 I'll share on my screen and we can get going. Here we are. So we're going to pick up where we left off last time. So the title of the lecture is proof of Hilbert reciprocity. But I will eventually remind you of the statement of Hilbert reciprocity. But I want to more remind you exactly where we ended last time. So I'll pick up a little bit of the discussion from the end of last time. So if f is a field, we associated to this an Abelian group called denoted k2 of f. And its definition was as follows. So it was defined essentially by generators and relations. So the generators are, well, they're parametrized by units in f cross units in f. So pairs of units in f. So for every pair of non-zero elements of f, a comma b, you're supposed to get an element of this Abelian group. And the only relations we impose are that abc should be equal to, it's an Abelian group written in multiplicativity. The group structure is denoted as multiplication. So little dot, say. So you impose multiplicativity in both variables. So abc is equal to abac. And then you impose the so-called Steinberg relation, that ab should be trivial if a plus b is equal to 1. So this is called the Steinberg relation. So formally, you can define it as the free Abelian group on these generators. Modulo, the subgroup generated by, well, the elements you get by multiplying by the inverse of this guy to move it over here, or the inverse of this guy to move it over here, and subgroup generated by these things here. And why did we make this definition of this Abelian group? So the key property was that giving a homomorphism, so for an arbitrary Abelian group a, so homomorphisms, oops, sorry, guys, from k to f2a are the same thing as symbols. So in the sense defined by Akelin that we talked about yesterday, so these are just maps which are multiplicative in each variable and satisfy the Steinberg relation. And this is a pure tautology just based on the definition of the Abelian group here by generators and relations. So and then we stated, oh, and there are particular examples of symbols which are going to be relevant for the discussion. So for any prime p, we have the tamed symbol, which I'll denote dot, comma, dot sub p. So it's defined by saying that a, oh, tamed symbol on q. So we take our field f to be q, and Ab sub p is defined to be equal to, well, minus 1 to the piatic valuation of a times the piatic valuation of b times, then you take a to the piatic valuation of b divided by b to the piatic valuation of a, you note that this has valuation 0 and therefore it makes sense to reduce this mod p. And you get an element in z mod pz cross or fp cross. So that's our target Abelian group for the tamed symbol is the multiplicative group z mod pz cross. And the tamed symbol takes this form. And this formula is not important, but what is important is the key property that Ap sub p is just equal to a mod p when a is relatively prime to p and that, well, Ab p is equal to 1 if both a and b are relatively prime to p. Well, what should I say? Yeah, if a is an integer of co-prime to p, if both a and b are integers, co-prime to p. So what I really should say is that if the piatic valuation of a is 0, and here I want to say that the piatic valuation of a is the same as the piatic valuation of b is 0. OK, so the tamed symbol exists also in p equals 2, but it takes values in a trivial group, a group of one element. But then there is also the two-out-of-kilbrit symbol, which is a symbol which I won't recall too much about for now, but it takes values in the group of signs plus or minus one. And then we ended with stating the following theorem of tape, which completely calculates k theory of the k2 of q. So k2 of q is isomorphic to the direct sum over p of an abelian group denoted a sub p, where a sub 2 is the group of signs, so the target of the two-out-of-kilbrit symbol. And a sub p for p odd is z mod pz cross, so the target of the tamed symbol. And the map is induced by, is given on the p factor, the two-out-of-kilbrit symbol for p equals 2, and the tamed symbol for p bigger than 2. And what do I mean by that? Well, to describe a map like this to the direct sum, the direct sum is a subset of the product. So if you want to describe a map from k2 to the direct sum, it's enough to describe the map to the product, and then it's enough to describe the map to each of the factors. And we said that maps out of k2 q are the same thing as symbols. And now I'm saying on the p factor, the symbol you choose is the two-out-of-kilbrit symbol for p equals 2, and the tamed symbol for p bigger than 2. OK. So now what we're going to do is prove this theorem of tate. So we have a dealing group given by generators and relations, and now we're going to just identify completely. So the one thing we should do right at the beginning, which is say why the map is well-defined, I described it as a map to the direct product, but why does it land in the direct sum? So first of all, this map lands in because of this property that we said earlier that the tamed symbol of x, y is trivial if there are no p's in x or y. So if the vp of x and vp of y are both equal to 0, then the tamed symbol is trivial. And so any rational number has only finally many primes occurring in it. So outside some finite set, any symbol will land inside the direct sum. And k2 is by definition generated by these symbols. And therefore, everything is a finite word in the symbols. And so again, there will always be only finally many primes occurring. And then outside of that, you're bound to be equal to 0. So this direct sum is the subset of the product given by those elements, which are eventually 0. So that's why it lands inside the direct sum. So now the statement is well-defined, and we can actually think about proving it. I should maybe have said in the general discussion of k2, so the generator is given by these pairs a comma b. Their image in this quotient group, k2f, is denoted by curly brace a comma b and curly brace. So again, it's not the set containing a and b. It stands for the element of k2 of f, which is the image of the generator a comma b. So that's just a bit of notation, which we'll be using. OK. So now, how are we going to prove this? We want to show that k2 is a direct sum. So it splits up into copies of these very simple groups AP. It often, when you're trying to prove that an revealing group is a direct sum, it often pays to first look at something, try to prove something weaker. So