 So it's a long title and those of you who are familiar with the theory of quadratic forms might think it'll be a very short talk because there's not too much to say about this. I'll try to convince you that there is actually. So let me introduce some notation. First of all, so let S be a three by three symmetric matrix. So I'll write my ternary quadratic form in terms of S. So X will be a variable. These will usually be integers. And so I'll let S of X be the quadratic form associated to S. So this guy. But it's not singular. So the determinant's not zero. So I'll call that capital D. And I'll assume it's positive. When you're dealing with ternary forms, it doesn't the sign of the determinant doesn't really matter because you can just multiply S by minus one. Now, if you're talking about, it's a more modern way to do this would be to talk about. Let us is and quadratic spaces in which case, you know, the, the S would be a matrix of inner products of basis elements and changes of variables would be encoded in changing the matrix. But for what I'm going to do it's more convenient for me to actually use matrices, even though it's more elegant actually to not do that. But anyway, so S and S prime two different matrices are in the same class or equivalent. If you can transform one to the other. And I'm going to use jail equivalence. I'm going to be using seagulls work. So this, this kind of constrained me in a lot of ways. And this is the most natural thing for what I'm going to be doing. So it means that there exists an A such that a transpose S a is equal to S prime. So this corresponds to a change of basis. But I'll say that they're in the same class. And then the same genius. If this happens over all primes, including infinity. And it turns out that there's only finally made classes in a genius. So we assume that a nice way to put it is to say that a is in GL three ZP, he had a cantors for all P, including infinity. And what that means is we're talking about GL three are in that case. So this is a weaker condition for you know if you have an individual prime, if they're in the same class they're kind of obviously in the same ZP class, and actually they're in the same genius if they're in the same class, but not necessarily in the other direction. However, there are only finally many classes in a genius is a theorem. Now, the Hassel principle. Well, okay, so first of all, we have a notion of isotropy. So, as is said to be isotropic over Z say, if you can find a zero of it, if there exists a primitive, meaning the GCD is one such that S of X equals zero. Okay, so this is you know a very well studied thing. And one of the early, early cases of the Hassel principle says that S is isotropic. If it only if it's us s x equals zero is solvable over ZP for all P. So S facts is isotropic. If it only if it's this question is solvable over ZP non trivial or primitive leaf you like. And again, you know, this is the opposite direction is is if it's solvable over Z it's obviously solvable over ZP, but the opposite direction is non trivial. And that's the, an early case of the Hassel principle. So, actually, in this case, the Hassel principle is equivalent to a very classical result. It's one of the actually hardest cases to prove if you're talking about over Z. It's a non trivial result. It's equivalent to Lejean's theorem. So it's a, it's a special case it looks like so this says that if I let S of X be diagonal so AX one squared plus BX two squared. You know, one of the things that makes this questions a lot easier is because we were talking about isotropy, the equation being zero. You can have a rational solutions and and factor out the denominator so it's really a question. It looks like it's completely a question about rational solvability. That's why the Hassel principle is an effect here. But anyway, this allows you to reduce the principle to Lejean's theorem where in a special case where a B and C are integers and you can assume that they're square free their product is square free so they're relatively prime. And you know they, in order for them to be isotropic it has to be indefinite so they have to be different signs. The Lejean theorem says that this is isotropic. So if and only if you can make the local conditions into a very nice form. So minus AB minus AC minus BC, these have to be quadratic residues over the sort of obvious things that's the opposite. The thing that's being left out. So that's famous Lejean theorem everybody's heard of this, and they also have to be not the same sign, ABC, not the same sign. So it's not trivial to see that this is equivalent to Hassel principle in general, you know for a general turn, but it's not to it's not really hard either. By the way, there's a, you're probably aware of this book by Andre Bay on the history of number theory, I think it's just called number theory. It's really a beautiful book, because he, he mixes the history with, you know, modern explanations of the results so it's a great place to learn about this. If you if you want to see it. So I think it's just number theory, not the basic number theory but number theory. It's really a history book. Okay. Now. So, again, you might think okay this is this is the end of the story, because we have a complete understanding of when a ternary quadratic form represents zero. Well, I want to convince you it's not the end of the story because what you can also look at orbits of solutions and there's a lot of interesting questions about orbits of solutions. So look at orbits. Well, what kind of orbits. Well, maybe the first way to look at it is