 Alright, so let me screen share. Okay, so. Yes, I want because I want to pick up. So today is going to be the last of my lectures that is focused sort of on the general theory of quadratic forms. abstract general theory that works over video for any field. But I want to start by, I think I was a little bit rushed at the end so I want to sort of recap the results from yesterday at the end. So, right, so let's, let's recall. So last time. So one of the outcomes was that we completely classified quadratic forms over a finite field. So we classified quadratic forms over fq. And so we had the following proposition. So, right, so the proposition is that any n dimensional quadratic form over fq is isomorphic to something of the format brackets one brackets one dot dot dot brackets one comma D. So you can put it in a diagonal form where you have n, at least n minus one ones, and then some other D which is going to live in fq cross. So you can well you can always put over any field you can put in diagonal form it over over fq it's even better you can you can make all but one of the diagonal entries, equal to one. And as a consequence. So you can also say. So, right, so as a consequence. So there are two isomorphism classes of quadratic forms, say of n dimensional quadratic forms over fq. And that's given based on D so D is an fq cross but it really lives in. It's only really well defined up to multiplication by square. So you can put it on the element D in fq cross mod fq cross squared, which is isomorphic does you want to as a group. So I just also want to make a correction from last time so last time I called D the discriminant so D can be, I mean, you can, you can define this as a determinant of the associated symmetric matrix. Last time I refer to this as a discriminant. I think that terminology is not quite right often the discriminant involves an extra sine factor. So D is called it the determinant and determinant of the quadratic form. And right I mean D is something that makes sense over any field very generally I mean you basically just take the associated determinant of an n by n symmetric matrix. And again it's called the determinant and the discriminant is basically going to be a modified version of that where you multiply by a sine. Okay, so this was this was a sort of sketch at least the proof of last time. And I mean right so what was approved well that the key point is that once you have a quadratic form of dimension at least to it automatically represents one at all it all automatically as a vector v such that v dot v is equal to one. And that means you can split off a copy of one. And then you can keep sort of doing that inductively until you get to something one dimensional and then that's why that's where you pick up this, this element D. Because any two dimensional quadratic form represents one ie has a vector v with v dot v equals one. Right, so that also follows because so any three dimensional so you can deduce us using any three dimensional quadratic form is isotropic so basically approve this fact last time using a using accounting argument and I think on the exercises, Explore some generalizations can be Chevrolet and warning. So let me also remark that this this particular proposition so at least the first part not not the two isomorphism classes part. It works over any field. Of you invariant at most two. So, for example, see Laurent series T. And, well, so this discriminant D instead of living in plus or minus one it always lives in f cross mod f cross square. Okay. Um, right. So today I want to sort of start to dig deeper into isomorphism so it's alright so this this work great if you're over a field of small you invariant have to give this complete classification. But we want to think about, you know slightly more general cases we want to, and we want to study. We want to try to classify quadratic forms up to isomorphism. And also as I mentioned last time we want to also study this actually more powerful question of when is a quadratic form isotropic. And right so in general what we're going to need is well so this this determined as a first invariant of the quadratic form but we're going to need sort of more sophisticated invariance of quadratic forms. And the idea is that you want to you know you want to write down some number for any quadratic form and maybe the number of this group. You can define that very easily if you have a diagonal form, but then you want to sort of be able to know when that's well defined so you want to you really want to be able to know when are two diagonal forms isomorphic over over some field. So today, well maybe the first order of business is that we want to really try to understand better, sort of how did, you know how to generalize this type of construction so it's pretty easy to see the determine to something well defined. But we want to study isomorphism classes. And invariance. I should say we want to start studying this quadratic forms more generally. Okay, so, right so maybe just to start with let me write down some examples of, you know how quadratic forms can be isomorphic. We can always put our quadratic form in a diagonal form so it's some string of numbers, or some string of scalars in the field, but then maybe it's not so obvious when one string of scalars and another string of scalars actually represents the same quadratic form. So let me just give some examples. For example, if we have just a one dimensional form brackets a, then that's, well that's isomorphic to brackets a times you squared, if you use any element of f cross, so here is an element of cross. So that's, that's because we can rescale the basis factor. So that that gives us one way in which we can modify. We get the same isomorphism class. Another example is if we take brackets a one brackets a two so we have this two dimensional quadratic form, then we can permute the factors so this is the same as brackets a two brackets a one, just by permuting the basis. Okay, and then maybe slightly more slightly more complicated example of an isomorphism of quadratic forms again of two dimensional forms