 It's a pleasure to introduce Danny Kraschen from UPN. So Danny did his PhD at University of Texas at Austin, and then he was postdoc, I think, at UCLA and was a member at the Institute for Advanced Study and then had positions at Yale and maybe UPN as well. And finally, he was a professor at University of Georgia and recently at Rutgers. And just now I think he's moving to UPN. So please, Danny, yes, right, right. So maybe I should say what's the name of the talk. Right, it is, sorry, it is the field arithmetic and complexity of algebraic objects. Okay, so yeah, thank you very much for the introduction. So let's just, I hear some background noise. Okay, okay, I think we're good. So yeah, so thanks so much. It's a pleasure to be here. It's great to be back at Park City where I did a program just after just as I left graduate school at Park City, which I enjoyed very much. And hopefully I'll get together again physically someday. It's great, you guys, if you haven't been there, you should definitely go. It's a really beautiful place. But it's also nice to be able to interact with people in this way as well. So I'm happy for the opportunity. So let me start by giving just a very brief summary just because I don't like suspense very much or I'm just not very good at it. So here's the punchline basically. So there are a lot of problems in field arithmetic and complexity of algebraic structures. And I'm going to really spend like, I don't know, a good chunk of both of my talks really just giving an overview of like some of like lots of open problems and the kinds of questions that we'd like to know about. Motivate comology has been helpful in the past, in the past. Maybe it will continue to help us. All right, so that's basically the talk, right? So now just the details. So today I'm going to primarily talk kind of from a field-centric point of view in terms of like field arithmetic and the kinds of things that we would like to know about field arithmetic. And tomorrow we'll shift focus a little bit and focus more on algebraic structures and the interactions between field arithmetic and algebraic structures. Okay, so I'm going to try to kind of continue just from a very kind of top-down perspective. So let's start with just the concept of field arithmetic. So this is kind of absurd, but why not just say like what is field arithmetic? Like what do we mean by that? What do we actually care about? Basically what I mean by field arithmetic in a very nebulous way is that, so somebody gives us a field. We want to ask the question of which systems of polynomial equations have solutions and why? And if you know that one collection of systems has a solution, then do you know that another system of equations has a solution? Stuff like that. So of course, really everything is about algebraic geometry is the subtext. There's the kind of eats all the other fields, but okay. So which systems of equations have solutions? And then moreover, we want to answer this based on measurements of our field. So we might ask, if one special system or type of system has a solution, has solutions, which others do as well. And this leads to all sorts of kind of boundedness questions and stuff like that. And we'll see a bit about kind of how that works. Okay, this is also like super vague, super nebulous. We will actually become specific as we go. So the basic kind of tool or set of tools that I'll kind of spend most of the time today discussing is Galois homology and its relations, which are things like Milner K theory, the bit ring and motivic homology. And so these and also tomorrow algebraic structures and from a some perspective, these Galois homology classes provide kind of natural systems of polynomial equations that if you can solve, tell us a bunch of other things about the field. That's kind of one perspective on maybe why they're useful, but they're kind of very, very special ones. So determining whether or not a class is trivial or the way in which it might be trivial or saying statements of that sort often lead to particular systems of equations which kind of have more importance than most other ones. Okay. And okay, so let's, yeah, so the, so to kind of continue to outline instead of actually talk about anything in particular, let me make one more kind of outlining thing and then we'll get into it. So kind of part one will be actually Galois homology and relatives. Part two is going to be kind of basic field measurements. So these are gonna be things like notion of dimension and things like that. And then part three is going to, after I've done all those things, I'll talk about kind of the basic problems, structural problems in Galois homology. Okay, so now let me go ahead shuffle my notes around. Okay, so let's start by just a quick overview of Galois homology. So the definition is pretty straightforward. So if you're given a field, then we'll let this be GK be the absolute Galois group. So this is some profinite group. And we can consider, I don't know, call it ABGK or something, Abelian groups with continuous action by the Galois group. So Abelian groups with continuous actions. Typically, we're going to consider such groups with a discrete topology on them. So the Galois group has this profinite topology and we'll consider these actions to be continuous with respect to usually a discrete topology on the Abelian groups. And so if we're given some Abelian group with such an action, then you can look at the fixed stuff, the invariant stuff. And this is a nice left exact functor. And then these groups H and GK blank are the right derived functors of this. And so, okay, done, that's Galois homology. It's important to mention that these so-called Galois modules have a natural tensor structure, just basically where the group acts diagonally. And we get cut products, H and GK and this M. Paris would say HM GK and some N into HN plus K GK and some N. And let's see, so these are useful because they just come up a lot when you study other kinds of algebraic structures. And a particularly important case is when M is the Elf roots of unity with, you know, to some tensor powers, in other words, tensored with itself some number of times. So this is a particular group with Galois action that comes up quite frequently. And in particular, these groups HN GK and MUL since they're in are particularly useful. These, but the cut product form a ring and you can describe kind of how they start. So I don't know what Mute of the zero is, but I guess by convention we just call it Z mod LZ. So, you know, if you take MUL and you tensor it with itself zero times, that's just gonna be Z mod N and the fixed stuff is just again, Z mod N and the H1 via Hilbert's theorem 90, one can show that this is just K star L by Hilbert 90. And so in particular elements in this H1 are represented by, let's see, what did I tell myself by the way? Okay, I told myself I would talk for no more than 10 minutes about Galois homology. I'm like seven minutes in. So I'm gonna finish really soon. Okay, so this is not gonna be like an hour and a half on