 Yeah, also, yeah, so let me just set up the screen share. Yeah, also I just wanted to encourage everyone again about the TA sessions. I think even if you haven't done the problem set so one is that you're very welcome to talk about previous, especially now that the Right. So I'm going to talk about the the worksheets from previous previous lectures and also if you even if you haven't actually looked at the worksheet and just want to, you know, learn stuff for or talk about math and generally, you should definitely be very much encouraged to go to these to these TA sessions. So I want to, yes, I'm sorry for a bit of a mess yesterday. I want to try to tie some of these things up today. And so yeah so let me just recall some of the we were yesterday. So, right, so let E be a local field of characteristic not equal to two. Then one of, I guess one of the basic objects that we're going to be interested in this course is the is the Hilbert symbol. So let me let me define that again. So the Hilbert symbol is the following function so it's a function in two variables, which goes from E cross modulo E cross squared times E cross modulo E cross squared to plus or minus one. So I'm going to write it as Hilbert symbol sub e just to emphasize that it's the version for the local field E. Right so it's going to be defined via the following formula. So the Hilbert symbol of a comma B. So five elements of the field E nonzero elements it's defined to be one. So E squared minus a x squared minus b y squared has a non-trivial solution. So in other words if this quadratic form one comma minus a comma minus B is is isotropic. And it's minus one otherwise. Okay, so, so it's a function on right so so, okay so right so I guess if you define it this way it's sort of clear that it only depends on a and B modulo squares. So you can think of it as a function on E cross modulo modulo squares times itself. And yeah it's just to find the squares that it takes values in in plus or minus one. Okay so right so let me write down some properties of the Hilbert symbol. So properties of the Hilbert symbol. So first of all it's symmetric so a comma B is equal to become a by inspection. And two is that, well there are certain sort of cases in which the Hilbert symbol is automatically plus one. So a comma minus a is equal to plus one, and sorry. A comma one minus a is equal to plus one so this this latter example is if a is in E cross. And it's also not one. Right so I mean so this is this is because and if you look at the quadratic form z squared minus a x squared plus a y squared, that's going to have a solution, given by 011. And similarly the quadratic form. Well, right so so namely we take z comma x comma y to be equal to. I think I just want to take it to be one comma one comma one. So that gives a non trivial solution. So in this case it's automatically equal to equal to one. And then there's one more relation that we get for free. So which is that a comma B is equal to a comma minus AB. One sort of natural relation we got on the Hilbert symbol. Right so so so I said if you take the quadratic form z squared minus a x squared, minus b y squared, well that has a zero if and only if, well you can multiply it by a so you get a z squared minus a squared x squared minus AB y squared. So we can multiply that by minus one again. So this has a zero if only the second one has a zero, if and only if. Well, right so we can rewrite this as ace what we can multiply by minus one and that's a x squared. So this is b squared plus a b y squared, and then rescaling x we get it into the form one comma minus a comma AB. So so we also have this relation, I mean I guess just by sort of manipulating sort of rescaling and manipulating the quadratic form. I didn't. In fact what I've said, so these these properties are, they haven't used the fact that he is a local field so so so right now. I mean this definition would have made sense if he is any field. And these properties so far would be true if he is any field. But in fact we're only going to be interested in this one is a local field, and that's because of the following sort of really fundamental property, which is that the Hilbert symbol is by linear. And in fact, right so it's it's it goes from E cross mod E cross squared times E cross mod E cross squared it to plus or minus one which I could think of as F2. And so it's actually a bilinear. It's a bilinear map of F2 vector spaces. And it's actually non degenerate. So it's a non degenerate pairing that we E cross mod E cross squared and itself, which is going to exist for. So this proper. Yeah, so for any any local field of characteristic not to. So this is really the key sort of fundamental feature of the Hilbert symbol, which is actually not at all obvious from the definition of the fact that, well first of all it's bilinear and to that it's non degenerate. In fact, right. So why is it in fact bilinear. I think that to my knowledge, this is not something that really has a so for for any local