is that brackets a one brackets a two is isomorphic to. typo a two brackets a one. Okay, so, so back over here so brackets a one brackets, sorry, brackets a one comma a two is isomorphic to a one plus a two comma a one a two divided by h h sorry a one plus a two. So there's an isomorphism like this. So you would be no you as a scalar here because these are these are one dimensional forms. So they're, I mean so one dimensional quadratic form is, I mean if you choose a basis is just given by a scalar. So yes, it really means squaring in the fields. Yeah, thanks. Okay, so. Yeah, so this is this is an example right so so to formulate this I should, I should assume that a one is not minus a two or otherwise this is not going to, going to make sense. Right, so how do we see this last one I mean this last one is going to follow, because, well it's it's a suitable change of basis. So, I mean if you have a one, you know a one comma a two and your basis vectors or e one comma e two. Well, then you choose the basis vector which is e one plus e two, and then you choose something in the orthogonal complement. And then if you sort of scale it appropriately you'll see that you've got something like this. So yeah so maybe I'll leave this to you. Okay. So these are these are these are three examples of isomorphisms. And it's a, it's a very nice result of a bit that that these are essentially the fundamental examples of isomorphisms of where isomorphisms come from so in other words if you have any isomorphism isomorphism isomorphisms given by strings of numbers, you can always sort of build. I mean you can always get from one string to another using these moves. So, so let me let me yeah let me be a little bit more formal. So this is called vitz chain equivalence theorem. And it's the following so suppose. I haven't an isomorphism between brackets a one dot dot dot brackets and brackets b one dot dot dot brackets b so suppose I have two diagonal. I mean so suppose I have two strings numbers such as the associated diagonal forms are isomorphic over over my field. So the statement is that we can get from the first string of numbers a one through and the second string of numbers b one through b n by using exactly the above moves. So, then we can get from the tuple. So I just think of this as a tuple as a string of numbers to be one dot dot dot b n in a sequence of steps, where at each stage we only change two entries to consecutive entries. And what we do is we use one of the above three moves, only two or maybe one in the first case, two entries are changed via the moves of the above example. So at each stage basically you're either rescaling an element by square. You're permuting to consecutive elements and from there you can permit any two elements, or you're doing you have two consecutive elements and you do this last particular step so you add the first two and then you do that you do this thing. In the third item in the example up here. So the statement is that you can, you can always, you can always go in a sequence of moves from from the first or in a collection in a collection of moves of such moves from from from the first to pull a one through and to the second to pull b one through b n. So, yeah, so before I prove this let me just sort of explain I mean so if you want to define an invariant if you want to define like an isomorphism invariance like a number that you can associate to a quadratic form. So this gives you a way of doing it will essentially you need to write down what the, the number is for for a diagonal form a one through and, and then you need to check that it doesn't change when you do one of these three types of weeks. So that's something that you can check pretty explicitly because it's yeah. So this is very useful. It's very useful to define isomorphism invariance and so we're going to use this later on in the course. Okay. So, so now I want to explain how to prove this theorem. So the proof of this theorem is is the following. So, so first of all, essentially by induction. Okay, so first of all let's say that. Let's say that a one through a n and some other tuple let's say a prime one through a prime sub n are chain equivalent. If you can get from one to the if you can get from the first to the second using these moves. Okay, so right so it's saying that if you have isomorphic quadratic forms and they're actually chain equivalent. So, right so how do we see that. So what we'll show is, we'll show. So we're trying to show that a one through a and that tuple is chain equivalent to the tuple b one through b n, and what we'll show what we'll show, which will be enough and actively will show that a one through a n is chain equivalent to something of the form b one, comma b two prime through b n prime. So, we don't need to quite show that it's chain equivalent to be one through b n. But what we're going to do is, it's, we're going to show that it's chain equivalent to us to a tuple that starts with the same entry that starts with b one, but then b two could be replaced by b two prime and so forth. Right. So, why is this enough. Well, by, so by the cancellation theorem is also due to that that was proved last time. This means that brackets b two prime dot dot dot brackets b n prime is isomorphic to brackets. So b prime, I'm sorry, brackets b two dot dot dot brackets b n, and by induction can can make that into a chain equivalence. So by induction on the dimension of the quadratic forms you can assume that that comes from a chain equivalence, and then you can compose the chain equivalence. So, so by induction on the dimension so we're assuming that isomorphism implies chain equivalence and dimension and minus one. So using that and using bits cancellation theorem, it suffices to show that a one through an as chain equivalent using these moves to some to some tuple that just starts with b one, we don't need all the other bias. It was going to vary. Okay. So, right, so how do we prove this. Well, we know that. So if a one through a n is chain equivalent to some tuple. Let's call it. Yeah, let's call it a one prime dot dot dot