 So welcome back everyone to Danny questions second talk on field arithmetic. Okay, great. So thanks again and so today, I have the following kind of a diagram of my of the lecture so this is the plot. Basically we've have what we might call arithmetic problems. Which is basically the kinds of things that I tried to set up last time these have the structure of like, you know you fix a field. Okay, and then we look at what we can say about complexity of co homology or of let's say algebraic structures. Okay, so this might be things like things about the group or the symbol length or the index of co homology classes things like that. So some examples where things like the diaphragm time dimension the co homological dimension. I think I mentioned the period index problem and the period symbol length problem. On the other hand, there is kind of a different kind of set of problems which you might call, I don't know algebraic structure problems. I think I probably should have like consulted a thesaurus before I like made this diagram up because like I'm oh yeah I don't know algebraic structure that that not really descriptive but what I really mean is maybe you fix a ground and then we're interested in like talking about properties of algebraic structures of some given type over all possible extensions of this of this field you know so or maybe just you know overall fields maybe you know like what can I say about, you know quaternion algebras over all fields quadratic forms of you know dimension 10 over all fields things like that. You know describe. Let's say structural features of algebraic structures of a given type over all field extensions. Okay over cannot so maybe incredibly transcendental infinite field extensions. You know but you know so maybe cannot is like a prime field or something like that, for example. I guess the so the plot today is what I'd like to do slide this down a little bit is describe. Kind of connection kind of told the story of a connection between these two things. By way of the notion of a central dimension. So I'll say just a little bit about what that is. And then, you know, say a bit about the things which which help us compute a central dimensions of some particular things of interest or I think compute is too strong a word but let's say, give upper and lower bounds for this. So, on the one hand, there's the notion of canonical dimension. Which can give an upper bound. Well, this is all like, kind of super vague and like almost nonsensical, you know but but whatever so like, in certain contexts, if we on, you know, canonical dimension will give us way if we compute canonical dimension will get a way of computing lower bounds for a central mentions for certain functions of interest for certain kinds of algebraic structures. And then on the other hand, you get perhaps lower bound sometimes from comological invariance. Okay, so this is like the story that I'm hoping to tell today. I'm hoping to, you know, kind of pretend although I'll remind you a little bit that that we, you know how to wonderful, you know, complete discussion about things like this on the left. And what I'm going to try to then do is motivate a couple of problems here on the right, describe a little bit about how to bridge this through and then we'll say a little bit about bounds. And, yeah, okay so let's actually start with a particular concrete problem. So, let's take a look at this particular comology group. So K is some fields. H2K mu L. So, this in this particular case, this actually is the L torsion subgroup of of H2 with coefficients and all of GM, which goes by the name of the brow group. And this is a particularly interesting comology group just because it has a lot of kind of nice concrete interpretations and it comes up in various contexts, but the problem that I'd like to present is, if we're given some alpha H2K mu L. And let's say we know the index of alpha is some number in so just to remind you, the index is going to be defined as the GCD of the degrees of field extensions, which, which make that comology class trivial. And in the case of the brow group, it happens to be the same as the minimum as well. So it's a little easier to think about. And I guess I should say, in general a nice open question is like whether or not GCDs and minimums are the same for other degrees of comology and other coefficients, this is pretty opaque in general. All right. So, so recall that that if K contains a primitive alternative unity, you know, and so we can identify mu L with just the constant group Z model or with, you know, itself twisted twice. We can identify this in turn with mu L tensor two, which we saw is the Milner K group mod L. And in particular everything is a sum of symbols because of this explicit presentation of the Milner K group. And so we know that we can write. We can also still we can write alpha as a sum of symbols. So here each AI is, I don't know like it's some BI comma CI for a BI and CI, or a K star. So we know we can we can find some number of symbols that we can represent that comology class. And the question is, how big does our have to be. Okay, so just to, just to back up a little bit then like, if we haven't, you know, if I haven't told you anything about the index if we had like, assume that we didn't have any information there, then the symbol length and principle can be arbitrarily long as the field gets more and more complicated. So this is so it's known that there's no kind of natural kind of that there's no kind of bound on the symbol length in the in absolute terms. But if we require that the index is is some fixed value. Well, then it turns out. So it follows from, let me say quote the literature. You have to string a few different results together that that there exists an absolute. Oh actually wait no maybe that's actually no I don't think that's true actually. I'm sorry I was going to say that we know that there exists an absolute bound that doesn't depend on the field, but actually, I mean, it does there. There exists an almost absolute bound a bound. Let's say depending only on the characteristic. I'm sorry, let me just think about this for a second. Does it depend on. No, no, no, no, no, there's there's just an absolute bound. I have this thing I like psych myself out because but whatever there, there is an absolute bound. We don't know, you know, so if we fix, if we fix L and N, then there is an absolute bound. You know the one that was depending only on L and N, but not okay. Okay, but, but we have, you know, no idea what this bound actually is. One second. Okay. So, yeah, let's see so. Okay. Yeah, so some cases are known. So, I'm going to mention a few of them I feel like I, there are a few others that one can put together from from some facts that exist but these