like what is Galois homology? Okay, I'm almost done. I hope, okay. So in H1, I just wanted to give, you know, just, it's algebra. So I feel like an algebra you actually like in principle can like define everything you wanna talk about and not like take the whole time doing it. So I thought it would be nice to make sure where we have a chance to try to do that. Okay, so if you have an element A in your field, then we write like little parentheses around A to represent the corresponding class in H1. And we use the notation A1, comma, dot, dot, comma, AN to represent the cup product of these guys, which would then live inside of HN, K mu L tensor N. These things are called symbols. And by the, by the, by the very famous block Cotto conjecture, maybe now, what should be called the norm residue isomorphism theorem proved by Vovatsky and many others. So norm residue isomorphism theorem tells us that HN, K mu L tensor N is generated by these as, you know, as an Avelian group. And in fact, I'll give, we'll say a little bit more about that. So I think it's a good idea to give a little bit of an example and I'll say a little bit more about that after we describe Milner-K theory, which is the next thing I'll talk about. One thing that's nice about Gaoua homology is it turns out to be a special case of atol homology. So let me just say notable. Gaoua homology is a special case, atol homology. The important thing to mention in that regard, so I'll kind of abuse notation a little bit and just use the same symbol M to represent this, the corresponding atol sheaf as opposed to a module acted on by the Gaoua group, but there's some kind of correspondence there. And the useful thing from this perspective is that atol homology kind of makes sense, extends on arbitrary schemes. So not just a scheme, which is actually the spectrum of the field. So formally speaking, one maybe should write is the atol homology of the spectrum of the field but typical abuse of notation just has us write a K then. But you know, but this is a, so that's that provides kind of an extra set of tools or interpretations in some sense, because we can relate this to this more geometric theory. Okay, so that's Gaoua homology. And now let me say, let me describe Milner-K theory so Milner-K theory is maybe a little bit longer to say, but it's much more elementary. So the definition goes as follows. So given a field K, we define the Milner-K theory ring, we define the Milner-K theory ring, which is a graded ring as follows. So on the one hand, the zero part is Z, the one part is K star. But the most confusing thing about this is that it's written additively, right? Because this is so additively. So to not be confusing, one typically writes this A for the element in here corresponding to the element in the multiplicative group of the field so that you have the addition structure in this ring is given by the multiplication structure in the field. And then it's generated by just this K one stuff. The products are written by concatenation. So this is really just not a definition but more of a notational point. Like we write A one comma dot dot dot comma A n to denote A one times in the ring, A two times up to A n just to reduce the number of squiggly braces. And again, these things are called symbols. And the kind of only relations or the relations are generated by saying that A B is zero whenever A plus B is equal to one. Okay, so it's like, it's easy to say it's perhaps a little hard to kind of justify in the abstract. I mean, you know, you can write down whatever ring you want to, but why is it useful? But it's like a truly magical creature the relationship you can see between these things in Gallic homology is, you know, which is hinted at by the use of the same word symbol is given by the fact that if I have, you know what I should probably say back over here instead of just A, maybe I should also if I want to write like A sub L so that we remember what the L was. Just to make it a little less ambiguous. So I'll also write A one, A n with a little L underneath we'll keep track of that. So if I have A B L, this is some element in H two, K mu L since we're two. And it turns out that this is equal to zero if A plus B equals one. And so what that tells us is that we get a natural ring map to this ring of Gaoua homologies just by, you know, just by taking some A in here to little parenthesis A. And the fact that really these are the only relations in here and the fact that they hold in there tells us this is a nice, totally legitimate ring homomorphism, the amazing and crazy thing which is really the, oh, there's all this stuff in the chat, should I be looking at that? Okay. We take care of it, don't worry. Great, great, great. Thank you. Thanks for taking care of that. Okay. Yeah, so the kind of truly amazing surprising thing is that, you know, of course this map, you know, factors through here, everything in here is L's version. And the real norm residue isomorphism theorem is the statement that this is an isomorphism which says that not only is kind of Gaoua homology generated by these symbols, that everything is a sum of symbols, but also that the only relations in Gaoua homology come from a degree two and things like this. So that's kind of remarkable, amazing and surprising. So this is the Block-Cado conjecture. Okay. Okay, let's move along. So the next thing I want to mention is the next variation is the bit group or the bit ring. You guys okay? I feel like I'm getting this feeling like I'm just reading a dictionary actually. Maybe I shouldn't be doing this. Or is everybody still awake? Okay, good, you know, I'm just like, just trying to read my virtual room. Okay, good. So bit ring. So my notes go right. So the bit ring basically as many of you probably have already seen in codes information about quadratic forms over a field. And this, okay. So just for the sake of convenience and sanity, let me just assume that K is a field of characteristic not two in which case one can naturally go between the notion of quadratic forms by which I mean homogeneous degree two polynomials and some number of variables and symmetric by linear forms. So there's like this kind of standard, polarization identity that lets you go between you have some by linear form, which you naturally would take on X, Y and if I let Q of X be B of X, X that gives me a quadratic form. And then vice versa, if I have some Q then I can define my value linear form on X, Y to be Q of X plus Q, Y, wait, Q of X plus Y. Minus Q of X minus Q of Y let's say divided by two and then we go back and forth that's why. And of course, these, if you have a by linear form this gives you a way of taking a vector and associating to it some dual vector. And we say that a quadratic form is non degenerates if this natural associated map is an isomorphism. And then we define the bit ring of K to be, well, it's, this is generated as an Abelian group by let's say classes non degenerate quadratic forms, isomorphism classes of non-degenerate quadratic forms. So I'll write the, I'll let Q to