an prime. Well, then this associated quadratic form is isomorphic is isomorphic to be one through b n. Which means that we can solve the equation. Some of, sorry, so you can you can solve the equation b one is equal to a sum of ai prime times Xi squared. Right so, well, because because b one is a is a length squared on the right it has to be a length squared on the left so you can solve this equation for some Xi. So what we're going to do is we're going to find the chain equivalent to pull right so so anything that anytime you have a tuple which is chain equivalent to the first one. Then you can you can solve this equation. And what we're going to do is we're going to find the, the, the chain equivalent to pull such that the number of non zero terms is minimized. The chain equivalent a one prime through a n prime, such that the number. So let's call this equation star, the number of non zero terms in star is minimized. Sorry. Sorry, so there's a question in the chat. A prime and be no I think this is right so, so a one through a and this is chain equivalent to be one through b n. And what we're going to do is we're going to work. A one through a and is isomorphic to be one through b n. And we want to say they're chain equivalent. And, or at least we want to say that a one through a and his chain equivalent to something that starts with b one. And we're going to we're going to show that by a minimization process. So what we're going to do is we're going to find something which is chain equivalent, we're going to we're going to look at all tuples that are chain equivalent to a one through a and that and we're going to minimize the quantity which is the number of non zero terms and star. So we're going to, we're going to do a search over all tuples a one prime through a and prime that are chain equivalent. And for each of those you can, you know, you can solve this equation star, and we're going to minimize the number of non zero terms. Okay, so we're going to find the chain equivalent to tuple such that the number of non zero terms in this equation star is minimized. And we realize that that's going to that's going to be that's going to actually solve the problem for us. So claim this solves the problem. Okay, so why, why is that. Well, so suppose. Right so suppose b one is equal to well so first of all I can always permute around my. I can also permute the tuple because that's one of the allowed moves. So let's suppose b one is equal to a sum from i equals one to our of a i prime x i squared. So suppose this is sort of with our minimal. Again I'm assuming I can I can write b one, the number of non zero terms in this equation star is this minimal. And let's say that number is our. And so in fact, so if I rescale. So by rescaling by squares, I can even assume that the Xi are equal to one. So I can assume that b one is a sum from i equals one to our of a i prime, because I mean that's going to follow because I'm allowed to rescale the i prime by by by squares. Okay. So what I want to claim is that actually R is equal to one, so that be one is just equal to a one prime. So, so what I'm claiming is that if I try to solve this minimization problem. It's going to be a space of all chain equivalent tuples to a one through an, and if I try to solve this minimization process. Then I'm automatically that's going to be exactly what I want, I'm going to get something that starts with a b one. Okay, so. So why is that. Well, so let's suppose b one is equal to a one plus a two plus dot dot dot plus a R. Yes, thank you so. Yeah, so the question so we're trying to minimize the number so so we know that be one is is a sum of the AI primes times some squares and we're trying to minimize the number of non zero terms. And so we're going to call that minimum are basically. Okay, so. So the claim is that if R is greater than one. Right so the claim is that if we don't already have be one or quadratic form, then we can simplify the quadratic form so that we're reducing the num the expression. So if be one is equal to say a one plus a two plus AI prime. Yes, thank you. All right, so by assumption we have some expression like this. And if R is greater than one. Well we can make the following move, we can take the tuple a one prime a two prime dot dot dot a n prime, well we can sort of move that to a one prime plus a two prime, and then a one prime a two prime over a one prime plus a two prime, and then a three prime dot dot dot a R prime. So that's one of the elementary moves we're allowed to do we're allowed to collect two of the terms. And then we'll, yeah, so we're allowed to do this. Notice that this actually reduces the value of our. So but then be one is equal to quantity a one prime plus a two prime, plus a three prime a three prime plus dot dot dot plus a R prime and so this should be a n prime. So by so essentially the idea is that if we didn't already have a be one in this minimal expression so in this minimal chain equivalent expression, then we could improve it we could improve it just by collecting the first two terms together by adding them because that's one of the allowed moves, and we would get something smaller. So and this is a smaller expression. So essentially the R minus one non zero terms. And this is a contradiction. So essentially the idea is that you have your you have your quadratic. Why the tuples are not the same length. Well because you've collected two of the terms right so so so we're saying that. So so so we're saying that be one is an expression is a sum of our of the AI prime. And what we're going to do is we're going to simplify the quadratic form by collecting together the first two elements of the sun, and we're going to make that into a single entry of a new tuple. And so then it becomes a sum of length R minus one, instead of a sum of length R. So, so that's why I put the parentheses around a one prime plus a two prime. So if you have a tuple. So, right. So I guess what what you could say is that for, so you could give this a name so for for every chain equivalent to pull a one prime through and prime. Let's call