about this counting function? Can you actually prove anything about this counting function? And in fact, can you do something similar if instead of asking about fibers that are conics, you have fibers that are curves of genus one? So we're gonna address this question in this talk, but we're gonna tackle a slightly easier question, more much easier question, where we sort of switch out the phrase containing rational points in the fiber or containing periodic points in the fiber. And I think both questions are interesting, but Peter's question is a lot harder. So before getting onto that, but that's where we're going, but let me just go over a couple of SIV problems which are quite classical. So if we're given a polynomial with integer coefficients and two variables, one of the basic proto-SIV questions one could ask is, can you show that this polynomial takes infinitely many prime values? And can you give an estimate how often it takes a prime value? A variant of this question, which will be more relevant in this talk is to try and ask instead that your polynomial is only divisible by primes that lie in some predetermined set of rational primes. So the set of rational primes that crops up here are these kind of Frobenian sets of primes. So these are just primes such that the Frobenius-Condruxity class lies in some determined union of contrast to the classes of the Galois group under some extension. Anyway, some set of primes coming from algebraic number theory. And so can you, so yeah, what the, I'd say that traditionally SIV methods of focus very much on question one there's a lot of very substantial important work on that. But one of the goals of this talk is to give a bit of agency to this second question because that's what happens to come up in this application. So let me just state it again. If, can you get an asymptotic formula or estimates for the number of times, sorry, the number of pairs of integers M and N whose absolute value is bounded by T and for which this polynomial is only divisible by primes lying in this, this Frobenian set of primes. And actually the particular set of primes that we'll care about are just the set of un-ramified primes that split in some fixed extension number field. So let me give you a couple of examples. So here's a theorem that will permeate this talk actually. It's a result of Landau from 1908. I'm sure you must have, many of you have heard of this before and it pertains to the univariate polynomial f of x, y is equal to y and we're going to take K to be the Gaussians. So the primes, this is a Galois extension and the primes that split in this quadratic extension are just the primes congruent to one mod four. And so the second question on the previous slide is just asking about the number of integers which are only divisible by primes congruent to one mod four. And his sort of complex analytic argument tells you that, yeah, 0% of them are of this shape. And in fact, you get an asymptotic formula of the shape t divided by square root log t. Also, I should maybe also reflect on question one. So here question one is just the prime number theorem. Here's a second example, found in the Operator-Creebro book by Friedlander and the Vagnettes. So we're taking a more complicated polynomial this time. f of x, y is equal to x to the six plus y squared and we're going to take K to be a Galois cubic extension. And then we're asking about, how often you get these integers m and n such that m to the six plus n squared is less than or equal to t and such that this polynomial is only divisible by primes that split in this cyclic cubic extension. I think getting an asymptotic formula is still out of reach for this problem but we do have upper and lower bounds and that's what's carried out using the beta sieve in this book. And you get an asymptotic formula of the shape t to the two thirds divided by log t to the two thirds. So yeah, question two has some sort of resolution for this particular scenario. However, I should point out that the corresponding question one is completely wide open. So I guess from the point of view of sieves, I'm trying to promote the idea of question two and here I'm showing you that it should be easier. In general. I mean, it soon becomes difficult though. So I think this would be interesting and useful but I don't know how to do it. So can you handle this same problem for cubic extensions which aren't cyclic? That's one question. A second question. So if you picked any binary quadratic form q for which k and sorry, for which little k and for which number fields big k and you handle question two for these kind of special polynomials which are q of x to the k comma y. Okay, so those are my sieve theory questions and I'd like to go on to say a bit about this question of local solubility and yeah, the goal is basically if you're given a family of diaphanetone equations whatever that means, can we say something about the probability that an element of that family is soluble of every piadic field? Okay, so this is not an entirely new subject. I want to motivate it by this, this I think very attractive example. So you take the family of diagonal conics and so they're parametrized by these three coefficients a x squared plus b y squared, c z squared equal to zero and we just take all of these diagonal conics we order them by