 Thank you very much, Mike, and thanks to you, Alina and Philip for asking me to speak in this wonderful seminar that you guys have organized. It's much appreciated. So what I'd like to talk about today is actually, oops, how do I advance to the next? Strange. It worked before. There we go. Oh, we've gone to 51. Okay, let's see if we can, there we, we're back now. Ah, wonderful. Okay, so what I want to speak about today is joint work with Stanley Xiao. Stanley is now at the University of Toronto, although he is on the job market this year. Uh, and, uh, let's set the stage here. So we'll take F always to be a binary form with integer coefficients. Assume the discriminant is non-zero, something non-trivial, and the degree is at least two, and we'll call it D. Okay, so what I want to look at is the, the set of integers, uh, which are represented by the form. So given any positive number, Z will let the set of, set of non-zero integers of absolute value at most said, for which we can find integers X and Y such that FXY equals that integer, we'll denote that by script RF of Z, and then the cardinality of that set, uh, by, uh, just capital R, I should say. Okay, so if we think about the, the simplest case, question, the simplest case is when F is a binary quadratic form. And in that case, in fact, that's, this goes back to the giants of the, uh, early days of number theory, Fermat and Lagrange, uh, the genre and Gauss. They investigated various properties in the set script RF of Z. As for the growth of, uh, capital RF of Z when F is a binary quadratic form, uh, that's, uh, that's a subtler question that, uh, was first taken up by Landau. Landau looked at the, uh, the case of integers represented as the sum of two squares. So FXY is X squared plus Y squared. And what he proved is that there's a positive number of C zero such that the counting function, the number of integers up to Z, which, uh, can be written as the sum of two squares, is asymptotic to C zero Z divided by the square root of the logarithm of Z. Uh, it's, it's striking, I think, that Landau's result was proved in 1908. And of course, that's already 12 years after the proof of the prime number theorem. Uh, now, in, uh, Michel Bauchman's lecture earlier in this, uh, seminar series, uh, he mentioned that, uh, uh, that extended, that Bernays was a student of Landau and Bernays four years later in 1912 in his thesis extended Landau's result on FXY equals X squared plus Y squared to, uh, any irreducible, um, positive definite quadratic form. And then, of course, you have to change C zero, C zero to another number that depends on the form you're looking at. Anyway, so, so not, uh, what Landau is showing is that, uh, that we don't get, uh, all the integers up to Z, we just get a very thin set that are represented. And, uh, how about this number C zero? Well, C zero is, uh, known as the Landau-Revenugian constant, and it's given on the screen here. So, what half of that infinite product over prime is congruent to three mod four, and we take the square root of the whole thing. Uh, how is, uh, how is Revenugian's name associated with this? Well, in fact, when Revenugian wrote to Hardy back in 1912 and, uh, in 1913, and he, uh, he sent Hardy that this incredible long letter with 120 amazing claims, some extraordinary integrals, uh, continued fraction expansions. One of the things he mentioned was that the, uh, the function R F of Z, when F was X squared plus Y squared, he said it was asymptotic to 0.764 dot, dot, dot, uh, essentially Z over root log Z. Well, he didn't phrase it like that, but that was the, the content of his message. And, uh, uh, later, in fact, he, uh, he specified that the 0.764 corresponded to this, uh, infinite product representation of C zero. And of course he, he didn't supply a proof and as typical, uh, his intuition brought him to, uh, to the correct, uh, order of magnitude for the counting function, and in fact with the correct constant. So even though this was, he certainly was not aware of landouts earlier work, but, uh, would give him credit for his astonishing, uh, insight here. Okay, but I really don't want to speak about binary, uh, quadratic forms. I'd like to go to forms of higher degree. That's the content of this lecture. And, uh, let me just take you through a bit of the, the history. The first result of, uh, substances due to Erdogan Mahler in 1938. And their result was that, well, they proved that if, if F is irreducible over the rationals, and D is at least three, then the function RF of Z, uh, grows at least as fast as a constant times Z to the two over D. So of course, uh, this is a different, uh, a different situation because if you take D equals two for binary quadratic forms, you just, you'd have growth like a constant times Z and we know that's not the case. Uh, so here's something else is happening. Um, when you see that result, well, the first thing you can ask, uh, the most natural thing to ask is, can one in fact, uh, get an asymptotic result here? What is the true order of magnitude? And in order to, in order to state such a result, figure out what the truth should be, we have to introduce the following region in the plane. That's the, the set of real numbers x, y for which the absolute value of f x, y is at most one. Uh, this is, uh, the fundamental region of F in the plane. And, uh, it's, uh, it's a region