 Okay, so right so before we start today, I just want to make a quick announcement. So there's a PCMI cross program activity today it's a panel discussion on career arcs, which everyone is strongly encouraged to go to it so this panel discussion is at 1230 mountain time and the link is in the email sent out. So this does conflict with the usual time for the TA session and I believe the TA session today is likely to be moved. Probably just sometime after that, but maybe just confirm, maybe I can confirm at the end or or check discord for to confirm the time of the session so okay. Okay. Hello everyone and welcome back. We're going to have some fun today. Let's see here. We go. So, yeah, we're going to be talking about reduction theory of. There's a lot of adjectives. There's a lot of definite binary integral quadratic forms I think I got them all in there but let's just just say what we mean so let's recall. Let's recall the context we had last time. So, we're interested in quadratic forms over the integers, and we're looking at the smallest non trivial case which is two variables. A binary quadrant that's called a binary quadratic form integral is a polynomial of the form. f of x y is equal to a x squared plus b x y plus C y squared, or a b and c are integers. So it's a homogeneous degree to polynomial in two variables. And we're also going to denote this as left brackets a comma b comma c right brackets, but be careful. This is not consistent with the previous use of this notation where this stood for a diagonal form in three variables, a x squared plus b y squared plus C z squared. And b here stands for the cross term x y. And in general, not every such guy is going to be diagonalizable, but actually see this as a consequence of something we do today. So there's no use in the diagonal notation it does it's not general enough so this will be the new notation, they'll just put this caution symbol here to not consistent with, with previous. Okay. All right, and the most important quantity in the most important invariant you associate to this. Such a binary quadratic forms is discriminant, which is just the usual expression b squared minus four AC. And we will assume, we will be assuming that the corresponding if you take the same polynomial but view it as a real quadratic form. And then we'll be assuming that that's positive definite. And as I explained last time that's equivalent to some other properties of this form, for example, you know, f of the x y is greater than zero, unless of course, x y is equal to zero. It's not negative and in fact strictly positive unless you're evaluating at zero. And it's also equivalent to a numerical criterion involving D. So, well the discriminant of f should be less than zero. And, well, to separate between the positive definite and the negative definite cases you can say that a and C, or, or definitely just one of them has to be positive. So both a and C are values, right, on one comma zero and zero comma one respectively so they have to be positive. And that actually is enough to guarantee that your positive definite. Once you have this condition as well, because you're either positive definite or negative definite right and that shows. Yep. Okay. So from now on. This form means such a guy. So it's a positive definite integral binary quadratic form. It's an app like this satisfying these equivalent conditions. I'm going to say form for that. And you'll have to remember that it's positive definite, for example. And we talked about the notion of equivalence of forms that we're going to be using. So, so also recall, we say that F is equivalent to F was strictly equivalent. Sorry, F is strictly equivalent to G. If and only if they're related by an invertible change of variables with determinant equal to one. So in general, invertible matrix over the integers will either have to turn one or minus one and we want to only look at those were making a finer equivalence. We're probably determined to be equal to one so concretely, this means that you can write that if you take F of like alpha x plus beta y gamma x plus delta y. Then this gives you G of x y or some alpha beta gamma delta in SL to Z. So the group of integer two by two matrices with determinant one. Okay. Now. So these are going to be our favorite characters from now on. So, well, also, but I just accidentally hit my camera. Okay. So also recall. That's if F is strictly equivalent to G, then, in particular, the discriminants are equal to each other. That was something we also saw last time. However, let me remark. There are many so that so well, well this is not invertible. So this is not a if and only a so discriminants being equal is also a coarser equivalence relation than being strictly isomorphic. And in fact, you can put a whole bunch of different rather natural equivalence relations sitting in between the strict equivalence and the equality of discriminants so let me just write a chain of implications here. So F is strictly equivalent to G implies F is isomorphic to G, and this I mean in the sense of, you know, using GL to Z instead of SL to Z so arbitrary invertible linear change of coordinates. This in turn implies that F is isomorphic to G over the chaotic integers for all primes P. Which in turn implies that's F and G or isomorphic over the rational numbers. Which in turn, I guess implies that that already implies that the discriminant F is equal to the discriminant G. And in general, all of these implications are strict. Well, maybe there's one implication that's not quite obvious I guess this one. So why if they're isomorphic over the chaotic integers for all PR the isomorphic over the rationales. Can anyone give me the reason. I think you can use. Can use what I'm sorry. Yes, exactly. So if you're isomorphic over ZP then you're certainly isomorphic over QP. But on the other hand, oh, G, sorry, since F and G are both positive definite they're also isomorphic over our, because there's only up to isomorphism there's only one positive definite quadratic form over R with, you know, plus plus in terms of solicitors law of inertia. So in Kowski, this implication holds. And, yeah. Actually wait does this implication or I think I'm going to end up confusing myself here. This only implies that these are equal up to squares. Oh geez I might have confused myself here. Yeah, let me let me put just I don't want to have to, I don't want to get something wrong so let me put this here instead. I might have been mistaken argument there. Anyway, in general, all of these implications are strict. And you saw some examples of strictness of implications in your problem set. From last time. Now, we're going to be interested in. Oops, be studying the set XD. So, for, you know, for D a possible discriminant so D a negative integer, the set XD, which is the set of equivalent strict