are the kind of the ones that kind of come to mind. If L equals N equals two or three. Then the answer is that you needed you only need one symbol. Okay. That's that's very small. If L is equal to N is equal to four, I think, it's certainly no I think I think the answer is two, at most two, but I have to. I should have like check that it's not like, but I am pretty sure it's to my guess. If L equals to an N equals four. You can do it with two. This is a kind of a famous classical theorem and central symbol algebras. And if L equals to any equals eight. Then a more modern famous theorem is that you need for this is a result of Rowan and Chignol. But other than that I don't, I mean I think there might be like something known about maybe like three and nine, you know, like three torsion classes of index nine I think, but I'm not, I'm not totally sure you know this is about as far as we know in kind of just absolute terms. But in the case that K contains some K naught, where, where you have a finite diaphragm time dimension, if you just contain such a field. And it turns out you can produce an explicit bound. So, you know, so that is so in particular, if you're in characteristic P. Every field of characteristic P contains a finite field. Those are those have diaphragm time dimension one. And therefore you actually have a result for that characteristic zero, not so lucky. So let me say a little bit about how this works this works by kind of a two step process on the one hand by by bounding this so called the central dimension and on the other hand by kind of arithmetic results in particular fields, knowing things about the periods of life. Okay, so so roughly speaking, so how does this, how does this work. Well, I think it shows that that if we're given some alpha in H2K mu L as above, then it turns out that we can find some field L sitting between K naught and K such that L has a has is L is a finitely generated over K naught with transcendence degree. Depending only on, excuse me, on, of course, cannot but also L and depending on L and in I called it this was the this is basically the period it's the, you know, the, at least gives a bound on the order of the class and it is the index. So, so this in a nutshell is the statement of the fact that central simple algebras or co homology classes of a given period, an index have finite essential dimension. I mean, this is actually a very specific number this transcendence degree that you can, you know, write down without too much trouble or at least write down an upper bound for it without too much trouble. All right. So, now, if if the diaphragm time dimension of K naught is bounded. Well, I guess I didn't mention this but it turns out that the that if you have a finely generated extension the diaphragm time dimension of that extension is going to be bounded in terms of the smaller field and the transcendence dimension of L right so the so the diaphragm time dimension of L is going to be no more than the diaphragm time dimension of K naught plus the transcendence degree of else was an important property that I probably should have mentioned. So overall, there is this nice result of a belly motsery, which says that we can bound the, the symbol length in H2K mu L in terms of the in terms of the comological dimension. Danny there was sorry not in terms of the sorry in terms, excuse me. I have a question in the chat which I'm going to interpret. Does this L the, I mean the transcendence this bound on the transcendence dimension doesn't depend on the choice of your class alpha, just on the L in the end but what about the field itself L. How much does that depend on the alpha. Oh, well, um, the right. Yeah, that's a that's a good question. So the idea is there is a how to say this exactly I mean the, there are different ways to kind of answer that question. I mean it kind of it kind of does depend. But, but the point is that there's kind of like a, at least a universal like a comology class on some on some ring for which everything is specialized from so you can so one way to say that is like, there's kind of like a universal field. I don't know why these things live, but the, but you, but it's not a matter of you know fields. It's it's a little bit awkward I'm talking in a field kind of way, but you don't have like you know, homomorphisms from a big field with smaller field right. So, so this L would be like the kind of like the function field of this, the quotient field is our. Exactly. Exactly. More or less it doesn't it doesn't depend too much on the alpha. Yeah, there's like, yeah, so there's some universal. Yeah, there's some kind of universal are. And all the L's are are homomorphic images of the R. That's really what's going on. Right. Great. Okay, thanks. Okay, yeah. I think L equals cannot. I don't think so because, oh wait did I. Okay, so it looks like what I wrote actually was not a sentence so let me turn it into a sentence and see if it makes sense. What I meant to say was, if I write some L finally generate a block such that I forgot to say such that. Alpha is in the image of the map h2l mu l to h2k mu l. The point is, you can find, you know, alpha arises, you know, maybe not from all the way down to K naught, but from some, you know, from some field that's not too much bigger. Okay, does does that. Okay, great, thanks. Yes, so what was I trying to say. Okay, so yeah what Mothry showed is that you can bound the symbol length in this group in terms of the diaphragm time dimension of K but I guess it's really L that is the relevant thing here. So, and yeah. So all if we if we glue all this together, what does this say it says that. Well, somebody gives us a particular Allen and in. And we say, well, we know that there is, I, you know, for a given alpha with a given Allen in, I can find some field whose transcendence degree is bounded in terms of Ellen in. Because this thing is bounded in terms of Ellen in this thing is then bounded. So if some bound on the diaphragm time dimension. And then Mothry says, oh, and then, you know, by, by what I was saying before alpha is in the image of this map so alpha really comes from here. And so it's coming from some field L with bounded diaphragm time dimension and Mothry says now we can bound the symbol length for an element in there. And then of course the homomorphic image, you know, has no bigger of a symbol length. And so, put that all together and you do get a bound on the symbol length. So, let me just say what this bound is just so you can see it, just for entertainment purposes and just so you know that these things. This is actually somewhat constructive all that sort of horrible. So the explicit bound that you got looks like it's fall so to make it look a little bit simpler. I'm going to assume that things are