that generally isn't non-vax. And in fact, it's indeed compact because if, for instance, f x one has a real root, then you're going to have, uh, you're going to have cusps going off to infinity, uh, which are close to the line given by x equals alpha y, where alpha is a real root of f. And so you're going to have all these spikes corresponding to the real roots. Uh, nevertheless, the, the area is well-defined of this, this region of the fundamental region. And, uh, we call that, we're going to designate that as A of f. Okay. And so it wasn't until 1967 that Hooli was able to give the first asymptotic estimates for a class of, uh, binary forms of degree at least three. And he did it in the, in the special case that f is irreducible, a binary cubic form. And in addition, he required that his binary cubic form had a discriminant, which isn't the square, not the square of an integer. And, uh, what he proved is that result that I've stated there. So rf of z is the area of the fundamental region multiplied by z to the two thirds. So that's a two over D, uh, D in this case is three. And you can see he just squeaked it out. He, uh, his error term is smaller than z to the two over three by this log log power of z. Nevertheless, that was a huge step forward. This is the first asymptotic estimate for, uh, for the counting function, uh, for a large class of binary forms. He, he finished the irreducible cubic case in 2000. So he was able to treat the case when the discriminant is a perfect square, perfect square of an integer. And, uh, in order to state his result, this was a, this was a subtler business. Um, we have to introduce a bit of notation. So suppose that fxy looks like that, b3x cubed, plus et cetera. Associated to fxy is the Hessian covariant of f. And that's a quadratic polynomial with coefficients capital A, capital B, and capital C, uh, where those coefficients are given in terms of the initial data of f. So the coefficients of f in that manner. So that's the Hessian. And now what Hooli did is he put m equal to the square root of the discriminant of f. Remember that is an integer because we've assumed delta f is a square, uh, divided by the gcd of the coefficients of the Hessian. And what he proved is that if f is an irreducible cubic, uh, you also assume b1 and b2 are divisible by three, delta f is a square. Then in fact, this counting function for the number of integers of absolute value of most z represented by f is asymptotic to the area of f times z to the two thirds with a further weighting factor. And the weighting factor is one minus two over three times m, that integer m, which we've defined about. Okay, so, so clearly something different is happening in the irreducible, uh, cubic binary form case when the discriminant is a square. We don't get the area of f times z to the two thirds. We get it weighted by this rational factor. Okay, well he was also able to deal with quadratic forms that had a particular shape. So he looked at quadratic forms that were as, as indicated here. So ax to the fourth plus 2bx squared y squared plus c1 fourth. So that's very particular shape. And, uh, he was able to prove quite a satisfactory result here. Uh, it depended upon whether the quotient little a over little c was a fourth power or not, fourth power of a rational number. If it's not the fourth power, then the counting function grows like the area times z to the half divided by four with an error term. Now here we're winning, uh, quite a bit on one over two. This is 18 over 37 not 18 over 36. So we actually win a positive power here. But if a over c is the fourth power of a rational, we write that as capital A to the fourth over capital C to the fourth query and see a co prime, uh, positive integers, then the counting function grows like the area times z to the half divided by four times again, we have this waiting factor. Uh, in this case, it's one minus one over twice capital A C. Okay. So when you see this pattern, we're getting the area times z to the two over D times some rational number. So he was also, who he was also able to treat, uh, the case when f is a product of linear forms with integer coefficients and a number of people looked at the, uh, the very natural, uh, interesting case of, uh, forms that looked like x to the D plus y to the D. So that's Browning, Graves, Brown, Woolie, Skinner, Woolie, and, uh, they obtained asymptotic estimates for the, the counting function. And then Mike, uh, Mike Bennett, Neil Dumbigan, and Trevor Woolie, uh, looked at a slightly more general binomial, uh, form A x to the D plus B y to the D with A and B non-zero integers. Okay. So that's a quick tour of, um, uh, results where asymptotic estimates for the counting function had been obtained. And this brings me now to our result with Stanley, which, uh, we have here. So the usual assumptions, f's a binary form with integer coefficients, non-zero discriminant, and degree at least three, epsilon's, uh, some positive real number. Then we can find a positive number, cf, depends on f, such that in fact the counting function is cf z to the two over D plus an error term, uh, big O of, uh, basically z to the beta f plus a little bit, plus epsilon. And this is only interesting if, of course, beta f is less, strictly less than two over D. And so I'm going to tell you what beta f is now. Uh, basically to determine beta f, uh, we