equivalence classes of forms of discriminant B. So we're going to be taking this course invariant the discriminant and asking to classify all of all possible forms of a given discriminant. And it turns out that this is sort of the best question to ask in this two variable case. I want to mention that outside of the two variable case, the classification by discriminant is what much too course to be kind of a useful way of classifying forms. And in that case it's this one that is the appropriate general notion. And this is called this in this case we say that F and G are in the same genius. And the thing you want to do in general is ask to classify equivalence classes of forms of the same genius. And that gives results somewhat analogous to the ones that we're going to be talking about here in the two variable case. However, in the two variable case because of special phenomena that appear which will explain. It's actually actually get a better result by classifying forms with the same discriminant rather than forms of the same genius so slightly coarser. And in fact you can understand forms of the same genius in terms of the results will describe about forms of the same discriminant. So, in generality you'd want to look at classify forms of the same genius but we're going to look at forms of the same discriminant because that's the nicest thing in this special case we happen to be considering. So that's that's just a general remark to orient you a little bit in the general theory. Right, so we're interested in studying a set let me check my notes and make sure I didn't. Right. So what's the first question you can ask about a set. Is it empty or not. So, for which D is XD non empty. In other words, given a negative integer D when does there exist a form of discriminant D. So, well, note that by definition D is equal to be squared minus for AC. So it's congruent to be squared mod four. In other words, D has to be a square mod four, which means that D has to be D has to be congruent to zero or one mod four. And conversely, if D is congruent to zero one one four it's actually not difficult to explicitly write down a form of discriminant D. So, if D is congruent to zero mod four. Then the form F equals X squared plus D divided by four Y squared. As discriminant F equal to D. And if D is congruent one one four, then F equals X squared plus X Y plus one minus D over four shows remember that he is negative yeah. I didn't remember D was negative. That's better. Plus one minus D over four Y squared has discriminant D. Sorry guys, my pen is failing me here. Has D equals D. So the answer to this first question of when the set is empty is the set is empty is non empty if and only if D is congruent to zero one four. So, so called, I mean these sort of obvious forms you write down actually do play a, they're not just random examples actually play a special role in the story so these specific forms are called the principle form of discriminant D. But or you might more generally call a principle form anything in the city, anything which is strictly equivalent to one of these two forms but we'll see that these two also have a special extra property which characterize them in their equivalence class. And the principle forms do play a special role in theory so it's not just an example, you can write down. Okay. Now I want to give you a first reason why this question is a natural question. So, why it's natural the group forms by discriminant. So, here's one answer, they're actually use several possible answers but here's one. And it has to do with this question we're interested in of which primes are, which primes for example are represented by giving quadratic form. Here's a theorem. So let D be negative and D congruent to zero or one or one mod four. So D is a possible discriminant in other words, let P be an odd prime, then P can be represented by some form of discriminant D. So, if and only if some condition you can actually check. So if and only if D is a square one. Now you can't say D is a square one P you can necessarily say that he is represented by some, I mean you can fix the form beforehand you can't necessarily a form of discriminant D before and you can necessarily say that he has to be represented by that form, but he will be represented by some form that has the correct discriminant. So here you sort of have to take all forms of discriminant D into consideration a priori to get the, the good result about primes being represented. And the proof is not too bad so let's actually give it. So first let's give the forward direction. So let's assume that P is equal to a x squared plus b x y plus C y squared. Well, P has odd petic valuation in fact petic valuation one it follows that it can't be the case that both x and y or congruent to zero mod P. Because if both of them were divisible by P then you'd be able to have you'd have a p squared on the right hand side but you only have a P on the left hand side. So that's pretty simple but let me reinterpret this. Let's, this means that if we reduce these numbers mod P. This is non zero. Right. So in other words it's a non zero vector in the two dimensional vector space over this finite field. So, any non zero vector over a field can be extended to a basis. So we'll extend this guy to a basis of the two dimensional vector space in Fp, right some two, and then calculate the discriminant in this basis, or the discriminant mod P I should say. So then, what's the matrix going to look like. Well, our basis vector annihilates the quadratic form because P is equal to our quadratic form. And therefore, when you reduce mod P zero is equal to our quadratic form so we're going to look something we're going to look like zero BB and then some C right. And then the determinant is going to be B squared. So we'll get B squared is congruent to D mod P, which is exactly what we wanted to show. So that means that D is a square mod P. Okay, so here I'm using that the determinant is the, I mean the discriminant is minus the determinant of the matrix associated with the quadratic form. Okay, so now let's do the other direction. Well, it's actually pretty simple. I mean, if D is, if D is a square mod P. So then we can write D is a D is equal to B squared plus four times. Oh, sorry, sorry, so D is a square mod P. It's also a square mod four, because it's congruent to zero one one four. And then P is odd so four and pure co crime so that implies that D is a square mod four P. So I'm actually going to use that I'm going to write D equals B squared plus four P times a random number which I'll call minus C. Right. I'm allowed to do that. I'm not supposed to give me what I want, because, well, this means that D is the discriminant of the form F where F is equal to P x squared plus B x y plus C y squared. So it's sort of, you sort of just do it. Just