field of characteristic not to be knowledge this is not something that one has a really direct and elementary proof of. I mean, so, so it sort of falls out from the general sort of local class field theory that's this bilinearity and non-generacy is contained in there. So let me try to sort of try to explain what's what's going on like why why is this this thing bilinear. So, so given an element a and E cross. Well if a, if a is a square then then then it's automatic the Hilbert symbols automatically one so we can. Right, we may as well assume it's not a square. But so then a comma B is equal to one, if and only if right so z squared minus a x squared minus b y squared has a solution. And right so if you sort of rewrite this this is the same thing as saying that B is a norm in the quadratic extension. From the quadratic extension e square root of a to E so B is in the image of the norm map, which goes from E adjoin the square root of a or the non zero elements and there and D cross. So because you have this quadratic extension. So you have this quadratic extension E adjoin the square root of a over E. And so then you get a norm map, which goes from the multiplicative group of that field extension back to the cross. And this this quadratic form has a solution well if it has a solution. And then well right so so then B is equal to the norm of z minus the square root of a divided by x over over y. Okay, so so just sort of writing out explicitly the formula for the norm then then you get this. And in fact, so what local class field theory gives you is that this this image of the norm is actually an index to so it's always a subgroup but it actually has index to. So the image of the norm from the square root of a is index to. And that's that's actually what bilinearity is giving you so it's it's telling you that if you fix an element a, and then as you let be very it's it's plus or minus one according to whether be belongs to this index to subgroup or not and so that's that's a linear right that's a linear map to F2. So the fact that it's bilinear is a is really is really a consequence of this fact that the image of the norm up this index to, and that's, that's, that's really part of this part of local class field theory I mean I guess that's going to be true, whenever you have an abelian extension. Sorry. Sorry, so there's something in the chat yes so if if y is equal to zero then is a square in which case it's it's automatically one yeah. Right. Okay, so, so at least that's bilinearity. So, right, but so if you want to prove it without sort of invoking these theorems then I think it's not so easy. What you can do is in the case when he is equal to QP, then you can work everything out explicitly, and then it's going to follow that it is bilinear. So, in general, so proving bilinearity directly seems not so easy, but you can you can actually do it just by hand for for equals QP. And you should also say that if P is odd, and you have any finite extension of QP essentially what works for QP is still going to work for for that final extension, but where you have to sort of work a little bit more as if you have like an extension of Q2. So, sorry, so there were some. Yes. Yes, that's a why thank you. Okay. And then you can, you can actually work out you can actually just work out explicitly so so if is equal to QP I mean these are finite groups, they're finite dimensional F2 vector spaces and you can actually just work out the bilinear, the Hubbard symbol by an explicit by an explicit computation. Okay, so right so for example, let's suppose P is greater than two. So I'm going to have elements x, y, which live in QP cross. So then I can write x as some power of P times a periodic unit, and similarly for why. So here a and b are integers, and UNV are periodic units so they're in ZP cross the periodic valuation equal to zero. Right so so so then the statement is that the Hilbert symbol of x comma y the QP Hilbert symbol is given by minus one to the a times B times epsilon of P times. Well the Legendre symbol of you over P to the beef power times the Legendre symbol of V bar over P to the eighth power. Right so so so here, sorry I need to explain this notation so here, epsilon of P is equal to P minus one over two. So that's some integer. And you bar and V bar are living in FP cross and they're the images of UNV. So they're the reductions mod P. And this this, you know, something over P is a Legendre symbol. In other words, it's the unique non trivial map from FP cross to plus or minus one. So it detects whether or not your the input is a is a square. Okay. Right so I think I've defined everything here so so in particular right so this gives a formula for for the Hilbert symbol for any two elements of QP. So this is something that we can prove pretty directly using Hansel's summer. But let me just write so sub example. And in fact the sub example is so so using the fact about the Hilbert symbol that have already been proved. You sort of reduce this