examine how f, uh, factors over the reals. So if f does have a linear factor over the reals, then beta f is defined according to this, this table here. Uh, I guess the thing to focus on is that generically, so in the case that D is at least nine, our error term, so let's go back, oops, our error term here is compared to two over D is essentially one over D minus one. Uh, but for smaller D, uh, we don't do quite as well. There's a different regime in place. In the case that D is equal to three, and f is irreducible over q, instead of 12 over 18, we get 12 over 19 for beta f. We can do better yet if we know that f has exactly one linear factor over q, and D is three, that's four over seven, and better still if it splits completely and then we get five over nine. Okay, so what about if it does not have a linear factor over the reals? Well, in that case, D has to be even, and, uh, generically, we're getting a one over D is our error term, that's at least for D, at least 10, and for D equals four, six, or eight, we get three over D root D. Okay, well, what's the proof dependent? Uh, basically, several things. We're making use of some absolutely beautiful work that has gone before. In particular, we need some very nice work at Solberger, who refined Heath Brown's periodic determinant method, and of course that turn can be taken back to the determinant method of Bombieri and Pila. We also need an argument of Heath Brown to control integer points in the cusps, and a classical result of Mahler. In addition, when D is three, that's when we have this, we're able to do better if f is irreducible, is reducible, excuse me. Uh, then we're appealing to a result of Heath Brown on integer points on non-singular cubic forms, and also to work of Hooli and, uh, and Schaub. Cam? Yes? There's a question in the chat room, Fabia Patsuki. Fabia, could you unmute and just ask away? Um, yes, so sorry for interrupting, Cam. I just want to know if you can say a word about CF. Ah, yes, yes, yes. In fact, I would say rather quite a bit about CF, but I thought I would put that off until, uh, closer to the end of the talk, but I will say quite a bit about CF. Uh, in a sense, CF depends on the automorphism group of f, and, uh, and so we'll go into that in quite a bit of detail later. Sure, thank you. Okay, so again, let me just give you a suggestion on how the argument works. How about this result of Mahler? So the result of Mahler that we need is from 1933, so it's a classical result, and, uh, it's, again, a counting function. So we look at the set of pairs of integers x, y for which the absolute value of f, x, y is at most z, and, uh, and I look at, whether it's zero or not, doesn't matter, and I look at the, the counting function associated to that set, n, f of z. So we're counting, uh, the number of integers for which not, not the integers represent, but just the pairs of integers x, y for which the absolute value of f, x, y is at most z. Okay, so what's Mahler's result? So he says, well, if f is a binary form of integer coefficients, non-zero discriminant to degree at least three, then with a f defined as before, so that's the area of the fundamental region associated to f, uh, the counting function grows like the area times z to the two over d with an error term of z to the one over d minus one. Well, Mahler actually only proved this result under the assumption that f is irreducible, but his, his philosophy works just find, prove it in general. In fact, Mahler's result does follow as a special case of some beautiful work of Jeff Thunder on, uh, decomposable forms. But, uh, let's, let's see what, uh, what's happening with Mahler. Okay, so if you look at, uh, at the, the region defined by the inequality absolute value of f, x, y less than or equal to z, then its size is, uh, essentially af times z to the two over d, its area. And you'd like to think in terms of geometry of numbers where what you're doing is you're just, uh, assuming that the number of integer points in the region corresponds to the area. And with this sort of, uh, mental template, it's no surprise that you get this result. That's what you'd expect. The only thing of course is that your region is, as I've already mentioned, non-compact. It's got all these cusps and potentially all sorts of integer points could hide in the cusps, could hide off in these spikes going off to infinity. Well, what Mahler used to, uh, deal with that possibility and so the deep part of this result is he used the work of Tuiz Eagle. I mean, he didn't have Roth, but he had the precursor, Tuiz Eagle, uh, and his advisor was Eagle. So he, uh, he, he was definitely aware of Eagle's work and he used that in this context to control the number of integer points in the cusp and in the balance of the, of the region, he was able to use geometry of numbers arguments to, uh, to get his result. Okay, so, um, uh, another thing that, so let's just go, go back here again. Uh, if, if we see this result and we say, well, wait a second, if it's the case that for every pair x, y, uh, we're going to get f since that pair x, y to a distinct integer, h, then in fact, the counting function is just going to correspond to the number of points. So we would get it immediately from Mahler's result, we would get the, uh, result we want. Well, we've seen that it's not so far from the truth. Um, we solved with Houli's result