height so a b and c run over integers. So we have a very nice criterion here for when this conic has a point everywhere locally and in fact, actually here because of the Hassell principle that's equivalent to having a rational point and that condition for piadic solubility at least at an odd prime is just that any of the primes, for any of the primes dividing A, the corresponding genre symbol minus b c over p should be one and similarly for the other two coefficients. So yeah, it's a very simple criterion and you can use that to get an upper bound for this density of conics which have points everywhere locally and this was done by Seher in 1990 and so he counted the number of primitive triple vectors, primitive triples of integers, a b c which are bounded in modulus by b and have this property that the corresponding conic has piadic points everywhere and so he was able to show that you, that actually 0% of these are everywhere locally soluble and he was able to make that 0% very precise in the sense that it grows at worst like b cube divided by log b to the three over two. Okay, so he proved this using the large sieve 1990 and in fact, you get a similar result if instead of just fixing sorry, just studying diagonal conics if you looked at the full family of conics with five coefficients, you get the same kind of result but with a different power of log b but still a negative power of log b. However, it becomes different if you look at instead of the family of plain conics you look at the family of plain cubics, you get different behavior. So this is now you have 10 coefficients defining your ternary cubic form and you order them all by height and then recently, Bargava, Cremona and Fisher were able to show that a positive proportion of these plain cubics defined over the rational numbers have local points everywhere and in fact, they were able to pin down that positive proportion. So I should say actually that the fact that a positive proportion of these plain cubics are everywhere locally soluble goes back to earlier work of Poonan and Voloch in 2002 as part of a more general discussion of hypersurfaces but the main thrust of this paper is to really work out what that constant is and it happens to have this oiler product so it's kind of nice that you get the same oiler factor for every prime peak and essentially what this is saying is that 97% of plain cubics when ordered by height are everywhere locally soluble. So what's, it's kind of interesting what's going on here, why are we getting a positive proportion here and yet for the family of all conics we're getting 0%. So this talk is really going to be driven by families of equations over the projective line. So it's very interesting to ask about high dimensional bases and I will touch on that but most of the motivation is around families of diaphanetone equations over P1. So we saw, we've not seen any examples of that before so in SES result everything was parameterized by P2 and in the bark of a Cromona Fisher paper it was parameterized by P9. So this is an example of I guess a threefold which naturally has a vibration over P1 and I'm just calling these Landau threefolds for the purpose of this talk and you can sort of see that they have the same sort of structure as the sort of basic problem that was considered by Landau. So I'll return to this example. Let me introduce a tiny bit of notation. So in this talk rather than writing all of these things out every time by using the word standard vibration I'm just going to mean a dominant morphism from X to P1 and I'm going to assume that X is smooth, it's projective, it's geometrically irreducible and it's a variety defined over the rational numbers and I'm always going to assume that the fiber above a generic point is absolutely irreducible that's going to be my setup throughout the talk and for these the counting function that I'm going to be interested in is this. So we are counting points in the projective line rational points in the projective line which have height at most B and so the height here is just the naive standard height on P1 where you just take the maximum of the components after you've taken primitive pairs of integers as represented in coordinates and the key thing is that we want only those rational points which lie in the image of the idyllic points of X. So that's completely equivalent to just asking about the rational points in P1 above for which the fiber above them has points everywhere locally. So it's exactly that kind of counting function that Swinnipe and Dyer was discussing at the start of the talk except that we are asking only about the local solubility everywhere rather than global solubility. Good, so we could think about this counting function again in the context of these landale three folds. So we know that S for example so S and T are co-prime integers in that example at the top of the screen and we require S to be a sum of two squares. So any odd primes dividing S have to be congruent to one mod four and similarly for T. So we essentially have like an application of landales result will give us that for this particular family, this counting function should grow like B squared divided by log B. That's the example that I want to just mention here. And sort of motivated by like a Ceres example and these kind of landale type constructions. There