this the B one length, the B one length is the, the number of non zero terms in this expression B one equals sum of AI prime times I squared. So let's call that the B one length. And what we're going to do is we're going to try to minimize the B one length overall chain equivalent tuples. And the idea is if the B one length is greater than one, we can reduce the B one length, and we can reduce the B one length by collecting together, sort of terms, but by doing this by taking the sum, and that reduces the B one length. So. Well, yeah, so I just sort of summarize the proof so proof strategy is look at all chain equivalent tuples a one prime through and prime that starts with a starts with the first one a one through and that you can minimize the B one length, and then sort of by this argument, this must necessarily contain a B one, because if it didn't contain a B one then you could collect terms and and reduce the B one length. And so then you've, you've sort of made your form a one through and chain equivalent to something involving a B one, and then you sort of conduct, and then you can continue induct by induction with the dimension. Again, yeah, sir. So to make this argument. Well, we must have this. We must not have at some point in time, even prime plus a to prime to make a zero, these things so because we are putting this in the denominator so I think then we must. We must make this argument for ionisotropic forms. Right, so thanks. So actually we don't need to do that but so thank you for putting that out. That's a that's a good point. Um, so so right so so send us question is right so what happens if a one prime plus a two prime is equal to zero then right so then I'm not allowed to do this move. But if a one prime plus a two prime is zero, then I can just cancel that from the sun so the sun is not even minimal. Right, so so we're looking at really the minimal expression of B one in terms of the I primes and if there's a subsum that vanishes and then our sum is not minimal to begin with. Okay so so right so by minimality in fact a one prime plus a two prime is not zero but I should have said that so thank you for. So, by minimality. Yeah, so that's that's a theorem and again this is going to be super useful because we're going to, because these are sort of very simple moves they only involve two dimensional quadratic forms and they're going to let us. So they're going to let us define, they call the Hassan variant a little bit later in the course, we talk about quadratic forms over the pediatric, the pediatric numbers. I mean it's going to show that it's well defined. Okay. So that's, that's vets chain equivalence theorem. And so the next thing what I want to do is to explain. So this is the concept which is the concept of the the groin dick bit ring and the bit ring. So this is sort of a really useful way of encoding this this this problem, right so you know we're interested in this problem of let's let's try to understand quadratic forms up to isomorphism over over a field, and actually, right so so maybe that's a little inconvenient to state by itself but there's a way of translating this into a really nice statement just involving ranks. So we can encode this into a commutative ring, and we can sort of reformulate this question, in terms of understanding the structure of this commutative ring so that's what I want to explain next. And sorry so the question. So is there a least number of moves required for this theorem to hold. Do we know what it is. Good question I think if you look at this proof it will give you. I think if you. Yeah I think if you look at this proof it is going to give you some sort of bound because. Well at each stage you're sort of reducing this, like this be one length, like, each time you're allowed to do a move and you're producing the be one length. So I think I think it is going to give you a bound. Yeah, I mean, yeah. I think, yeah probably if you really want to make a quantitative probably it's better like to allow all permutations for example and count that as sort of one, one, one move instead of just allowing transpositions and so forth but anyway so I think you can make it explicit. But I don't know offhand. Okay, so the next thing I want to do is to define the growing the equipment and bit ranks. Yeah, so as I said this is, this is going to be sort of a really nice way of sort of encoding this problem of let's classify quadratic forms up to isomorphism, just by producing a commutative ring and you know we can then try to study this commutative ring via various techniques. Okay, so this is called the growth of it ring. So, right. So, yeah, so f is f is my field throughout so so the ringed. Okay, first let me define it as an abelian group and then I'll explain why it's a break. So, the abelian group. GW of f is, is, is the following. So it's all formal differences of isomorphism classes of quadratic forms over f right so so what does that what does this mean. Well, I mean this is saying that. If you look at isomorphism if you look at quadratic forms over f say up to isomorphism, and you can add them, you have a notion of a direct son. So there's a there's a well defined sort of addition law on them, but you can't quite subtract them I mean you can't subtract vector spaces, for example. So what you do is you consider all formal differences of isomorphism classes of quadratic forms over f. So, you know, in other words all expressions, let's say, packets vq minus brackets wq prime. And, well you say that two expressions are the same. If, well, you move everything onto the same side you move everything onto the same side and then take that those are equal. So, I mean sorry so this is exactly like so if you think about how you, if you know what you know if you define the natural numbers, then you define the integers are the non national numbers are non negative integers, then you define the integers as formal differences of natural numbers. And it's exactly that procedure that you're doing here. So this is now analogous. Okay, so I put more details about this on the problem set which isn't