that in fact that was essentially the case, that had to have been essentially the case. We know in general, of course, uh, a binary form of the Tuiz equation, f x, y equals h can have many solutions. And, uh, and so what one has to show is generally there are very few, uh, solutions to the associated Tuiz equation. And so each pair x, y gives essentially a, uh, a new integer h, uh, which would be counted by the function r, f is it? Okay, but there are, there are definitely instances where that's not true. And that's why we introduced the notion of an integer being essentially represented by the form. And we say it's essentially represented if it's represented, first of all. And whenever we've got two pairs of integers x1, y1, x2, y2 that are sent to the same integer h, then there's a reason for it. And that reason comes from an automorphism, uh, so a gl2q transformation that leaves the form unchanged. Okay, we'll come back to this. Um, but let me just finish off by remarking that if there's only one integer pair for which, uh, fx1, y1 is equal to h, then h is certainly also essentially represented since the identity is in the automorphism group of all this. Okay, so that's this notion of essentially represented. So we've got Mahler's result, we've got this notion of essentially represented, and we'd like to show that generally, uh, things are essentially represented. So we want to control when they're not essentially represented. And for this, uh, that's where the work of Solberger comes in. What we do is we look at the, uh, the surface x given by fx1, x2 equals fx3, x4, and p3. That surface is smooth since our assumption is that the discriminative of f is non-zero. And what we do is we count integer points on x in a box which don't lie on a line on x by means of, uh, a result of Solberger, the bigger the box, the worse our estimate. But we're going to play off the size of the box against, uh, how, uh, how far up we truncate the cusps in Mahler's perspective. And that's going to give us our error term. And, uh, so Solberger allows us to count points that on x in a box which aren't a line of x. And now by examining the lines, uh, we're basically able to, uh, count the associated integers h which are not essentially represented and of absolute value at most set and show that there aren't too many of them. Okay, but now we come down to these integers which are essentially represented and to figure out how our count should go, we need to study the structure of lattices associated with the automorphism group of f in its subgroups. Okay, so I am going to return to that, but I would like to get, uh, the, the topic of k-free integers represented by f also dealt with. So let me switch gears a little bit and now we're going to, in some sense, go back to, uh, to our first theorem. So let k be an integer with k at least two, an integer is k-free, this is the standard definition, if it's not divisible by the k-power of a prime. And what I want to do is I want to introduce the, uh, first of all the set, rfk of z, so that's the, the set of k-free integers h up to z in absolute value for which, uh, there is a representation, so for which there are integers x and y such that fxy is equal to that h. So here instead of rf of z, that, that set that we looked at initially, we're imposing the further restriction that to be in that set you have to be k-free. And we look at the, the counting function associated with that set, so that's capital R, fk of z. Okay, so well, in fact, the first results on this, these counting functions rfk of z, so they would correspond to the result that I mentioned earlier, uh, uh, Erdisch and Mahler and the, in the case where one's not imposing this additional counting, uh, this additional k-free condition. Uh, well, the first result that I'd like to mention here is due to Gouvet and Mazer, they proved it in 1991, and they extended work of Hooli again. And what they looked at is just the square-fee case, so they assume that there's no, uh, no prime p such that p-squared divides fab for all pairs of integers a, b. In other words, one's, one's, uh, we're going to count, uh, square-free, uh, integers represented by f, but of course if there's some universal square divisor of that, of all the values of f, that, that'll be an uninteresting count, that'll just be zero. Uh, so you impose that restriction. Uh, you also assume that all the irreducible factors of f over the rationales have to greet most three, this is imposed by sitting constraints. If epsilon is a positive real number, you can find positive numbers c1 and c2, which depend on epsilon and on epsilon, such that if z is big enough, then the number of square-free integers represented by f up to z is at least a constant times z to the 2 over d minus epsilon. So you're, you know, you couldn't do better than to constant times z to the 2 over d. You're getting said essentially the right order of magnitude. Okay, well, Yaptop and I, uh, looked at this and we considered more generally, uh, the case of, uh, k-free integers. So let k be an integer at least two. Again, assume that there's no, uh, p to the k that divides fab for all integer pairs, a, b. There's no universal k-powered divisor of all the values assumed by f. And, uh, we made use of a lovely sitting argument due to graves and also the result of Beardish and Mahler to show that if k is at