 Welcome everybody to day eight of PCMI USS. We now have a new speaker, Dustin Clousen, who will continue our Kiehl's lectures or maybe it will be completely different. We'll see. And let me remind you that today we have another cross-program session, Modular Origami by Glenn Whitney. And I again encourage you to go for that. That is at one o'clock to PCMI. All right, Dustin, go ahead. Thank you. Okay, thank you. Hello everyone. I'm going to, yeah, in some sense continue where a Kiehl left off, but also in another sense start a different thread of topics. Can everyone see the sheet of paper where it says lecture eight Hilbert reciprocity? Yes. So, well, let me just give a brief recap of what we've seen so far in the course and then I'll say what we're going to do. So, up to now, what kind of stuff have we been talking about? Well, one major topic was the theory of quadratic forms over general fields. We've had things like Witt's theorems and so on. You know, every quadratic form can be diagonalized and then there's a certain set of moves for going between the diagonalizations of the same form. So about classification of quadratic forms, what you can say over a completely general field, but then we also saw some things for certain more specific fields. So we saw for finite fields. For example, Fp, we gave the classification, and that was fairly simple. And we also saw for pediatric numbers or real numbers, or more generally local fields. We also gave the classification. And so what about from now on, what are we going to be talking about? So from now on, we'll be doing number theory. And that, well, the simplest way of saying what that means is more or less our main object of interest is our field of interest is going to be the rational numbers. And we want to explore the features that arise when you take the rational numbers and in fact later on we'll take later on in my lectures we won't take the rational numbers anymore we'll get even more basic so to speak. We'll take the non-field. It's not a field, right, but we'll be looking at quadratic forms over the integers instead and then there are some quite interesting phenomena that arise, which distinguish it greatly from the case of the rational numbers. And but the work that we did in the course in the first half will not be in vain, and in fact there's an important general principle, a general principle in number theory, which is roughly speaking that says to study the rational numbers, you should first study the q, you know, it's non-trivial completions. So the periodic numbers and the real numbers, then then see how they fit together. And what we're going to be doing in these first few lectures is sort of this general principle applied to the case of quadratic forms. But more specifically, we're going to take up the discussion of Hilbert symbols that Keel left off with. So, well, that's something that we had for QP and R and now we'd like to see how they fit together over the rational numbers. So let me just give a brief recap of Hilbert symbols. So, if you have A and B, two non-zero periodic numbers, then you can associate to this a sine denoted A comma B sub QP. So it lives in the two-element abelian group of signs, and it's defined by, it's plus one if, well, a certain quadratic form is isotropic. And just to write that concretely, it's A x squared plus B y squared equals z squared has a solution, which is not equal to the trivial solution 0 0 0. And it's equal to minus one otherwise. So that's the definition. And then, and then you do the exact same thing for R replacing QP. So you do it for all the non-trivial completions of Q. Those are parametrized by prime numbers P, and then the usual absolute value given the real numbers is a Keel discussed in one of his lectures and the definition of the Hilbert symbol is uniform. Okay. But then there's the question of how to calculate it, and that's something we also saw, and you break it up into three cases. One, the first case is the case where P is odd. Then I'm going to write the formula for it in a slightly different way than a Keel did. Then a B QP can actually be calculated in terms of the pediatric valuations of A and B so it's minus one to. Well, not just in terms of that but using that valuation of a times the valuation of B multiplied by, and then you take a funny thing so you take a raised to the power given by the valuation of B and divide it by B raised to the power of the valuation of A. And you can note that these two things here A to the valuation of B and B to the valuation of A have the same valuation, because in both cases, it's the valuation of A times the valuation of B. So, when you take their quotient, it has valuation zero, which means it's a pediatric integer. All right, well, pediatric unit in fact. And then particularly you can reduce it mod P and ask whether or not it's a square mod P. So you take the Legendre symbol of the residue of this thing lives in ZP cross. You take the Legendre symbol of the residue of this thing mod P, and that's that gives you a sign plus or minus one and then you multiply it by this sign here. So the formula for the one way of is the equivalent way of writing the formula for the Hilbert symbol when when P is odd. And I should just recall that I here is a VX. This is the pediatric valuation of the pediatric number X. So in other words, if you write X as so in other words you