do it by hand you can easily solve the equation discriminant. Well, anyway. Okay, so that was that theorem. So that says that yeah prime is represented by some form of discriminant D if and only if D is a square mod P. Let me make a remark. I didn't understand the other indication. The second one or the first one. So our assumption was the D is a square mod P. However, we also have the hypothesis that D is convert to zero or one mod four which means that D is a square mod four. The hypothesis that P is odd. In other words, P is relatively prime to four. So by Chinese remainder theorem it follows the D is a square mod four P. Are we so far. Yeah. So let me just write what that means. It means we can find a B such that B squared is congruent to D mod four P. So D is equal to B squared plus some multiple of four P which I'll call, which I'm calling minus C. And that equation just says that D is the discriminant of this form. Yeah, so it's very, it's very elementary. I'm not doing anything fancy here. So let me make a remark. So this condition we had that D P equals plus one can be rewritten as a congruence condition, modulo D. A congruence condition on P modulo D by quadratic reciprocity. So let me pick up into cases of, according as to D is zero one one four but in the end you find that indeed, in all cases, you get a congruence condition what D so there was an exercise, sort of similar to this a couple of problems that's back. So let me give it just us just a very simple example so. So D was minus four, which is the discriminant of this x squared plus y squared. So D is equal to four on P is equal to plus one if and only if P is congruent to one month for. So it's a congruence condition mod minus D, which is the same thing as one. So, you know, so because of this observation, this condition is actually it's very easy to check it for a given prime. It's just a congruence condition of the discriminant. Okay. Now let me explain a corollary of this theorem. So, if it just so happens that the size of this set of equivalence classes of forms of discriminant D is equal to one. Then if F is any form of discriminant D, we have that P is represented by F, if and only if well yeah just the same thing. Again this nice congruence condition on P. Well, you have any form of discriminant so we know that this condition in general is equivalent to saying you're represented by some form of discriminant D. But when the size of this set is equal to one that means any two forms of discriminant D are equivalent to each other and the equivalent forms always represent the same set of integers. Because you can get back and forth by just do linear transformations in the variables. So, if it's represented by some form it has to be represented by F. Yes, there is. In this argument that we just gave in the case where XD has just one element in it. Well one class basically of strong equivalence. And we don't we didn't use anywhere the fact that the equivalence is strong right because the fact that they represent the same elements is from regular isometry. That's right. So you could say something a little more refined here. Absolutely. So that is, it's not the strongest possible statement of this form and that's a very good remark. Yes. And so. Yeah. Okay. But anyway, this is going to be sort of our general strategy. So, so now, what's the main theorem about this set XD. So, which we're going to prove today. So for all D less than zero, well, I might as well add the common zero one month for all the theorems true without that. XD is finite. So may not always have just one element. But it is always finite. And it's cardinality is denoted HD. And it's called the class number. But actually the theorem is the term that we're going to prove is actually better than just this abstract statement that you have a finite set. In fact, in fact, given D, one can explicitly produce by a very efficient algorithm. So you can actually make this completely explicit and algorithmic. So given D, you can actually figure out what the class number, you can calculate what the class number is and find representatives. So you can actually make this completely explicit and algorithmic. So in D, you can actually figure out what the class number you can calculate what the class number is and find representatives for all equivalence classes of forms of a given discriminant. So before giving the formal proof, let me say what the basic idea is. So basic idea is to, well, try to make a and B as small as possible. So you maybe you start with a random form ABC, right, and then you apply you want to apply changes of variables in order to make aim a or smaller or be smaller. And then either one you win because eventually it'll get down to something really small and then they're there only be finally many possibilities. So that's the basic idea. And to do it, we don't have to use. So priori use arbitrary elements of SL to Z. In fact, you only use two specific kinds of elements of SL to Z. So you use two kinds of change of variables. So the first kind is where you want to sort of swap X and Y, but that actually has the term that minus one so you have to swap them and change the sign of the other. So this corresponds to some matrix like I guess zero one minus one zero. And what does this do on the level of forms. Well, you can just plug this in so plug wine minus accident here, take f of y minus X and rewrite it in standard form and you'll see that what happens is you get ABC. You can see that you can transform this into C minus be a. In other words, you're allowed to switch A and C as long as you change the sign of the. Okay. So that's the one kind of variable change. And the second kind of variable change is, so let me make sure I get this right so you can send X to X plus K times why. And you can just keep y fixed. So it's kind of a sheer. And this corresponds to, I might get this slightly wrong, but I guess it's this one. I could, you could make it for trying to use the upper triangle instead. So the guy here is going to be an integer. But actually this is just the, you only need this one because this is just the cake power of this element here in the group SL to Z. I'm sorry to ask another question but for the first kind of changes, should it not be the first column zero one and then minus one zero because X goes to why and why goes to minus X. Yes, indeed. Thank you. Although, there is a question of whether it's the matrix or it's inverse the kind of, I love what I mean by this really. You see what I'm saying. Yeah, I get it. Yeah. Yeah. Yep. Right. So, if the notation he's using in for a certain matrix is correct, then the first matrix is correct. Yeah, I think it's consistent whereas before, there is this comment that wasn't being consistent so that's, yeah, so that's good. Anyway, in a case we're not going to think about