 Thank you. It's a great pleasure to be here. And I didn't know what kind of audience I'd have, so I'm going to be a bit introductory today. My goal was at the end to explain, to give an approximate definition of universal mixed elliptic motive. I mean, it'll be more or less complete, except there's lots of things to be explained, and I'll do that in detail in the next lecture. So, let me start. So, the first thing is history in context, and maybe a lot of people know this, but I thought I'd start here. So, very brief comment is, so the 1960s, whoops, Grotendick discussed the theory of pure motives. And so that's an effort to make smooth projective varieties into an abelian category and find a universal cohomology theory. So, basically pure motives over K, and this guy here would be a field. And basically the objects are varieties, and the morphisms are correspondences, and so on. I think this theory is still not well understood because of all the problems of all the standard conjectures. So, basically objects, smooth projective varieties, and the morphisms are correspondences. So, you get more objects because you look at correspondences that are item potents, and they give you projectors, and so you can break up these varieties. So, the second thing is maybe about 1980, there was Balanson wrote a paper on regulators, and I'm not sure exactly who introduced mixed motives, but the idea appears in this paper, and so he defined motivic cohomology. And I'll just recall that in a second, but he defined it, he did an n-runner bound, the problem of finding co-chains, which was later solved by Vojavodski. But it turns out, at least for me, it was a very good conceptual way of thinking about this. So, how does this work of X? And just to be safe, I'll take X to be, let's take X to be smooth, but I don't think this is necessary. So, you have the k-theory of X. So, the algebraic k-theory of X, which was defined by Quillen, and you have Adam's operation. So, what do I, I can't remember how they're written. Yeah. So, Cp, which take k dot X into itself, and these can be simultaneously diagonalized after tensuring with Q. So, you have that k dot X, tenser with Q, is equal to a direct sum of the, I'll just write, I'll leave it when I put this guy here. And so, I think this is n greater than or equal to zero, and this is where Cp acts by p to the n. So, it's, so you can break up the k-theory here. And basically, the projection of this onto this piece is the churn character. So, the churn, the, the degree n part of the churn character is just projection onto this, this factor here. And this is compatible with maps to all cormology theories. And so, what he did was he defined, I should mention, I don't know how interested you are, but I will, I've made notes in this way and I will put them on the web somewhere. I'll let you know in the next lecture. But anyway, so, the motivic cormology of X with coefficient, say, in Qn is equal to, is defined to be, you look at 2 to n minus j Xn. So, this may look odd. So, let me take us, say a little word about it. This is a universal cormology theory for the following reason. Because there are maps from k-theory. Say, if I took k, let me call it km, of X, and I, I can map it by, say, the degree n part of the churn character. You think of this as being k upper minus m. So, it's in degree minus m. And this will go into, into any cormology theory into degree 2n minus m of X with coefficients in Qn. So, this is for any theory. So, you can put ordinary cormology, you can put deline cormology, you can put tau cormology and so on. And you always have this map here. And this will always factor through h, m, Xn. But this is, this is just equal to, if I did it right, this is the motivic cormology in degree 2n minus m of X, Qn. So, it's showing that the, that this is really the universal cormology theory. And so, if you want to produce, so this is the old-fashioned way. And I guess I'm still old-fashioned, sort of, pre-Voyov-Odsky, is that, you know, there are questions about how do you produce consistent sets of elements, compatible sets of elements in all these different cormology theories. The idea, they all come from the same element of k-theory. Okay. And so, then the idea was that, again, I didn't run to the library and check it. So, I apologize for that. But, so this may be a little bit revisionist. I should also point out here, it's, it's not clear that this cormology theory vanishes in, in negative degrees. So, there's Baylinson-Soulet vanishing conjectures that this guy vanishes in negative degrees. Is, is that h, h less than zero, a motivic of X, Qn equals zero. At least one X is smooth. And you can also, later block showed that these, these pieces here rationally are the same as his higher chow group. So, there are various things that one can do here. That gives you an integral version of this theory. Right. And so, another comment is that if X is defined over Z, for example, then all of these things should be finite dimensional. And their dimensions of all of these guys are predict, you know, supposed to correspond to the order of vanishing of certain L functions that may not be known to exist. So, so let me give an example. So, the first one is, this is really, I'd say, block is that X smooth says that H to n, motivic. So, this is just a restatement of an old result. This is chow. Because these, the chow groups are the, the, these I graded quotients