pull out as many factors of P as you can so to speak, although the X one it can be positive or negative so this is an integer. So that what's left is a, a pediatric unit, so a unit in the pediatric integers. So it has no P's left in it. Right. So that's the general formula for odd prime P and now I just want to make a couple of remarks about this so that the two important special cases of this formula. So, so let me give an example, example, which Achille also talked about but it's going to be important so I'll remind you of it. So if, if a and B are indeed units in ZP. So they're unit pediatric integer units, then the Hilbert symbol is is equal to plus one. Because, well, then this is the same thing as saying V of A is the same as V of B is the same as zero. So the pediatric valuation is zero, and then you just see that all the relevant terms just drop out. Everything becomes trivial. And Achille also give a direct argument for this lecture. So that's another example. And another example is that. So if, if A is in ZP cross, and then we take B equals P. So in other words if we're looking at a comma P QP, then this exactly gives the Legendre symbol of the aim, a module. Well you take a modulo P a lives in ZP cross you take it modulo P you get something in FP cross, and then you take the Legendre symbol of that. And that's a plus or minus one. This is a plus or minus one and you're the same. So in some sense the Hilbert symbol, can be written in terms of the Legendre symbol, but it also recovers the Legendre symbol when you input this simple input here. So, yeah, okay. And then, now let me make an important remark. So, this this thing here is not true for P equals two. So this property that if they're two if they're, if it's a pediatric integer unit, then the Legendre, the Hilbert symbol is trivial is not true for P equals two. In fact, if a and B are in Z two cross, then you can you can write a formula for a B comma q two and it's not complicated but it is also non trivial. So minus one to the a minus one over two times B minus one over two. And also if we can also give the analog of the second formula, but for people's to some just slightly more complicated so you take a squared minus one. This will be divisible by eight, so you can divide it by eight and you still get a tool. And then you take minus one to that power. So, oh no, I realize I the sun just came out in Copenhagen a very rare occurrence which I did not anticipate. So I'll have to just close the blinds. So, and in fact there's also a general formula you can write down for the Hilbert symbol, but it actually it'll be enough for us to know these two special cases. They're not a combination of these two in general. Okay, and that's that's all we'll need to know about the two edit Hilbert symbol and then let me also just recall for the real numbers that a br is equal to minus one if and only if both a and b are negative. And otherwise it's it's plus one. Right. Okay, any questions about that. In the recap of what he'll talk about a little bit tough for me to keep an eye on the chat, but I assume the key was doing. Yes. Thank you so I was I was wondering in the formula that you wrote for the for q2 and for the to add a unit you in the exponent you have a minus one over to, for example, what what would this expression mean for for it to what do you mean by that for like for a to add a unit. What would that mean, the same minus one over to the very good question I'm glad you asked it because it does require some commentary to say what this means. So, A is if you have a two added unit, then what you know, but well you have the two added integers map to Z mod to Z, by reduction mod to a string homomorphism. So if you have a two added units, so if you have a unit in this ring, it gives you a unit in this ring, Z mod to Z cross. So that's just saying the units units mod to where it's just one right. So in other words, being a unit and it's actually if and only by Hensel's a version of Hensel's lemma if you're a unit mod to then you're a two added. So, saying you're a two added unit is exactly the same thing as saying you're congruent to one module or two in this ring Z two. So it's the same thing as saying that if you take, you know, X minus one you can divide it by two and you still get a two added integer. And then we use this homomorphism from Z two to Z mod to Z. So we can extract a sign, not just out of any integer but also out of any two added integer, so or even odd criterion, so to speak. And then that so this makes sense as an element in Z mod to Z and that allows you to make sense of all of this nonsense here. Was that a coherent explanation. Yeah, I think I get it. Thank you. Yeah, you're welcome. Yes, and, you know, so similar, slightly more elaborate version of the same thing makes sense of this experiment as well. Any other questions. Okay. Let me make another corollary of a visitor. This is an important remark I said, let me make another corollary of this remark before we move on. Because if a and B are non zero rational numbers. Then a b qp is equal to plus one for all but cliently many p. So when you look at this equation and a x squared plus b y squared equals z squared for any and B, but if you fix a and B then for almost all the p addicts it's it's going to have. a trivial solution. And what's the reason behind this well. Well, the claim that I claim that in