to think about orbits under automorphs of the of the quadratic form. So, let me say that. Let me let me denote capital C of X capital C of S sorry to be. So these are the primitive solutions. I'm a to with S of X zero. So these are the these are the solutions that we're interested in. Now suppose you have two solutions. So if you have one of it. If you can go from one to the other using an automorph. So I'll just write out what that means so if there is an A in G o 3 Z that first of all takes x to x prime, and is preserved and preserves S. So it's an automorph of S or isometry however you want to look at it. So let's up the solutions assuming that they exist into orbits and there's actually finitely many is a theorem. I mean this is a classical result to it's true in much greater generality. So there are finitely many orbits, Z orbits, maybe I'll say Z orbit. So let's go back to that by little C of S. Okay, now you can be interested in how many orbits are. It's not trivial question it's not obvious at all. How many orbits there would be. And you'll see why I guess eventually, but you can also look at the orbits over the pediatric integers. So the parallel question. So, it's the same Z p orbit. If the corresponding relations hold, or you know a and G o 3 of Z p, including infinity. And again there's only finitely many Z p orbits for each P. So let's see P of S, he's a number of Z p orbits. The result I want to describe is the following so given an S that we've been talking about. So for S non singular ternary. There's a relation between the number of Z orbits and the number of Z p orbits. It does involve the genius, which may not be consistent just one class. But the theorem is that if you sum over the genius forms in the same genius, and you take the number of orbits, the forms in the genius. That's the same thing as the product over the primes of the number of local orbits. And this is a finite product. So CP of S. It's usually one. So it's not equal to one for most finitely many. So, this is a kind of a quantitative version of the House of principle. It's not exactly. It doesn't imply the house principle, as it stated because the hospital would say that any yes prime would be isotropic provided the right hand side is not zero. And you sort of I would call the supplement of the House of principle so it tells you a bit more because it gives you some indication about how many orbits you have. Excuse me, do you count orbits with one over the size of the centralizer. No, so that's that so that leads me to the next comment because it looks like Ziegels main theorem. Yeah, because Ziegels main theorem has looks looks kind of like this but for the sum over the genius you would have measure of representation, which is basically a measure of the size of the isotropic group. And on the right hand side, you would have periodic product of periodic densities. So this is kind of a simpler version. It looks, it looks similar but it's actually simpler because these are all integers. And there's no waiting going on at all. So I got into this because I was reading Ziegels and he there's some cases where his theorem doesn't apply and this is one of them. Because the both sides of his identity are actually infinite in this case. So he didn't he didn't treat this case and I got very curious as to what was going on here. After many struggles and many sort of missteps this is this had finally emerged as sort of a version of the Ziegels theorem. In a way, it kind of looks like you can think of it as being, if you know about Ziegels theorem. It always has a mass formula embedded in it. And kind of what's happening here is I'm moving the mass formula from one side to the other. And maybe that'll become a little more clear when I sketch the proof. Now, there's a corollary if, if H is one. So if the class number of the genius is one, then you have a very nice statement, because it just says that to L two solutions are equivalent over Z, if not only if they're equivalent over ZP for all P. So that that really is a quantitative version of the, or a refinement of the house of principle. So XX prime in CFS are equivalent. So I like I like this statement a lot. I say more of it equivalent and Z orbit if only you are in the same ZP orbit will be I have one more question. Does it realize as a zero Fourier coefficient of Eisenstein series. If you look at the Ziegels. Yeah, that's a good question. I'm not. Yeah, I'm not I'm not sure. I have to think about that. Yeah, that's a good question. Thanks. Yeah, let me let me think about it. Maybe it'll pop into my head. So, I want to give you some examples of this. So first of all, the, you get nice examples just with the Legendre equation with the diagonal equation. So if you take the case that's used in the proof of the house of principles of you assume AB and Sears a times B times C is square free. And it's isotropic. Let me just always assume that then the class number is one and the number of orbits is one. I'll give you some indication at the end how this is how this is proved I mean, you can compute the right hand side of the theorem in different ways but you can also do this by reduction. And this includes the case of the Pythagorean triples and that's kind of an old theorem from the 30s but as far as I know people haven't really looked at this question much in general. So, if you take a more general case of Legendre theorem, which over the rational these are not interesting questions because everything's equivalent. So it's really over the integers that the, and the orbit question is really an integral question. You