our prime power so let me assume that L is p to the T and in is p to the s. So, you know, one knows that the period divides the index same prime powers and just focusing on one prime power. So here T is less than or equal to us. We get that the length is at most. So it comes out to T p to the power p to the power to s minus two plus one plus the diaphragm time dimension of cannot minus one. So, good that you have the minus one there that makes I know that really saves it right. So, if you, if you look at that like for the particular example like this kind of like L equals to in equals eight case, for example. So here you know so in other words p is to see is one and as is three. So we know that the length by the T null row and stuff is at most four. But the bound that we get instead comes out to two to the 17 plus the, the diaphragm time dimension of cannot minus one. So this is not sharp right, but, but you know it's less than infinity right which is which is something. And it's it's explicit. And so this is kind of like kind of a somewhat annoying you know if like if we were to say cannot is equal to q. In this case, because q doesn't have a, a finite diaphragm time dimension, we don't know, you know we know that there is a bound one can show there is a bound. But, you know, there's no way of, you know, we don't know have any way of writing it down. It's, you know, I mean, even if it's like a horrible number, we don't even have a horrible number of some sort of question of alphas with large symbol length. How to say it's very difficult to for somebody to write down an algebra. And, you know, it's so it's easy to write down an algebra. And very difficult to show that you can kind of write it as a sum of very few symbols, you know, I mean, so like, you know, like, oh, sorry, a co homology class in this case these correspond to algebras but But yeah, I mean, you know, the real question is somebody writes down a symbol, that's the, you know, sorry, a homology class that's a sum of 20 symbols. And then you happen to know for some reason that it should be rewritable as a sum of three symbols but how would you do it it's there's not it's this is this is hard. I'm not sure if I answered that question but So, you know, so one has Oh, I guess I should say like so one has upper and lower bounds, you know, in the you have a lower bound to on this symbol length. The lower bound the guy must have scribbled it down somewhere. Yeah, so In this case, the lower bound looks like the ceiling of S over T Plus one. It's really like the plus one is very hard. So like, you know, like, so it's, it's very difficult to show that you that you know that that you can't write it as a small number of symbols. You know, so this is So the real answer is somewhere between these two numbers. It might be that the real answer is S over T plus one. It's pretty horrible right. Okay, so let me now mention Let's go. Ah, okay. So an analogous problem for quadratic forms. So this is the problem of fister numbers. So the question, it has a similar kind of a flavor so question. Maybe I'll just let's just recall just for a moment. There's been a lot of information for you guys in the past, you know, 24 plus hours or whatever. So, given a field. Okay, we defined this ring called the vet ring. The elements are let's say isometry classes of quadratic forms. That is to say they are degree to polynomials up to change of basis. And non degenerate, which means that the associated bilinear form sets up a perfect pairing between, you know, the vector space and it's an isomorphism with a vector space and it's dual. Perfect pairing with the vector space in itself. And let's see. So the addition is via perpendicular. The product is the so called chronicle tensor product. Which maybe, you know, is. Yeah, so like a right and then there is a notion of equivalence. Or, you know, the, which is that we just via declaring that hyperbolic plans are trivial. So the hyperbolic form depending on how you write it is the form x y or it's the form. The form is squared minus y squared, you know, characteristic not to the same thing. Or the way we've described it in this kind of diagonalize notation, we would have written this form as one minus one. Okay, so. So the VIT group and inside there we defined this thing called the fundamental ideal, which consists of even dimensional forms. A very special kind of form inside of here are these things. These forms to look like one comma minus a, which we use the notation. Okay, I have no idea where that phone is or how to make it stop ringing, but I'm sure it'll stop soon. Okay. So, yeah, so this so we have these very special, even dimensional forms. And if you, we can take products of these guys. And these are called infold fister forms. And these things live in the power of that fundamental ideal. And in fact, they, they generate this, this empower. Okay, so. Right, okay. So that's the, so that's the, that's the setup slash reminder. And the basic question is the following. Given Q, a quadratic form of let's say, dimension D in I to the end. We know that we can write that Q is, let's say equivalent in the bit ring to a perpendicular sum of some number of infold fister forms. And the question then is how many do we need. Okay, so that's the basic question. And really more precisely even is it even bounded. Yeah, so like, so if I said, you know, I have like, I don't know, like a 80 dimensional quadratic form in I to the fourth. Is it possible that I, you know, the, or, I don't know, just reforms to write me, or is there, or maybe there's like no bound maybe there's for every, you know, integer there's some form of that size where you need more than that many, you know, first reforms to write it. So, there's, there's some aspects of this, you know, where we can at least get some, some kind of bounds on where the question kind of makes sense so like a remarkable result of of Sasha visex is that if the dimension is less than that. Well, the answer is you, it's only going to be one fister form. You know, I mean you'll, you know, well, actually maybe that's that's not exactly true, because I guess you could be just. Well, okay, okay. Okay, well, our is definitely because you can be a multiple right so but our, you know, our is bounded, certainly, and like either one, two or three or something like that I should have to think about it. Or is there anybody else who can just save me and tell me what the answer is. Sasha, are you there. Yeah. Right. It's a multiple of a fister form so it's a multiple of a fister form so what does that mean that like. Well, that means it's a difference of n plus one full fister form and then full fister form right. It's a difference of okay okay so three then I guess is well okay well, or I guess because I didn't