general assertion to a couple of special cases. So for example by multiplying X and Y by squares, you can assume that a and B are either zero or one. So a sub example is that the Hilbert symbol of you comma V, the P adic Hilbert symbol is equal to one. And so here you and V are just P adic units. So why is that because well you you want to show that the quadratic form z squared minus you x squared minus V y squared has a solution. And while this is a quadratic form over QP, P is greater than two and it has three P adic units on the diagonal. And while this is a quadratic form over QP, P is greater than two and it has three P adic units on the diagonal. So this is going to have a solution if and only if it has a solution mod P by Hansel's Emma. And we already know that any quadratic form in three variables or P has a root. So because this quadratic form has is isotropic is isotropic by Hansel's Emma. So I guess we did this example on Friday, but yeah so since it's isotropic might be okay. Right so another sub example. Again you sort of reduce the general case to these two sub examples using general properties of the Hilbert symbol in particular that it's invariant under scaling by square. If you take the P adic Hilbert symbol of P times the unit time and and a P adic unit, then this is exactly the Legendre symbol of V over P. And so why is that well because if you look at the quadratic form z squared minus P you x squared minus V y squared. Well in general if you have a quadratic form over QP where P is greater than two, then you sort of break it up into two pieces where the P adic valuation is even in the P adic valuation is on. And it's isotropic if and only if either of those pieces is isotropic. So this is isotropic. If and only if z squared minus V squared by squared is isotropic. Because because of this again this fact about a classifying isotropic and isotropic forms over QP. And that's if and only if the sitting inside ZP cross is a square. So by Hansel's summer that's the same as the bar over for P equals one. Okay. So yeah so these are basically the two examples to keep in mind and similarly you can do something where both the p adic valuations are equal to one. And, and then that that actually implies the general case by just by sort of rescaling. In fact, so once you've actually computed it so so once you've computed it then it follows that it's a homomorphism, and you can also check directly that it's non degenerate. So, by computing directly, it follows that it is a non singular binary map. So we're not proving, you know this is not an abstract argument this is just you compute it and then you check it. Okay, sorry so actually I think I should start with this example so another example that's worth keeping in mind the simplest example is a case where he is the real numbers. So this is the simplest example of the Hilbert symbol. So in this case, then the Hilbert symbol of a comma B over the real numbers is equal to minus one if a and B are both less than zero. And it's one otherwise. Right so just looking at looking at signs in the in the quadratic for the z squared minus a x squared minus b y squared. Right and so finally I should state the answer for for q two so for the two adic numbers it's a little bit more complicated to work out the Hilbert symbol. Sorry, non singular by linear. Yes. Right sorry so this is so in particular. Non singular. Yes, thank you. So in particular, while you have r cross mod r cross squared times r cross mod r cross squared to, you know, plus or minus one. And these are both isomorphic to Z mod two. And the map is well it's the unique non super. Okay. Non singular. Yes. Okay maybe I should say non degenerates. Non degenerates. Okay. Okay. So for p equals two. It's a little bit more complicated. Right so for p equals two. One needs to work a little bit more because it's a little bit trickier to determine what the squares are in q two. So we need to use the classification of squares in q two. And so recall that for example I mean if you have a square in q two then it's two adic valuation has to be even. So, well, first of all, then that let's just reduce to the case where it's just in degree zero. And so it's recall that x in z two cross is a two adic unit. So if x is congruent to one modulo eight. So that's that's something you can prove I think maybe it was on. I don't know if I did it in lecture but maybe it was in in one of the, the homeworks and that's something you can prove using the sort of either the refined version of Hansel's level. That's a little bit more refined than the one I explained in the lecture, or by sort of changing coordinates. Oh, thank you. Yes, not not a tragic unit. It is a tragic unit, but it is. It's always a tragic unit, but it's a tragic square. Thank you. Okay. Right so so instead of just working my P now you have to work my P cubed. Okay so again. Right. So what was my notate sorry. What was my notation again. Yes. Right so sorry so suppose. So you have x equals two