the major season so it's just going to be about how a B and C change that's all we care about so let me write down how a B and C change here. So a B C. You can plug it in and check for yourself. I did it in my notes so I'm not, you know, bull crapping you. It goes to a, and then B plus to a K, and then whatever the third one is it's not going to be important for us. And third one was actually determined by the other two right because we know what the discriminant is. So we can always solve for the other one so we don't really need to keep track of C and we can just focus on a and B that's also part of the general strategy. Right. So what is this move do it keeps a the same, but it replaces be by a multiple of two a. Okay, now let me state the most precise form of the theorem explaining exactly how to find this set of representatives, and this is called reduction theories this is the very topic of this lecture. So, the definition that you use is this. So a form. C is reduced. If. So, if. So, I am C are always positive right, we said that the B can have candy negative it's possible for B to be negative. But but now we're going to call the form reduced if the absolute value of B is less than or equal to a. A is less than or equal to C. So we're trying to make B and a small and this is the precise way in which we'll be able to do it. And more over, if we're in a boundary case. So if either a equals B. Sorry, a equals absolute value of B or a equals absolute value of C then there's actually going to be some freedom and we want to fix that freedom. So we're trying to make sure that B should be non negative. So we just have to handle the boundary cases. We didn't throw this in there would be two in equivalent forms. Two forms to reduce forms which are equivalent if we're not going to want to have so we have to choose one of them and we do it by fixing the sign of the end if you're in a boundary case. Okay. So, that's the definition of what it means for a form to be reduced and now the most precise form of the theorem is the following and I guess this is Gauss's. He actually, I mean he, he did everything we're going to be talking about not not everything we did a lot of the stuff we'll be talking about and more. In his book, this position is arithmetic thing. Actually, there's a fun story about that book. So maybe I should plug this book. So it's definitely the most famous book in all of number theory, one of the most famous books in all of mathematics. And it relates to the subject of our, our lectures in more than one way. So first of all, he discusses a lot of this theory of binary quadratic forms integral binary quadratic forms that we'll be talking about here in the second half. But he also was gave the first proof of quadratic reciprocity. And we also gave a proof of quadratic reciprocity in the form of Hilbert's product formula. What I said was due to Tate, but then here's a funny story which Tate likes to tell that he came up with this calculation of K to a few. And then, once he had done it he realized that he'd seen that argument before somewhere, essentially. And it was in the various for first proof, the Gauss ever gave the law of quadratic reciprocity which is one of the first things he does in this book this position is arithmetic thing. And if you read this book and look at Gauss's proof of quadratic reciprocity the first one that he has in there, then it is, in essence the same as the proof of his calculation of K to a few and the deduction of a bit reciprocity from that so it could be fun for you guys to. I well this book you can't go wrong reading this book it's just got so much amazing stuff in it. But just, that's just a little plug for one of the most famous books in our subjects history. Anyway, I digress. So, every form is a is a strictly equivalent to a unique form, a unique reduced form. So, in other words, yeah, so you make a list of all the reduced forms. And then you've got up to strict equivalence you've got exactly all of the forms. So no no to reduced forms are strictly equivalent and every form is strictly equivalent to some reduced form. So, yes, there is. And if we were to remove the condition of strict equivalence and replace it by equivalence would we lose uniqueness. Yes. I think thanks. Yeah, a lot of things become less nice. Okay, so let's give the proof. It's actually not too hard. So let's first do existence. So suppose given up. So existence of a reduced form in the. So, among all of the possible ABCs which are strictly equivalent to F choose the one for which. A is minimal. So it is a positive integer. So, you know, it's, it's got a lower bound right so just take the least least lower bound, but the greatest lower, I don't know, the minimal. Yeah. I can't find more ways to say it. Choose one for which is minimal among all possible options. So then the first claim is that. That's right. See, it has to be greater than or equal to a. So otherwise, you use the first move. And you can switch a and see right you change the sign of evil we don't care about that at this point right now we're making sure a is less than or equal to see. So that was one of the conditions in being reduced a is less than or equal to see. So that's good. So we can use the. So now, then use the first, then they use the second move to make sure that absolute value of B is less than or equal to a. And in fact you can do better you can do minus a strictly be strictly greater than minus a and less than or equal to a. Well, recall that the second move. It doesn't change a, but it moves be by multiples of two times a. So by shifting over using this you can always make it land between minus a right that's an interval of length to a. And you can choose which edge case you want to keep and we want to keep the one where B is equal to a and not the one where B is equal to minus a. Okay, so we've already satisfied the first condition in the definition of reduce so we just need to make sure the boundary cases that we can have the boundary cases be okay. So if a is equal to see them, I'm getting confused about these case distinctions. Okay. All right, so, so that if these, okay, minus B is equal to a. No, wait, wait, sorry. Actually, if absolute, yeah, absolute value of B equals a means that either a is minus B or a is B, a can't be minus B because we explicitly ruled that out by our choice here. If a is equal to be. If a is bigger than zero B is also bigger than zero so we're okay. So the only thing we need to worry about is whether is if a is equal to see, then we need to be able to change the sign of B, but we can do that using the first new again, which switches and see that doesn't do anything because and see or equal, but changes the sign of B. Yeah, well I won't write all that down, but you can arrange this by just using the first move again. The only case