of k theory in degree zero. So, this is chow group. And then, another example is two, I'd say, this is more or less Borel and Balanson, is that, so let's take f to be a number field. And so, you'd like to know what the motivic cohomology of qn is. So, so let me write it down. So, we'll need this later on. And this basically is the ingredient for understanding what mixtape motives are. Is that the first one is that hjm, I'll just write fqn equals zero and less than zero. Two is that h0 equals zero and not equal to zero. So, this is sort of expected. Three advantages hjfqn equals zero if j is bigger than one. So, the interesting case, and four, I'll just put hfq of one is just f cross tensed with q. So, that leaves the remainder, which is, I'll just write like this, dim h1fqn. So, this is going to be n bigger than one is equal to r1 plus r2, n odd r2n even. And here, the degree of f over q is r1 plus 2r2. And as usual, that's the number of real embeddings and this is the number of complex conjugate pairs of embeddings, complex embeddings. Alright? So, in Borel's, basically Borel computed the rational k theory of a ring of integers in a number field by computing group homology, double-colomology of SLN and Baylinson interpreted Borel's map, what did it raise to go? Am I going crazy? Ah, there they are. Okay. There. And Baylinson show, you know, constructed Schoen, Baylinson and other people constructed Schoen classes. There's something that I, okay, so the next part I'm not an expert on. And I think Levine did some as well, is that, so, ah, and so I think there's another part here is that Baylinson also, I think, conjectured the existence of a category, and so this is what's important for us of mixed motives. And such that the motivic homology of say xqn is equal to x in this category of say q, and then you take some motive associated to x, whatever that is, and so that motivic homology would be annexed in some category. And so, Vojavodski and Levine, ah, and this category should be, ah, this guy here should be Tynakian, right? So basically it means it's an abelian category with tensor products and duels and a faithful function, forgetful function to say finite dimensional vector spaces over q. So, ah, so Vojavodski and some, due to Deline, I guess, they constructed a triangulated tensor category with these properties such that, ah, so this is not, this is incredibly good, but it's not a, ah, it's not a Tynakian category. So, ah, and I should say something about the objects q of n, at least in all of these theories, roughly, ah, these are the Tate motives, these q of one is roughly, I think of it as h1 of g of m, and q of minus one is equal to h1 of the multiplicative group, and, ah, then q of n is equal to q of one tensor n, this is n greater than or equal to zero, or q of minus one tensor absolute value of n, right? So that, that's, so I'm old fashioned I think in terms of compatible realizations. So, so I have a question, do people take a break after one hour or is this a marathon? Okay, so, ah, let me say something about mixed-taste motives. So I'd say this is, again, I, I did not make, I did not carefully check the literature here, so I may have the history slightly screwed up, but let's take, ah, k to be a number field, and we'll just take s contained in this ring of integers, and, ah, so the upshot of this work here, Levine first for fields, and then Dylene and Gontroff for rings of s integers, as there is a Tanakian category of, of mixed-taste motives, I'll call them over OKS. And I should say here these guys have, maybe I'll say a bit more later on, but the fact that you've got maps from k theory, say, into extensions of hodge structures or Galois modules tells you, ah, what the realization of this guy should be. So the hodge realization of this, it's just the hodge structure of, ah, dimension one and type minus n minus n, and the ellatic et al is just, ah, it's a one-dimensional Galois representation, so gq is acting on ql by the, ah, nth power of the cyclotomic character. Yeah, so, ah, so, yeah, so qn is going to be given by the nth power of the cyclotomic character. Okay, so, ah, and, so this is only mixed-taste motives, so these are successive extensions of these guys here. So, ah, and then you have the property that hj, yeah, I want to underline this here, ah, so they have the correct x groups is the point. So the x groups are given by k theory. So, and objects here have a weight filtration, ah, I'm going to start doing something slightly non-standard here, ah, yeah, ah, and the weight filtration is by sub-objects of this category, and I'm going to denote it by zero in, the pieces will be called MR of, maybe my object is V, MR of V containing MR plus 1 of V, and so the union should be all of V and the intersection should be zero, ah, later on it will become clearer, I hope why I'm using M, ah, rather than the usual W, because basically the objects we're going to be dealing with later on in the elliptic case have two weight filtrations, and the thing we normally think of a weight filtration will be denoted by W, and then we'll get a second weight filtration which will be this one here. And so this has the property, ah, such that, so I'll use this GRM, ah, to R of V is isomorphic to a direct, so, ah, Q of minus R to some power, and, and GRM odd of V is equal to zero, so it makes, Tate modives are of even weight, so you, if the graded quotients are Tate