fact if x is in q cross, then the valuation of x is equal to the pediatric valuation of x is equal to zero for all but finally many p. And this is because you know you write access a portion of integers and then there's only finally many problems appearing in both of those integer in the prime factorizations of both of those integers. So if all the primes different from those primes, the pediatric valuation will be trivial there's no keys to pull out either in the numerator or the denominator. And then that then applying this to both a and B you'll get a finite list of primes occurring in both a and B and when you're outside of that you're a pediatric integer so by this remark here. The Hilbert symbol will be true. And this is the, this lets us state the Hilbert reciprocity law with the Hilbert product formulas it's also called. So, so this is the theorem, maybe I'll keep this up as best I can. So, called Hilbert reciprocity or Hilbert product formula. So, so that's a and b be rational numbers. So we're trying to see how the different p out of Hilbert symbols are related to each other. And, but for that to make sense we have to start with something that maps to all the periodic numbers for all p so we start with a rational number. And then it turns out we get a non trivial relation between the p out of Hilbert symbols for the various completions of q. So, take the real symbol and mark multiply by the product overall p of the p out of Hilbert symbols, then you get plus one. And, well, so note that this internet product reduces to a finite product by the previous corollary. So you just have to, if as long as you if you choose a prime p which is not in the prime factorizations of a and B and is not to then you're guaranteed that the symbol will be trivial so it's really only a finite product. In fact, you can say what this means very, yes. Please tell me if I, please tell me if I'm mispronouncing your name by the way, I don't know. No, no, that's, you're saying it correctly and thanks for asking, I wanted to ask. So with this, I mean, we're, I guess we're going to see the proof in a minute but I was wondering, would this formula be related to the product formula for the norms that if I have a, if I have a rational number than the product formula over all the norms qp where we also consider p at infinity to be well the usual absolute value is equal to one. Indeed it's very much related. They belong to the same family of theorems, there's no logical connection between them, but they belong to the same family of theorem this one is much more complicated. We were not going to see the proof in just a moment in particular it's going to take us a while to I mean, we might get to it at the end of the lecture but it's not something we can just state and then automatically prove there are some build up. So this is it is exactly the same general nature but it is on the other hand, a level of complication above. Thank you. Sorry, I don't know how to make my hand raise but could I return to a more mundane question related to the various previous question. So if you just scroll up a little bit. So to speak, yeah. So you're a minus one over to a is a pediatric integer. So if you have the two addicts expansion of a, that means that constant term is is now zero when you take a minus one. By two, do you just shift that expansion, one to the left. Exactly you shift the digits one to the one to the lower power of two divided by two in the pediatric expand in the two which is just to shift. In terms of whichever direct practice. All right, thank you. Yeah, you're welcome. And you guys please do feel free to interrupt me with questions because I've been in graduate classes I've taught recently have been calculus for chemistry students. So I'm not used to. I'm not used to this so please it helps me out a lot to know how to how to explain things with the more questions you guys ask. Okay. Where was I was in the middle of saying something. I asked, so I was going to say that, well, I mean this looks really fancy right an infinite product but we could think back to the definition of the Hilbert symbols. And they were phrased in a very concrete way. So, concretely, doesn't help proving it, but at least you can phrase it helps phrasing it. So concretely, if you can look at this equation a x squared plus. View wise bird equals z squared. So, the set of completions of you, in which this equation has a non trivial solution is well first of all finite that's the remark we made earlier about the Hilbert symbol being plus one for all the family many p. The Hilbert reciprocity set saying is saying that this finite number of completions of q in which this equation has a non trivial solution is an even number. Do you mean does not have a non trivial solution. I mean this, the thing I said in the definition of the Hilbert symbol so a solution which is different from 0000. So, x equals y equals equals zero is what we call the trivial solution to this equation. Because, you know, any equation like this you write down is going to have any homogeneous equation is always going to have 0000 as a solution so we call that one trivial we say we're not interested in that only interested in non trivial ones. And so I meant to ask if you meant to say that there's an even number where there is only the trivial solution. It's like non trivial one but then. No. No, I think I said it