yet considered that. So analogous to the definition of the integers as formal differences of natural numbers. Yes, right so maybe I should just, right so maybe I should just be a little bit more explicit so, for example we're going to say that the formal expression brackets v comma q minus brackets w comma q prime is equal. So in the growth in the event ring to let's say brackets v. Well, let's say v one w one, let me give these names is equal to brackets v to q to minus brackets w to q to prime. If. So these are sort of formal differences of elements and we're going to say that they're equal in the growth in the event ring, if. We move everything to the same side so if the one direct some w to his isomorphic to brackets v to direct some w. Yes, exactly so thanks for yeah so this is exactly sorry. So the question is, is this like the group completion of a monoid this is exactly the group completion of a monoid. What, so you expose a little bit on the exercises. So if you if you look at isomorphism classes of quadratic forms over F, it's an abelian monoid, because you can add things but you can subtract them so it's not a group. And it's exactly the group completion where you so whenever you have a billion monoid so you can add things but also subtract, there's a universal way of form group. And, while we formally add differences and that's exactly this construction. Thanks. Right. So, in general, if you do this construction you have to be a little bit more. So in general. So, I should also say that if I if I really define things like this, then I'm sort of implicitly also using the cancellation theorem so I'm sort of implicitly using vets cancellation theorem. If I want to define things like this. So, so this is, this is correct. And so the nice thing about this is that right so this is actually a commutative right. So, sorry, so first of all this is operates an abelian group but this is actually a commutative ring. And it's a commutative ring via construction that I think maybe hasn't yet been defined in the course, which is a construction of tensor product. So via the tensor product of quadratic forms. So if you have symmetric, so if you have symmetric bilinear forms or inner products on vector spaces v1 and v2, then you get a symmetric bilinear form on v1 tensor v2 by tensoring the symmetric bilinear forms. So this is because can tensor symmetric bilinear forms. So there was a question why are we implicitly using vets cancellation and yes that's exactly right so the answer in the chat is exactly right so so this this definition of of when two formal differences are equal so we're trying to put this on the problem set. This, this is a, in principle what you should really say is that these two formal differences are equal. If, instead of saying that v1 plus v2 v1 plus w2 is isomorphic to v2 plus w1. So I think that they're stably isomorphic to become isomorphic after adding the same thing to both sides. And, but the, the content of vets theorem is cancellation theorem is that that's the same thing as just being isomorphic. So, yeah. Okay. Um, right so so this is actually commutative ring and I mean maybe the simplest example is if right so if f is a complex numbers. So the grown G fit ring of F is is just given by Z, because all dimension and quadratic forms are isomorphic. So isomorphism classes of them of quadratic forms are as an ability to monoid the non negative integers and when you do this procedure you get the you get the integers. So GW of C is is is given by Z. And, right so so GW is, it's sort of great because if you want to say that two quadratic forms are isomorphic. Well so any quadratic form just gives an element of this ring GW, and to say that two quadratic forms are isomorphic is just to say that the two elements of the ring are equal. So now you've translated the problem of one or two quadratic forms isomorphic to one or two elements of the commutative ring equal. Two quadratic forms are isomorphic. If, and only if they yield. Equal to any quadratic form again gives you an element of the growth of that ring. And the statement is that to protect forms or isomorphic if and only if they equal if they yield equal elements of the growing the growth of that ring of F. And so this is great because now, you know we can try to study properties of this commutative ring and like we can try to study linear functional on this commutative ring that for example will detect isomorphism classes. Okay, so. So I also want to explain that there's a nice presentation for this ring. So the presentation of GW of F as let's say as an abelian group, and then I'll explain the ring structure. So namely, you have generators. Right so any quadratic form is a direct sum of one dimensional forms brackets a. So that means any element of the, the groin dick fit ring is a formal difference of brackets a's. So it's generated as an abelian group by these classes brackets a as a ranges over F cross. So so this this groin dick fit ring is has has generators which come from the one dimensional forms, and any element of the groin dick bit ring is a is a formal difference of sums of these classes. So then you have some relations and the relations well the relations are just the ones that we've already seen. So you have relations. They come from well brackets a times you squared is equal to brackets a for you and F cross. And then you have the relation brackets a one plus brackets a two is equal to. So this is equal to brackets a one plus a two plus brackets a one a two divided by a one plus a two. So this is actually, I guess this is a presentation of the groin. I mean this is a presentation as a as an abelian group. So you get exactly the, well, essentially by bits chain equivalence theorem I mean these are these are exactly the ways in which you can get from one isomorphism class to another isomorphism class of quadratic forms. So so you get exactly these generators and relations. And while the multiplication is given by brackets a times brackets B is equal to brackets AB. So there's a very explicit presentation of the of the groin dick fit ring