mean a pair of a vector space together with a symmetric by linear form or quadratic form on it or a symmetric by linear form on it but I'll just write that as Q. Isomorphism classes of non degenerate symmetric by linear forms or quadratic forms. And then we modulate the ideal which is generated by a couple of different things. We want to just say what it means to add these guys together. So, if I take Q one and I add it to Q two then I want to identify that with the orthogonal sum of Q one and Q two. So that's, you know, a way of kind of saying what the addition is or kind of saying how it differs from kind of the free object. And then also saying that this particular two-dimensional quadratic form is identified with zero in the bit-ring is just so called a hyperbolic form. Okay. And the, so the ring structure is given by the, but a tensor product. So this is sometimes called the chronicle product of bilinear forms. You know, you just, well, you know, I'll, you do the obvious thing, which if I write down this guy, I'll probably, you know, just have too many indices around. Okay. Okay, so this is the, this is the bit-ring. And the, in some sense, the, you know, the, this is, so I guess I should have said with the Milner K theory probably it would have been good to point this out. Kind of what's, there are many things that are great about kind of the relationship between Gauss-Comology and this Milner K theory. And on the one hand, you have these generator and relations kind of presentations. But on the other hand, you also have this kind of group that ties together all the different L's, you know. So it's kind of like some integral version of this Gauss-Comology. It's not just, you know, you can look mod L and you can look at the various prime cell but you also have this thing which is kind of living over Z, you know. So that's kind of an interesting kind of feature that Milner K theory gives you. The bit-ring gives another kind of like enrichment of like a part of Gauss-Comology by kind of trying not the different primes together but kind of like the different degrees together in some sense. So let me kind of say a little bit about that. So the relation to Gauss-Comology. Well, what you can do is you can define this, this ideal and the bit-ring to be the ideal of even dimensional forms. And we can look at just various powers of these even dimensional forms. So this thing is called the fundamental ideal and you look at its various powers. And it turns out that there's a map from Milner K theory, let's say, I don't know, the ith bit of Milner K theory, maybe I'll say the int bit of Milner K theory, the consistent to the int part, to the int fundamental ideal minus the int plus first. So it'll be defined by taking these symbols to very special quadratic forms. And let me just, I guess I need to say a couple of words to define those quadratic forms. So a little bit of an aside. So Graham Schmidt, because two is invertible, tells us that every quadratic form can be after a change of basis written as, let's say a one x one squared plus dot dot dot in this kind of diagonal way, a and x n squared. And we denote that particular form like this. Also another very special kind of form is this thing with the little double brackets, which is a shorthand for one common minus a. And then we define double bracket with a big long list to be the product of these guys. And these are so-called infold fister forms. And it's not totally obvious, but it's not super deep to show that in fact these infold fister forms generate I to the n of K, everything in here can be written as a sum of infold fister forms. And so then the map, it goes like this, you take this to these infold fister forms. So these are kind of the natural repository of our symbols and now the next kind of truly amazing and ridiculous forms. And then the next kind of infold fister forms is that if you consider these guys factoring through, so everything inside the right is turns out to be true torsion. And if you look at this natural factorization, then this as well is an isomorphism. And this goes under the heading of the Milner conjecture, again by Levotsky and various other people. And so together with the part of Blockada, which was known a few years prior, this is a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a sort of a string of form. And so this tells us, well, I guess I should write this down here. We also have the, this is H and K mu two, mu two to the end is the same as mu two. And so we actually then can relate these quotients Since like, part of the utility of this is that if we would like to understand some comology class, then this allows us to say that it corresponds to a quadratic form, some special quadratic form, at least modulo, you know, some notion of equivalence. But somehow it's like, you know, we can use, we can reason with these quadratic forms to say non-trivial things about the comology classes because of this kind of concrete interpretation, kind of, you know, mod to the comology is kind of like caused by quadratic forms, one of the places to think about. Okay, these, you know, in some sense, they like one of the, yeah, so one way that this is kind of often thought about is that, you know, this gives rise to these natural kind of maps going the other way, these guys, which are, which are just a sequence of comology classes then that can be associated to certain quadratic forms. Okay. And, you know, so the real point is that these, well, one of the many points is that these guys are then surjective, that any comology class can be related to some quadratic form. Okay. So we're almost, we're almost done with, with, what was it, with part one, we have like, we're, I just want to, so we're still in the, we've done Galois Comology and I'm going through the relations, the related theories. There's one last one, which is Motivic Comology. And then, you know, and actually, you know, so, so, so this I was guessing would take about a half hour and, you know, I might, we're going to be pretty close. This just going to take maybe 10 minutes and this is going to be most of the time. Okay, to talk about the problems. So that's, that's where we're at right now. Okay, so I'm going to say a little bit about Motivic Comology. So, so as I said before, like one of the kind of key features of the, of the Milner-K theory was the fact that it kind of like simultaneously related all these Galois Comology groups for different Ls together, right? These, they provided some sort of like integral version of, of these, of these Comology groups. And in some sense, you know, Motivic Comology kind of continues to do that for, for other twists. You know, so like, so, like, so maybe one, one kind of selling point of it is that it provides integral version, you know, I should say like Motivic Comology is like, is like a mini headed beast, right? That goes all over the place and has various interpretations and relationships, other things. You know, everybody, everybody likes it for their own reasons, you know, and some of those reasons agree