right. Okay, it's okay it's possible. So let me double check. So the Hilbert symbol, let me go back to the sheet of paper I define the Hilbert symbol. Yeah. So the Hilbert symbol, we want to know when it's minus one. Oh, wait. Good night. No, you're right I got it wrong. Thank you. Yeah, okay. Has has, wait, okay, where am I okay now I'm getting myself all confused. So the Hilbert symbol is plus one, if this has a yes, does not have a. Okay, that's it. It's a double negative which is going to be my excuse for getting confused but thank you very much for correcting me. And sorry for giving the wrong answer first. Yes. Does not have a, that's a mouthful isn't it does not have a non trivial solution is fine I didn't even. That's just because if you have a bunch of signs and if you want their product to be plus long then the number of minus signs has to be even right. Okay. Yes. Okay. And by the way, it's, it's, I mean it's really annoying to have to always separate out the real numbers and the pediatric numbers. So I want to also use uniform notation and let me switch to all right new for what's called a place of the rational numbers and a place is either a prime number P, or this another thing which you can think of as corresponding to the usual absolute value on the rational numbers so and, and then for every place we can write a Q sub place new we can write Q sub new for the corresponding completion. So Q sub new is QP if new is a so called finite place of place P from number P and it's are new is the other place, the so called infinite place. So, yeah, I just. Is there an analog of this Hilbert reciprocity for general number fields. Indeed there is. And it works exactly the same. Well, yeah, so you can you can phrase it, you can make a yeah yeah so the every number of field also has a set of places completions you can define a Hilbert symbol with values and plus or minus one and the same statement is true. So it's much more difficult to reduce that statement to the case of the rational numbers so there's not that much more content, but on the other hand. Well I'm about to explain that Hilbert reciprocity is equivalent to quadratic reciprocity law from elementary number theory, and actually so well, while the translation from Q for general number field is in some sense trivial in the Hilbert reciprocity formulation and the quadratic reciprocity formulation is very much non trivial. And this and to phrase in a similarly an elementary way what the quadratic reciprocity law is not an arbitrary number field is quite difficult and this is one of the virtues of phrasing things in terms of the p addicts and so on, is that it's easier to add to other number fields. The statements become easier to make and then just as easy to group, but it really the first part is the most crucial part, the statements are easy to make when you use this language of completions and so on. And also I should say when you move to a general number field you can upgrade the Hilbert symbol to, you can always make it take values in the group of roots of unity in your number field, which is a finite cyclic group which is oftentimes bigger than just plus or one. So you can also get a, yeah, so there's also a slight refinement. And there are a million other refinement, you know, Hilbert reciprocity is refined by art and reciprocity is refined by Langland's correspondence I mean the story never ends the quadratic reciprocity law is a well spring. Much of modern number theory. Can I, that sentence that starts with concretely some things are written below but there can you read it one more time or clearly mark what goes where. Yes, absolutely. So that concretely the set of completions of q actually I just now introduced a new name for that so play I'll call them I'm calling them places. This is a equivalence plus of absolute values. So of course wants either a prime key or the other one. The set of places up to in which this equation does not have a non trivial solution. Yeah, it's fine item. Does not have a non. Well anyway I get confused trying to read the sentence I wish you guys the best luck. Maybe another way is just say, if P is large enough, prime is large enough it always has a solution in QP. It always has a non, non trivial solution beyond some point and have the ones that, and then you have an even number before that where it doesn't. Where it does not. Yes, exactly. Okay, I'm going to move on now. I want to explain now that the Hilbert reciprocate about is there or the chat is just going wild a but he kills kills on it. Awesome. Okay. Okay so I want to use this public reciprocity law I want to show that the Hilbert reciprocity law implies the quadratic reciprocity law. So, let me remind you of the quadratic well actually I'll just come out so let's, let's specialize. The non zero rational numbers, let's specialize to a equals P and B equals to where P and Q are distinct odd primes. And let's figure out what this Hilbert reciprocity is saying. So, first of all, the Hilbert symbol P comma Q sub new is always going to be plus one, unless new is equal to either. P q to, or you know the usual absolute value, which corresponds to the real numbers. All right, what I was calling it q mu. Yeah, sorry. Okay, now I'm, now it's just embarrassing. If you is not in this set because then it then new corresponds to an odd prime number, and P and Q, it'll be an odd prime