to the a times you and y equals two to the b times V. So then the assertion is that the two adic Hilbert symbol x comma y over q two is equal to minus one to the epsilon of you times epsilon of V, plus alpha omega of V times beta, oh sorry. Not alpha. Sorry, a times omega V plus B times omega of you, where. So first of all, epsilon of you and epsilon of V is you minus one divided by two. So my two. And omega of you is you squared minus one divided by eight. So my two. Right so for example omega of three is going to be. It's going to be one, but omega of one is going to be zero. So yeah so these are these are giving you sort of two attic characters on Z to cross with values and plus or minus one. And it's, it's given by the following formula. I'll put this on the sort of problem set in more detail, but right so the example is that if you just have two two attic units so if a and b are equal to zero. Right so so so then the two attic Hilbert symbol of you comma V is given by minus one to the U minus one over two times V minus one over two. So you minus one over two here is, I guess it's taken mod four. It's right it's not an integer, but it is a two attic integer so you can say whether it's right it's it's divisible by two or not. Right, use Z two to two. Right, so if you take the Hilbert symbol of two comma a, or two comma you, or use a two attic unit then this is minus one to the use, use square minus one divided by eight. So, so these are formulas for them. The two attic to add a culprit symbol. And again I guess you can check using this that it's actually a non non degenerate bilinear bilinear pairing. So yeah so these, this might sort of start to remind you of quadratic reciprocity these conferences. And in fact the Hilbert Hilbert symbol turns out to be closely related to quadratic reciprocity and I think that's going to come up in in one of the, one of the sense lectures. Okay. Right, but in any case one can sort of work everything out pretty explicitly in the case of QP, or the real numbers. And, yeah, so you got these formulas and you can check the bilinearity and so forth. So, let me also. So I guess in this lecture I want to wanted to explain a little bit more about how you can use the Hilbert symbols to write down sort of invariance of quadratic forms over, over a local field and actually give a sort of the classification of them. I also want to continue with something that I explained that I sort of explained last time, which is the relation of the Hilbert symbol to the structure of the bit ring. So the Hilbert symbol. So it's closely related, or it's sort of encoded to in in a structure of the bit ring of the local field. So it's closely related to the structure of the bit ring w of E is our local field. Right and basically there's the following observation, which is that the form. So the Hilbert symbol of a comma B is equal to one will by definition that's if and only if brackets one minus a minus B is isotropic. And the observation is that this is true if and only if brackets one minus a minus B, AB, which is equal to brackets one minus a tensor brackets one minus B is hyperbolic. And so this is kind of fun to check and I think I maybe put it on yesterday's problem set so if you have a four dimensional quadratic form of determinant equal to one so of this form. Then it's actually hyperbolic if and only if it's isotropic. And that's because if it's isotropic it always has a hyperbolic piece. And then the complement is going to have under these assumptions is also going to have discriminant determinant minus one, and hence that complement is also going to be forced to be hyperbolic. So actually this is a special case of a more general statement of involving so called fister forms of quadratic of dimension power to but in fact let's check directly. Um, okay. So, right, so in particular, this Hilbert symbol is also somehow encoding the multiplication or it's closely related to the multiplication in the in this bit ring. And so the fact is that right so the bit ring of E contains. So if you sort of online this that right so the bit ring contains this ideal I which is the even dimensional forms fundamental ideal. And, right, so a consequence of this theory so I can, I guess this is a consequence that, again, if you're working over like some final extension of q two. I'm not sure if you can just prove this directly by hand in all cases. But you can sort of, well, by the work that we've done here it's, it sort of gives it you over qp for any p. So first I cubed is equal to zero. And you can form this, you can form this filtration where you have w of E containing I containing I squared. And you can form the associated graded and then that contains IQ which is equal to zero. And then you can form the associated graded terms so girl zero is equal to w of E mod I which is given by Z mod to to girl one is I mod I squared. And that's given by E cross mod E cross squared. It's just true actually for any field this doesn't require it to be