with each other. You know, there's like, there's a lot of perspectives on like kind of why, you know, why it's a great thing. And because it just relates to all sorts of things. So like, I'm giving like a very narrow biased perspective on like, kind of what, you know, one thing I like about it. Okay, so one thing I like about it is that it provides integral versions of like, of these kind of natural, naturally arising at all Comology groups. These, let's just say, H, I, K, Mule, Tinser, J, you know, for a given I and J, you know, these, these, these might come up in various ways. And, you know, and what we, you know, so, so the, you know, so we've seen that the, that the Milner K theory kind of gives an interpretation when I equals J. But, you know, these other much of the groups among other things do, you know, do the other ones. So, so the, yeah, so the, so this, this works by works by constructing these so-called motivic complexes, which is a, which are, so these are written like, maybe Z of M. So these are complexes of pre-shaves on, let's say, generally described on, on smooth schemes over some given field. Okay. Um, you know, usefully considered in, in the Zarisky or etol, or maybe even Nisnevich topologies. So the, the Zarisky hypercomology, so I'm not gonna, I'm not actually gonna like, so, so it's not so horrible to actually like write down what these complexes is. You know, there, there are kind of concrete ways to write them down. I did kind of look ahead at other people's notes. And I think so, at least somebody's gonna write, write these down in some, in some way. So, for example, you can, these, yeah, so, so, so anyways, I'm, I'm, I'm not writing them down, but somebody will. So the Zarisky hypercomology of these are the, are the Motivate Comology groups. So H, N, X, Z of M, which, which is also written H, N, M, X, Z. These are the Motivate Comology. I mean, I guess we'll also just write H, N, X, Z of M, just without putting the double, you know, hypercomology bars, you know, but that's what we mean. You know, of course, if we're, you know, so, so if X, if X is, is spec K is a field, then, you know, of course, there's not that much going on with the, with the Zarisky topology. So we're really just talking about the, the Comology of some particular complex of, of dealing groups and all that. But, you know, with the Atol Comology on a field, there is, you know, obviously that, that extra layer going on. So, so these relate to, to the other, to the other groups described to the, I should say, to, to Gawa Comology, et cetera, as follows. So on the one hand, there is a quasi isomorphism of complexes of sheaves between just the kind of constant mu L, since they're N, you know, or kind of the single, you know, the, the one term complex that and, and Z mod L of N, yeah, yeah. There is an isomorphism. You mean, you mean in the etol topology, I guess, no. Oh, oh, yes. Yes. Sorry. Sorry, I'm sure I'll do that kind of thing again. So there's an isomorphism in the kind of diagonal case with Milner K3. So this was, this was, this was kind of shown with the kind of earlier versions of, of, you know, of, of motivic complex of the, before the actual kind of modern. So there, you know, there were, there were many incarnations of these kind of motivic complexes before they were known to have all the right properties and all that. So this was kind of from the point of view of the, of blocks, higher child groups where you, which were, which are now, you know, known to be to also give much of the homology, but you know, so anyway, so there is an isomorphism, you know, we can identify this, these particular much of the homology groups with Milner K3. And let's see. And then I guess I should say so blockado, you know, roughly, roughly said identifies the, the Zariski and the atoll of this, of this particular guy, right, of this, of this diagonal guy, which, which then, you know, from the other things. So, so the point being that this thing, by this isomorphism gives you, gives you Gawa homology, you know, identifying, you know, the atoll homology of this metabic complex with the atoll homology of this, which then is the Gawa homology of the field, with this thing, which coincides with, with the Milner K groups. That's one way to describe what's going on with the, with the blockado conjunction. Okay. Right. So, yeah, so the, so part of the utility of the, of the metabic homology is, you know, it, it, you know, it provides kind of a glue to relate kind of these, these different perspectives together in a, in itself, a coherent theory with a lot of extra structure. So, you know, you have these, there are kind of comology operations and various other tools that, that motific homology has that, that, and then the fact that they relate to these other things like that they relate to Gawa homology and these other things lets you then, you know, kind of do more things. Okay. So now, okay, so I'm, so I'm done with that. And now I'm going to very briefly go back to, go back to some, some basic invariants of field arithmetic that we would like to use to, to kind of formulate our problems. So, okay, so the first, the first kind of notion, which is very problematic is the notion of the dimension of a field. So there are, there is not one kind of like standard definition of dimension of a field that there probably shouldn't be. There are kind of a series of competing notions which don't necessarily agree with each other. Let me, let me give a few though. So definition. If K is funnily generated over a prime field or an algebraically closed field, K0, you know, we say the, maybe this is like the naive dimension of the field is, well, it's the transcendence degree of K over over K0. If K0 is algebraically closed, it's the transcendence degree plus one. If K0 is finite and plus two, the prime field is characteristic zero. So this is, this is, this is kind of like the, the kind of naive and dirty definition just based on the properties that we've, that we've seen that such fields exhibit kind of in nature. So, so this is some sort of naive notion of dimension. Let me give another notion of dimension based on Gallic homology. So we say that the homological, homological dimension of the field is the, well, let's see. I'm sure I said this. So we say the homolog, okay, we'll say the homological dimension. I know I'm going to like mess this up because it involves inequalities. Comological dimension is at most N if HM GKM equals zero for all M bigger than N and for M torsion. Okay. And then, so the homological dimension is then what? It's the smallest thing so that you're not less than that thing. It's just a matter of grammar that I'm getting confused on, right? This is the minimum N such that the homological dimension of K is at most N, let's see, right? This is like the Lewis Carroll definition or something. But okay, I hope I said that correctly. So in other words, what I'm trying to say is the homological dimension is N means that sometimes, so maybe I should have just said it in a more positive way. Comological