was a lovely paper in 2016 by Dan Loughran and Arnais Schmetz where they made this conjecture about what should happen in general. So we're going to be working with standard vibrations again. It kind of makes sense to assume that the total space has points everywhere locally otherwise this question is not going to be very interesting. And there's this kind of technical assumption which asks that if you pick any of these points in P one and look at the fiber above it among the components that exist in that fiber there should be at least one which has multiplicity one at least one irreducible component of multiplicity one. So that's some sort of technical assumption. And if you have these two assumptions then this counting function should grow like B squared. So B squared is like the total number of rational points in the projected line of height B but it should die off by a log B to this power of this power of capital Delta. So let me define capital Delta quickly because it's going to be kind of important. Oh yeah, before doing that I should say that they give evidence for this so they prove an upper bound which matches this prediction using the large sieve. And there are some lower bounds in the literature but I have to say I think that's a much harder problem and it's a little bit more patchy. So what's this Delta? So it turns out it's a rational number and this is how you construct it. So you pick a point in P one, a closed point in P one and you look at the set so you look at the fiber above this closed point and you look at all of the sort of irreducible component absolutely irreducible components of this fiber which have multiplicity one. Okay, we know that there's at least one of them by this assumption in the conjecture. And then you pick a finite group gamma through which the action of the absolute Galois group on these irreducible components factors and you define this rational number. So it's just the proportion of elements of this subgroup with the Galois group which act with a fixed point on this set of irreducible components. Okay, and with that to hand, you can now define Delta. So Delta is just the sum of like one minus Delta and one minus little Delta as you go over all of the closed points of P one. So maybe a couple of comments here. So firstly, if you have one of these standard vibrations so one of the assumptions was that the generic fiber is irreducible. So that means that there's only going to be finitely many D for which the fiber is not irreducible. And as soon as the fiber is irreducible that means that any element of the absolute Galois group is going to act with a fixed point on this set of fibers. There's only one irreducible component in the fiber. So for all but finitely many closed points D this little Delta is equal to one which means that this sum over Ds over closed points D in the definition of capital Delta is really just a finite sum. Maybe I could give an example of its calculation. So let's go back to this Landau example. So if you, so that there are, the fiber is a, is a quadric surface, is a quadratic surface. It's, you know, for most, so it's only going to be reducible when S or T vanish. So when S and T are non-zero then you, the fiber is irreducible and so this little Delta is equal to one. If S is zero, so that corresponds to the closed point zero. So if S is zero, then, you know, the, this fiber is, is, is, you get these two components which are conjugated under the Gaussians and quadratic extension. And so therefore you really only have to look at, yeah, there's really just this C, Z mod two Z which acts on this, on these fibers and only one of them is going to fix the components. So this little Delta is equal to a half. So it's just the identity which is going to fix the, fix the component. And so it's the same for the closed point at infinity. And so that's telling you that Delta is equal to a half plus a half, which is one. This is, this is, this is consistent. If we just go back to this, this conjecture. So the conjecture is telling us that we should expect for this example B squared divided by log B and that's indeed what we, what we saw at the top of the screen. Okay, so I think this is a really nice conjecture. It's sort of generated a fair bit of research, you know, and anytime you have something like this you can immediately think about how you might, how you might extend it. And that's what I want to do today and I'll just mention a couple of ways that you can extend it. So firstly, you could try and make some predictions for the leading constant in this asymptotic formula. And that's now been done. This was done last year to do some beautiful work of Dan Locke from Nick Rome and Epithemius Sophos. So firstly, they make a conjecture for the leading constant and they back it up by upgrading and says up a bound to an actual asymptotic formula. So they calculate the leading constant here and it's kind of interesting as it has this structure as a product of local densities but there's these other two factors like the number two and the number pi over three, pi to the three over two, which take a bit of explaining and then that's what they end up doing in their paper. So I recommend taking a look. You could also ask about moving away from P1 as a base and maybe looking at some higher dimensional things. And I