by by generators and by these by the relations which come from comes directly from the from it's right right sorry so there's a question it. Can you realize any commutative ring is GW. I think the answer is no, because for example one thing that's really special so I'm going to say more about a variant of this color bit ring but one thing that's really special about the G this ring GW is that it has all these generators brackets a and all these elements square to one. So brackets a times brackets a is brackets a squared which is one. So most commutative so in particular it's it's generated by all these elements it's square to one. So, yeah. So I mean in fact there are lots of restrictions on on the groin dick that rings. But so for example, as you sort of see this on the exercises groin dick that rings never have odd torch. So all the torsion is to torsion. In fact it's sort of closely related to more drinks at the field. Okay, so. Well, so I guess I'm going to mostly focus on a variant of the groin dick that ring I just want to mention that. So the groin dick that ring itself. So, I guess it's, it classifies these quadratic forms. I, you will also so if you if you go to a Kirsten Wickelgren's lectures at the graduate summer school, then you will also see some applications of the of the groin dick that ring to to enumerative questions for a very different form. But so in this course what I. Sorry, what I want to focus on though is actually not the groin dick that ring, but, but a variant of the groin dick that ring which is, which is going to be a little bit more convenient. Right. So observe. And this is this is related to this phenomenon that we have, we have this notion of hyperbolic forms. Okay, so. Sorry, so let me say it this way. So, in the ring. GW of F. There's going to be a very nice ideal. There's an ideal generated by the hyperbolic plane. So generated by brackets one minus one. Well, I guess maybe I should say brackets one plus brackets minus one. So hyperbolic plane. And what's really nice about this ideal is that this ideal looks the same for every field. So this ideal is just Z. Why is it, why is it Z. Well it's because if you take a hyperbolic form and tensor it with anything it becomes a hyperbolic form again. So, hyperbolic tensor with any form is hyperbolic. And I think this was some version of this was on the exercises if you take a hyperbolic form so brackets one minus one is in particular it's isomorphic to brackets a minus a. So it's on the exercise. G brackets one minus one is a some more fit to brackets a minus a. So you always have so for any field F you always have this ideal. Which is actually just a free a billion. It's a Z on on the hyperbolic plane, which sits inside F. So we always have an ideal, which is given by Z times the class of the hyperbolic plane. And it sits inside the ground exit ring of F. And because it's an ideal we can form the quotient. And then W F is the quotient by the hyperbolic forms. And so again this is this is kind of nice because hyperbolic forms, sort of they look the same over any field and any question involving hyperbolic forms is sort of easy to answer him or this is a sort of linear algebra. The questions that are most interesting are really the anisotropic forms. So if we want to sort of think about these questions it's, it's, it's often nice to quotient by by the ideal and again it's an ideal that's a copy of the integers and produce the produces which is which is defined as the quotient. So in the bit ring. So question from last year. So basically when we just to make sure that whether I get this right. So when we quotient out by this ideal generated by the hyperbolic plane. What we do is that by the splitting theorem we have about isotropic forms breaking into well splitings of hyperbolic planes and something anisotropic. So we actually just focus on the anisotropic ones right because we kill all the, the isotropic bits, the hyperbolic plane summons, is that correct. That's exactly right yeah thanks that's that's that's exactly right and that's why the bit ring is really convenient. So in fact, in the bit ring. Right so so one thing that's nice about the bit ring that's that's maybe nicer than what happens in the groen dieck that rain. So in the groen dieck that ring not everything not every class is coming from a quadratic form because you have these differences. But in the bit ring, in fact, any class is represented by a unique. So any element of the bit ring is is is represented by a unique anisotropic form V comma q. Whenever you have an element of the bit ring there's there's always a unique anisotropic form that you can associate to it and conversely. And that's because well any form is is anisotropic plus some hyperbolic forms, and hyperbolic forms go away in the bit ring. And for example, if you want to form the negative in the bit ring so so this is an so if you have a form V comma q that has an inverse in the bit ring. So, the additive inverse in the bit ring of F of V comma q is V comma minus q. It's the same vector space with a negative of the quadratic form. And that's because if you add these two quadratic forms together so this was on a earlier problem set. Well because V comma q direct some V comma minus q is is hyperbolic. So this is a form of copies of the hyperbolic plane, and hence a zero in the bit ring. So this is really that the really nice feature of the bit ring, which I think one one feature which makes me a little bit nicer than the road to that ring, which is that every every element of the bit ring, actually comes from a form and that form and a unique anisotropic form. Up to isomorphism. And so for example if you want to add elements in the victim. In the bit ring. Well you add the anisotropic forms. And well it doesn't have to be anisotropic anymore, but then you just sort of peel off the hyperbolic part. Okay. So yeah, sorry so there's a question what did yes so I'm going to say something about that in just a second thanks for the question. Okay, so, so let me let me also comment that in the bit ring. I mean basically you have