local. And go to is again given by Z mod to and go to is given by Z mod to because there's a unique anisotropic form of dimension four. So, and that's, and because again because I cube the zero in this case there's no higher, no higher terms. And right so you always have a pairing for from girl one cross girl one just a multiplication pairing to girl to which is the mod to. And this is a pairing E cross mod E cross squared times E cross mod E cross squared to Z mod to and that's exactly the Hilbert symbol. So the Hilbert symbol is exactly encoding the multiplication and the associated graded of the bit. And so you might say okay well this lets you prove the bilinearity but then we have to sort of back up a second because actually seeing the girl to is Z mod to you have to do some sort of computation to do that. So either I think either I think to see this I think you either need some sort of computation or you need to invoke some machinery like a little question through. Okay. So that's a little bit about the Hilbert symbol. I guess what I wanted to say next is right how to use the Hilbert symbol to define the invariance of quadratic forms of a local fields. But in fact it'll be convenient to say this not just for the Hilbert symbol, but this more general notion of a symbol. So there's the following definition which is somehow axiomatizing properties of of the Hilbert symbol, and this definition turns out to be well. So let, let F be a field, and let a be an abelian group. So a symbol on F with values and a is a bilinear map, which we'll call fee, which goes from f cross times f cross and today. And it satisfies this basic identity which I essentially stated without comment a while back. So it's not fundamental which is that fee of a comma one minus a is equal to zero for a not equal to zero or one, so it's well defined. So a symbol is, is just a bilinear map from f cross times f cross into a so right so to be careful here by by linearity. I'm considering f cross as an abelian group so only say bilinear it's with respect to addition, like that. So it is a multiplication in the field F. So we're considering f cross as an abelian group. And in addition it satisfies this equation which I guess is coming from addition in the field somehow a fee of a comma one minus a is equal as automatically equal to zero. And so right so this axiomatizes some of the properties of the Hilbert symbol, right so bilinear map of the modules yes thank you. So axiom, oops. Okay. So you. So this is a basic axiom this this this feature here. But in fact, so I wrote down some more properties of the Hilbert symbol that. But in fact, many of those properties are actually consequences of this, this one axiom that fee of a comma one minus a is equal to zero and bilinearity. So for example, you also have the property that fee of a comma minus a is equal to zero. Which is equal to. Because minus A is equal to one minus a divided by one minus a inverse. And so that's going to give me if a comma minus A is equal to feel A comma one minus a divided by one minus a inverse, which is equal to fee of a comma one minus a minus fee of A comma one minus other verse a comma one minus a plus p of a inverse comma one minus a inverse, and that's equal to zero plus zero zero, using the symbol axiom. Okay, so starting that with a simple axiom you also get that you also get some other identity such as fee of a comma minus a zero again that was more or less immediate from the definition of the Hilbert simple, but in fact it follows from this one property fee of a comma minus a equals zero. So another fact that you get whenever you have a symbol is that it's automatically anti symmetric so fee of a comma be is equal to minus fee of b comma a. So you automatically have anti symmetry. In the case of the Hilbert symbol we were just taking values in an F2 vector space or anti symmetry is the same as symmetry but in general if you consider the definition of a symbol, what you'll get is something that automatically has to be anti symmetric rather than symmetric. And right so how do we see that well we can consider fee of a b comma minus a b and so by definition that's equal to zero so fee of a b comma minus a b is equal to zero as we've just shown. Right so so now we can sort of expand this so we get fee of a comma minus a well that's zero but fee of a comma b plus fee of b comma minus a plus fee of b comma b. So we can cross out the first term because we already showed that zero. And we can also rewrite this as fee of, we can also write the second what we can write this term right here is fee of B comma a plus fee of B comma minus one, then plus fee of B comma B. And then we get fee of a comma B plus fee of B comma a plus fee of B comma minus be using by linearity, and this vanishes. And then what we get is that fee of a comma B must be minus fee of B comma a. So anyway just sort of playing with these identities you get you get some additional consequences for free and in particular you get this anti