should say that in their paper, Locke and Schmetz actually do work with five vibrations over PN and you get the same sort of picture arising. But this was something that I spent a bit of time with when Roman Serpent was in my group and we looked at an example. So the basic question is, what can you say about other vibrations from X down to Y? But now you just assume that Y is some fan of variety. So it's not P1 anymore, but just some variety maybe for which you know how to prove Manning's conjecture. So you know how to count the number of rational points of bounded height on the variety. So can you say something about, the number of rational points on your variety, the bounded anti-economical height such that the fiber above the point is everywhere locally soluble in this vibration. It's a natural enough question. So if you kind of naively conjoined the Manning conjecture and this Locke and Schmetz conjecture, it would suggest that this counting function should grow like B times some powers of log B. And so you should get log B to the row minus one. So this is what you would see in the Manning conjecture where row is the rank of the Pickard group of Y. And Locke and Schmetz would be suggesting that it decays by this log B to the delta. So delta is this invariant that's associated to the structure of the vibration and the behavior of the fibers. So we were, sorry, yes, we explored this in a very particular case. So if we take a split quadric in P3, so this is a smooth Fanno variety. It has higher Pickard rank. So it has Pickard rank two. And we took a particular X. So this was the X that we took lies inside P3 cross P3 and it gives you this morphism from X to Y, the fibers of which all these diagonal quadric surfaces. And it turned out that you get something slightly weird happening. So as part of the summer project, we calculated that delta for this problem is two. But the counting function is bounded above what basically has order of magnitude B. So I, yeah, so just to point out that the expected exponent of log B according to that conjunction of conjectures on the previous slide is row minus one, minus delta. As we saw, delta is two. So we were expecting to see like B divided by log B for this counting function. So that's kind of slightly surprising to us. We've looked at it a little bit and it seems like there might be some thin sets inside Y, thin sets of rational points inside the rational points on Y, which if you remove them, you would get an asymptotic formula, which is a little bit more in keeping with the expectation we had on the previous slide. So I think there's some forthcoming work in the pipeline here of Cameron Wilson, a PhD student at Zimios Sophos, who's been studying this. And I think that's the sort of co-thin set that you need to take to make these work. Okay, well, that was two ways in which you could think about generalizing this conjecture or extending the conjecture. The one that I want to focus on, actually, and the one that this previous on the archive is about is this technical condition that they had to think injecture. So they had this condition that above every closed point, there should exist at least one irreducible component, which has multiplicity one. So what happens if you drop that condition? Yeah, so that's what I've written there. Well, it turns out that it seems to have a very strong effect. A very strong effect indeed. And in some sense, our paper is trying to make more quantitative, an observation of Koryo Telen, Skorobogatov, and Swinness and Dyer in the late 90s. So they played around with these situations where you have vibrations with multiple fibers and they wrote down a number of examples. This is one of them. So you take basically a standard vibration that is built from taking a proper model of this affine variety. So T is the variable for the affine line. And so C and D and F, they're polynomials with coefficients in the rational numbers. And you can see that for fixed T, this is an intersection of two quadrics. And if you sort of think of it projectively, it's an intersection of two quadrics in P3. So it's a genus one curve. So this is a vibration in whose fibers are all genus one curves. So it's a bit more complicated than the examples that we were looking at before where we had conic vibrations or quadric surface vibrations. We have higher genus curve vibrations. What they actually proved was, well, two things. So firstly, they proved geometrically that you get double fibers occurring above the roots of F inside X. So this is an explicit example where double fibers actually occur. And moreover, so what they actually proved was that there are only finitely many integer points inside P1 in this situation that, sorry, in the situation that the degree of this polynomial F is at least six. Then they showed that basically the rational points upstairs, the image of the rational points upstairs lie in, you know, finitely many points downstairs. So that is, well, sorry, yes. And I should point out that in fact, something stronger follows from their argument. This was something observed by Loughran and Livian Madison, was in fact, even this local solubility counting function is finite. So whatever the value would be, there's only finitely many elements of P1 which lie in the image of idyllic points on the total space. Provided