to consider is a will see and then you just use the first move to change the song. So that's it. Well for existence. Okay, and actually that's all we're going to need. But so I'll go quickly through a sketch for the uniqueness. So there were two right so F were strictly isomorphic to G and G prime, and these are both reduced. Well, then G is strictly isomorphic to G prime. So it's enough to show that no to reduced forms can be strictly equivalent. Right. So the key claim for this, which is actually an exercise on your problem set is that if F is a reduced form. And it's equal to a comma B comma C, then a is the least positive value of F. So, since strictly equivalent forms always represent the same set of values. This means that if you have, you know, two strictly equivalent forms then at least the a's have to be equal right. And then you have to do a little more argument to get the things themselves have to be equal you have to see that if you have two strict forms of the same a, then any variables which goes from one to the other has to be a move of type two. And then you can see that when they're reduced this K actually has to be zero because otherwise you'd be moving the out of the window that you're looking at. So, the main claim is this and you need a little extra thing about what possible moves there are when the a's are the same but it kind of falls out from how you prove this anyway. So if we go too much into this uniqueness. The existence is kind of enough for the applications we're going to go into anyway. Okay, any questions about that group. So let me make a let me make a remark. Well as I presented it it's not quite an algorithm right so how to give an F, how to find F prime equivalent F with F prime reduced. You can explicitly find the change of variables. So there's an algorithm, and it's very simple it's a so the algorithm says you can either make a smaller by a move of type one or be smaller by a move of type two. And you can add you can so in practice you can always just do it you try to make a and be smaller and apply moves of type one and two and you'll succeed. So you have to alternate moves of type one and moves of type two. So you have to go back and forth between them. And that's actually kind of it's a kind of it first it seems kind of remarkable the only need to use these matrices, essentially, to make any two forms equivalent. But actually, if you look closer at what's going on here you can use this kind of stuff to see that these matrices generate SL to Z. So what are the coincidence that we only needed to use those two kinds of chains of variables and in fact you can kind of read that off from the argument, plus a little calculation of what the automorphisms are reduced for which is also on the problem set. So, yeah, so it's you know a good way to prove facts about groups like this fact is to make the group act on something interesting, and then you learn things about the groups this is an example. So you have SL to Z on the quadratic form so you can learn that this fact for these two things generate SL to Z. That's those are the only things you need to make a form reduced. So it's not quite obvious that that implies this fact but it's, it's on the problem set. Right. What else. Now let's talk about the main corollary, which is the claim we had so XD is finite. And that, well, so we need to see that the set of reduced f of discriminant D is finite, because every form is equivalent for a reduced one so that set is fine at the next day is fine. Note that let me do a little. So the absolute that D is negative right so if I want, I mean, I could say negative D oops, I used a small D when I went to use a big D absolute value of D is equal to forest AC minus B squared. Now because the form is reduced. So I find that a is less than or equal to C. So this is greater than or equal to for a square so C is greater than or equal to a right. So this is greater than or equal to for a squared minus B squared but absolute value of B is less than or equal to a. So I can also turn that any quality around and make this an a squared as well. Because it's reduced. And that's equal to three a square. So this gives you a bound on a in terms of the discriminant so a is less than or equal to the square root of a negative D over three. But it feels weird to write square root of negative you and negative is positive so I'm writing absolute value of D instead. I mean, negative D when D is negative. Anyway, you know what I mean. I hope. Right. So we have a bound on a. So it's also important then, by definition of reduced again B is bounded by a so absolute value of B is also less than square root of D over three and recall that as positive yeah. So that means that both a and B. So the possibilities for a B are finite. And then but D equals B squared minus four AC then fixes C. In terms of a and B. So here's a corollary. Well, just sort of reading things. So if F one dot the dot F H is the list of primitive forms of discriminant D, then well, H is equal to HD. And every forms of discriminant D is equivalent to a unique fi. So this is how you calculate class numbers. You just look at all the primitive forms of discriminant D there's going to be only finally many of them look at all the possibilities for a and B and so on. And then that's your list and then the number of them is the class number that's it simple as that. So let's do an example, our favorite example D equals minus four. Yes, there is. What do you mean by a primitive form. I cannot recall that them. Wait, what's the word I used. And I say, you mean reduced reduced. Sorry, primitive is the wrong word to primitive means something else. Ah, sorry guys. I didn't know that earlier. No, yes, I'm not reduced. Thank you so much for catching that primitive actually means something entirely different. So, yeah, just. Okay, so D equals minus four. So let's put this into practice so we saw that. So we know that zero is less than a is less than square root of four thirds right less than less than or equal to but it's an integer so whatever. So that that means that a has to be equal to one. And then absolute value of B is less than or equal to a and then if that means that B is either minus one, zero or one. But actually, since we ruled out, I mean, the border case condition gets rid of the minus one unless that's not going to be so relevant. On the other hand, we have B squared minus four AC. So core times one times C is equal to minus four. And that tells you that B squared is converted to zero for which and the only thing is either zero or one that tells you that he has to be equal to zero. So C are uniquely determined and therefore so is C that tells you there's a unique reduced form of discriminant minus four, namely, x squared plus y squared. A corollary of this is an odd prime P is of the form can be