symmetry. Okay. Right. For the purposes of quadratic forms. The main interest is actually in symbols that take values in F2 vector spaces. But for example, it is sort of interesting and fun to think about symbols that take values in more general groups. So, for example, there's a symbol on QP. And it values in FP cross called a tame symbol, which actually is a refinement of the Hilbert symbol. But for for thinking about quadratic forms it's actually somehow. Somehow it's just symbols that that take values in an F2 vector spaces, such as the Hilbert symbol. So yeah so this definition of a symbol. I think maybe maybe it first appears or was first really made systematic in this paper by Milner on algebraic case here in quadratic forms. And it, I guess, when you look at it for the first time, I mean it, it seems kind of it seems a little ad hoc. You're just tax imposing this axiom fee of a comma one minus a is equal to zero. But somehow it turns out to be somehow it turns out to be sort of really fundamental. And particularly it turns out to be well for the purposes of course one of the first ways it turns out to be fundamental is that you can, you know you can try to classify symbols on on a field. And for example the classification so we've seen that the Hilbert symbol gives you a symbol on on QP. And actually it is like sort of the universal symbol. It's sort of the only symbol on on QP with values in plus or minus one. And when you try to classify symbols on Q it naturally sort of that's naturally going to produce quadratic reciprocity. So as I think Dustin will explain. Um, but what I want to explain, I guess in the rest of today is given a symbol, there's a way of naturally extracting sort of numbers from a quadratic form. So, given a symbol, there is a natural way. Given a symbol with values in an F2 vector space to extract invariance of quadratic forms. Okay, so that's going to be the following following construction. So let the from f cross times f cross into some some a billion group a be a symbol, and suppose also that is an is actually an F2 vector space. Okay, so then there's a way of assigned then then you can use fee to define an invariant of every quadratic form over F. So then can use the symbol fee to define an invariant of any quadratic form over F, which I'm going to call epsilon sub fee. So that's going to be an invariant that takes values and this F2 vector space. So if you have a symbol with values and after F2 vector space a, then for every quadratic form you can you can write down. You can write down an invariant of this after space a. So here, what by construction epsilon sub fee of a quadratic form is well if you have a quadratic form brackets a one through an, it's going to be defined as the product over all I less than j of fee of a a comma a sub i comma a subject. Well, sorry, I guess, I guess I'm writing a as an ability in groups I should really say it's a song. We're sort of adding them. Yeah right so in practice fee is going to be plus or minus one so think about the product but let me write it. Okay, so so this is this is so for every every quadratic form of the looks like brackets a one through and we're going to write down this element in a and that's going to be the invariant of the quadratic form. Okay, so so this is what we want to be our definition but we need to show that it's well defined. So what we need to do need to show that this is actually invariant of the choice of the diagonalization. So this is how we want to define the invariant. What does it say at the end of the line. epsilon sub fee in a sorry that's so I'm confused. It's way above any quadratic form over F. Yes, sorry for my handwriting. Okay, so what we need to show is that this is independent of the diagonalization. And so it actually is an invariant of a quadratic form. And right so so in fact this is this is right so we saw sometime last week that if you want to go from one diagonalization to another diagonalization. So we sort of a set of moves you're allowed to do for a given diagonalization. And you can always get from one diagonalization to another diagonalization or diagonal sequence using this sort of sequence of moves, and each of these moves is only going to change two terms. Okay. So recall from last week that there's this finite set of moves. Well there's this, these three different types of moves that one is allowed to do on a diagonalization that you can get from one. I mean that you're if you're within the same isomorphism class you can get from one to another. And so all we really need to do is, is we need to show that epsilon fee is invariant under these these moves. Right so. So brackets a comma B is isomorphic to. Sorry so for example brackets a comma B is isomorphic to brackets B comma a brackets a comma B is isomorphic to brackets a comma B times some element squared and brackets a comma B is isomorphic to brackets a plus B. So maybe over a plus B. So, okay, so these are the three moves as I explained it last