say you could like subtract I guess so. I can introduce probably constant slam the one lambda air lambda. Yeah, okay. It's, it's small. It's, it's, it's some small number. What's on. In any case it's bounded. It's bounded. Yeah, it's totally bounded. But for for for D, bigger than that. It's, you know, it's not clear. I guess I mean that was that's my impression. I mean certainly like if you're bigger than like to the in plus one I mean I feel like there's no there's nothing to be obvious to be said I don't know I wonder if you can say a little bit about things in between but anyways, okay, but for for for D large. It's does not there's no obvious bound, nor is there. Yeah. I mean it's entirely possible for all we know that it's not actually bounded. So, on the other hand, maybe what should I say so. It is bounded. But first of all, and for for and less than or equal to three though. So, beyond, you know, higher not so clear. So let me say a little bit about, you know, kind of why is in less than or equal to three okay. Yeah, maybe I should actually define a central dimension. Let me let me go ahead and and do that. Yeah, maybe I should actually define a central dimension. Let me let me go ahead and and do that. So, let me. So I want to do it though. Yeah, okay, okay. So, definition. Okay, so yeah, so I guess I should say like a central dimension of what the so the idea is that we fix. Let's let's just kind of preface this. Let me just say a central dimension. So we preface this by saying, we're fixing a type of algebraic object. So this might be, you know, for example, classes in here, or we might think of these as Brower classes we might think of these as elements in the bit ring or elements in here or we might think of these explicitly as quadratic forms of, you know, dimension D. These classes are an IDN or what, you know, we fix something like this some concept like that. And all of these, you know, we can think of as functors from the category of field extensions of some field of some fixed field cannot to well at least sets. So, then the if we're given such a functor above. Then, and let's say we're given some alpha. So if we're given half of, I don't know, okay, for some K over K naught. Then we define the essential dimension of alpha to be the. Let's see if I want to say this. Yeah, so this is the minimal. Transcendence degree of some L over K naught. Such that alpha is in the image of F of L to F of K. So, you know, it's kind of how many parameters you really needed to define this alpha. And the essential dimension of the functor is the largest of the essential dimensions of alpha for alpha and F of K overall, you know, K over K naught. So, you know, it's basically how many parameters do you need to define stuff. So just by by way of imagination, like how should we, you know, think of these things kind of like, you know, the so kind of what a natural way the one typically bounds the essential dimension is by coming up with the versatile families. So that's probably worth, worth mentioning that as well. So definition. So let's say if we're given a functor script F from, let's say from K naught algebras to sets. Then we'll say that some I don't know alpha in F of R is versatile. I'm not sure if this is exactly the perfectly standard way of using all these words but it's, I think it's pretty reasonable, but we'll say that something in here is versatile. If for any, you know, beta, let's say in F of, you know, K for any K over K naught. There exists some R to K homomorphism, such that beta is in the image of alpha under the induced map. Okay, so the idea is like, you know, if you have some, some versatile animal this this alpha. So here's my, here's my picture, you know, of a, you know, so so like, how do I draw a picture of this. Okay, so here's our. And here's like F of R. You know, sitting above it. Maybe I should draw it like a cylinder, right? F of R sits above it. So then the idea is like, you know, if I, if you, you know, if I, if I pick a point in R, you know, like a prime or something like that, right? And like mod up by, you know, that that prime, then I get something in F of R mod that prime, which is this, you know, the fiber over that or some sort of like family thing, right? So the idea then is, you know, if I have any like L, and I look at, you know, some, some beta over here, you're going to run into trouble you need the alpha first. So here's alpha. So here's alpha is my special guy here. Thank you. So like alpha. It's like a horrible picture, I think, but okay, the thing is like for every, for every beta, there's some way of like, of course, specializing is really like kind of geometrically going the other way. It's like associating L to, you know, some point over here. And it's like saying everything, you know, like this, it's saying that this beta really is like kind of coming from restricting to some. Okay, this is not a, not a great picture. Do you feel it anyways? Is it okay? I like it. Thanks, Mark. Okay. Yeah, it's yeah, okay. So anyways, let's see. Yeah, I mean, so it's just like that. So it's your mentor. Oh, yeah, yeah. So like, I mean, so basically the, the essential dimension of a, of a, of a scheme is basically the dimension of the scheme at least of the scheme is like a variety and, you know, I mean it's the central dimension of what would it be? It would be the essential. I guess it would be the, I guess it would just be the dimension of this of the scheme, right. The, the variety itself. It gives you an upper bound anyways. Oh, I'm sorry. Yeah, an upper bound. That's that's what I should have said. Yeah. Sorry, let me close this door, which magically opens. Okay, sorry. So where was that right? So, okay, so this is this idea of a versatile family. And so the basic observation is that if there exists a versatile family, a versatile, a versatile family, I just go and look at the picture. If there exists some versatile alpha in some F of R, then the, the crawl dimension, let's say, and let's say where R is, let's say finite, finally generated over K naught, then the crawl dimension of R is going to be an upper bound for the, for the essential dimension of the functor. So some, some insightful member of the audience asked about the example of the functor of points of a scheme X over K naught. And then I guess the answer is then the essential dimension is just the dimension of X over K naught. That right. Um, F of L is because you need to at least, because like the generic point is at least only going to be defined over right that point of that right transcendence degree. So yeah, it would be that it would be that in that case. Exactly. Yeah. So, yeah, so if, so if your functor was the functor of points of a, of a scheme. Then, then the essential dimension will just be