written as x squared plus y squared, if and only if P is congruent to one one four. So we did it we proved one of our desired results. And it fell out of the sky I guess but the proof is that we just saw that h minus four is equal to one. So there's only one equivalence class of forms of discriminant minus four on the other hand this was the condition for Peter we represented by some form of discriminant minus four. So, that's good news. I'm going to make another remark which is just a small remark that, you know, we worked on the standard examples of forms of discriminant D so this x squared minus D over four y squared, and x squared plus x y plus one minus D over four y squared, these are reduced. And they are indeed very small, so you can believe it but also you can just check that that is true. So, that's another justified that's so they are the reduced forms representing that equivalence class of principle forms. So that was good. And let me make one final remark. The class number being one is rare. In fact, there are only nine values of D for which it holds. And I don't know it off the top of my head so minus four minus two minus three minus seven minus 11 minus 19 minus 43 minus 67 and finally everyone's favorite minus 163. So this isn't there. This is the theorem of a beginner, who was an amateur mathematician and when he presented his result, nobody could follow what he was talking about and they thought his proof must be wrong. I think 20 years for someone and people came up with another proof for a similar proof and then they finally went back and realized oh wait this finger guy who's actually making sense we just couldn't understand him because he wasn't really speaking our language. He proved this amazing theorem. It was an open question since Gauss. You know, what were the Gauss conjecture there only finally many values, which ones that they are exactly. Yeah. It's pinpointed by Hager and it led to his technique also led to a huge amount of advanced mathematics and important stuff in number theory. So, that's a pretty cool story. All right. So that was what I wanted to talk about for today. And tomorrow we're going to switch gears slightly because we're going to be looking at the same subject from a new perspective. So this perspective of quadratic fields and rings of integers and quadratic fields will have to do a little bit of an introduction to that and then we'll be able to make the connection between the two things, binary quadratic forms and quadratic fields. Yeah, something to do with the fact that they both have the word quadratic in them. Okay. Any questions. I have. Sorry, is this following the following. What's it called Cox's prime to the form x, y squared, or is this just a common argument. I put that book in the list because I know that book has a huge amount of stuff. I put that book. I actually haven't ever opened that book. So, I mean, it's about the very subject we're talking about so I assume everything I'm saying is in there. Like, I remember, like the same argument center so. Okay. I mean, but I have read Gauss I have heard this because you want to share with medicaid so that might be where I'm getting this from and Freddie is confirming that it is a good book. Sorry, can I just make a quick announcement before we can request, but so I think the plan is for the TA session to be right after the cost program activity. So 130 PM mountain time so. I might not be able to make that. But the TAs are so great that I feel superfluous every time I show up. So, if you look at the equals, so we are looking at the different binary quadratic forms, non-equal and binary quadratic forms up to action of SL2Z. That will be a bigger. Now if I look up to GL2Z, that will be as that will still be fine. In fact, the smaller than the class. Yeah, this the court, the set XD would get smaller if you use that. Yeah, why is that not given primary importance why is this class number given a separate name etc why is that not so important. Because if I want for example just to give an example if I wanted to state the analog of this so I said every form is equivalent to exactly one reduced form. That's only true with SL2Z. I would like to say for GL2Z you'd have to, I mean it's not a big deal to make them modifications. So maybe that's not the real reason okay the real reason will come later in that there's a beautiful formula in fact for HD called Dirichlet's class number formula. And it doesn't look as pretty. Right, the formula for the analogous thing with GL2Z. So there's just, it just turns out that when you work through the theory that if you understand the SL2 equivalence theory first then it's it you then you can it is possible to deduce what's going on with the other one, but you can't go to SL2 equivalence and the result for the GL2 equivalence just doesn't look as pretty as the result for SL2 equivalence so it's definitely smart to do SL2 equivalence, I guess for those two reasons. One of what's prettier and two you can deduce the other one from this but not backwards. Yeah. That's that's what I would say at least. There was also a lot of theories you had your hand up at some point. I wanted to ask the same thing that Sundara did. So my question is covered. I can ask another question. Please just so will so this result of what's it, what's his name, Hagnar if I'm not mistaken. I can find where I. Oh, there it is. Yeah, Hagnar. So will this result follow from the class number formula that we're going to derive. It's that class number formula is really amazing. I mean, it's hard to use. I mean, first of all, I don't think I. I don't know any argument why a priori argument for why the value of the formula is even a, you know, it's an integer when you look at the formula but that it should be a positive integer is actually not so easy to prove. And that the only way I know how to prove it is by proving the class number formula and so it's saying it counts something. And it's actually not so easy to make a direct analysis of the class number formula so that's not how it goes. The argument is much more interesting than that. Yeah. Thank you. It uses, I mean, at least in the modern understanding of what Hagnar was doing uses some sort of advanced topics in arithmetic geometry. For example, so in these rings of integers and imaginary quadratic fields they also play a special role in the theory of elliptic curves. Because they can act on elliptic curves and the morphisms ring of elliptic curves can be rings of integers and imaginary quadratic fields. And that's about all they can be. If they're not busy. So there are some special elliptic curves and this gives special points in a modular elliptic curves and these points are now called Hagnar points and it's by studying the arithmetic geometry of these modular spaces that you eventually whittle down the