think about it. This is a case where you don't have square freeness. And in this case, if Q is a product of prime of distinct primes distinct primes congruent to three mod four. And you can prove it's a classical result of Meyer that the class number is one. But the number of orbits is not. In this case, it's the product of P minus one over two over the primes. It's not always true that there's one orbit. I mean, you could think that that might be true, but that's definitely not true. And also the class number is not always one even though it often is if you have simple S is usually the class number is one but when you have high powers of primes dividing the coefficients. It's often that it's not usually true that the class number is one. Another case would be if the Q if the Q is prime product of primes distinct, which are congruent to one mod eight. And also if you have a condition that they're all quadratic residues of each other, which is kind of an interesting condition. So the P quadratic residues of each other. They have a pretty big class number. H is two to the number of primes. Class numbers always a power of two this is a theorem of Meyer and then laser actually worked on this much later and gave much more general nice results. But the class numbers always a power of two. In this case, the number of orbits of each class of each S in the genus. It's all the same. It's P minus one over four. So the, well, basically, lots of different things can happen here I just wanted to illustrate that now. At the end if I have time I'll tell you how how you actually prove these from. It's actually better not to use the theorem to prove these in fact there's a different approach that you can use some more elementary approach. Okay, so what really led me this question of Ziegels what led me to this problem but also I was motivated by a proof of Ziegels theorem that was due to ask in Rudnik and Sarnac from many years ago. Where they, they used an idea of comparing asymptotic counts in an orbit to asymptotic count of the of all solutions. So I wanted to see if you could apply that method to this problem and some and somehow get some substitute for Ziegels theorem and that's really what happened. So, but you need to know first of all you need to know the mass formula, you can't get around this. So this is part of it. And to formulate this one very beautiful circumstance here is it if I take the group of automorphs. So these are the things that came into the orbit. So these are just the transformations that leave us alone. So this is a nice group of subgroup of the orthogonal group of s. Now, if I think of it as being in the or thought the orthogonal group over our, it has a subgroup, which is what you get when you restrict everything in a vast to the connected component of the identity of the special orthogonal group. So that's, that turns out to be the group that you want to look at. So it's the spinner group. And this turns out to be isomorphic to a fuchsian group acting on the upper half plane. This is the entry of analysis for me into this. And there's different ways you can do this, but this is a fuchsian group. This is a subgroup of SL to our PSL to our, and it's co finite. And this is due to Frick a client. Originally, I think although you know instances that were known even all the way back to Gauss. But there's an explicit map that you can write down, which I actually need. It tells you how to go from an element and O plus s to I did I didn't write down what this is but this again it's it's the subgroup kind of in the connected component the identity. And this, this group has many nice properties for example it's has cusps if and only if s is isotropic, but it's always co finite and the volume. This is the volume hyperbolic volume of h pod gamma s. So that's finite as well. And that's what enters into the mass formula. So the mass formula compares the sum of the volumes over the genus to product of local densities and the local densities of the are the local densities of the orthogonal group, restricted to ZP. There's lots of local densities that show up here. So this is Eagles theorem. So you compute the number of elements in the orthogonal group over the pedics, but taken modulo p to the end. So you can think of this is counting solutions to a congrats by p to the M. So this is OP. So OP of s is the, I'm not going to write out the definition so this is the ZP version of the orthogonal. But anyway so you count the number of solutions, modulo p and then divide by p to the 3M and take this limit. And it's a result of zeal that it stabilizes. So this gives you a rational number. And then you take and the product is actually convergent. So when you take the product over here. This is actually convergent and gamma infinity is the corresponding thing for infinity, and it's explicit gamma delta delta infinity. In this case is pi over 4d squared. There's lots of little subtleties that you have to deal with in dealing with the subject because the whole zeal uses volume on the orthogonal group and the volume on the fuchsian group is is twice his volume. And there's all kinds of little things that have to match up exactly so it's kind of a it can be quite a tricky, tricky thing to deal with zeal's theorems. Okay, so what, what are the asymptotics I'm talking about. They're really well known. So one of them is counting all the, all the points x primitive solutions with some restriction on their size there's definitely many so you have to put some restriction on the size. The usual way to do this is to use a norm. What you're really doing