instead of saying that it's a direct sum of copies of this group, you should produce a filtration where the associated graded of the filtration is given by the APs. So we're actually going to write down a natural filtration of this group and identify its associated graded to the AP. And then we'll see as part of how we argue for that. But in fact, the filtration is split and it's actually a direct sum. So now let me go into details about this. And you don't have to know what a filtration is a priori. I'll just be saying everything concretely. So a filtration is just a sequence of subgroups. But normal subgroups in general, we have an appealing group sum. Okay. So for n greater than or equal to one, we'll define a subgroup, Ln subset K2 of Q as the subgroup generated by the symbols X comma Y where X and Y are integers which have absolute value less than or equal to n. Okay. So what do we have? Well, we have the trivial group sits inside L1, L1 sits inside L2, sits inside L3, et cetera. And all of them sit inside K2 of Q. So this is our so-called filtration of K2 of Q. And there are a couple of remarks I want to make about this filtration. The first remark is that it's an exhausted filtration. So if you take the union of all these subgroups, Ln, you get the full group K2 of Q. So every element of K2 of Q lies in some Ln. Why is this? Well, it suffices to show that every symbol X, Y lies in some Ln because K2 of Q is generated by symbols. So everything is a word in the symbol. So if you just take the maximum of n for all of the symbols occurring in the word, then your element of K2 will live in that subgroup. But then again, so these are non-zero rational numbers. But by the bimultiplicativity, we can reduce to the case where they're actually integers. And then, yes, it is true that any two integers have absolute value bounded by some n, right? You can just take n to be the maximum of the absolute values of the integers occurring. So that's why the filtration is exhaustive. So the next remark is that we can understand pretty, without too much work, we can understand what L1 is. So we're gonna do some kind of inductive argument eventually. And so we should start by understanding what L1 is. So by definition, L1 is generated by, well, I should say non-zero integers, right? Like zero, you're not allowed to have a symbol where one of the elements is zero. You're supposed to be Q cross. So non-zero integers of absolute value less than or equal to n. So L1 is generated by non-zero integers, pairs X, Y, where they're non-zero integers of absolute value less than or equal to one. So there's only four of those, right? We have minus one, minus one, minus one, one, one, minus one, and one, one. However, part of bimultiplicativity is that you're a homomorphism in each variable separately. And this is the identity element in the multiplicative group. So since you have a homomorphism, this thing has to be trivial because of this one and this factor. This one has to be trivial because of this one and this factor, and this one's trivial for two separate reasons. So all of these are trivial. And now I will ask you a question. You are dear participants. Is this symbol trivial? And I won't continue until someone says something in response to this. It doesn't have to be an answer. It doesn't have to be correct, but you must, someone must say something. Yes, Elephorius. To be frank, I'm not sure I understood your explanation right before about why the three symbols below it are trivial. So could you please repeat that? Sorry. I'd be very happy to. So the, so we have a universal symbol with values in K two of Q, right? So saying that this is trivial in K two of Q is the same as saying that for any symbol, phi, phi, we have phi of minus one one is equal to one, right? So certainly this implies that we could just take A to be equal to K two of Q, but it's equivalent by the universal property. And this is because if you fix, if you fix one of the variables and let the other one vary, you get a map from Q cross to A, right? But it's actually a homomorphism by the multiplicativity in the second variable. Now every homomorphism of between groups necessarily send the identity element to the identity element. So this, when you evaluate on X equals one, you have to get one. And that's the explanation. So if you wanna write it out, you could say that minus one, minus one, and minus one one times minus one one by by multiplicativity or by one of our defining relations is a brackets minus one one times one, which is one. And then we're in a group, so we can cancel this and we find that minus one one is the identity. That's also just repeating the proof that if you have a homomorphism, then it sends the identity to the identity. Okay, then thank you for, yes, now someone said something, would anyone like to say anything else about the question? About whether minus one, minus one is trivial or not in K2 of Q. Any activity in the chat? I can't, I don't quite know how to navigate and read the chat without stopping my share. So, oh wait, here we go. You have to click more. Yeah, so there's a claim that it's trivial, but then it was realized it wasn't correct. So how could we ever hope to prove? Let me ask this question, a more general version of the question. How could we ever hope to prove that an element in K2 of a field is non-trivial? It's a group defined by generators and relations. You could find the group and the symbol on it on which negative one, negative one stop one. Indeed. By universal property. Indeed, that is the clever way to prove that an element is not trivial, is to find a symbol on which its value is not trivial. So, are there any symbols on which the value of minus one, minus one is non-trivial? Yes, Althea? If we take the symbol that we defined on R, which was, which gave the values one, unless both of them were negative, in which case it was minus one, it was an isotropic, we can restrict that to Q. And then if we plug in minus one, minus one to that, we get minus one. So it has, so it's non-trivial by that and it has order two. Wonderful, yes. You answered both the question and my follow-up question. So yes, so indeed, this is not trivial, no. And the reason is that if you, yeah, you can calculate the real Hilbert symbol and you get minus one. So if this were trivial in K2, it would necessarily be trivial when you make a homomorphism to any of the alien group or in other words, it would be trivial