dimension is N means that HN GKM is nonzero for some M torsion. There exists an M such that and HM GKM is zero for all M torsion and M bigger than N. That's what I'm trying to say. Anyways, so the homological dimension then, homological dimension of K is equal to the dimension of K if K is finitely generated over either a finite or algebraically closed field, but if finitely generated over Q if also has no real orderings. Say if you contain an imaginary number field or something like that, then it's also going to give you the right number. Dimension and the homological dimension will be the same. But more generally, if you're finitely generated over Q and you have orderings, then your homological dimension will be infinite. If K has orderings, homological dimension of K is going to be infinity. And that actually is just coming from this Miller conjecture thing, for example, you know, you, well, okay, if you have, well, maybe that's, maybe that's like, I'm supposed, you're guys are supposed to have things to think about, but I'm not telling you. So maybe that's something to think about. Okay. Exercise. Okay. So, let's see, what else should I mention? Okay, yeah. Another kind of related notion is that we say K is Cn or has the Cn property. If for positive integer D and m, at least in bigger than D to the n, every homogeneous polynomial f of degree D in m variables has a non-trivial zero. So, the Cn property tells you that, that if your variables, if the number of variables is larger than a power of the degree where the power depends on that, you know, this n, you know, then you definitely just have a non-trivial solution. Enough variables gives you a zero. So, definition, the diophantine dimension of K is the minimal of the n such that K is Cn. So, you know, so again, examples, if K is finitely generated over finite or algebraically closed, then the diophantine dimension is again just the dimension. And that's also the homological dimension. There's a variation of the Cn property called the Tn property where you are allowed to look at not just one polynomial but a collection of polynomials and you're looking for kind of simultaneous things and that lends to a notion called the Cn rank, which I guess I'm not really going to need. But there's a whole bunch of other things of this sort that exist. Let me, maybe it's worth mentioning because it's like, because there's like a nice natural open question about it. So, definition K is Tn if for d1 up through, let's say, dr positive numbers and for m bigger than the sum of the di to the n, every system of polynomial equations, you know, f1 equals f r equals 0, where the degree of fi is di and in variables has an entrivial solution. And then the minimum n such that K is Tn, this is called the sin rank of K. And so, let's see, so if you're Tn, then you're a Cn because, you know, if all, you know, this is a seven, you know, r equals one. And so, the sin rank of the field is bigger than the, did I get that right? I think? Sounds right. Okay, good, good. Then the, did you say that these are all homogeneous equations? Oh, I should, yeah, I should, homogeneous of that degree. Yeah, homogeneous of degree di. Yeah, thanks. Yeah, so the, so, I mean, the converse is just like not, maybe these are actually equal. I don't think that I don't, I mean, I don't know of any fields where they're known not to be equal and I'm not 100% sure, but I think that I'm not sure if anybody else knows any such examples. So, you know, so they might be the same. And this is just like, you know, a nice illustration of our ignorance, perhaps, or at least mine. I think, though, that I'm not sure if it's none of these are the same. I don't think it is. Now, the relationship between the, the C DEM and the D DEM, the comological and the diaphanetine dimension, also has some open questions associated to it. So there's a very famous example of, of axe, which is a field which is C one. Well, let me say, which, so let me say it's, it's a field, sorry, of comological dimension one, but, but no diaphanetine dimension. But on the other hand, there is a question of Sarah, which is, does, if you know your, you know, maybe the diaphanetine dimension has to at least bound above the comological dimension. You know, so, so you know that you can have a small comological dimension and a big diaphanetine dimension. But if you have a small diaphanetine dimension, does that force you to also have a small comological dimension? So this is still, this is still open. You know, there, so you can, so it was, it was observed by Sarah that, that, so Sarah observed that the, that the Milner conjectures tell us that that the diaphanetine dimension is at least as big as the so-called two comological dimension. So this is where you're, you know, this is, this is only considering the two primary parts of, of comology. So the, the two comological dimension is the biggest, it's the biggest, I guess I should have defined this before, the two comological dimension is the biggest one such that this thing has, has some two primary part, you know, has some two torsion, but these ones never have any two torsion. You know, so the two torsion only lives up to degree in and every, in every column. And so, you know, so you really, you know, you know, Sarah really, Sarah's question really, you know, we can kind of split it up and say, you know, is this thing true, let's say p for all primes p? And, you know, we know, and so the Milner conjecture by giving us this concrete interpretation of elements in Gaoua comology in terms of quadratic forms, which are specific single polynomial, you know, homogeneous polynomials kind of relates this diaphragm in a comological dimension. In general, you know, it's, this is still open, but, but let's say together with Ellie Motsry, a few years ago, we showed that, that the comological dimension was linearly, admirably in the diaphragm time dimension, but the slope isn't one that we were we couldn't get a slope of one. The, the slope we got was roughly the log base two of p. So that kind of gives it for p equals two, but in general, this is the best that we could do. So this kind of like not uniform, it's not uniform and p is the problem. Okay. So let's see. Yeah. And, sorry, I'm going to jump in with a question being rude. What's the variable in this growth is p? Or dimension, diaphragm dimension of k? What's what do you mean? So what I mean is something like the, we get a bound that looks like, you know, on the order of, you know, off bias, a small constant, the diaphragm dimension times the log base two of p. Oh, okay, great. So in p, I see, it goes like a log of p and p and like the diaphragm dimension times a factor in, in what, in k or something. But okay, I got it. Okay, okay. Yeah. Yeah. And the constant is such that, like, you know, that you get, you get the correct conjectural values for some particular small things, but then it just starts growing, you know, so it's like, it's kind of sharp in the bottom. But then it, it grows a little bit too big. So let's see. I guess