list of possibilities with class and the ones given by Hagnar's list. Thank you. It's a completely different kind of argument from the class number formula. I guess they're called CM points and then Hagnar points or if you map that modelized space to something more explicit than the image of them. CM as in complex multiplication. Exactly. I think I have even even one more question. I kind of come up as I'm looking at my notes questions are great everyone's questions are great. So please just feel free. So we saw in this in today's lecture that we get this nice generating set for SL to see which it's just two matrices actually, given by this action of SL to see on on the forums that we were considering today. So it's generally a good strategy for some groups, if you want to find a generating set that's nice have them act on some set and look at that action. Do you have any other example in mind of like a group action that gives me information about the group itself. There are too many to name. Let's see. This is kind of the classic example or you can make SL to the act on all sorts of related objects the upper half playing a tree can write down. But yeah, I. Oh boy, there's so many. Okay, I mean, there's a kind of generalization of SL to the SL to Z generalizes to a concept called an arithmetic group, which is more or less something like the integer points of some matrix group. So, it could be SL and Z G L and Z or an orthogonal group over Z. These are all arithmetic groups. And for every arithmetic group you can write down a really nice topological space. In fact, Romani manifold on which, on which the group acts, and the question and finding fundamental domains for this action. So that kind of like a notion of a reduced point in this topological space. So, you know, finding a representative for each orbit, oftentimes gives you by techniques of so called geometric group theory gives you an understanding of the group, for example, generators and relations or co homology calculations or things like this. And yeah so one of the best ways to understand arithmetic groups is through this action on this. So called associated symmetric space. So that's a big class of examples but I mean they're really just all over the place if you have any group, one of the main techniques in studying it is just devising some some nice topological space on which it acts. Yeah, what are some other good examples. Are there any popping to your mind. Well, I guess a lot of, I mean maybe maybe right now I guess many groups are sort of defined through like they act on or sort of automorphisms of something. But I mean I guess like. Yeah, I don't know. I think that I think there are like lots of examples of I guess things like mapping class groups and all these things are sort of studied by making them act on some, some object. But I guess maybe very concretely. Yeah, sorry maybe I don't have a better answer right now but I guess in general, I think one studies, one typically studies groups by in terms of their actions on. I guess on a more elementary level isn't there some proof of the silo theorems by making G acting on like subsets. Yeah, I don't know. Yeah, it goes but there's something that's act on some sense of giving cardinality yeah so that's I guess that's a minor example of this kind of phenomenon. Yeah, you have your hand up again. Or even because it just came to me like as a probably very elementary example, maybe the first example we see, could it be Cayley's theorem that even if you have a group acting on itself. You get Cayley's theorem and you can view it as a, well as a subgroup of some SN and maybe you can do some computations more easily if you represent it as a subgroup of a symmetric group. I don't know. I don't really see the action of a group on itself by translation I'm not sure that maybe maybe it's just because it's so basic that I don't notice it being used but I was going to say, sorry, maybe one comment is like if you look at this, I guess there's this book by Sarah trees and thoughts of sort of fun examples of like the compositions of groups and so forth that you get there. When you look at the group apps on like a tree or a graph or something and so. Right. Yeah, actually, um, you show that groups are free in many cases, like if you're a portion free subgroup of I guess I'll, and yeah, anyway, sorry. Yeah. Yeah, this is how you can prove, give a, we said you can give generators for SL to you but if you're a little more refined in your techniques you can actually give a presentation of SL to you to figure out the generators and the relations. You can also calculate co homology and all sorts of stuff. And this technique of having an act on a tree in this case this is a just the one dimensional case of a sort of geometric theory technique of having groups act on a lot nice topological spaces and works for certain low groups such as SL to see And it will be finitely presented SL to see so finally many relations to. So I mean, I give the presentation in the problem set I'm not asking you to prove it I'm just telling you for your information. But if you, well, maybe I'll just say it in words so I mean they're, I think the nicest presentation is, you know, they're, they're, they're two special reduced forms. One is x squared plus y squared and one is x squared plus x y plus y squared and, and what's special about them is they have more automorphism and all the other forms. The first one has automorphism group cyclic of order for the second one has automorphism group cyclic of order six. That was actually already on your previous problem set that particular example. And all the other ones have automorphisms or order to automorphism. And but it's always the same order to automorphism. It's just, you know, minus one zero zero minus one that's an automorphism of every quadratic form just change the sign of X and Y. So I thought that SL2Z is just the so called amalgamated product of cyclic group of order for with cyclic group of order six amalgamated over this common subgroup they have, which is the cyclic group of order two. So, and, and from this amount, I mean amalgamated product is some, you know, push out in the category of groups but what it means concretely is that SL2Z is generated by an element to elements subject to the relations that the first one has order four or the first one has fourth power equal to one the second one has sixth power equal to one, and the second power of the first one is equal to third power of the second one. So that's a presentation of SL2Z by generators and relations. And it's kind of fun that you get this big, you know, almost torsion free group generated by two small cyclic subgroups right. There's non commutative groups that you start with something of order four and something of order six they can generate like an infinite cyclic