least r minus one over two, or in the, uh, the case of square-free, if k is two and r is six, um, that, uh, that we get r, f, k of z, the counting function is at least a constant times z to the 2 over d. Okay, so these results, so both the Gouvet and Mazur and Yaptop and I, uh, made use of these counts, these estimates for the counting function in order, uh, to, to study twists, quadratic twists of elliptic curves. So if you've got elliptic curve defined over the rationals, you can ask, uh, how many twists of it have, uh, large rank where large is at least two. And in fact, uh, both Gouvet and Mazur and Yaptop and I applied these results to, uh, take advantage of certain constructions which showed that there were many such twists by, uh, using these estimates. Okay. Uh, Stanley was able to extend the range, uh, for which this result holds, Gouvet's 13 holds. Um, he did so by generalizing the determinant method of, uh, he's Brown and Solberger to the setting of weighted projectives weights. And as a consequence, he was able to, to sort of, uh, widen the range over which these results hold. So essentially what you do is you, you strengthen the sieving side of the argument. And, uh, he showed that if K is at least, well, essentially seven times, uh, R over 18, um, and K arms not three eight, and then the previous result holds. So this is pushing it up from R minus one over two to seven R over 18. All right. But now, um, we're able to get an asymptotic estimate for this counting function by sort of building on the approach we used to, uh, to get an asymptotic estimate for R F of Z. So here's our statement. So as always, F sub binary form integer coefficients, non-zero discriminant degree D at least three, R is the biggest degree of an irreducible factor of F of the rationals. If you want to just think of it as D, that certainly works as well. Uh, we assume there's no universal P to the K divisor and we assume that our condition, 14 here on K and R holds. Then in fact, the counting function grows like CFK Z to the two over D. So we're not, uh, this is maybe not so surprising. We're not dropping down too much. The truth is that some number CFK Z to the two over D plus an error term. Well, again, I better have that GK R of Z going off to infinity to make that a true error term. And here, here we don't win nearly as much, essentially win, win a factor of log Z. Uh, we have to be, it's a little more delicate in the case that K and R to six or three eight, we don't win quite as much in those two cases, but nevertheless, we do get the asymptotic estimate. Okay. So, so how does this, how does this go? Well, it goes in a similar pattern. What we want to do is we want to just as the previous result built on Mahler, we need something like that. So we look at the set of pairs of integers XY for which FXY is K free and below a given bound. And we define the counting function as just the cardinality of that set. And so we're, we're getting rid of, uh, of integers which are K free. So we want to sieve out and the, the natural thing to do here in this setting is to sieve out the following way for, for each positive integer M, I want to define row F of M as the set of pairs IJ for which F IJ is congruent to zero mod M where IJ is in zero M minus one. And, uh, we introduce lambda FK which is this infinite product over primes of one minus row F P to the K over P to the 2K. So this is going to take into account uh, divisibility by P to the K. The product converges, no issues there because they're discriminant, it's non-zero. And lambda F of K is exactly zero whenever we have some universal P to the K divisor. And we put little CF of K to be just lambda FK times the area. So we're doing the natural thing. We're weighting the area by this sieving factor. Here's our result with Stanley. So if F is a usual assumption, we're going to require that R is the largest degree of an irreducible factor of F over Q. That's what the sieving sort of lets us take. So as before K is an integer where 14, so that's that condition of K and R. With that CF of K that we've defined, the counting function grows like CF of K times Z to the 2 over D plus a smaller order error term. So little CF of K is is just the area weighted by that sieving factor. And so what we're getting is basically this is not a generalization of Mahler, but it's an extension of Mahler to the case of K free values assumed by a binary form. So let's see again what we're doing is, so Mahler is counting NF of Z and we're counting NFK of Z. And we're introducing this weighting factor, which is the natural weighting factor coming from SIV. Okay, let's just see. Sorry, sorry. Okay, so let's just see. Let's go forward. There's our weighting factor, the result that extends Mahler. And let's recall the first result and Fabian's question was about C of F, capital C of F. Having seen the Mahler result, you'd expect that capital CF of K would be equal to just lambda FK, that weighting, that sieve weighting factor, multiplied by CF. Now, in general, that's not the case. And I give a very simple example here, GXY is 8X cubed plus Y cubed and K is either two or three. Then in fact, CG of K isn't lambda GK times CG, but four thirds of that. So, what's happening here? So what's going on? Well, what we have to do is we have to go back, basically what's going on is that this idea of being K free