it's, it's, wait a second, I only have like 20 more minutes or something, right? Is that right? Yes. Yes. Okay. Okay. Okay, well, so, so where we are is, so my plan was, go back to the plan. It's supposed to be, you know, 30 minutes, this was supposed to be 10 minutes, and this is supposed to be 50 minutes. So right now, we're here. So, yeah, so, okay, so we're, so it looks like some of this stuff will be for tomorrow, but let me at least start because I do, I really do want to talk about these, some of these problems as, you know, which are kind of the main focus. So let's see. I mean, it's probably worth just saying a little, like some very brief philosophical thing about this question of Sarah, because I think it really like, it really points out a very like central question, you know, in the, in this basic philosophy. So the, so the point is, you know, assuming some particular bound on the diaphragm time dimension, we're saying just a very broad blanket statement saying like every equation, every polynomial equation of a certain type must have a solution. You know, if you have at least this many variables, etc., then you have a solution. And so in order to say that, that then you get a bound on the comological dimension, like why would you even expect that to be true? The idea would be that maybe associated with particular comology classes, you could write down an equation such that if you can solve those equations, then the comology class is trivial. Right. The real kind of, you know, the, so to show this, we want to say that if we're given some alpha and some comology group, then there exists some, some, some particular polynomial equation or maybe some system, some single polynomial, let's say F alpha. So that's that F alpha has a zero implies alpha is, is trivial. You know, if you had some, if you had some way of like constructing these things in a very efficient way, then you, you could then say, well, because I know a bunch of things that automatically have solutions, if, if I, if I crank the degree of alpha enough, these things will all have to have zeros or something like that. That's, that's what we want to say. And, you know, in, in like kind of historically, in the beginning of these things, it kind of seemed kind of hopeful and that like, you know, for very specific comology classes of very special types, let's say if, you know, star was equal to two, or if we were looking at mod two coefficients, then really we were able, and in a few other cases, we were able to really come up with some, some polynomials of that sort, you know, in some cases at least. But, you know, so, so the, you know, the case where alpha is a symbol, this is very close to the idea of looking for these so-called norm varieties. So, norm varieties are supposed to be these varieties that have points, well, it's supposed to be, whatever, there are these varieties that, the ideas that they should have a point if and only if this symbol is split. And I think, you know, like kind of small number of circumstantial evidence was pointing to maybe these things were described by hypersurfaces or something like that, you know, eventually we kind of gave up on that, you know, and kind of the, these norm varieties ended up being constructed more kind of abstractly, you know, without kind of sticking them, you know, they're not hypersurfaces in the way they're described. I mean, kind of like, you know, but somehow, you know, if they were hypersurfaces of the optimal degree somehow, you know, then this would kind of follow. The problem is they don't have such a concrete realization as that. Okay. Okay. So, now let me maybe talk about structural problems in Gaouac homology. So I'll maybe do this a bit more briefly than originally planned or just not quite get as far. But so, okay, so the, so what I just like to talk about is, you know, what would we like to know about Gaouac homology? So this is, you know, this is basically like, I don't know, a cry for help or something like that, right? So I'm just, I'm, you know, you know, I'm, you know, these are things that we really would like to know. You know, we I'm just hoping that like, you know, one of you guys will like have a bright idea and like put us out of our misery and maybe use some, you know, some insight from, from motivic homology eventually, that will make this more clear. But here, let me just say, what are the, what are the problems that, that we need to help with? So the first one is what we might call the period index problem. So let me just first off define if alpha is, let me just like, just to save a little bit of time and notation, write h i of k when I don't adorn it, just to mean h i k mu l tensor j, where I'm just not going to tell you what, what l and j are. And we'll, we'll kind of add them in if we need them. Okay. But so if I have some homology class, you know, we say l over k splits alpha if it becomes trivial when you, when you extend it up to, up to l to this, you know, this factorial property of Gallup homology that you can kind of just change the field like that. So that's called restriction. And we define the, the index of alpha to be the GCD of the degrees of extension fields such that l over k is finite and splits alpha. Okay. And, you know, the period of alpha is just the order of alpha, but, but, you know, we, we might typically just kind of, yeah, so at most l, right. Okay. So, so one can show basically using some of the kind of functorial properties of Gallup homology that the period divides the index and they have the same prime factors the index divides some power of the period. So, you know, the, the period is smaller, but they have the same prime factors, great, potentially smaller. And the, the period index problem is for a given k what is the minimum m such that index of alpha divides period of alpha to the m. I guess I should say for a given k and for a given i and j and l, you know, so this is taking place in h i k mu l since we're j. You know, so how big, you know, or we could similarly just ask, you know, because l is a good substitute, you know, you know, what, how, you know, how big can the index be in terms of some power of l. Okay. So, an explicit conjecture due to Kurtzmann, I think he stated it as a question, though, is if k has a diaphanetone dimension in, or let's say dimension in, if you're talking about a, a finely generated field and alpha is in this particular one, then the index should divide the period to the power n minus one. So we specifically think that there should be, you know, these, these kinds of bounds. But in general, like, you know, we have like, we're very, very far from this. So, you know, even in this case, no known bound for, for, for k equals, let's say q of t, you know, for, for, for any else. So, so, you know, we, we might believe that everything, so this is a dimension three field and everything should be like, you know, the index should be the square, at most the square of the period, but there's no known bound, actually. It might be infinite for all we know. So this is a question