the relations in GW. So you have the same generators you have the generators that come from the one dimensional forms, and you have the relations in GW. But also, you have this relation that brackets a is negative of brackets minus a. Well, because if you add these two you got the hyperbolic thing just zero. So this is the additional relation that you're imposing in the bit ring that you don't have in in the growth. Okay, so it's sort of a really fundamental question now in the theory so like if you want to classify quadratic forms over an arbitrary field. So if you have some field and you want to try to classify quadratic forms over it. Well, as we've seen it's equivalent to classifying anisotropic forms because everything has a unique anisotropic component. So the problem is really like, how do we determine the bit ring of an arbitrary field. And what can we say about the bit rank so this is really one of the. This is really a sort of a central question in this theory of quadratic forms. So, yeah, so in fact this is very, very well studied question and I think you'll explore it a little bit on the church explored a little bit on the problem sets on the problem set today, but maybe let me just give some examples. So an example is if the if you look at the bit ring of the complex numbers. So, we saw that the growth in the ring of the complex numbers is is just Z, right because it's all quadratic forms or isomorphic. And you're modding out by the hyperbolic forms which are the ones with even dimension. So the bit ring of C is Z mod to, and the map is coming from take the dimension mod to. So in general, for any field F. A dimension map. Well, if you think about it from the growth in the exit ring, it goes to Z. Take the dimension of the quadratic form. It's not quite well defined on the bit ring because the hyperbolic form has dimension to. So if you pass to the bit ring, the dimension is only defined mod mod to so it goes into Z mod to. So for any field, you always have a dimension map from the bit bit ring to Z mod to. And if your field is a complex numbers or if it's algebraic reclosure even if it's quadratically close any how much is a square root. Then that map is an isomorphism. So that's the first example of a bit right on the next example is right is is the real numbers. And in this case if you look at the bit ring of the real numbers. Well the claim is that it's actually isomorphic to Z. The claim is isomorphic to Z via the signature. So, so if you have any quadratic form over. So if you have any quadratic form over the real numbers. Then what you can do is you can you can send it to its signature. So that's the safe you diagonalize a quadratic form you have a bunch of plus ones you have a bunch of minus ones and you take the difference number plus ones and the number of minus ones. And that signature is well the signature of a hyperbolic form is zero because hyperbolic form has a signature zero it's one minus one. So the signature actually gives you a map on the level of bit rings, which goes from w of r into the integers. Okay. So in general, if f is a field, which has an ordering. So what is an ordering on a field I mean it's like it's a relation less than it there's there's an ocean for one element of the field to be less than another element, and that relation satisfies the the usual. I mean all the usual properties I've said the usual ordering on the real numbers. And another way of saying is that you have a notion of positive elements so. So let me not maybe spell it out but let's say you have a notion so if you have a field with an ordering you have a notion of notion of positive and negative elements. Right so so then you can define sort of a generalized signature with respect to this ordering, which goes from w of f to z for so whenever you have an ordering on a field, you can define some version of the signature for quadratic forms, and you got a map to Okay. So, so I'm just about out of time so maybe I'll just say that yeah so you'll expose a little bit. So yeah I'm going to try to say more about the structure of the bit ring a little bit later in the course. But so for example you'll on the problem, you know on the exercise, there's a really. I mean so there are a lot of like general structural theorems about about the structure of the bit ring. So, so if you have an element of the bit ring, you can always look at these sort of generalized signatures on the for every ordering of the field. And it's a, for example, it's a beautiful theorem of fister so that that these these sort of generalized signature elements detect all the torsion in the bit right and in fact all the torsion is to to power torsion. So it's a theorem of fister, which will be on the problem set that any element in the kernel is torsion. I mean so that's kind of cool because it's saying that if you have, if you have some element, if you have like an anisotropic quadratic form representing some element of the bit ring. Then, if all the signatures for every ordering or zero, then it's torsion in the bit ring which means that some direct sum of copies of it is hyperbolic. And that relates to the question of like, you know which elements of your field or some squares and so forth. So, there is a lot of sort of general stuff you can say about the bit ring and it's relation to field orderings, which you explore a little bit about in the problem sets. And later in the course we'll see some sort of relations of the bit ring to Milner case theory, and that's so that's going to be relevant when we talk about the bit ring of the periodic numbers. Yeah, so I think I'm out of time so I will, I will stop, stop here and I'll post the exercises soon. So, okay yeah sorry so there's a question which is what do the. Yeah so please ask questions now but let me start by answering the questions in the chat. So, yeah so you can ask, you know what are the units in the bit ring for example. I mean and this is a type of, I guess sort of course sort of structure at the arm about the bit ring. So, yeah, it's going to have to map to plus or minus one under each