doesn't play nicely with the automorphism group of the format. Okay, so how does CF depend on that? Well, okay, so I've got to give you a tour here. So if we've got an element A in GL to Q and FA of X, Y is identically equal to F of A1X plus A2Y comma A3X plus A4Y, then we say that A fixes that. And the set of A which fix F is the automorphism group of F, and we're going to denote it by that. Okay, well it turns out that there aren't too many options here for our automorphism group up to conjugation. So if two elements, two subgroups of GL to Q are conjugated, we can find some element T and GL to Q such that G1 is conjugated to G2. Okay, so it turns out that the positive number, as we've seen in sort of the Hooli case, Hooli's examples, is a rational multiple of the area, and the rational multiple depends on the automorphism group. And there are essentially, it depends on the equivalence classes of automorphism groups, then there are 10 equivalence classes of finite subgroups of GL to Q under GL to Q conjugation to which the automorphism group might belong. And let me just show you in this table. So they're just C1 up to C6. These are the generators, D1, the dihedral group up to D6. Those are the generators. And so that's it. That's the total number of possibilities. Well, I mentioned lattice is associated with these guys. So let me introduce the lattice of Z2 lambda. So that's the set of all integer points, UV and Z2 for which A of UV is in Z2 for all A in the automorphism group. That's a lattice. That's an ambience determinant. M is one of the automorphism group is essentially trivial, so either C1 or C2. Things get complicated when we have say D3, D4, D6. If the automorphism group is conjugate to D3, it's got three subgroups of order two, the generator's A1, A2, and A3, and one of order three, generator four. And I have to associate a sub lattice with each of these guys. So let lambda I be the sub lattice consisting of the UVs for which AI UV is in Z2. And let's put mi equal to d lambda I for one, two, three, four. Well, we have to do a similar thing for d4. We look at subgroups of the automorphism group modulo plus minus the identity. We're going to define sub lattices in a similar way and the determinants of those sub lattices. We do it again for d6. Again, we've got three subgroups of order two and one of order three, and we have to study how everything interacts with these sub lattices. So now I can answer Fabian's question, and this goes back to the very first theorem I mentioned, and that's this positive number Cf in the statement of theorem one. So it's equal to Wf times the area where Wf is some rational weighting factor, and it's given by the and it's given by the following table. Well, it doesn't quite fit in, but here I'm giving you the representative of the conjugacy class or the conjugacy class on one side here. The weighting factor is given by these rational numbers. The Mi's correspond to the determinants of the lattices I've mentioned. Let me just remark that in the case of d3, d4, and d6, the integer M is the LCM of Mi, of the Mi's say M1, M2, M3, and M4 in the d3 case and M1, M2, and M3 in the d4 case. Okay, so these are the weighting factors that we have to put in front. So I would just like to finish off very quickly by showing how Hooley's result from 67 fits into this framework. So if we've got a binary form and we've got some element in our automorphism group of F, that element acts on the roots of F by sending a root alpha to this expression. If it fixes a root, well, then it has to be a root of this quadratic, but if F is an irreducible cubic, alpha has degree three, so those coefficients have to be zero. And that allows us very quickly to show that, ah, in that case, a has to be just the identity. If it doesn't fix a root, it's going to permute the root cyclically, and thus must have order three. And any element in the automorphism group of order two would fix a root of F. So in that case, the automorphism group has to be gl2 conjugate to C3, the cyclic group of order three. So automorphism group of F looks like T, C3, T inverse with T and gl2q. And you can calculate, you can determine what forms with integer coefficients are invariant under C3. They have this special form with A and B integers. Okay, but let's take the discriminant there. We can work it out. In that case, it's this square. So that tells us that F is equal to G of T for some G invariant under C3. So in this case, the discriminant is the determinant of the six times the discriminant, which is a square. So in particular, it's a square. So what we conclude from this, that if F is an irreducible cubic, we need this irreducibility condition, and this discriminant is not a square, well, then the only option is the automorphism group is just the identity, just C1. And we go back to our table we see in that case, the weighting factor is one. And that's why we get, and that's why Hulu's result follows with the counting function like the area times the weighting function of one. So the area times z to the two thirds. Okay, well, I haven't defined, I haven't told you how capital CF of k is defined in the square free case. That becomes, that's a little more cumbersome that has to be defined to take into account both the weighting factor and the automorphism. So I'm just going to leave that out and I'll finish by thanking you.