in some sense of like, of how complex, you know, how hard is it, or the, you might ask this as like kind of how complicated are cohomology classes of a given period and degree. The, the, there's this rough idea that, that, you know, the, the Galois cohomology kind of resembles some kind of like exterior algebra in some way, and that if you get close to the cohomological dimension, so rough idea for h i of k, i close to, to the dimension of the field should have small index kind of classes should become more simple as the eyes get approached the kind of biggest they can possibly be, you know, for example, maybe, you know, i equals the dimension, maybe period equals index, something like that. But, but really like all that we know is based on like just a very small handful of circumstantial kind of evidence and some very small examples related to this is the symbol length problem. And this is probably where I'll end. So, so if you, if you recall, you know, it came out of the, of the block auto conjecture that the Galois cohomology, we know that h n of k mu L tensor n is generated by symbols. So these are these these cut products, a one through a n. And the, so the problem is, you know, given a given k and n, find the maximum, well, I guess find the, find the minimum number m such that every alpha in h n is a sum of no more than m symbols. So, you know, every particular cohomology class is a sum of symbols, but like, how many do you need? And this is another kind of measure of complexity of this cohomology class. And it's kind of describes how easy it is to write down in some sense, you know, kind of like how easy is it to write a class alpha. Now, so we want, you know, a bound in terms of maybe the diaphanthine dimension or, or the dimension or something like that. You know, so you can certainly construct fields that don't have a finite dimension or diaphanthine dimension where you need a arbitrarily long large number of symbols to write cohomology classes. But for kind of finite dimensional fields, there is maybe the feeling that there should be some sort of bound. Actually, I don't think there is a bound for, so that's my personally, my guess is that there isn't a bound once you get to like maybe n equals four or so. But, you know, even for finite dimensional fields, but I don't, you know, I mean, maybe, maybe we'll talk a little bit about that, about that tomorrow. It's just my personal feeling, but I'll try to explain my feelings a bit. So, let's see. Yeah, theorem that from number field, everything is one symbol or something? Maybe that's worth saying. Yeah, absolutely. So there are, yeah, so maybe I'll say a bit about what's known. Yeah, so what's known. So, for number fields or global fields, so these are kind of fields with a reasonable notion of dimension, dimension actually two fields, these are dimension k equals two. You know, over, you know, finally generated over prime field. We know that every class in H2 is a symbol. Well, wait a minute, what am I talking about? So, I guess I have to say if we have to include roots of unity for me to say that. Um, with an elf root, with a primitive elf root of unity, then we can say that. Kind of the, the distinct, you know, so the kind of finer points of that, I guess I'll have to, you know, the statement that, yeah, we'll all wait for next time to add more details, but okay, so number fields, you know, um, you know, everything just is a symbol if you have roots of unity. Um, what else do I want to say? So actually, um, you can get, um, I wanted to write, oh yeah, um, just because I only have time to maybe say like one thing about it, I'll, I'll write this kind of, kind of surprising thing, which is like, um, so if, um, so this really, um, great result of Montserie, um, showed that, um, that for, um, for H2 assuming roots of unity, um, you know, uh, what do you need? If, um, let's say L is a prime power, for example, if L is, let's say, um, p to the t, um, then we need, at most, um, t, uh, times p to the diaphanetide dimension of k minus one, uh, minus one, uh, symbols. So if you have a finite diaphanetide dimension, this is the, the kind of, uh, strongest thing we know of this sort. So, you know, if the diaphanetide dimension is finite, then we get a bound. Uh, conjecturally, the bound, um, shouldn't depend on p, you know, I don't know, not conjecturally, like we suspect that the bound shouldn't depend on p. Um, but you know, I guess it's really hard to say, you know, um, for an, you know, for a higher, uh, dimensions, a degree, let's say, for a higher degree comology, um, we know almost nothing except, um, you know, so, like, for some very specific fields, like there is this, um, like for some specific, um, kind of, uh, in some sense, three-dimensional fields, although in a sense that's somehow somewhat of a variation of the things that I've mentioned, we know that things in H3 are a symbol. There's a nice result of, uh, Suresh, for example, um, for H3 of, um, of function fields of periodic curves, um, which are in some sense three-dimensional, um, uh, are all length one. You know, they're just single symbols. Um, but, you know, basically we know nothing, you know, about these symbol lengths for higher-dimensional fields. Um, and let me, um, let me, I guess, finish by, you know, just like it was, you know, in the, um, in this case of Suresh conjecture, kind of the punchline for morally how to, why this should happen should be the existence of kind of some nice forms. Like, um, like why should we expect these kinds of bounds, um, to happen? We expect these kinds of bounds to happen when it's, when it's somehow possible to write down, like, generic co-emology classes of some type, and kind of our, our, our inability to, you know, like, the, the point is, is like, if you have a bound on the symbol length, then you have a uniform, you know, you can kind of, like, by making all the, um, kind of constituent symbols, kind of variables, you can kind of write down a generic element in co-emology as a sum of, you know, at most 28 symbols or whatever you have. The kind of inability to kind of, like, write down, like, a general form of a co-emology class for a given field is what makes this, this problem, you know, difficult. Kind of the description of co-emology classes isn't a priori bounded. Okay. Um, I've, I've, I've set enough for today, or at least I've said what I've said for long enough. So, uh, let me go ahead and stop there. Thank you very much then. So any questions? Hello professor. I was thinking maybe I could ask a question. It says, uh, do we have analogs for this field arithmetic to more general communicative rings? Um, well, I mean, of course, you know, you can, you know, absolutely ask the, the same kinds of things, um, for, um, for, you know, for commutative rings. It's, I guess I should have maybe mentioned that like, um, yeah, I mean, so like, you know, so as I, as I mentioned, you know, atal co-emology, um, kind of generalizes Galois co-emology in the, in the context of, of, of schemes. Motivate co-emology is also