is you're counting rational points on a chronic. So this is a special case of a very general set of theorems. Due to fraca many and shinkle, and also pair who counted who gave the constants in the asymptotic. What I needed to do was not to use the norm but a slight variation. So, to describe this. So this would R be the ad adjunct of s so this is just the determinant of s it's basically the inverse made integral. So I take the inverse and multiply by the determinant. And if I take a Y in our three. So that this adjunct form has the value four times to the determinant, which is positive, then there's only finitely many. conditions that have an inner product between X and why so why is general with this condition and that turns out to be important it's that it's general but so this would be the X is for which the inner product of X and why is between zero and T say. It's finite. And even though this condition is you're talking about something lying between two planes. And, but you can see that it's finite with this condition because basically what you're doing is you're taking the lattice points that lie, you're cutting this cone, the condition that s is zero is a cone. And what you're doing is you're cutting the cone in an ellipse instead of a hyperbola. You get my pictures pretty bad here but you're counting points on the cone between two planes that cut it so that it's a compact set. And what their theorem implies is that this has an asymptotic formula, not only is it finite has an asymptotic formula. And so what is it what's the constant. It's a product of local of local densities again. One half, there's a density infinity product of Sigma P's. And these, these Sigma P's are the local density for counting the solutions, sort of the natural thing. So this is an actually pretty non trivial theorem. In fact, I approach it from the point of view of the Hardy Hardy little with circle method which is where this kind of asymptotic was first given is given for usually for quadratic forms and many variables, in which case you're getting Gauss sums, and these products of local densities converge nicely. And I've introduced some notation which I haven't defined this CP of S of the solutions, the periodic solutions. So, I'll just, I won't write that down, but that's what they are. It's sort of the net everything's kind of natural here, which you'd expect. Again, this is non trivial because we're talking about ternary forms and the, the Hardy little with method becomes difficult because very difficult actually to do it directly. Because you get non convergent products. So there's there are ways around it he's Brown found a very nice way to deal with the local densities in the ternary case, when you have weights. So I won't go into this but you can prove this. There's this, there's several different proofs that you can give of this there's the original proof of Franca chinko and many and pair. You can also give a sort of a more classical proof but it's highly non trivial. Okay, so the new thing that I did was to try to do the corresponding count when you're restricting to an orbit. So if I let's, if I take a fixed x, which is a solution, and I define the orbit of it, which is what basically what we define before. So these are the points in here, which are in the image of something in the orthogonal in the group of automorphs. So just basically what we did before. And we also have a local version of this. So, we have this local orthogonal group. Well, it's it's the isotropy group. So, this is this, I have to look at the group, the subgroup of the of the automorphs to fix this point x. So this is the difference here. So, then we can define local the local densities. So, I'll call Delta P because it's really part of the group of automorphs. So again, it's sort of the obvious thing, you have to get all the little constants in front right. So again, you take the number of points in the isotropy group module P to the M and then you compute that that's a finite number to group finite group, and then you take the limit and again this stabilizes. Okay, so what is the theorem. So again, it s is going to be isotropic. And I'll fix this why, which is going to give me the norm type thing. Then it for a fixed x in a fixed solution, we can count the points in the orbit of x little x with this constraint on the size. And again, this is asymptotic to a constant times T. And let me write down what the constant is. It does involve the volume. And then it involves the product of the local densities of the isotropy subgroup inverted. And this actually converges and for most P it's, it's easy to see these are actually powers of P, except for primes to don't divide two times the discriminant so it's actually pretty, pretty simple in some sense. And how these, what these look like, not so simple to compute them exactly but there's no problem with convergence of the of the product over primes can be given. In fact, let me let me write down. Okay, so anyway this, this is the orbital asymptotic. You know, this I view this kind of as the main theorem for me. This is the hardest thing to prove the thing that's hard to prove is not that there's isn't asymptotic but to compute the constant. This is actually turns out to be kind of kind of challenging thing to do. And it's what you need to prove the first theorem and everything has to be exactly right. So both and factors of two, the bane of the number theorists show up in every way you can imagine sure. So it's a anyway, that's the key result and for me anyway that's the key, the key result. So you can, you can ask questions