on any symbol, but we have a symbol, the real Hilbert symbol on which it's minus one. In fact, we also have another one. Minus one, minus one of Q2 is also equal to minus one. And in fact, we know by the Hilbert reciprocity that we're trying to prove that if you're non-trivial on one of the places, then there also must be some other place in which you're non-trivial. And it can't be an odd prime because these are co-prime to P so the value of the tamed symbol there would be trivial. So this must be true, but you can also see from your formula for the tamed symbol at Q2 that it is indeed non-trivial there. And my follow-up question was gonna be, what is the order? So now we've shown that L1 is generated by a single element. These ones don't do anything. So it's a cyclic group. And now Eleftherias has told us that it's a cyclic group of order two. Why order two? Yes, go ahead Eleftherias. To be fair, I mean, I'm probably gonna repeat an argument that I think was mentioned in the chat by Garrett. So I have to be frank here, but the idea is that if I do minus one, comma minus one times minus one, comma minus one, by multiplicativity, this is equal to minus one, comma one, because I can, I mean, I can, and then minus one, comma one, we have already said it's trivial. So it's not the identity, but it squares to the identity. So it's an involution. Yes, indeed. So another way of saying it is that this element has order two. So under any homomorphism, the value will also have order two. And again, we fix this variable and we get a homomorphism from Q cross to K two, Q cross. And so its value on minus one must also have order, well, order at most two, I should say, but we already said it's non-trivial, so that's order two. Anyway, the conclusion of all this is that L one is cyclic of order two, and it's generated by minus one, minus one. Very good. Thank you guys for participating. So, right. Now, the next remark is gonna be that, so we have this filtration indexed by natural numbers N, but in fact, something interesting only happens when N is a prime. So if N is not prime, then there's no difference between L minus one and LN. So there are no jumps in the filtration only occur at prime numbers. So why is this? Well, we have to show that everything in LN minus one lies in LN, but LN minus one is generated by these X Ys for the absolute value of X and absolute value of Y less than N. Oh, sorry. No, sorry, we have to show the opposite thing. I have to show that we have by definition that everything in here is in here. We have to show that if something is in LN, then it's in LN minus one. So that's something in LN is generated by a symbol X Y where absolute value of X and absolute value of Y are less than or equal to N. But if N is not a prime, then that means that, well, so that we can write it as a product of numbers less than N by just looking at it as prime factorization. And then using bimultiplicativity, we'll find that this thing indeed lies in the subgroup generated by X, A, Bs with A and B absolute values strictly less than N. So, like if you had, for example, NN, you just write N as a product of primes and then use bimultiplicativity and you'll find that you lie on the previous subgroup. So the only way you get something new is at a prime number. Okay, so we know what L1 is and we know that jumps can only occur at prime numbers. So the basic question that remains is what is the quotient of Lp mod Lp minus one? That tells us how much the filtration is growing as you increase N. So that'll tell us how big the group K2 of Q is. And now comes the next remark. Well, if you look at the tame symbol, let me write Pp for the map from K2 of Q to Z mod Pz cross for the homomorphism corresponding to the tame symbol. So on the generators X comma Y goes to the value of the tame symbol on XY. So Pp of brackets XY is equal to XY sub P. That's not a remark, that's a notation, but the remark is that Pp kills Lp minus one. So in other words, if you have anything in Lp minus one, then the value of Pp on that will be the trivial element here. And the reason for that is that everything, every integer X with absolute value of X less than P is coprime to P. And we said that the tame symbol kills anything with any integer, any pair which you have a pair X comma Y and they're both integers coprime to P then the value of the tame symbol is trivial because their VP is zero. So we have Pp here, we can restrict it to Lp and then it kills Lp minus one. So that means that we get an induced map from Lp mod Lp minus one to so Z mod PZ cross. And now the key lemma, this is an isomorphism for all primes P. I'm sorry for the, I see that there's some sunlight affecting that. I'll see if I can set up some sort of obstruction. I need to place something right there. That's also going to disrupt my writing so I have to make it taller. There we go. Right. So let's for the moment take this key lemma for granted and use it to prove Tate's theorem calculating K theory of two. So assume for now. Now what we're going to do is prove by induction for n bigger than or equal to two, the map from Lm to the direct sum overall primes less than or equal to n of AP induced again by the due to the Hilbert symbol of the tooth factor and the tame symbol on the P factor. So basically the same map but just restricted to Ln and restricted to the direct sum and projected to the direct sum key less than or equal to n of AP is an isomorphism for assuming the key lemma. So this will suffice because if this holds, Tate's theorem follows by taking the union over n basically. So when you take the union over n on the left by what we showed originally you get K2 of Q and if you take the union overall n on the right you get the direct sum of all APs. Okay. So we're supposed to prove it by induction and therefore we should start by looking at the case n equals two. So we want to show that L2 maps isomorphically to A2 which is the group of signs via the two out of Hilbert symbol. But let's look at what the key lemma says when P equals two. When P equals two, again, this group is trivial. So what the key lemma is saying is that L2 is actually equal to L1. And we already saw that this is cyclic of order two generated by minus one, minus one. And we also saw that the value of minus one, minus one on the two out of Hilbert symbol is the non-trivial element in A2. So we have a homomorphism between cyclic groups of order two which matches up the generators. Therefore