modives you don't get anything in odd weights, let's see, ah, I got lots of experts here and some non-experts, let me just say, let me do, I've got to do this at some point, so I might as well do it now, quick review of, ah, mixed hodge theory, this is a very quick. So, ah, one is what's a Q hodge structure, ah, a Q hodge structure is of weight M, consists of, the first thing is, ah, a finite dimensional Q vector space, and I'll call it VQ, and B, a decreasing filtration, ah, F dot of VC, so VC is just the complexification of VQ, and so you'll have VC, and, ah, it contains some FPV, and so again, the union of these is all of VC and the intersection is zero, and so it has to satisfy, VC is the direct sum of, ah, HPM minus P, ah, so summing over P, where HPQ is equal to FP intersect F bar Q, so you've got complex conjugation because you've complexified a rational vector space, ah, right, so, ah, and so let me, one very quick example is that, ah, one QN, and two, ah, this is weight minus 2N, and HM of a smooth projective variety, alright, so now I want to say what, ah, most people, let me just say what a, so what's a mixed hodge structure, a, ah, Q mixed hodge structure, so again, ah, V consists of, of A, ah, a finite dimensional Q vector space, VQ, B, a filtration of that, so, a weight filtration, and here I'm going to use, ah, the letter W just to confuse you, so, we'll have WM minus 1, V contains WMV, it's defined over Q, and again intersection zero, the union, the whole thing, and, ah, these satisfy, ah, each GRWM of V, so this is defined to be, ah, WMV mod WM minus 1 of V, ah, with the induced hodge filtration, okay, so, which means you, for example, restrict to this by intersection, then you project to the quotient, ah, is a hodge structure of weight M, and so, yeah, ah, oops, I forgot C, thank you, ah, a hodge filtration, I knew I left space at the bottom of that board, of VC, and hodge filtration's a decreasing, so this FPV contains FP plus 1 of V, again intersecting zero, union, the whole thing, and I got a stress that, ah, little slogan is, a mixed hodge structure is more than the sum of its graded quotients, there are non-trivial extensions, and that's what we're mainly concerned with here, and so, a hodge structure will be a mixed hodge structure, again, if you've got a hodge structure of weight M, you just define WM minus 1 equals zero, WM equals V, right, so, ah, every hodge structure is a mixed hodge structure, all right, so there's something I want to stress here, so there's some important facts, all this you can find in, in Jolene's paper in Public Assembly, yes, yes, ah, important facts, and so, ah, and to me these are, there are, there are many, but here, that GRW dot is exact, the associated graded of the hodge filtration is also exact, but, ah, and 2, again all of this, the definition to do Jolene and so on, the least trivial thing here is this, Jolene is at the cosmology of every complex algebraic variety, ah, has, ah, has a natural mixed hodge structure and, you know, compatible with morphisms and so on, okay, maybe I'll skip that, all right, so, ah, yeah, so let me just do a few simple examples, okay, let me just think, I just violated the order of my notes and so now I've got to think about how I want to say something, ah, all right, so let me, what I've got here, let me just try to jump, of objects of mixed tape motives over Z, so let me say something about this, so again, I'm sure this is well known around here, but let me say it just for completeness, so, ah, so I was saying before, yeah, so when I, when I say realizations, there are various kinds, they come under the name of Betty, sort of hodge, or maybe I'll put say Q Dharam in this case, hodge and say eletic, let me just first of all discuss, so realization mean you've got this abstract motive and then you can realize it in all these different categories and these different categories are compatible in strong ways, so V equals Q of 1, I already mentioned before, so Betty, again, apologies for those who already know all this stuff, but, all right, so it's just equal to H1 of C star, Q is just equal to Q and I'll think of it times sigma, so here's my copy of C star, here's my loop sigma and then let's do Q Dharam, I'm going to do the dual, so if I looked at H1 Dharam, I'll say GM over Q, it's just Q times say DZ over Z, if Z is the coordinate in GM, so H1 Dharam, okay, we've got to take the dual of that, it's going to be Q times DZ over Z dual, right, and now you have this an isomorphism, so we can, let's call this, this will be V Betty equals this, so this is going to be V Dharam equals this and then you have V Betty tends to C, this is basically the Dharam theorem, V Dharam tends to C and let's see, so it should be that DZ over Z check should be equal to, should correspond to 2 pi i sigma, is that right, put it the wrong way, did I get that right or have I got it upside down, let's see, that's right, I think, so and then this has a Hodge filtration, V Dharam equals F1 V Dharam and so then you've got Hodge as you put these two together, this guy has a Hodge filtration and this gives you a, this gives you a Hodge structure of weight minus 2, so sorry, this should be equal to F minus 1 because when the dual contains F minus 2, sorry, contains adding one F0, right, and then Aladdin Katal, so I'm going to put VL, it's just going to be, I'll take it to be say, it's equal to pi 1 of say GM over