number distinct from P and Q, and therefore both these numbers will be, you know, new attic integers. And so the Hilbert symbol is plus one. So, we only have to calculate what the Hilbert symbol is for these four possibilities, and then read out what comes. Well, so first of all, P q r is equal to plus one. Well, because P is positive or Q is positive. And what about in for two. So what is the two attic Hilbert symbol well by the formula I wrote earlier so P and Q are two attic integers, two attic units. So this is the same thing is minus one to the P minus one over to q minus one over to and then what about P q q p. Well, by another formula doesn't it's going to be symmetric in p and q that the p, you know, the Hilbert symbol is symmetric in both variables. So, the same as q p and I wrote down the formula for that it's just q on people a genre symbol, because q is a p attic unit, or no p's in it. And similarly P q q q, excuse me, is a P on q. So then the quadratic, sorry, the Hilbert product formula is saying that minus one to the P minus one over to Q minus one over to times q on P times P on q is equal to plus one. And now these are all signs. So, I can sort of when I cancel them by dividing by them is the same as multiplying. So I could, if I want I can move this over minus one to the P minus one over to minus one or two and then we get one of the standard formulations of the quadratic reciprocity law. So in other words, well what is this, what does this function look like this is equal to plus one. But unless p and q are both congruent to three month for, in which case it's minus one. So concrete and this is these are signs which tell you whether q is a square mod p or p is a square mod q so concretely what this is saying is that he is a square mod q, if and only if q is a square mod p, unless they're both three square mod four, in which case it's opposite. So he is a square one to it and only a few is not a square mod p. And in the problem sets what the first problem will be about using this to extract sort of concrete criteria, for example for when 13 is a square module and hard crime, which is kind of fun, fun elementary number theory sort of stuff to work out in quadratic reciprocity you've never, never thought about that before. So if you actually get more than this, this doesn't just have quadratic reciprocity it also contains the two so called supplementary laws for quadratic reciprocity so there's something quadratic reciprocity doesn't tell you anything about the crime to doesn't tell you anything about signs so plus and minus one. So, so Hilbert reciprocity also gives that minus one on p is minus one to the p minus one over two. So again, so p is again an odd prime. And two on p is equal to minus one to the p squared minus one over eight. So also the criterion for whether minus one is a square mod p and whether two is a square mod p. And since it also tells you when q is a square mod p for an odd prime few it can be used to tell you when anything is a square mod p, because those are simple as a homomorphism so you understand it on any number if you understand it on prime numbers and minus one. Right. Yeah, so this is for this you can plug in, you know, a equals minus one and B equals P and for this you can plug in a plus two and B equals P. So I'll leave that also as an informal exercise. No, I'm the exercise sheet. Okay. So any questions about the statement of the Hilbert reciprocity law or the deduction of quadratic reciprocity from it. Now I'm going to present to you a proof of quadratic reciprocity of a Hilbert reciprocity which I kind of dislike, but it's quite beautiful. I'll tell you I'll sketch the proof and I'll tell you why I dislike it. But it. So we're going to give tapes proof of Hilbert reciprocity. I mean I like it too. But there's one reason I dislike it for. And I'm going to just, it takes, it consists in two steps and I'm not going to go into details about either of these two steps. This is just to give you a preliminary idea of how of the logical structure of the proof. So, for step one, we have these Hilbert symbols right but and but in Achilles lecture he also introduced the general notion of a symbol. We have a certain bilinear map on F on this on a field a certain bilinear map on f cross to a target of helium group. And takes proof the first step in takes proof is to classify all symbols for the for the field at equals q. So, you can think of these, well, this Hilbert symbol started life as a symbol of a qp but you can restrict it to q and just as well view it as a symbol on q. So all of these things are symbols on q. And if you understand all possible symbols on q, then you'll actually be able to deduce and we'll see this in just a second that there must be some relation between this symbol and these symbols the real Hilbert symbol and the real Hilbert, Hilbert symbols for various prime p. So I guess maybe step 1.5 is deduce some relation of the form. So, product overall places mu of a b mu to some epsilon mu is equal to plus one where epsilon mu is either zero or one. And when we have to classify all symbols, you're going to deduce that there must be some relation between the various Hilbert symbols for the various places we don't you can't deduce what the relation is they just from the classification. But then in step two, you, you see that you prove that the only possible relation of that