that dimension. Let's see. Yeah. Yeah, we'll just be the dimension of the scheme. And I guess, you know, if I had, if I had said this a little bit differently in this second definition, you know, if I had had a functor maybe not just from cannot algebra is but maybe from schemes over cannot, then you get, you know, then, then that would also kind of work nicely with that example right you so you get that the essential dimension would be, you know, you could see is going to be bounded above by the dimension of the scheme. But in the particular case, asked where the functor is the functor of points. If you look at the generic point you can also see that as a, as a lower bound as well. Okay, so anyways, this is the notion of a central dimension. And so, let me. So just as a very silly example. You know, let's say where we start with a quadratic form of dimension in maybe that's our quadratic forms over a cave dimension in will be our f of K let's say. So that's one example of such a such a fun term. And the essential dimension of this guy is just going to be in of course assuming that we're not in characteristic to so don't have to think too hard. But the point again is that you know every quadratic form can be diagonalized and written. Like that and so you. You know so ever so this is defined. Over, let's call it L being K not a joint a one through a and which has transcendence degree at most and alternatively, the, we could say that the form q. equals, I don't know called like x one through x in over cannot a join the x one of the plus minus one x into the plus minus one is versatile. But there's another way to kind of see that upper bounds like every quadratic form comes from this one by just specializing some particular values some non zero values for x one through x a. Okay. Okay, and, and you know it turns out, you know that you can, you know, construct. You can play some, you know, nice. Versal spaces for first central simple algebras which give of a given index, which gives you kind of a bound on these essential dimensions as well. Okay, but now, but now how about the. The fundamental ideals. The fundamental power of powers. Well, I mean you know there's not that much about being about being an eye one that's just being even dimensional. But, but if you if you recall, you know we have this. There's this isomorphism given to us by the by the Milner conjectures, which identifies this. And so, what that says, you know, is that we have then a short exact sequence. And, and so, I'm going to say this. So, so, and this was this thing that we called Ian. And we say so we can say that this, this tells us that quadratic form of I don't know dimension D in i to the end is the same as a quadratic form of dimension D. And I do the end such that Ian of Q is zero. Right. And so, so you know to you're missing. There's one of those ends has to be an end plus one I would think. Yeah, yeah, yeah. Thank you grammar grammar, finished sentences did not finish sentences of dimension D. In i to the end plus one. Wait, oh wait, other other way, wait, wait, what am I. I'd, this is a plus one on this side. Yeah. Awesome. Thank you. Thank you. Thank you. Okay. I'm glad you're here mark. I'm glad you're here. Okay good. Okay, so. Right so so then to kind of proceed what we what we need is a way of kind of like kind of what it takes for comology classes to be trivial. So kind of the tool. The tool that's kind of built for that is the is this notion of canonical dimension which is kind of a special case and formally of a central dimension. So, so let me just define it so canonical dimension. So, let's say. So, the idea in canonical dimension is we want to. We want to cut that like variant is trivial. You know, the, you know, it's some some kind of like binary decision like a yes no decision about about elements of some other functor typically right so. So what we're going to do is we're going to suppose that that F is a functor from field extensions. K over K naught or maybe for K naught algebras or something like that, not to sets but to pointed sets. So where the point is going to represent kind of triviality in some sense. And for for alpha in some, you know, it's defined over some field. We define a new functor, which I'll call I don't know. It's going to be defined by alpha just because I, you know, that's the only decoration I could think of. And it's going to be defined by by F check alpha of artists some field extension is, well it's either going to be. It's going to be empty. If, if when I take alpha, and I extended to L. This is not equal to the point. The special point in my point and set. And it's. I don't know and it's not empty it's just some singleton set. So this is a new functor that deterred that describes whether or not I'm trivial. My classes trivial. And then we define the canonical dimension of alpha to be the essential dimension of F check alpha. So if we parse that, what we're, what we're trying to say is like kind of, what does it take to make alpha identified with the point. You know, so we want to say like, you know, so we want to say that. So, so canonical dimension of a equals, you know, is it most are means that if alpha L is the point, then there exists some. Maybe I'll say alpha K, because that's how I was using before alpha K is the point. So alpha was defined over F K, okay. Okay, so now I realize that I need to say that so we define a new functor F alpha on extensions of K, as opposed to K naught because they all have to be bigger than that, the one where alpha is actually defined right so I should have said that. So, sorry, I just, I have to come to terms of the fact that I need a new letter for a field, it'll be a crisis averted. Okay. So, come on, sorry, canonical dimension of alpha less frequent are means that if alpha. So, where was that alpha, if alpha L is the point. There exists some E sitting between K and L with transcendence degree of E. No more than our such that alpha E was already the point. Okay, so like it's basically how many parameters do you need to say that this thing is split. I just said that right now I feel like I didn't say that. Yeah, it was right. Sounds good to me. I feel like I said something backwards actually. I feel like I said something like this, this, you know, I mean, let me just. It needs that you say it needs at most that canonical dimension less than or equal to our means you can do it in less than or equal to our parameters. Oh yeah that's right. Okay, I did right. Okay, thank you. Yeah. Okay. So, great. So this. So the typical approach to bounding to bounding these things. This is the concept of generic splitting schemes. But I guess maybe like, you know, before, you know, before I even say about like kind of how where these come from we can at least kind