that the degree of F is at least six. I think there's some conditions on C, D and F as well. I think F needs to be even degree and square free and maybe co-prime to C minus D, something like that. So yeah, so that technical condition that they had in their conjecture, might have seemed somewhat strange, but if you remove it, you get drastically different behavior. And that's what we're seeing here. And yeah, that's what we wanted to do was to try and investigate this a bit further and put it on a more quantitative footing. So maybe I should, I mean, one of the things that we struggled with in this, I mean, it's always nice to have sufficiently explicit examples so that you can kind of see what's going on. Like in the Landau example, we could very easily see the kind of multiplicative constraints that were placed on S and T in order that the fiber is everywhere locally soluble. And it would be nice to have some examples in this setting too, but it turns out that there aren't the same kind of conic bundle type examples in this world. So there's a famous theorem of Grave Harrison Starr, which implies here that as soon as you have double fibers or multiple fibers, the generic fiber cannot be rationally connected. So in particular, it cannot be rational. So in particular, it cannot be a conic or a quadric. So they are necessarily more complicated like this. So yeah, so having these fibers that are higher genus curves is the norm in this world. And it can be a little harder to see what's going on explicitly. So here we closely follow some ideas of some very influential ideas of rhetoric campana. It was a geometry, but explored this kind of questionnaire, ethnically as well. So if you have a standard vibration, you can associate an overfold to it. And the way you do that, so an overfold is just a pair of a curve and a Q divisor. But here we're having a curve is just P one. And we build this Q divisor in a specific way. So for each closed point in P one, we just write down M, M sub D. So it's a positive integer. And it's just gonna be the minimum multiplicity that happens in the fibers. So above D, you've got a bunch of components, irreducible components. Some of them are gonna be, sometimes there's not gonna be any multiple fibers in which case you take MD to be one. Sometimes there are gonna be multiple fibers. And you just take MD to be the minimum multiplicity that occurs in these fibers. And then you just define this, it's like strange looking Q divisor, D sub pi by taking a sum over these closed points of one minus one over M as a coefficient of that divisor, of that closed point. So again, generically these things that the fibers are absolutely irreducible. So generically that means MD is gonna be one. So again, this looks like it's an infinite sum but really it's just a finite sum. Good. And then the basically key idea, which again goes back to Campana really is this idea that we can interpret this local counting function, sorry, local solubility counting function as a problem about counting Campana points on this orbit fold. Okay, so that the topic, so independently of this story, the topic of counting Campana points on orbit folds has sort of received increased interest and attention over the last few years. It's sort of a world where it kind of lies in between the world of rational points and the world of integral points, which I think everyone agrees are too difficult, but Campana points are still expected to exhibit the same sorts of things, properties and behaviors that you would see in the world of rational points like a man in type conjecture, some kind of like a brown man in obstructions and all of this stuff. And in fact, there's a nice paper by Piro Pan, Smith's Tenemoto and Arie Albrado where they phrase or they articulate what a version of the man in conjecture would look like for counting these Campana points of bounded height for much more general things than P1 and this divisor really just for, in a much more general setting. So anyway, there is that story, which I think is important, but what's gonna be key here is that, yes, we certainly need to be counting Campana points, but in order to really detect this local solubility business, we are also having to throw in some extra conditions, some extra multiplicative constraints, which are gonna come from some kind of application of the Chebotarov theorem. So with that in mind, so these are associated to any of these vibrations, we can associate this all before base. And then we have this, this is the first conjecture. So if you assume that the, that this divisor written there is ample, then this counting function should grow at most like B to the two minus the degree of D plus epsilon. So actually the condition that minus KP1 plus D is ample is equivalent to two minus the degree of D being positive. So it's equivalent to that exponent being positive. So this kind of general topic actually goes back like this sort of bounding points on, campana points on varieties goes back to a beautiful talk by Bjorn Poonan in 2006, which I remember seeing at the, there was an MSRI program. And this pertains to a particular base. So you take the protective line and you just take three, three divisors above which you have some higher multiplicity. So you take a half times the close point zero, half times close point one and