type is the desired one. So all of these places have to be contributing so all of these epsilon these epsilon us have to be plus one. And then the relation that you abstractly have necessarily. So I think the only possible relation is the one that gives over reciprocity and that's the proof of over reciprocity. Very, very, very weird deduce there must be some relation and then by various number theoretic trickery, you prove that the relation has to be this one. Oh, thank you. So I was wondering, looking at that Hilbert formula, are there like, it's kind of a general question maybe in a bit vague. Are there like any cases where we know the value of the Hilbert symbol for say all places except one and we can use this formula to like find that that place where we don't know the Hilbert symbol. That is, I would say, perhaps the major application concrete application of over reciprocity indeed. That is what it's telling you, yeah, if you know the value that that's a very good remark if you know, another way of saying this silver reciprocity is if you know the value of the Hilbert symbol at all except one place then you then value that that missing place is indeed determined and in a precise manner. Yeah. Thank you. And we will be using it in exactly this way in one of the coming lectures. Other questions. Yeah, Paul. You're muted. Still muted. Now. Yeah, sorry, at the bottom of the page that you have displayed. It's where Hilbert's reciprocity is. So where the statement is or where the statement of Hilbert rest. Yeah, that page. At the bottom of that page you seem to be saying that the real Hilbert symbol of A and B is one plus one, but I thought it was minus one at both are negative. No, no, no, no, no, no, no, no, this is a change in notation. I'm changing it from. And includes includes includes is something called infinity, or whatever you want to get for which you need. Yes, you knew is equal to R. I'm just restating it. I'm not adding any mathematical content. Okay, sorry. Thank you. So now I want to make a, so I can tell we're probably not going to get through the proof today, but I left some wiggle room in my lecture finance that's okay. So let me just remind you again about the concept of symbol, and then deduce the interesting book keeping device of the universal symbol, which turns out to have quite a lot of significance. So, well, let me just recall the definition that Achille gave. So if F is a field. And a is in a billion group, which we're going to write multiplicatively. Somewhat unusual. Somewhat unusually, but in all the examples it's going to be naturally a multiplicative group so it'd be too confusing to use additive notation for the general case. A symbol, or otherwise it's sometimes known as a Steinberg symbol if you want to be very precise about it on F with values in a is a function. And that theoretic function system map of sets from f cross cross f cross to a such that there are two properties satisfied. And one is that it's bilinear in each variable where you remember that all, all these are being groups under multiplication so I'll call this map as such that Fee of a B comma C is Fee of a comma C times Fee of the comma C, and then in the same other variable as well, a comma BC equals Fee of a comma B times Fee of a comma C. And the second property is the so called Steinberg relation. So Fee of a comma B equals one. If a plus B is equal to one, and an A and B are in. So, and the examples that we had. The Hilbert symbol, so on any q mu, the Hilbert symbol is a symbol with values in plus in the group of signs plus or minus one. But note, so let me make a remark. If, you know, f goes to E is a homomorphism of fields. It's called this f little f and fee is a symbol on E, then well fee composed of F, F, F applied to each variable separately I mean is a symbol on F. And that's just because we only use fields, I mean we only use addition and multiplication to phrase all these properties and so it's a routine check to see that this holds. So we can also view the Hilbert symbol, the qv Hilbert symbol also restricts to a symbol on q. So all of these different Hilbert symbols for the different places can be used as symbols that they start life as symbols on q and you, but you can risk, you can do them as simple justice symbols on q and that's what we're going to do for now on. But let me also make a slight elaboration of the Hilbert symbol, at least that odd prime so let's go back to the formula for the Hilbert symbol. Yeah, for odd primes. Nope, can't find it. So let me write it again. So, and so another example. For P and odd prime. Well, it doesn't have to be odd. So for P a prime. There is this, there's a tame symbol so called tame symbol, which I will denote. Tame symbol on q a B sub P, and this will be defined to be minus one to the VP, pediatric valuation of a times the pediatric valuation of be multiplied by you take a to the pediatric valuation of be divided by be to the pediatric valuation of a, and then you reduce that mod P. So it's the same thing as the formula for the Hilbert symbol, and it's what it where does it take values, it takes values in FP cross. So it's the same thing as the Hilbert form symbol for an odd prime P, except that you don't take the Legendre symbol at the very end, you just reduce mod P and that's it, and you remember it as an element in FP cross or Z mod P Z cross. And the same, the