time. Um, but, in fact, maybe instead of just writing them out explicitly the main thing is that they only depend on two terms at a time. So they somehow they only depend on isomorphisms of two dimensional quadratic forms. So, I guess if you think about this it's it's going to imply. Yes, maybe a plus B is not zero for the last one. Yeah, thanks. So if you want to show this construction of epsilon fee is actually well defined. What you really need is, is just a fact about two dimensional quadratic for so if a comma B is isomorphic to C comma D so via, for example, the above moves. Then the symbol fee of a comma B, because I said or it's not isomorphic it's equal. It's just equal to fee of C comma D. So, so, so what you need to do is, you need to essentially show something for two dimensional quadratic forms. So this is the basic thing one has to check. And right so. So, in fact, what's what's sort of nice is that the definition of a symbol is sort of precisely, it just sort of works perfectly for for something like this to be true. Right so if a comma B is isomorphic to C comma D, then we can write to see as a x squared plus be y squared for some x and y in our, in our field, and in fact sort of by rescaling well so it's sort of clear that if you. Right so if you rescale by a square, then that's not going to change the value of the symbol because the symbol is taking values in an F2 vector spaces. So for, so for definiteness and in fact without loss of generality. We can just say that C is equal to a plus B. So in fact we can sort of work and in the case of the last example. So C is equal to a plus B. Okay, and so, so then right so what do we have we have a divided by C plus be divided by C is equal to one. So for in this kind of type of situation. So, for example, x is equal to one y is equal to one. Sorry, so I didn't I think I didn't say that well. You want to see that it's invariant under the following three moves. And, well it's, we're assuming we've seen that fee is anti symmetric, or and which means symmetric since we're working with active after spaces so it's, it's invariant under the first and it's also a very second move because it's bilinear and it isn't changed, and we're taking values in F2 vector space so we really just need to see the last one. So we can let C equals a plus B, we have a divided by C plus B divided by C plus one which gives by the simple property fee of a over C times B of a over, comma B of a over C is equal to one. And now we can sort of expand out all the terms that's fee of a comma B times fee of a comma C times fee of again we'll use symmetry fee of B comma C times fee of C comma C is equal to one, and that's going to give fee of. And then going back between additive and multiplicative notation but hopefully it's not too confusing so fee of a comma B times fee of ABC comma C is equal to one, and ABC equals D up to squares. So, so this is also equal to fear. So this implies that fee of a comma B is equal to fee of C comma D. Sorry, so I said this a little bit quickly but essentially the point is that this, this last move that you're allowed to do, when you, when you change, you know you go between two two dimensional quadratic forms is exactly sort of governed by this, this A comma one minus is equal to zero. And it lets you show that you have a well defined invariant for two dimensional quadratic ones. And then essentially that gives you it in general because any isomorphism between diagonal forms is is going to sort of be obtained by composing these isomorphism two dimensional forms. Okay, so, so that's a proof. For example, if he is a local field. So then what we can do is we can define for each quadratic form and invariant in plus or minus one by the above construction applied to the Hilbert symbol. And epsilon of B, if V is a quadratic, each quadratic form, each quadratic space V, we get a sign plus or minus one by taking the. Yeah, thanks. Right. Yeah, so epsilon is not to be confused with E and my handwriting essentially you can't believe. Okay. So, so we got assigned by taking essentially the products of the Hilbert symbols along the diagonals, and that gives a sign plus or minus one. So this is called the the Hassan variant. And it comes from it comes from from the Hilbert symbol. And just sort of a warning is that it's not quite additive so it's not true that if you take epsilon of V direct some v prime. So you have to epsilon of V times epsilon of V prime, but then there's also a cross term which is the Hilbert symbol of the discriminant the determinant of the, and the determinant of the prime. So, so you have this construction called the Hassan variant. And it, I guess just the only thing sort of warning is that it's not quite it's not a home. It's not a map on broken you can that brings it satisfies this, this extra. And this is the sort of additional on. Okay, and so then there's the following theorem, which is that it this is a complete well this together with the discriminant and the