it must be an isomorphism. So that handles the base case of our induction. So now assume the claim for n minus one and try to derive it for n. So if n is not prime, both sides are trivial well, sorry, well, we have, sorry, I mean, there's no jump. So we have Ln minus one equals Ln by a previous remark but also clearly direct sum over all primes less than or equal to n of AP is equal to the direct sum over all primes less than or equal to n of AP. That's okay, well, it's the direct sum over the same set. So it follows, so if n is not prime, it follows immediately from the inductive hypothesis that the claim, yeah, that the claim holds. If n is prime, then let's look at, I'm gonna use the language of short exact sequences. So, because I don't know, because that's how I think, and well, I don't know. So, well, so we have a, we have Ln, we're interested in going from Ln minus one to Ln. And the difference between them is controlled by the quotient, Ln minus, mod Ln minus one. So in short exact sequence terms, we have a short exact sequence like this where this map is the inclusion and this is the projection onto the quotient. On the other hand, we can write an analogous kind of short exact sequence for this direct sum here. So we have zero going to the direct sum over all primes less than or equal to n minus one of AP going to the direct sum over all primes less than or equal to n of AP. And then what's the quotient here? Well, we've just added one more factor, namely an, n is a prime, you remember. So I put an here with the knowledge that n is a prime. So we have this short exact sequence and that in fact, we can make compatible maps between these short exact sequences. So the map we know is an isomorphism here extends to the map we want to be an isomorphism here. Exactly because of the claim that phi P or phi n, I guess annihilates l n minus one. That's what makes this diagram commute. That's what gave us the induced map, which was an isomorphism by the key lemma. And now we just use the general fact that if you have a map between short exact sequences where the outer terms are isomorphisms, then the middle term is also an isomorphism. So, or in other words, if you have a map between, yes, there it is. About what you just stated, isn't this, would this be like one of those, what do they call like three lemmas or five lemmas or like those things in McLean's homology basically, those things at the start? There's, yes. There's something called the kernel, co-kernel. I mean, there's a, you should probably just prove this thing directly, honestly, I think, but there's this more general fact called the co-kernel, kernel, co-kernel, long exact sequence or the snake lemma or something. Anyway, it's, if you have any map between short exact sequences with no hypotheses about anything being an isomorphism, you get a long exact sequence which goes, zero goes to kernel, goes to kernel, goes to kernel. And then there's a non-trivial map going to the co-kernel of the first guy again and then co-kernel, co-kernel, I'm going to zero. That's the snake lemma, right? That's the snake lemma. And you can read off the snake lemma this statement, but I mean, so it's just saying that you have a map of a billion groups and then you have a subgroup that maps isomorphically, this guy which maps isomorphically to a subgroup here and the induced map on quotients is also an isomorphism. Then this map of a billion groups you started with is an isomorphism and that maybe one should just prove directly, I don't know. Thank you. That said, I think rather than in terms of the snake lemma, actually, as even worse, I think about it in terms of drive categories in my head. So that's how I encode this. Oh, it's distinguished triangles, yeah, okay, fine. Yeah. But yes, right. So that's the inductive hypothesis is exactly, in other words, oh, sorry, the inductive step is exactly given by the key lemma here. So we need to, so this is what we really need to prove. So how are we going to, I'll restate the key lemma over on another page. So, oh shoot, the sun must have moved. Who knew that would happen, right? So we need that LP mod LP minus one maps isomorphically via PP to, oops, to Zeeman PZ cross. Wow, I'm sorry guys, my pen isn't, there's some connection broken between my brain and my hand. Okay, so let's first argue surjectivity. So we need to know that if you have anything here, then it's hit by something in LP, right? But the mod LP minus one doesn't matter for showing surjectivity. So everything in here is represented by, so Zeeman PZ cross by an integer X, which is strictly between zero and P. Okay, we know we can do that, right? The least positive residue mod P is always between zero and P and if it's a unit, then it's also strictly bigger than zero. Then we claim that the symbol X comma P, which does lie in LP, well, it maps to a class of X and Zeeman PZ cross under this purportive isomorphism. And well, this is the first property of the tamed symbol that I recall. So X, P is always equal to X mod P. So that's a, so this claim is just a restaining of this property of the tamed symbol that I recall. So that shows surjectivity. Now you might think that I'm about to prove injectivity thereby concluding the argument, but it's actually a little tricky to try to directly attack injectivity. So instead I'll show something a little stronger or seemingly stronger. So next step will show that these things that we just used to prove surjectivity, they give all class, all of the classes in LP mod LP minus one. So every class, everything in LP mod LP minus one is equal to sum X comma P. Or in other words, anything in LP is congruent to X comma P, sum X comma P mod LP minus one. And that would be enough because well, here's one possible argument. This means, for example, that this quotient has size less than or equal to P minus one because that's how many of these guys there are. But because of the surjectivity, we're also known as size greater than or equal to P minus one. So therefore these are a billion groups of the same size and we have a surjection between them. So it must also be an isomorphism, but or you could also just say that this is explicitly giving an inverse to our purported isomorphism here. So yeah, so it's enough to show that these guys are exactly giving us all the classes in this quotient group. And we'll do this in two steps. So in the first step, so the XP form a subgroup of the quotient group LP mod LP