of go back and kind of think now about kind of like how one might have shown, you know, like kind of how this feeds into what we were talking about before. Right. So, like, so let's say somebody gives you a quadratic form in, I don't know, let's say an I squared. Okay, and equals two of some fixed dimension. And we're trying to show you know why can you tell me why is it. So what I guess. Okay, let me let me. Going backwards and forwards is going to hurt me a little bit so let me let me just finish and go linearly actually. So I know pedagogically I should have like made those decisions beforehand. And I did I'm just second guessing myself. Okay, let's let's keep going forward. The next approach is generic splitting schemes if you want to bound this this canonical dimension so the idea definition. If we're given some F as above. In other words a functor to pointed sets. Alpha and f of K. We say scheme. X over K is a generic splitting scheme. Or alpha, if, if alpha L equals zero. If and only if X of L is not so it's a scheme which has a point over a field exactly when that element is trivial so really what we're saying is like the criteria that a particular you know object is trivial is described precisely by some collection of polynomial equations having a solution. And you know so and easy exercise again is that X is a generic splitting scheme for alpha of finite type over K implies that the canonical dimension of alpha is no more than, than the crawl dimension of X. So just call that the dimension of the dimension of X. Okay. And the, I guess I would say the big open problem, or one big open problem. So does there exist. finite type. generic splitting schemes for co homology classes in hi. Hey, you well answered right. So we know, you know so I mean can you bound, you know the. You know, so these so these co homology classes that is to say these things are, you know obviously pointed. These are groups of their pointed sets associated to field extensions of, of K not you know K being some arbitrary field here. And, you know, so is a co homology class being trivial described by a finite set of algebraic equations we really polynomial equations we don't know that. So we do know, we do know this in special cases. So, I guess I should start with I equals zero when it almost doesn't make sense right because then the end because then it's filled extensions don't help you, it's like yes or no. So I guess the answer is yes, they exist or something. Okay, but for for I equals one. This is very straightforward. These are basically just a tall algebras particular finite schemes over K. For, for I equals to. I showed that it works for all twists. So these, these exist there. But for, I guess for I equals to j equals one this is very classical. And these are called the severity of our varieties for. And then you know, I guess we have just a few other cases so for example, for, for. Let's see what other cases do we know we know for for I equals three j equals to L a prime. These are the mercury of susan varieties. Yes, only for symbols of that sort for symbols in here and for symbols in I equals for j equals three L equals three. You can construct these via the so called Albert algebras. And these are for symbols. And then for for symbols. In general, for, for L prime. Can do this, let's say, up to. L prime to L extensions. So I really kind of should have given some more precise definitions to make to make this this clear but the, you know, the deal is like if you're, you know, often you can kind of work with one prime at a time and assume that your the fields are let's say prime to L close where L is a prime number and then some good model come all to there. And so in, you know, from the point of view of that problem. You know, you can construct what might be called L generic generic splitting varieties in this case so these are the. These are rosts norm varieties. You can play a big role in the in the proof of the block auto conjecture. But this is only for symbols. I guess I should say for for L equals to symbols. You could use the fister quadrics. Without this kind of issue of prime to illness. Yeah. So let's see. So, you know, I, so I haven't connected it back up I now have to like strategize you know we're getting to the end of the talks I have to decide what what to do exactly these last 13 minutes. I want to want to want to bring things back to the beginning. Let me just say that kind of the, the, the, the main one of my, you know, the one of the main points that I wanted to come up with is if there exists. Generic splitting. Schemes for classes in, in like, hi came you to for I heard the fister numbers. Really because, you know, we could balance is central of quadratic forms of a fixed dimension. In on to the end, you know, for, for, for larger ends. So, so I should say, I should at least kind of outline kind of how that, how that goes a bit maybe just a very, very brief. You know, very brief outline so like, so if I were to let, I don't know, let's say squiggly I upper in lower D of K is going to be quadratic forms of dimension D in I to the end. So, so if the essential dimension of ind was finite. Then. Oh, okay. Yeah. It is, it is good that I'm saying it like, okay, so. So if the essential dimension was was finite then. So if we're given, you know, some ground field. K naught. And Q is some quadratic form of this type over some field containing K naught, then Q would be defined over some L over K naught with bounded transcendence degree. Okay, so now we wouldn't we would need to know something about our K naught in order to then actually get a bound on on on these lengths. So, for example, if the if the diaphragm time dimension of K naught was bounded, for example. Maybe if K naught was. So this is to say like if our field, maybe contained an algebraically closed field or obtained to contain the finite field or something like that. Again with with Q, there's, I really don't know what to say. But, but if the diaphragm dimension of K naught was was finite. Then we'd get a bound. On the diaphragm time dimension of L. And. And, so I mentioned that before that we know how to go from a bound, it's this kind of thing involving the logarithm, etc, whatever, but you get some bound. Okay, and then. Oh, let's see. So then we would. Oh, what am I, what am I. Okay, this is what I should have actually written in my notes because I'm going on like a tangent and I'm actually not. Okay. Sorry, let me let me let me reform. What I was gonna say, hold on a moment. The argument I actually wanted to say was was actually pretty more straightforward. Let me give it. So, okay, there, there is something of this of this sort, but let me give actually a more straightforward kind of statement by assuming something a little bit stronger