half times the point of infinity. And when you unravel the definition of, campana points, what is a campana point? What's a rational point in this setting? It just turns into this very charming and concrete problem, which is asking about counting positive integers, A, B, C in some range. They have to be square full. So if, you know, the prime divides A and P squared should also divide A. And they satisfy this linear equation, A plus B is equal to C. So Bjorn was popularizing this back in 2006. And I think we still don't know how to solve it, sadly. In this case, the sort of prediction for the exponent is two minus the degree of this divisor, which is three over two. So the expected exponent is a half. So we're really kind of expecting that this counting function grows like B to the half. You could think of this different, I mean, square full numbers have the same approximate density as squares. And we know how to handle this problem for squares. It's just counting Pythagorean triples. So certainly B to the half is a reasonable thing to believe in this setting. I think the current record is still B to the three fifths plus epsilon. So you're not even able to handle this, this very simple looking problem. However, what we'll see is that it is much easier to handle problems when you have just two divisors, sorry, yeah, two close points above which there's some higher multiplicity. This problem of three, three multiple points is much more challenging. Yeah, so anyway, I want to try and, you know, the motivation in this conjecture, well, the motivation next is to try and make this conjecture a bit more explicit, you know, is this epsilon acuding some explicit power of a logarithm and what is it? So the first ingredient that we had to sort of develop in this setting was a sort of more general criterion for testing local solubility. So if we pick a point in P1, a rational point in P1, we take a sufficiently large prime P where sufficiently large is just relative to the data of this vibration. And we suppose that, you know, the fiber above that point has peatic points. Okay, so what can we say then about X is there's some condition on X that we can write down? Well, this is the condition that comes out of the work. So if you pick any close point above P1 and, you know, a close point in P1 is basically the same thing as a binary form, H. So this close point we're thinking of as a zero set of a binary form. Then there are two possibilities. So either the peatic valuation of H, this binary form evaluated at this point, X is bigger than MD or it's equal to MD, but then you get this extra information that the Frobenius element at P has to fix one of these components in the fiber. So this is a slightly different kind of time. This is a slightly different set to the one I defined before. It's very similar, but we are, you know, above this close point D, we're looking at all of the geometrically irreducible components, but we're only gonna look at those which actually have multiplicity equal to this minimal multiplicity in M sub D. Tim, there's a question in the chat. Sean, maybe you wanna unmute and just ask away directly. Yeah, on the previous slide, could you show the previous slide? Yeah, so, I mean, you mentioned that you were, are you counting a container points or this N log by B, what does it count? Yeah, well, that's literally the definition I gave before, which is counting points in P1, which lie in the image of adelic points. So there are no questions, there's no question of container points at that point. Not, there isn't a question of container points at that point, but it is gonna follow from the sparsity criterion that I was just describing that you are basically reduced to counting container points. On the top, on the total space? On P1. On P1, on P1, okay, okay, thank you. Yeah, thank you. Thank you for the question. Yeah, so that's, well, that's kind of what you can see here. So I guess, you know, to illustrate this, if we were looking at the close point above, the close point at zero, so the point zero comma one, then the binary form is just X naught and you can sort of see that it's saying that, well, basically that, at least with respect to sufficiently large primes, X naught has to be M naught full. And, you know, if the prime divides X naught exactly to the Mth power, then you get this extra information that this Provenius element fixes one of the irreducible components in the fiber. Yeah, so it's basically telling you that you are reduced to a container point on P1, but you have this extra kind of Provenium condition. Good, so let me say a little bit about what happens for one fiber, one multiple fiber, because that's a little easier to see what's going on and we prove an upper bound. So we sort of refine that epsilon, you get B to the two minus this expected exponent. And then you get the same log B that occurs in this Locke and Schmades conjecture. So here, I'm really just taking the, assuming that there's one multiple fiber at the origin, sorry, at zero. So yeah, just rapidly, I guess I already kind of discussed this, so the sparsity criterion tells us that we're just counting now pairs of integers, which are co-prime, which lie in a box. And I get that X naught, so the first integer has to be M naught full. And for any of the other divisors that occur in this