minus one. So they are closed under multiplication, which is something that has to be true if they're giving all the classes. And then step two will be the XP generate LP mod LP minus one. And so if you have a group, which is generated has a group by a subgroup, then it's equal to that subgroup, right? So the subgroup is closed under multiplication. So you don't get anything more when you generate by it. So it is indeed enough to handle steps, steps one and two separately. And okay, so for step one, well, we need to look at, so take zero less than X less than P and zero less than Y less than P. We need that, well, we need to understand X comma P times Y comma P. And we need it to be equal to Z comma P for some Z or not equal to, but a congruent mod LP minus one for some zero less than Z less than P, right? So then it will be a subgroup. So how will we do this? Well, note that this is equal to X Y comma P. So let's, now if we take X and Y, they're between zero and P, but their product need not necessarily lie between zero and P, right? It could be much bigger. Like if they're both equal to P minus one, then of course it's way bigger than, but we can always reduce it again, mod P, take the least positive residue. So we'll let Z be the least positive residue of X Y mod P. So then certainly it lies in the required range, but we haven't verified this congruence yet, but let's write out. Well, in particular, Z is congruent to X Y mod P. So this means that we have X Y is equal to Z plus P times some Q, I don't know, or Q is an integer or a natural number, I guess. And now I want everyone to pay very careful attention to the next step because it's in some sense, the crucial step, and it shows the power of the Steinberg relation. So let me go back to the very, we have to use the Steinberg relation at some point, right? I mean, it was in the definition of our group K2 and here's where it comes in. So I'll remind you that the Steinberg relation is this one that says that the symbol A comma B is trivial if A plus B is equal to one. So anytime you have an equation A plus B equals one, you get a relation in K2, but it's better than that because we have a field. So anytime you have a relation A plus B equals C, any additive relation whatsoever or everything's non-zero, you're gonna get a relation in K2 because you can just divide by C and then you'll get A over C plus B over C equals one. And then you get a relation in K2, right? So that's what the Steinberg identity is doing for you. It says any additive relation tells you something about K2. And here's an additive relation, X, Y is equal to Z plus P times Q. So let's apply this general principle to that. So let's rewrite this as one is equal to Z over X, Y plus PQ over X, Y. P times Q over X times Y. It's weird that I'm using dot here and not here, but it's just multiplication in both cases. So now from the Steinberg relation, we deduce that Z divided by X, Y comma PQ divided by X, Y is trivial. Okay. Now, we will be done if we can show that this is congruent to Z divided by X, Y comma P. Sorry, comma P mod L, P minus one. Why is that? Well, so if this is congruent to that mod L, P minus one, then that means this is trivial mod L, P minus one. But by multiplicativity, that's the same thing as saying that Z comma P is congruent to X comma P Y times Y comma P mod L, P minus one, which is exactly what we wanted to show. So if you restate this using by multiplicativity, it's saying the same as saying Z divided by X, Y comma P is congruent to one. So we do just need to establish this congruence here. But now look, X and Y are smaller than P. So any symbol and Z is also smaller than P. So all of these guys have absolute value less than P and these guys have absolute value less than P. So by by multiplicativity, if we want this to be equal to this, it's gonna be enough to show that Q also has absolute value less than P. So we just need absolute value of Q less than P. Yeah. So, and now this is just very elementary. We can just write the formula for Q and check it. So Q is equal to X, Y minus Z divided by P, right? So, well, Q is a positive number, by the way. I mean, everything's positive here. So I just need to show Q is less than P. But this is less than X, Y divided by P and X and Y are both less than P. So this is less than P squared divided by P, which equals P. So indeed this Q, the thing you're dividing by, and when you're doing this division algorithm thing is less than P and that finishes the proof of the key lemma, which finishes the proof of Tate's theorem. So QED takes calculation. All right. Any questions about that? Can you go through this last, why does Q less than P imply the last equivalence? Yes. So I'm gonna ask you to do something in your head, okay? So we're gonna fix the left-hand side, the left-hand side, the left-hand entry of the Steinberg symbol here. And we're gonna use bimultiplicativity to write the Steinberg symbol. So this Steinberg symbol is equal to Z divided by X, Y comma P times Z divided by X, Y comma Q divided by Z divided by X, Y comma X divided by Z, oh no, divided by X, Y comma Y, right? And now all of those things, if Q is less than P, then all of those things that I just said, except for this one, are gonna line LP minus one because all of the terms involved in both variables have absolute value less than P. So the only one that doesn't is this. So that's all that remains when you use multiple, you use multiplicativity in the second variable is what I'm saying. And I'm looking at the clock and seeing that, well, there's not gonna be much time for the second half of the proof, but that's all right. We're gonna take things at their natural pace here, not gonna try to rush anything. Indeed, yes. So we've established Tate's theorem. K2 of Q is isomorphic to a direct sum of P of AP. So there's another way of rephrasing this, I guess. So what does this mean in terms of symbols? So a symbol is a homomorphism out of K2 of Q. And now K2 of Q is isomorphic to this direct sum. So a symbol is the same as a homomorphism from direct sum over P of AP to A, which is the same as separately specifying a homomorphism from AP to A for every prime P. So, yes, yes, please. Did we do step two in this previous proof? No, thank you. I forgot to do step two, thank you so much. Yes, yes. Step two, I will add to the exercises. Yes, thank you for that. Now it's your problem. No, yep, okay. Oh no, now I'm out of focus. Let's try that again. So