thing when you combine them in the right way. I mean the group won't I mean it'll contain an infinite cyclic subgroup right this one one zero one, for example, as the magic of non commutative words. So SL2Z you said it's amalgamated product of cyclic group of order six and cyclic group of order four over a common subgroup of order two, if I heard you correctly. Thank you. So you can see Sarah's book for the for a proof of that. There should be some way to do it using the action on binary quadratic forms of all but it's much more vivid to use this geometric language. I guess concrete also more generally like all these so and they're all like finally presented groups, and I don't know if that's something that's sort of obvious to see directly but I guess you see it by like making it act on something that's just going to be fine. Even if you don't have an explicit presentation. I guess for SL2 you can probably sort of do it by hand, right, but for SLN and around. Yeah, I mean it right because to do it by hand you have to solve this problem of giving a fundamental domain. I mean that is there the reduction theory has been generalized to more variables but it's some to make it explicit is not so easy. And when we say that SLNs are finally presented like SLN of Z or SLNs of SLN of Z. Yeah, well, you can generalize a bit, but yeah. Definitely. So, just now you've completed a review before the end of the lecture that study of binary quadratic forms has some relation with what extension of Q. So, is there. And I'm not similar to this for a binary quadratic forms and QE extension so on. No, this is so special. It's this is so special. There's also not an analog over a ring different from Z. I mean, really, if you if you prefer to set up slightly this thing just breaks it's a really special phenomenon. Yeah. I mean, there based on the question that the fact that you get a degree to extension as Dustin had kind of hinted at is more related to the fact that you're working with degree to forms, not that they're binary. Does that make sense. I mean, I don't know. I think you need both of them to be equal to, I think you need both that it's binary and that is degree to, I mean, yeah. That's, that's right but I don't. Well that could be a degree six extension right because it's not, I mean degree to extensions are always galore, but does that matter. I don't know if that matters. Sorry, ignore me. There's it's just really, it only exists in this case so it's kind of I don't think you can point to any one specific aspect of it. Right. So this is just a special phenomenon for the integers and binary quadratic forms. Yeah, I think this, I think this phenomenon extends at least a bit to binary cubic forms, where there's like some correspondence between binary cubic forms and then like cubic rings like rings that are degree three over the integers. Let's just figure out which of these cubic rings are actually like domains, and then look at the like corresponding fraction fields. I think it's not. Maybe it's not like quite as nice as this binary quadratic case, I think you can try and do something similar in this binary cubic. I didn't know about this so what is. What is this. Where is this done. So I don't know like, like the return from same thing but I think in like, I give us people. And I think in this like Davenport Halbron stuff where they're like counting like cubic number fields. And they're like the questions like how many like cubic number fields are there discriminates like bounded. I think the way they get, they get at this is by like comparing with counting binary cubic forms. Yeah, so not at all obvious to me why these would be related okay thank you for the okay I guess this is also what you were saying Freddie more or less. This is new to me is all I can say. Thanks for the information guys. Weird okay. Yeah, I think the correspondence is like, not so straightforward. Jesus. Yeah. That's interesting. I guess weird that. I mean, like in the binary in this case like it really is you're using the fact that you have two variables and the degree is to in some sense I mean it. So I guess there's some reason why you can get rid of one of the variables in the cubic number field case. Oh, you don't let it vary or something you don't let the coefficient vary. Like the way you do it in the quadratic this case is you use the norm function which gives you a quadratic form right in this case. What's going on. Yes, I think in the chat of left area is linked to paper that talks to this in that second section that maybe. Okay. Yeah, so this is giving asymptotic results. I guess it's probably not as exact correspondence. What you're saying. Yeah, very similar to the correspondence is like less with like the number fields themselves and more with just like rings that happen to be like rank three over Z. Right, and then you have to like, yeah, figure out you know which of these are domains and what happens in the in this case as well as just that almost all of them are domains and then I mean like there's discriminant zero or something the only thing you need to worry about. So I can see this. Okay, so they interesting. There is an actual bijection between the set of GL to Z equivalent classes of integral binary quadratic forms and the set of isomorphism classes of cubic rings. Wow okay it's some weird thing it's not the norm anymore. Okay weird okay whoa. Okay remarkable. All right. I stand corrected, but I don't quite see. It doesn't seem like the same construction to my eyes, but it does. Yeah. Okay, any more questions. I think I think I have one maybe can to verify something so when you give me that example of of that arithmetic group action on a on a remanian manifold that remanian manifold is called the associated symmetric space, you said. Yes, that's right. It's nothing. I mean it's nothing obscure I mean the, you just take the same matrix group and take its real valued points. You get some lead group and then you mod out by its so called maximal compact subgroup which you can make explicit I mean for SL and it's s on. You get some contractable space which is nice on which your group acts and. Well, it acts almost freely if you pass to a finite index subgroup and be free and then it doesn't the big, the big subtlety comes because it doesn't act co compactly. And that, if you get when it does act co compactly, you're very happy but oftentimes it doesn't for example for SLM, and then you have to work to like compactify things and like all sorts of stuff. The modular curves themselves give examples of this with SL to z acting on the upper, SL to z acting on the upper half. I mean it's like the, the first example of this. Yeah. Thank you. You're welcome. Okay, so I, as I said I don't think I can come to the TA sessions today but I'll see you guys tomorrow at the lecture. Bye.