so like let me instead of saying like if the essential dimension was finite, let's just say like, actually, let's suppose that if there exists a versatile element, which in I am. So let's let's actually. So the one reason that the essential dimension might be bounded is because of the the existence of a versatile element. This is supposed to be bounded. Maybe I'll continue that train of thought if I, but let me do an easier train of thought first. If there's if there's a personal element inside of this inside of here, then. So, this is actually. So, this is actually a very rough argument with with with some technical details emitted. It sounds very obvious what I'm going to say but it's actually, there's actually a problem with what I'm going to say, but there's a solution to the problem. But, you know, so okay, if there's a versatile element in ID, then it only needs that particular versatile element needs some particular finite number of fister forms to write it, you know, if there's a versatile element. Then, you know, it has it has some given length. And, and everything else is a specialization of it. And so that length is going to give some kind of upper bound. This is actually like not not a correct argument, but it looks right doesn't it. I mean, I'm easily convinced. That's a good problem you don't know when you specialize that you don't know when you're here or they might have denominators but so you have to deal with that and exactly. Yeah, so there's kind of like, you know, so there's kind of like some kind of, I don't know, like no theory and induction feeling kind of argument where you go to smaller loci and get some sort of bound, so you can, you can bound it. Right, so that that particular work on some big open set, and then that element or and you just eventually exhaust the whole. Exactly, exactly. So it might get worse on certain specializations, but you kind of go down by dimension and it can only get worse a finite number of times. No, that sounds like a complete argument. That is a complete argument. Yeah, that's the complete argument. But so the. So the deal is, yeah, there's, there exists a bound. It turns out, you know, because of this almost the bound that you get here wouldn't be explicit actually, you know, because of kind of, you know, but but one could hope that you actually get a more like, like quantitative bound in some cases, if you know something about comological dimensions and stuff like that. So this, I, you know, like, so maybe just to make this argument easier just so you can see how a few of the pieces fit together like, if you knew that these essential dimensions were finite and you had a bound on them. Well, let's just say, and if you knew, like the kind of the, let's say the general some general period index conjecture. Then these would give bounds on symbol length in these co homology groups. And then, you know, there's a there's a an inductive argument where you could basically push it into higher and higher parts of the fundamental ideal by approximate by by taking these symbols, matching them with fister forms because these forms map the symbols and this correspondence which then knocks them into higher and higher degree co homology, and eventually, they all disappear. And then, you know, then you would get some explicit bound assuming that you had some explicit bound here. Really, this is just saying like if we knew this conjecture and that conjecture and then we would know something else. It's not maybe but it's nice to see kind of how these things fit together I think. Now that I'm now that I'm like out of time. Let me say something about motives. Honestly, I didn't have very much to say about about motives but if if they kind of come into this picture at all like, I mean they do actually come into this picture in various ways in terms of the tools that people have actually used to to make progress on many of these things, but. But on this part, which I didn't actually get to talk about. You know that one way to. So I should say, let me just let me just say that I think that that this is that these things are actually false. In fact, I think that that these essential dimensions are actually infinite. For that's my feeling at least in the last in the last couple weeks, I think that's how I feel. But I think I think these essential dimensions become infinite and how would you actually figure out if something is infinite, the usual way. To figure out the things maybe aren't bounded is to come up with so called comological invariance so these would be some comology classes associated to elements here that are non trivial that live in relatively high degree If things live in high degree comology, then they have to be defined over fields with big enough transcendence degree over your canots. And if those get bigger and bigger than there's kind of no bound on the transcendence degree is the idea. So. And you know I think one of the more kind of hopeful ways of kind of constructing kind of more interesting homological invariance would be. You know, following. You know some very kind of nice observations. These norm varieties kind of identify kind of a corresponding to symbols and comology. And it turns out that you can kind of see. And so I think that's one of the things that I think is really important. And I think that's one of the best classes in, you know, various varieties that that come up by these. Actually, I'm like way I'm this is like a bad. I'm already like two minutes over time and I'm like, just starting to talk faster and faster the feeling of like, it's like this feeling of like guilt for like, not having said anything about this but then it's like it's already too late right. And you know if people feel like talking about these things like during breaks and stuff like that we can. But let me like, let me be a better citizen and just stop now actually. Well, thanks very much Dan. So any any questions, comments. I had a quick question. I have a number of presentations of different dimensional quadratic forms in terms of sister forms. How does that grow with dimension. Is it exponential. Kind of like lower bounds like how many do we think we, we should need. I'm pretty sure it should be exponential but I don't actually, it's not something that I know off the top of my head I guess I mean, I don't think about that. So, yeah, you know, I think, I mean, I think what we, I mean very little is known about these actual fister numbers I mean what we have is really kind of circumstantial evidence based on things like how the essential dimension of some of these things seems to be growing. And, and I guess the. So how does, how does that go like in once you're in I cubed the essential dimensions start growing exponentially with the dimension right is