same reasoning for why the Hilbert symbol at an odd prime is a symbol will prove to you that the tame symbol is a symbol so that the Legendre symbol had no plate no role in that argument and you're actually going to see an abstract version of this in your exercises so you'll go over why this kind of thing always defines a symbol. Yeah, so indeed it doesn't find a symbol with values in this group of units in this finite field. Now of course, for people's to this is trivial, because F2 cross is a trivial group so you just get a trivial symbol. So this is only interesting when P is an odd prime. But it exists of course also in people's to nothing preventing you from taking people's to here. So this is a but when he is odd it's more refined than the Hilbert symbol right. Um, so now. So this this gives a now a wealth of symbols on the rational numbers and takes classification is going to say that essentially we've written down all examples up to up to a trivial modifications but how do we make that precise so I think there's a there's there's this artifice of the universal symbol which so to some weird formal game that you can play that actually turns out to be useful. So, so set a to be the free a billion group on symbols. Well now that's a that's a bad word to use on on f cross cross f cross on this set. So this is a subject to the following relations. So our modulo the subgroup generated by everything that makes the Steinberg property and the bi multiplicativity hold. So, you know, ABC. Yeah, unfortunately the standard so at the free a billion group on this set. So I should be writing it, come at AB, C, but unfortunately the standard notation is to use these. Things instead, when you're talking about, well you know what I'll write it in terms of the standard things that I don't shoot sorry guys. AB comma C is equal to a comma C or so. ABC equals a B inverse AC inverse, and a B was one for a plus B. Okay, so you simply take. The second relation, do we mean to have an equal side there for. No sorry it's supposed to be times yes thank you. So product in your group. Yeah, so I want to make the AB comma C equal to AC comma BC so I have to mod out by the mod out by the subgroup generated by these things. And then I want to also include the subgroup generated by these things thank you I was getting myself confused in the way I was writing it down. Yep. So what does this give you. So then, well this, this, this, this particular a is denoted by. So it's, it only depends on the field F right so it's an a billion group you can define for any field, created in group generated by this set subject to these relations. So it's a functor of the field. So it depends only on the field, and it's denoted by K2. So we're hinting of course that they're also KIF for all I, but we're not going to go into that so we're just, we happen to be interested in the K2, it's the one that's relevant for us so that's the only one we're going to talk about. So, this is this K2F is it is the target of the universal symbols so what does that mean. So, so there is a symbol on F with values and a values and K2F. Defined by, so a comma B maps to the image of a comma B in its quotient in the quotient, which defines K2F. And this image is usually written as curly braces a comma B. It doesn't mean the set consisting of A and B in this context it means the image of of this, this generator a comma B, and you're generating set for your being when you model by the relations which you, which you used to define it. And this symbol is is universal. So moreover, between symbols on F with values in a, an arbitrary dealing group A and homomorphisms of a billion groups from K2F to A. And so I was yes and a is no longer the specific thing, which I now call K2 about a is now a general of you. So for now an arbitrary dealing group us giving a symbol on F with values and a is just the same thing as giving a homomorphism from the specific. And the bijection is given by composing with the universal symbol. So, so if you have such a homomorphism, then you have your symbol here with rising K2F. So if you have a homomorphism, you can pose with this homomorphism you get a symbol of values and a. So that defines the map in this direction. And to go backwards, you just use the definition of K2F. So by definition it's the free of getting people on f cross cross f cross subject to some relations. So you can give a homomorphism out of it when you just specify the value on the jet values on the generators. And then the relations are satisfied and the relations exactly say that you have a symbol. So in the end, this is one big total. Yes. So this universal thing in the name refer it doesn't come from like this universal property kind of thing that you mentioned for the homomorphism. Is this the universality it refers to. Yes, indeed. So it's, it's the initial object in some category. You can make a category of symbols on a fixed field at where the homomorphisms are sort of given by homomorphism, but where the morphisms are given by homomorphism to target a billion groups. And a to f is the initial object in that category so it satisfies a precise universal property if you like to category theoretic terminology. Thank you. Yeah, I'm just noting that there's no statement about symmetry of the symbols in the no indeed so a keel showed and when he was talking about symbols that the version of symmetry