rephrasing. So for every symbol, P, Q cross, cross, Q cross, to an arbitrary dealing group A, there exists unique homomorphisms. I don't know, Phi P, no, I already used that notation. Well, I'll just write them AP to A, maybe I'll call them FP for all P, such that Phi of X comma Y is equal to the sum overall primes P of, well, let me write, I'll write the two addict term first. So X Y comma Q two, F two of X comma, plus sum overall P greater than two of FP of X Y comma P. So in other words, every symbol can be expressed in terms of the symbols occurring in the statement of taste there. I'm the two addict Hilbert symbol and then the tame symbol at odd primes P. Now you might have, now there was one symbol that we already knew that was missing from the statement of taste theorem. And that's the real Hilbert symbol. So, but now we know it must somehow be accounted for because I mean, every symbol is accounted for when you calculate K two of Q. So let's apply this to, yeah, P of X Y equals X Y comma R, the real Hilbert symbol. So that is a symbol whose target group is plus or minus one group of order two. Now what this rephrasing tells us is that there are for every prime P, there must be a homomorphism from A sub P to the secret group of order two, such that the real Hilbert symbol is expressed in terms of the tame symbol for all primes P. But now, you know, AP when P is odd is Z mod PZ cross and every map from Z mod PZ cross to the group of order two, factors through the Legendre symbol, which is also a map to plus or minus one because the kernel of the Legendre symbol is exactly the squares, which always die on map only map to a group of order two. So, instead of specifying the FP, we can just specify a map from plus or minus one to plus or minus one. And instead of saying we factor through the tame symbol, we can actually say we factor through the P out of Hilbert symbol. So in the case where the target group has order two, we can replace the tame symbol by the Hilbert symbol. So, A has order two. So plus minus one has order two. So we get, oh, and I used additive notation. I'm sorry guys, I said I was gonna use a multiplicative notation in A all the time. And I got myself confused. So we get that, well, there exists epsilon P in zero one for all P such that the real Hilbert symbol X, Y is always equal to the product overall prime as P of the P out of Hilbert symbol, X, Y, Q, P raised to the epsilon P. So, the real Hilbert symbol, yeah, for all X, Y and Q cross. So the epsilon P is just encoding the possible whether this homomorphism can plus or minus one to plus or minus one is trivial or not. And so now the game is going to be to finish, Hilbert reciprocity says that epsilon P is equal to one for all P. Or in other words, all the primes are contributing. We try to express the Hilbert symbol real Hilbert symbol in terms of the piatic Hilbert symbols. So that's an equivalent form of the statement of Hilbert reciprocity, just this product formula here. So how are we going to show that? Well, well, we're just gonna plug in a bunch of rational numbers and see what the relation has to be as best we can. So plug in minus one, minus one, then we deduce that minus one is equal to the, well, you know, the two added Hilbert symbol to the epsilon two times, but then all the other odd primes are the value of the piatic Hilbert symbol on minus one, minus one is gonna be trivial for all odd primes P because it's a tamed symbol. So that one's not contributing there. So we deduce that minus one is equal to minus one to the epsilon two, which means epsilon two must be doing something. Yeah, so it has to be plus one. So the two added Hilbert symbol has to be contributing. Otherwise, this formula couldn't be valid when you plug in minus one, minus one. So that's good news. Now, if P is a prime which is congruent to one mod four, we can plug in minus one comma P and then over here, the real, you know, the real Hilbert symbol is gonna be trivial here. So we'll get plus one is equal to, and then the two added Hilbert symbol is equal to minus one to the P minus one over two. Wait, just like I'm getting myself confused. I better check my notes. Oh no, sorry, P is congruent to three mod four. Yeah, that's why I was getting confused. Minus one to the P minus one over two times. And then the only other thing for similar reasons which can contribute is the piatic Hilbert symbol. So all the other values will have to be trivial because again, it's the tame symbol and it'll be co-prime to every other odd prime. But this is equal to minus one. So this must also, oh, sorry, add to the epsilon P. So this must also be equal to minus one which shows that the epsilon P is also there. It's also equal to plus one. So we've shown that epsilon P is equal to, is there, is equal to one for two and for any prime congruent to three mod four. Now if P is not congruent to three mod four, it's either congruent to one mod eight or five mod eight. And for five mod eight, you can do a similar deduction by looking at the value two comma P and you'll be able to deduce that epsilon P is equal to one. So the epsilon P is there. So the only remaining case is P congruent to one mod eight protecting that the piatic Hilbert symbol has to contribute. And there you need to do some extra work which is actually quite difficult. So there you need to use a lemma which says that if P is a prime congruent to one mod eight then there exists a prime Q less than P such that P is a quadratic non-residue modulo of two. So P is not a square mod Q. If you have this, you can look at the symbol P comma Q and use induction to show epsilon P has to also be, has to also exist. Because the only thing, the only other symbol, the only, the piatic Hilbert symbol is trivial unless you're P or Q here for this P comma Q here unless you're at the prime P or Q here and then at Q you're gonna be non-trivial. Therefore you have to be non-trivial at P by the product formula, which will mean that which means that epsilon P has to be contributing. But proving this lemma is actually quite difficult. And I had originally thought we'd actually have time to do it, but that's really not the case. So maybe I'll actually put it in the exercises with the, you know, just so you guys see at least a sketch of the argument. And if you want to work it out, you can. It's a very clever elementary argument but it does take a lot of work, right? So that was it for Hilbert reciprocity. And