not. Yes. Yeah. And so I don't know, I mean, I don't know kind of what. The lower bounds one even has for kind of essential dimensions for for higher powers of the fundamental ideal but worst expectation is not better. But I don't know, really don't know. Thank you. Sure. Some other questions. In cases where they're known to exist. Do these generic splitting schemes parameterize some natural factor. Um, do they present. Well, I mean, you know, in some sense, they're, oh, well, okay, yeah. So I'm not sure how I would answer that in terms of like, kind of in terms of like, like strictly speaking, as you phrase that in terms of functors but I guess like they they typically have. I mean, I guess, you know, what are they, what are they. So they have like, they do kind of have this kind of describable inductive flavor of how they're built up by looking at, you know, the, the, the flavor of it is, you know, you have. You kind of think about your symbol like that. And you are looking for. For things that you're looking you're kind of parameterizing extensions. That that that's that split. What is that. It's the, it's the statement that that a n is the norm of a kind of, you know, degree extension, which splits this thing. So you're kind of parameterizing degree and extensions that split this and elements in these things whose norm is that that is that an accurate description. Yeah, I think that's that's what they do. Okay, good, good. So in the, in the, in the cases where I mean for certain kind of small symbols, you know, like for, you know, where in minus one is like, you know, is like one or two or something like that. You know, you're just basically saying like elements in a field extension of a given norm, or, or subfields of a central simple algebra, or elements in a central simple algebra with a given norm. You know, in the, in the, in the kind of prime in L equals three. And let's say n equals four, where this is three you can identify these guys as cubic subfields of an Albert algebra and so elements in an Albert Albright with a given norm, things like that. So that's about as like concrete, I would know how to describe these things as. I mean, the case n equals two is very concrete there you're just taking, taking the L route of a one and asking that a to be a norm from the L through a one. Exactly. Exactly. Right. Right. Makes the symbol divisible by L. And that makes the symbol divisible by exactly so the kind of, so like kind of primed L make sure it's split. But this is only an example of pure symbols right for non pure behaves completely differently right for example. Interesting example probably is n equal n is arbitrary but L equal to and some of two symbols. When you get middle dimensional quadratic grass money and for the Albert form, which is sort of different from sister quadric right it's not a quadric but grass money instead. So for some of symbols clearly it will be completely different. Right right right kind of yeah so. Right right right right this this yeah that's a very good point right because so this, these these statements on like, you know we're really saying something very special and very non trivial about like, kind of, like the only way that this thing could be split is because there's this very special thing happening like this being a norm of something over here that's what's that or whatever that's, that's a very special kind of statement that's about symbols. But like if you have a sum of two symbols and you want to net trivial or some of more than two symbols, God forbid, you know, like, to show that something like that is trivial that's, you know, that is a whole different kind of set of equations right. So number two, you could imagine there should be like more hope in general by kind of saying that two things should be isomorphic somehow or something like that but. But again, you know, like only in special cases, can you kind of make that work for you. Some other questions. Any comments, questions. I mentioned, you know, the classical case of for H two of mu L was a very broad varieties. Yeah, and then you you had done work constructing similar things for H two mu L tensor J. Right or really for, you know, H two mu L, pretty much with whatever coefficients. Yeah. Other a tall group schemes if you want. Yeah. So how, yeah. Well, not that great exist right. I mean the, the, here's the point is really just like, you know, like how do you. It's, it's, it's just this very awkward thing, you know, like, how do you describe how to split a comology class right. I mean, you know, the so, you know, you imagine that you have some comology class. You know, I mean, let's let me just say like this, I mean every comology class is like, you know, it's like presented as like the inflation of some comology class from some, you know, from, from some kind of big Gallo extension, you know, it's like, you know, something in HN of like the Gallo group of E over K with wherever you know you are, you know, it's coming it's coming from some comology class like that, but you know, the, the part of the difficulty in dealing with like how comology classes become split is that, you know, if your class. classes is split by, by this extension, you know, you have so you have this HN, let's say the absolute Gallo group into wherever you are. If you know you're split by some L, then that means that you become that you go to zero over here. You have some alpha over here, but your alpha kind of like to work with your actual alpha it's going to come from some kind of finite thing over here. And then you might say well, did it actually get split over here. And the answer is, you know, maybe not zero here, because for, for general in this kind of inflation map isn't injective. You know, the kind of discrepancies you have to use basically something like, what is that spectral sequence called, you know, the normal group comology spectral sequence. And so, you know, which means that there's basically these kind of correction factors coming from kind of smaller comology of intermediate fields and stuff like that. And if any is equal to two, that's just some H one stuff that you have to deal with, then that's not a big deal right, because H one kind of is like inherently really bounded and, but, but if you're anything above in being to then the kind of correction factors in H two, and then, you know, kind of all bets are off as to kind of how to, you know, kind of because things in H two can have kind of arbitrarily large, you know, index and be very hard to split and stuff like that. And so, that's a nutshell about why the problem is kind of awkward I guess. Okay, thanks. Okay, so maybe we can continue the discussion in the problem session. I think it's about time for that. So let's thank Danny again for a beautiful talk.