that holds for an arbitrary symbol it's not quite obvious but it's also but it is elementary is that fee AB plus PBA inverse. So if you're taking values in the group where every element has order to, or in other words, the same thing is giving an F2 vector space then it's the same symmetry but otherwise it's different. Thank you. Indeed you can see that these tame symbols in general are not symmetric but they are anti symmetric. Right. To finish up, let me state. So, you know, it's a pate's calculation of K2 of Q. So, yeah, your own date. K2 of Q is isomorphic to, well, here's the way Milner writes it a to direct some a three direct some a five so it's a direct sum over prime numbers of certain finite of getting groups or a to is the group of signs. This is a Z mod PZ cross for P odd. And, you know, this isomorphism sends a B to the two addict Hilbert symbol, then the three addict Hilbert symbol, sorry, then the tame symbol AB three AB five, etc. So, you know how to get a Hilbert symbol when the tame symbol is trivial and then otherwise for odd crimes you just use the tame symbol. So, and this classifies all symbols right, because so if you know what the universal symbol is then you know what all the symbols are because giving now a symbol on Q is the same thing is giving a homomorphism out of this group. The correct sums that the same as giving a homomorphism out of each factor. And that's these are all cyclic finite cyclic groups so it's just giving an element of a certain order in your target being. And in the next lecture we'll prove Tate's theorem and explain how to deduce Hilbert reciprocity from it. So, sorry. So, once we show Tate's theorem, can we then say, sorry, if I'm asking a simple question, I just forget my answer sometimes. Is the, is the only homomorphism from K two into like a P that the one that sends AP like the AP to itself and everything else to zero. No that no it's not so you. It is a very good question. So it's I said, Tate's theorem says, you know, these are basically all the symbols that is not correct litter as a literal statement so you really have to analyze. Okay. For example, you know let's look at symbols with values in a two. Oh sorry I stopped my screen share well let's do let's do in our brains right so let's look at symbols with values in a two it's the same thing is a symbol with value and plus or minus one. So you really can't have a lot of cross terms. So you really have to just have to take your card being group as an abstract to being group and asked for all homomorphisms from a to all homomorphisms from a three all homomorphisms from a five and you can choose these and combine them together to get an arbitrary symbol with values in a. So I'll explain this in detail in the next slide you're welcome. In Tate's proof is he also going to vary a no we're just going to now for now on all the complexity of the possible target group is just contained in the definition of this one K2 K2 group. So we're only going to be now on in taste there and we're just going to be looking at K2 defined by generators and relations. And we'll prove that that group defined by those generators and relations is isomorphic to this direct sound. And we won't talk about. What I intended to ask was, so we have the definition of symbols, we have this map from F cross F to a. So, and in the first step of Tate's proof you wrote down date classifies all of the symbols possible so in that thing is equal to that's what I needed to ask. Yes, indeed so that's step one does correspond to this theorem of Tate's giving describing this group K2 of q up to isomorphism. So, describing all symbols on a field is exactly the same thing as describing the group K2 of the field. Yeah, because we said there will be symbols will correspond to homomorphism from K2. Precisely. Yeah. So I said the same thing in a different language and step one classifying all symbols is the same as this step one determining the group K2 of q. That's the that's the virtue of this kind of universal nonsense. Thank you. Yeah. Yeah, you're welcome so yeah just. Can I can I ask a question. Yes please. So I was, I was thinking a bit about this with regards to the question I asked you a while ago about this universal property. So what would this like I'm just curious about what this category of symbols would look like I'm not sure that I, I got what how you explained the the morphisms in that category, because I'm a bit curious about what that category of symbols over a field off a field would would look like. I mean, I just made up a category I'm not thinking you I'm not thinking really about this category I'm just, I just made one up on the spot I guess I don't know. I mean a symbol symbol the objects are the symbols right. And I think this is what I want to say a map between two symbols is just a map between the target appealing groups, which makes the diagram commute. Oh, I see yeah that that would. Yeah. I see it I get it. I can, I can think about that. Yeah, okay, thanks. You're welcome. I'm not saying you're going to want to go and study this category I mean it's just the. I just wanted to give a formal meaning to saying the universal. Thank you. You're welcome. Thank you Dustin. And then I've also there's a question. My pleasure. And so there, I have office hours right now, and then there are TA sessions and there's a cross program. And then see you tomorrow.