that's it for today's lecture. Although, of course, I'm willing to take questions. Marianne, that's correct. Yes, I was responding to the chat. Eleftheres, you have a question? Yeah, I was wondering because I'm hoping to kind of go through this proof again because it's a bit lengthy. Is there any specific textbook that you're following for this particular proof or? Absolutely, I'm very closely following. I'll put it in the chat. Now I forgot what the book's called, just a second. It's a book by Milner on algebraic k-theory. Now I have to, when we see what exactly it's called and exactly what chapter. Chapter, oh darn. I think it's chapter seven of Milner's book on algebraic k-theory. Is it the introduction to algebraic k-theory? Yes, yeah, that's what it's called, thanks. I'll have a look, thank you. Chapter seven, I think. I don't quote me on that. It's something about, you'll see if the one that's about k-two, yeah. Yep, and the book is freely available online, I believe. So if you Google it, you'll find it. Yes, another question? Eleftheres? Yeah, it just came to me. I think I wanted to ask it probably from the last time because we've mentioned this construction of a universal symbol on a field, which I guess if we think of it as an initial object in the category, it would be unique up to some isomorphism of symbols. In case our field is local, because we've also constructed the Hilbert symbol for local fields, would the Hilbert symbol be the universal symbol in the case of a local field? That's an excellent question. The answer is no, but for bad reason. So oftentimes when you consider local fields, QPR, you really have to consider them as topological fields. So you have to take the topology into account to get clean results. And it turns out that if you just take the bare definition of k-two of QP or k-two of R for that matter, you can prove, this is actually a good exercise. Thanks for mentioning it. You can prove that you get an uncountable abelian group just because, well, yeah, I mean, there just aren't enough relations to collapse all of the pure symbols down into anything reasonable. That's actually a very nice exercise. Not too easy, I suppose. Well, anyway, I don't have to put everything in the exercise sheets. But if you ask for the analogous topological abelian group classifying continuous Steinberg symbols with values in a Hausdorff topological, you have to add Hausdorff, with the values in a Hausdorff topological abelian group, then the answer is yes, the Hilbert symbol is the initial one, is the universal one. So once you add the continuity into the statement, then it becomes true. But if you're just looking at QP as an abstract field and Steinberg symbols as apt to an abstract abelian group, then it's actually not true. Because basically, and I mean, it follows from what you said, because the Hilbert symbol maps into Z mode 2Z, whereas if K2QP and K2R are actually uncountable, then obviously they're not isomorphic to Z mode 2Z, right? So the symbols can be isomorphic. Correct, yeah. So there are some exotic symbols, but they break the natural structure of QP and R. They're not continuous. That's kind of, so there are other symbols, but you don't want to look at them, in my opinion. Thanks. Sure. And that, by the way, the statement about continuous Steinberg symbols is also in Nohner's book. Yeah, same chapter. I have a question. Less time you told us that you didn't like this proof. Yeah. Why is that? Oh, yes, yes. It's because what this proof really gives you is some relation between the real Hilbert symbols and the chaotic Hilbert symbols, right? And then you have to use clever tricks to see that it's the correct relation, that all of these epsilon p's are equal to one. And in particular, this lemma I didn't prove, I mean, yeah, the proof is just a very clever, I mean, it's somewhat natural, I guess. Anyway, I feel like if you're giving a proof of Hilbert reciprocity, your proof should somehow directly explain why that relation holds. It shouldn't say that some random relation has to hold and then by tricks figure out that it must be the correct one. That's why I don't like, I didn't really mean I don't like it. I do like it. It's just, for me, it's not a satisfying proof of Hilbert. Does there exist such a satisfying proof? Yes, there are at least two. So the whole formalism of global class field theory as structured by Arton and Tate and Iwasawa is, you know, it makes it into a statement of a similar form as Hilbert reciprocity and gives sort of natural proofs for it. And when you see there's an incident, some insane specialization of it. The most non-trivial, the simplest non-trivial case is Hilbert reciprocity. So, but that involves this whole co-emological formalism for novel fields and so on. There's also another proof that directly explains that, which is really different in nature, which is based on, well, there's this subscript two and K two, right? And what does that stand for? Actually, so the correct definition of K theory in general was given by Quillen and he defined that, you know, KN for any N should always be the end homotopy group of some space called the K theory space. K of N. And you're writing down right now? I'm writing down for my own benefit. You're free to do the same on your end. So, yeah, so I'm writing down just what I'm saying. I'm not writing down anything more. So the KN group is the end homotopy group of some space K of F and there's actually a proof of Hilbert reciprocity, which, you know, does the reciprocity at the level of these K theory spaces. And then it's much harder to cheat and give tricky proofs because, you know, to prove something about spaces, you really have to pin it down and understand why it's true. But that proof is actually uses a lot of algebraic topology and it's quite complicated to understand. Although in my mind, it's quite, it's very natural and it has good explanatory power. But yeah, so those are the only two proofs I know that I consider the good proofs. Yeah, Ilarphaeus? So with regards to what you just said about the, you know, Quillen's, you know, idea that K theory can be thought of as, well, KN is some N's homotopy group of a space. Would this apply to like K theory of, okay, can this idea be applied to, because I mean, I've heard of K theory of like a ring or an additive category or a topological space. Would this apply to all of those K theories, like for all those cases? First, two, but not