dimension. Yeah. Sorry. Yes. Thank you. Yeah, so I think I was. Yeah. Okay, so, in fact, maybe if we go back up I should. Maybe since we're using everything so I didn't do this right so let's let's just write it multiplicatively let's write because a is going to be plus or minus one. Let's write everything multiplicatively. Yeah, so thanks. Um, yes, I'm sorry about that. Okay, so, so here's the sort of main theorem. So if he is any local field of characteristic not to then quadratic forms over er isomorphic. So if and only if, well, first of all, they have the same dimension. Okay, so that's pretty necessary. And then they have the same determinant, the determinant of the quadratic form lives in E cross modulo E cross squared. And if he is QP so imagine he is QP then that's that's a that's an F2 vector space of dimension two of peace order three is P is equal to two. And then same has to invariant. Okay, so this is the main theory. So this is the main sort of classification theorem. This is also very closely related to the statement about bit rings that I that I also mentioned earlier so the bit ring of is basically seeing, I mean it's seeing up to associated graded it's basically seeing this E cross mod E cross squared, and, and the Hilbert symbol pairing to plus or minus one. When you define it this way it's actually a slightly different statement it's sort of off by some sort of exponential. Okay, this is also true. Okay, so I guess I should just say say something about the proof. So the proof of this most of the ingredients so again so we're not going to be right so we don't have the ingredients to prove this, as I've stated, but we basically have the ingredients when he is equal to QP. When he is equal to QP. We basically have the ingredients. And it essentially is going to follow from well, essentially the main point is a quadratic forms of dimension five or isotopic. Um, so I guess I'm about out of time so I should just sort of say, say why this is true. So the main point of why this is true is really that quadratic forms are. The main point is that quadratic forms of dimension five are isotropic and quadratic forms. Well, there's a unique and isotropic quadratic form of dimension for. So basically if, if you have quadratic forms let's say, or quadratic spaces v and v prime have. Sorry, so let's say. v comma q, and v prime comma q prime, have the same invariance. Then what you want is that they're isomorphic. And that's the same as saying that v direct some v prime with the quadratic form, Q, Q direct some minus Q prime is hyperbolic. It's the same by that cancellation for for for this to be hyperbolic. And now what we want to show is that right so you have you want some criterion for a quadratic form to be hyperbolic. And what's going to happen is that if, if you have two quadratic spaces with the same invariance, this is going to have the same invariance as a hyperbolic form, hyperbolic form of the appropriate dimension, because we know how the determinant and the house invariant behave and direct sums. So, you reduce to showing that if you have a quadratic form with the same invariance as a hyperbolic form, it's hyperbolic. And you prove this by now you prove this by induction on the dimension, because once you're dimension at least five, you can anyway split off, well here it's dimension six because it's even, you can split off, you can split off a quadratic, you can split off a hyperbolic form because, because quadratic forms of dimension at least five or hyperbolic. So you can keep doing this and essentially you then just have to prove this in dimension for. And in dimension for if you have a quadratic form of discriminant one, or if determinant one. Then it's either isotropic, then it's either an isotropic in which case that you can detect that from the house invariance sort of by construction, or it's completely hyperbolic. So maybe I'll put some of these details try to put some of these details on the on the homework set, but this is essentially sort of a restatement of the classification in, you know, the main sort of results about quadratic forms over, over these local fields and it's really it's really sort of a restatement, just in terms of a slightly, you know, slightly different ways of normalizing the variance. Okay, so I guess I will end here so yeah beforehand I just want to also thank PCMI and the organizers for the invitation to, yeah, to this summer school. And they also want to thank the, I want to thank the TAs, so even Morgan and Freddie for all of their work. So Dustin will take over for the lecture starting tomorrow I will still be around and so feel free to email me, or feel free to ask me or email me or ask me questions and so cocoa or discord and so forth. And yeah so thanks. Thanks a lot for your time and attention. I also suggest that we all unmute and give up your applause for a wonderful series of lectures. All right, thank you very much so I guess I should stop the recording but I'm happy to stick around for