about this you can say what about the constant does it change as the orbit changes or are the is a distribution uniform across orbits. And the answer is sometimes it is and sometimes it isn't. So the examples that I wrote down earlier, it turns out that they are all the same, but you can come up with examples where they're not all the same. So it's kind of an interesting question which I haven't looked at in detail as to understand, you know, more in terms of s, how the orbits themselves are distributed. Okay, but for here I'm going to concentrate on how you can use this to prove the main theorem about counting orbits. What's the error term in T. I didn't. I didn't computed it. What this does is it comes from counting orbits of an Eisenstein series. So if you know something about the poles of the Eisenstein series, you know, you can, you can definitely give an error estimate. The group is an arithmetic group. But I, I don't think it's congruent subgroups so what I use is I use a general result about Eisenstein series, the meromorphicity. That's another interesting question, which I haven't, but I haven't looked at this. I'm concentrating mostly on the constant. Okay, so a corollary of this which is, is where the mass formula comes in is the sum over the orbits. So, if we sum over the. Peter can ask a question. You say it's not a group but I assume that that's because you went to the spin group and then it's metaplectic congruence I guess. So I think it's sort of fundamentally non congruent just in terms of the polls. That's my guess. Okay, yeah, so that would mean that you could, you could definitely give a good error term. Right, I think so. Yeah. Thanks. Okay. Also, you got asking the chat. You should be X dash y t in a way here. Here. Yes. So that's just an inner product right. These are vectors so that's the that's a dot product of the vectors. You're looking at X. Oh, oh. Thank you. Why am I writing is pretty bad sorry but that's correct. Yeah, so X is fixed and then everything depends on X. All right, so what you can do here then is is sum up over the orbits and you, you are going to get first of all you're going to get an identity, because you have these two different asymptotics. So when you sum over the orbits, one way to write the corollary is a formula for the volume. There are two ways you could write this but one way is to write it as a formula for the volume, which is kind of kind of nice because it doesn't involve something over the genus, but involves something over the orbits. And the local density is product is actually a finite product if you write it this way. And then this also involves the sigma from the first asymptotic. And so that's kind of a nice little formula. By the way, I didn't do this but I think you could play around with this with the gasp on a formula to try to say something more about the structure of the group, because we basically the number of orbits is a number of cusps of this group, and the volume in the gasp on a formula also relates the number of cusps and all and the number of points of fight order. So it might be kind of interesting to use this to say something more about the geometry of the group. I haven't done that really. So to get to the, to use this in order to count orbits, you need to relate the isotropy group and the size of the group to the size of the orbits, but that's something you can do. So, how I say this I won't write down the definition but you can you can probably extrapolate what they should be. So I can talk about the Zp orbits and their densities. I'm being a little bit imprecise here but then you can use the orbit stabilizer theorem to relate to relate these on a finite level. So this will give you an identity between these local densities. So here's basically the sizes of the orbits, the Zp orbits and the size of the group, the whole group divided by the size of the isotropy subgroup. So this varies with X. That's going to tell you the size of the orbits. And, and that's what the other, the first asymptotic really did for you it related the count and the of all the solutions to the sizes of the orbit of the sum of the orbits, right. So, by combining these, then you get, you can get a statement which allows you to compare the orbits, the sizes of the local orbits and the global orbit. So, but we still have this problem with the fact that the genus consists of several classes. This is a headache that you can't get rid of. So, but there's only finitely many so let me list them. So I'll take representatives of the genius. And then I'll take orbit representatives for each of these guys. Okay, so what you can then get from this. And this is not an immediate corollary but is that you take the product of the sigma piece from the, from the full asymptotic. And relate that to the sum over the, over the genus, and then a sum over the orbit representatives. This little C, and then the product of the local densities of the local orbits. Okay, so that's what comes out from the these two asymptotic formulas. And that's the thing that's useful for proving the first theorem. One of the problems that arises is, when you're doing the local orbits, they can, they include solutions to your S of X equal to zero, they're simply local solutions. But the theorem, the main theorem was talking about global solutions and their local representatives. So we have to know that every local solution actually comes from a global solution to use this. And for that I learned this from Reiner Schultz Appelot, who helped me with this because he, he was able to show that you can deduce this from a