the last. Not the topological space one. Yeah, that requires some additional machinations. I mean, that's a little bit of it. Ironically, yeah. Like you think that that'd be the easiest to put into this framework, but no. Yeah. And K zero would be the connected components. It would be the pi zero of that space. That's right. Okay. I see. Thank you. Yeah. And by the way, also this relates to your previous, your comment on the previous lecture of Therios where you said that the product formula is analogous, the product formula for absolute values is analogous to the Hilbert reciprocity law. Well, in this K theoretic approach, it is just two different specializations of the same space level result. Like you can take pi two and you get the Hilbert product formula and you can take pi one and you get some form of the product formula for absolute values. So there you can really see them as part of a family in a quite precise way. I don't know. Maybe Achille knows more proofs of Hilbert reciprocity. I mean, oftentimes you prove it by proving quadratic reciprocity, but that destroys the symmetry of the situation as well. Like it doesn't involve all contributions from all friends anymore. So there are actually very few direct proofs of reciprocity. Yeah. Yeah, we're also just gonna comment. So like I guess, for example, this proof doesn't directly generalize or end this computation, doesn't directly work for other number fields, right? Right. So this is a, yeah. This is a very interesting story. Yeah, you can ask what the analog of Tate's calculation is for an arbitrary number field and it's very, very interesting. So there's nothing as exact, but there's something quite beautiful going on. Maybe I can't resist telling you guys don't stick around at least. So let me share my screen actually because it's a bit of a story, but trust me, it's worth it. So you can, so in K2 of Q, there was this weird thing that we took the tame symbol at odd primes and we had this extra thing we had to take up at the prime P equals two. We had to put the two out of Hilbert symbol in by hand. But let's rephrase Tate's theorem just in terms of tame symbols. So we have K2 of Q mapping to the direct sum over all primes P of Z mod PZ cross. And then the kernel of this is actually in Quillen's language can actually be expressed as K2 of the integers. So Quillen's definition works just as well for arbitrary commutative primes as it does for fields and it's some general thing tells you that, yeah, well. And so Tate's calculation is equivalent to saying that K2 of Z is Z mod two or plus or minus one. So it can be equivalently phrased as a calculation of K2 of Z, but now it's a much simpler abelian group you're talking about just the abelian group of signs. Now for a general number field, you have something similar. You have K2 of F mapping to direct sum over all non-archimedean places. So you're not dividing infinity is I guess how people you would do it and then the residue field, the units in the residue field. And again, it's surjective and the kernel is again K2 of the ring of integers of your number field. And then the question is, what's this guy? And it's not known in general what this guy is as an abelian group, but it's known that it's finite and it's almost known that it's value, its order is up to some trivial factor, trivial and easy to understand factor. The same thing as the value of the Dedek and Zeta function of F at the point minus one, which is only well defined because the Zeta function has analytic continuation and happens to be a rational number. And then that's some trivial factor you multiply with cancels the denominator and makes it an integer and it's exactly the same as the order of this group. So this is at least known up to powers of two by work of wiles and it's expected that the powers of two are correct as well. Well, I didn't tell you what the trivial factor is but it is quite easy to understand. So for example, when you take F equals Q then you have the Riemann Zeta function and Zeta Q of minus one is minus one over 12. I don't know how well known that fact is but and then K2 of Q is two and you can work out the trivial factor and it's minus one over 24. Oh, no, sorry, it's minus 24. And it's not, I mean, you might think this is a little bit mysterious but this is a 24 as a very, I mean, the reason it's 24 here is basically, ah, nevermind, it's too long a story. It's very easy to calculate the 24 for your. Yeah, it's cause 24 is the smallest number such that everything in every unit mod 20 is the largest number such that every unit mod 24 square is to one. Every element in the unit group has order two. That's a good little exercise in number three. 24 is the largest integer with the property that every unit has order two. And that's essentially why the 24 appears there. And then so you're really relating to mysterious quantities with each other in this isomorphic, in this equality for a general number field but it doesn't tell you, it doesn't tell you about the structure of this as an abelian group, right? It only tells you its order. Yeah, but anyway, this is a whole part of a whole big relationship between special values of zeta functions or L functions and orders of various L abelian groups that are naturally attached to number fields or more general objects. And it's one of the most mysterious and kind of fascinating aspects in modern number theory. Because there's no reason why should they be related, right? I mean, it just comes right out of the blue, right? Yes, so there is. So about what you just wrote. So the first exact sequence that you wrote for with K to Z, that would be a split exact sequence, right? Yes. By Tate's result. Yes, indeed. Because it's, and is it like, and for the second thing that you wrote with a dedicated zeta function, is it like, in the general case for number fields, does a similar thing happen as in we're looking for some quotient that is just, I guess, two torsion or something? Because in the case of Z mod 24 Z star, you said that every element has order two. And so, it's like a two torsion thing. The fact that it's Z here and Z here and Z here is actually a bit of a red herring. So, the analogous thing, it doesn't involve looking at OF mod something or another. I mean, it has to do with how F sits in a relation to cyclotomic fields. And then, I mean, it's a bit of a long story. Okay, well, we'll hear more tomorrow. And don't forget other things today. Yes.