something that, you know, that is not just exclusive to fields, of course. And even, you know, the bit, um, the bit, uh, the, um, the bit rings as well. But the way in which, you know, kind of like these things relate to each other is not, is not so neat, you know, as it is for fields. Like, um, I mean, you know, kind of the objects are still defined, but the, but a lot of the tools break down a bit. So, I mean, you can ask the same questions. Um, but you know, it's, you'll have like less to say about them, I guess. Okay. Thank you very much. Sure. And we already kind of have very little to say. It's pretty hard already. Any other questions to Dan? Uh, what were the original motivations for defining dimensions of fields? Like, you started with one definition and then you, you know, added some new ones, but like the first one, where did that come from? Yeah. Well, I mean, so, um, I guess, you know, in, um, I think different people will give different answers to that, but like, um, you know, I guess, I mean, the way I think of it typically, uh, is like, um, in terms of acting like the function field of an indimensional variety in some sense, perhaps over the complex numbers or, you know, some, um, you know, like, so for example, you know, a finite field, you know, of course, um, you know, isn't naturally the, um, you know, the function field of anything in particular, but, you know, you know, the, but because the absolute Galois group of it, you know, um, is the same as, you know, if you did like, Laurel series over the complex numbers, it, um, that is some sort of like a hint as to it kind of behaving in some, in some ways, like something that's like one-dimensional, you know, um, kind of like, uh, some geometrically one-dimensional thing. And, you know, I don't have a good answer to that question, I guess. Um, you know, there's, um, but it's, you know, the motivation is, you know, these things are behaving like geometric objects in some sense of the same dimension. Um, hello, professor. So I was wondering, uh, are there any relations between these dimensions of fields? I mean, various dimensions with homological dimensions of rings, like projective dimension, injective dimension, or like some related notions like tall dimension. Well, I mean, you know, you know, I mean, I guess only in some of the, the nicest cases, I suppose, you know, like, I mean, just in the sense that like, you know, if you actually have the function field of a complex variety, then it's the fraction field of a ring of the same curl dimension, right? But, um, but in some sense these kind of like, um, some of the kind of field dimensions can be kind of like invisibly wrapped, you know, in the arithmetic in the way that like kind of the, like a finite field feels, you know, I don't know any kind of like, um, any kind of like, um, algebra geometric sense in which a finite field should be kind of dimension one or any kind of commutative algebra justification for that. But in terms of the Galic homology, it kind of behaves one dimensional and the arithmetically it behaves one dimensional. Okay. It's not unrelated, but it's, it's kind of, um, yeah. Okay. I was just curious. So maybe it's not a relevant question. No, no, no, it's good question. Yeah. More questions to them? Um, I was, I was wondering if there are computational tools for approaching the symbol length problem kind of analogous to like Gromner-Basey's or something of that nature where, you know, given a certain element, can we, how, how computationally feasible is it to decompose it into symbols? Right. Right. Right. Um, yeah, I don't, I don't know a good, a good answer to that. I mean, you know, they, like, um, you know, in some sense, it's like, you know, you can think of it as like a search for some notion of like dependence, right? Like, if you have like a list of symbols that's too long, is there some way of rewriting it and like kind of, you know, um, and kind of like, uh, contracting it, you know, somehow? I mean, I think, you know, in some sense, like, um, Matsuri's proof kind of contains some, um, some very nice kind of arguments that have that kind of feeling philosophically, although I don't, I don't know, um, if it would actually be possible to kind of go from that to something particularly algorithmic. Um, you know, um, the, it's, it's, uh, it's very, it's very tricky, you know, because we have, you know, there's these particular generators and relations for Milner-K theory, which then are for Gallic homology, but kind of the implications of these very simple relationships are pretty subtle and it's kind of, uh, it would be wonderful to have like some more procedural way of kind of like working with it, but, um, but I don't, I don't know of anything like that, you know. Thank you. Can I raise a comment? Please. Is there, is there like a trivial statement if you look instead of Gallic homology, you look at Milner K theory, mod P and characteristic P to something by the P rank of the field or something like that? Oh yeah. Yeah. Yeah. Yeah. So I, I guess that, that actually reminds me of something that I should have probably said, which is like, you know, there's all this like mod L stuff and roots of, L roots of unities and all that and like, um, and L is prime to the characteristic throughout, you know. That was clear. But I'm just saying as a, as a toy case, sometimes the mod P case and characteristic P is, is useful. That's all I'm saying. Oh, absolutely. Yeah. No, I, I, there, I maybe somebody else knows the answer to like, to how that, or do you happen to know the, there are, there should be something more straightforward with just the, the P rank and, um, and bounding the mod. Yeah. Yeah. I just, is there something like that? I suspect there might be just because that's, that's the sort of the original proof of block Cato was in that case. Right. And it's very, it's very closely related to these questions of P rank. So it was just, I really don't know, but I imagine, I mean, in other words, if you take just the vector space of dimension N and ask what's the symbol length on the R tensor power, isn't there just some formula in terms of the dimension and R or something? That's all I'm saying. But anyway, maybe it's a good, it's, maybe we can talk about it later. I don't want to waste everybody's time with this, but just a thought. Right, right, right. Yeah. Sorry, but, so I'm feeling just, it's not my, my lecture either. Please. But, but the right analog of, of KS3 mod R and characteristic P for, I think for what you are doing is not category mod P. This is this Cato Galois Comergy group, you know, this is the H1 of this one shifted by, you know, because we have this Galois Comergy Cato groups, okay, which plays a role in characteristic P of the Galois Comergy. And, and that, and that, yes, once again, this is H1 in this, in this K, K theory mod P. So I guess those are the right analogous groups. Okay. So this is my contribution. Any more questions? Well, actually in two minutes, the problem sessions should start. So, so we should round up.