signature. And well you could say okay so maybe you're, you don't have any orderings in that case you should also look at the dimension map to mod two. So I think the statement is that that's exactly the criterion for an element of the bit ring to be a unit. So for example you can classify, can classify the prime ideals. So I mean this is really fun because, sorry, right so you can classify the prime ideals of the bit ring w of F will basically in terms of ordering so the fields. It's actually really fun because you can, I mean, you have the bit ring which is the thing you define in terms of quadratic forms, and you can show for example that. Well the prime ideals essentially all come from order I mean if you have a prime ideal of the bit ring you can use that to actually produce an ordering on your fields. So there's sort of a converse to the signature construction, which is that whenever you have a map from the bit ring and to see it it always comes from an ordering. So yeah so in fact I think that's that's going to be on the problem set, and it's yeah it's fun. Okay so. Okay so the next question is, is there an analog of Sylvester's law for an ordered field involving the generalized entry. Yes, exactly so I mean so Sylvester's law is saying that some the signature is well defined up to isomorphism so another way of saying Sylvester's theorem is that it's Yeah, so the signature is a well defined isomorphism invariant of a of a quadratic of a quadratic form. And in fact, Sylvester's theorem is true over any anytime you have an ordered field. And actually, maybe it's a fun exercise like if you want you can. You can you improve it for example using the same way that you would prove Sylvester's. Yeah, so I guess you can prove you can prove the sort of generalized Sylvester theorem using using the same methods that you would use to prove it over, over the real, over the real numbers. But I think it's also kind of fun just to see that, like to really see that the bit the bit ring is really encoding, like the orderings of the field and, and, and yeah, and signatures. Okay, so the next question was, what is the map from the bit ring to Z mod to it's the dimension mod to. So if you have a quadratic form, you send it to its dimension mod to and writes the dimension is a priori is defined in the integers. But if we're saying we're working with the bit ring, then we have to send hyperbolic forms to zero, hyperbolic forms of even dimension so we work on to. So there's always a map from the bit ring to Z mod to, and that map is an isomorphism if you're on if your field is algebraically algebraic closed. Yeah, so okay so the next question is what information can we get from the spectrum of the bit ring. I think the answer is basically you get the information that you get is, is what are the orderings. So, so the prime ideals of the bit ring they, they all come from, well I mean, right so you have this dimension, you know dimension mod to. So that's always there. But then the other prime ideals essentially they come from orderings of the field that's a. So, so in fact you can make the collection of all orderings on a field into a topological space it's a profiling space. And so you're going to see that. I would also say that if you have a field that is not ordered. So that mean, or that cannot be ordered so that that by a tier of iron and shire means that minus one is a sum of squares. In that case the bit ring is all to power torsion I mean it's, it's, yeah, so then it's, in general it's not, you know it's not going to be simple to torsion but it could be sort of to power to power torsion. So, it's a yeah it's also kind of fun, which is that. Well, I will try to put some of the exercises at some point. So if you have a field, which is not orderable. So it does not admit an notion of ordering. Then that's going to happen if and only if minus one can be written as a sum of squares. And right so I guess you can what you can do is you can ask for the minimal number of squares needed to to write minus one as a sum of squares. And that is always going to be. Sorry, I guess that is always going to be a power of two minus one thing. And, and that's going to be exactly the order of. Yeah, that's going to be exactly the exponent of the power of two that annihilates them. Yeah, so somehow the general structure of the bit ring is that it sees a lot of the, the sort of yeah I guess the sort of course structure of the bit ring is going to see. It's going to see sort of information about like orderings in your field and you know when when our elements like sums of squares and so forth. I think when you, if you want to see sort of more detailed information, then that's going to, so I think the most detailed information about the bit ring is going to come from its filtration by the kernel of the map from w of r to Z mod to the function on to. And so, so, so, so that's. So, that that bit ring modulo, well the associated graded terms of that filtration were the subject of a celebrated conjecture of Milner that was proved by Orlov, which I can provide ski that that's the same thing as no more K theory want to. So, so, so if you work on this associated graded terms, then you can really relate this, it turns out this is basically the Galois co homology of the of your fields want to. And so I mean, I guess in this course we're I mean so in these special examples that we're going to say it's going to be cases in which the bit ring is somehow relatively tame in that this this filtration so this station which I'll maybe try to say more about like on Monday or something is is basically finite or it becomes a torsion freedom at like after three steps. And in those cases, you can really sort of classify quadratic forms in terms of fairly sort of sort of simple invariance and the discriminant as a determinant and this has a house and various signatories. More questions. Thank you. Thank you. My office hours now. Okay. Yeah, I guess I will sign off then. Yes, I will try to post the problem set in Chicago shortly. Otherwise I'll see you. See you tomorrow. We're seeing officers.