Knazer, in Knazer's book on quadratic forms. But you have to translate his theorem into, for me anyway, into the language of forms. And also, his theorem is about representations of one lattice into another, but that translates nicely into into representations of an integer by quadratic forms. And the, the advantage of that approach is that it doesn't require that you look at different classes in the genus, it combines all the classes of a genus into one thing. So, what he, what he explained to me is that what the, what theorem, the first term actually is equivalent to is a statement that representations of zero that fall into, into genera. And that's that theorem says that each genus consists of one class of representations. This is only for people who really know the subject already, but that's what I was able, I was able to then apply that theorem of Knazer together with this result to get theorem one. Now the way I actually did it was to go to, instead of having pediatric densities to actually go to counting solutions to congruences. And so that's, that's actually what I did. Okay, so, you know, if you're interested in this I put a preprint on my website and you can see more details of this. Okay, now I do want to say something about what's involved in the proof of this theorem here. Because it's, to me it was kind of, kind of fun. So, again, we use the Eisenstein series. Okay, but there's a subtlety that happens because how are you going to get these product of local densities of isotropy groups from the Eisenstein series. Well, the volume is natural because that comes from the residue of the pole of the Eisenstein series at one that just shows up very naturally, but these guys maybe don't. Maybe they do, maybe somebody knows how to do that, you know, immediately, but the way I did it was to actually compute the first of all you do a reduction of your s. This being isotropic has a nice has a nice effect. So I can, I can basically do a reduction on it and make it be equivalent to another s which has a simple form. So first of all I can put zero here because it's isotropic. You can also make these, these terms zero, and then you get this nice form. So this is basically Gauss, where a b and a and b are positive integers and c and d can be not they can be their integers. But what this is doing at the level of the group is you're taking x, which is the solution to the equation. You can think of this as representing a cusp. And what it's doing is it's taking that cusp to infinity. And this change it's really a change of variables which takes the cusp to infinity. This has a nice result that you can compute the parabolic subgroup. And then infinity, of course, it looks like this. There's some positive kappa. So this to be in. And the problem is to compute this kappa, depending on s. It depends on A, B, C and D. And this is the fun part. So that's where the relation to the local isotropic group comes in. So what you do is you use the explicit isomorphism to compute kappa. And this is going to be what it is. So kappa turns out to be square root of B. This B is in this reduced guy. And there's an integer F. And this is where F is minimal in C plus such that I realize I'm running out of time here. So A, B and C. So this is an integer. This, this come, it's kind of natural. So this is, you compute the pair, you compute the group element that corresponds to the parabolic transformation. And then the, to be in our Fuchsian group means that the entries have to be integral. And that turns out to be this condition. These are integers. So then what you have to do is you have to compute the isotropic group of this special guy. And by taking the cusp to infinity, the point is simple. It's 100. And then you actually compute them. You count the number of points in the isotropic group modulo p to the m. In different primes, everything is related to A, B, C and F. You know, because they actually compute the isotropic group, you look at, you look at S one of X equal to X. And you write down what the, you write down actually. I'm saying this wrong. You have to, you have to take the transformation of S one by an element in your group and count the number of elements in the group modulo p to the m. And I was quite amazed that it actually worked out because everything depends on A, B, C, A, B and C, and these change over the orbits. But it actually works out perfectly when you do this. It tells me there's probably a much easier proof in the background because usually when there's a highly computational proof. And usually when a highly computational proof works out beautifully for no apparent reason, then there's something behind it which is making it work. And I suspect somebody will find this and tell me about it at some point. But just before I stop, I had to do this for p equal to two, which those of you who have experienced with Ziegels theorem know that that's something that's usually not done for good reason. But here it's actually not so bad. It turned out it was by doable. And again, everything worked out perfectly through the theorem. So one last thing to get the examples. My my original approach to this was was element my original approach was like this but in a special case but then I did an elementary approach, which is to take this reduction further, and actually it's an elementary proof of the theorem, but it only works for special s, but those special s are actually the ones in the example. So you can, they can actually be derived by an elementary method, which is just taking the reduction technique further. Okay, so I guess I'll stop there.