vibration, this binary form H evaluated this integer, X naught, X1 should lie in this multiplicative span of the primes that split in some appropriate extension associated to the divisor. So that's just what that notation means. I just mean that any prime dividing this binary form evaluated at X naught and X1 should be a Frobenian prime for some predetermined Frobenian set of primes. And yeah, it turns out that this is something that you can also use the large sieve to estimate. Good. So here's another example. So these are the health and surfaces, sort of nice family of surfaces. I think they were introduced in the late 19th century as I think they're classes of surfaces which admit vibrations to P1, the fibers of which are genus one curves. And I think they, so they're elliptic vibrations, but they're actually all rational. So they're kind of a very special class. And so I'm gonna assume that there's a triple fiber now above the point of a closed point at zero. And I'm also gonna assume that above the point one I, there's a reducible fiber, but I'm gonna assume that it's split by a cyclic cubic extension. So I just mean that yeah, it's not geometrically irreducible. And if you work over this cyclic cubic extension then all the components are defined over it. So again, if you go through that sparsity criterion, you get a similar kind of conclusion. So firstly, X naught, yeah, so now I, yeah, sorry, I think that I mentioned a sparsity criterion. I just mentioned a necessary condition, but you can also, so part of the work was to get a necessary and sufficient condition under suitable assumptions. And that takes a bit of work. It's a bit above my pay grade. It uses sort of log geometry and things like that. But you can use it here to actually determine that X naught is not just cubed full, but actually a cube and the binary form associated to this closed point is just the sum of two squares. So we're asking that X naught is a cube and that X naught squared plus X one squared lies in the multiplicative span of primes that split in this cyclic cubic extension. But that's a problem we saw before right at the start. So that was one of these sieve problems that I was trying to advertise. And this was sold for us by Feedlander and Appanards. And we get upper and lower bounds, which are matching. But yeah, it'd be nice to have more success stories with the sieve, which could then be used to investigate this particular problem. So I'm on my last slide. So I just mentioned it's very natural to try and think about this epsilon when you have more than one multiple fiber. So yeah, let me just go back. So when we had one multiple fiber, we were able to prove this upper bound and it was a kind of a predictable upper bound. It was kind of B to this exponent, which can be less than two. And we've got the familiar exponent of log B in the denominator, just as we saw in the Locker and Schmetz work. But yeah, but it turns out that if you have more than one fiber, that all goes out the window and we sort of get this kind of slightly surprising and crazy looking conjecture. So we are gonna assume again, that we've got the standard vibration, that the relevant divisor example, obviously it makes sense to assume that the total space is everywhere looking soluble. I haven't defined all of these terms. They are, if this peaks your interest, you can look in the paper. But we basically construct a morphism or polymorphism and the conjecture is that this exists and it has the structure of a G-torser, the finite tail group scheme G. And actually this counting function should grow like it's the same thing on the top. So B to this exponent, two minus the degree of the divisor. But on the bottom, you get something slightly wild. You get like an exponent of log B which is a minimum of all of these different rational numbers, capital Delta, associated to different elements, different sort of twists, V in this parameterized by this H1. So yeah, these, if you, so the V's parameterized these twists of this morphism and the pi V that occurs in that exponent is a standard morphism you get by normalizing the pullback of this morphism pi along this twist. Okay, so I'm not gonna go into this in any detail whatsoever. I just wanted to make a couple of comments verbally. So if it turns out that if you are in the situation that there are no multiple fibers occurring, this immediately recovers the original Locke-Rensch-Maitz conjecture. If there's one multiple fiber occurring, then it turns out that this orbit fold is simply connected and so this H1 is trivial. And you actually just collapsed down to the exponent that we saw, sorry, on the previous slides, so just the Delta of pi. And we, I guess in the paper, we try and prove some evidence for this slightly weird looking thing and we prove an upper bound which matches this when you have two multiple fibers over zero and over infinity. And we also provided an example which shows that you definitely need to take something smaller than what the Locke-Rensch-Maitz analog would suggest. So you definitely need to take this, it seems like you need to take this minimum. So we proved and provided an example which shows that this is correct and different from the kind of naive analog of the Locke-Rensch-Maitz conjecture would suggest. Good, thanks very much for your attention.