Q, say 1 and this is just going to be isomorphic to the L power roots of, whoops, sorry, tends to be QL, so now what about a general V, it's a similar story, so we're going to have V Betty, we're going to have a V Dharam which is going to have an F dot and this is a Q, these are defined over Q and it will, and both of these have a weight filtration and we're going to have that the Hodge is going to be that V Betty, V Betty W dot, tensed with C is isomorphic to V Dharam, W dot tensed with C, this guy here also has a Hodge filtration and this together gives us a mixed Hodge structure, whose weight graded quotients are, this is a mixed Hodge structure with weight graded quotients, say 2N, weight dot of V equal to a direct sum of Q of N's, various Q of N's. Because I'm thinking of this as defined over Q, so in the elliptic case, eventually I'm not going to do it in the beginning, I want to do the Q Dharam case because that's more or less Leven-Racine and Q Dharam gives you splitting, so I mean I think even in the theory of mixed tape motives, the fiber functor is the Q Dharam, I mean the one over the field of definition, so what should I call it, just Dharam? I think of Dharam as being of a C. So let's just do a simple example here. So this is an example of a mixed tape motive, I guess I could have done that there over Q, so let's take V to be equal to H1 of say gm with say lambda in k cross minus 1. And so we can draw a picture of this guy here, here is say 0, here's infinity, here's the sigma in our, that we used to the generator here, we're going to have a 1 and a lambda and we'll take a path, say from here to here, and you can see here that you've got an exact sequence, 0 into H1 gm into V and then here you get say the reduced H0 of 1 and lambda into 0, it's not too hard to imagine that this guy here is Z of 0, so okay, so here maybe I should take everything's Q coefficients. So I mean this example works perfectly fine over Z as well. And so here we can, we still have the form DZ over Z makes perfect sense, we'll have a dual in here, so this will be in H1 gm 1 lambda which is the dual of this, so we're going to have the dual class in here DZ over Z dual and here we have sigma and here we have the path gamma which is a lift of the generator 1. And so the, so what you see, so to see where the, so the hodge filtration F minus 1 is spanned by here say DZ over Z check and in say the complex cohomology and DZ over Z dual, so to know what the mixed hodge structure is, there's an extension, you have to know where the hodge filtration is inside this with respect to say some integral basis and it's 2 pi i sigma plus log lambda, so this is really just the integral from 1 to lambda DZ over Z times gamma. And so the exact branch of this depends on the gamma you choose but F minus 1 is determined by, so yeah gamma is well defined mod Z times sigma. And so it's just defined by lambda in C mod 2 pi i Z, the ambiguity here or here, whatever and this is, this just maps by the exponential map into C star, sorry by log lambda and so this really corresponds to lambda in C star and if you think about it, this is actually X1 hodge, yeah I've ended up doing this integrally so that sequence there may perfectly sense over Z. So this is an example of a non-trivial extension and you can see we're getting, maybe I should have said here, K is contained inside C. So it'll actually be in K cross, the extension is good enough to recover the point. So let's look at a much less trivial example is, okay, so 2, I'll do the details actually in the elliptic case, but details of various things like the definitions and so on, this is due to Delin and Gontroff. So let's look at, so we can look at P1 minus 01 and infinity over Z and we have to make sense of this, but let's, we can talk about, let's let pi be equal to say the, well I'll just call pi unipotent equal to pi1 of P1 minus 01 and infinity, say X at first, take unipotently completed. And I'll define, maybe I'll say a few words now about what this means and so we'll take, let's just take an X to be an element of Q minus, Q cross minus 1, you know, for some of this I can just take it to be any complex point of this. And so let's first of all just think of this as the pi1 in the complex case. So I'll give you a brief explanation. So if I have a group, so what's unipotent completion? I'll do it later, but here's the quickest definition. So I have a group gamma say a discrete group and I can look at a category C. Let's call it say, I'm going to say C gamma is just equal to the category of unipotent representations of gamma. So this is a Tanakian category. So say over F. So this is a field of characteristic zero. And so this is a, this is Tanakian. And so you've got a map from C gamma into a forgetful function into vector spaces over K and you can just, so what's pi1, what's gamma, the unipotent completion of gamma over F is just equal to the fundamental group of C gamma with respect to omega. That's the quickest and most abstract way to say it. I'll say it in a concrete way here. Sorry. A little K. Vectors over F, right. So it's just the automorphisms of this fiber function that are compatible with tensor products. Or I can do it in, here's the most banal way I know how to do it. I can look at all, I just look at all representations of gamma into a unipotent group. So this is, so unipotent over F. And I look at, so this is rho. And now you can look at the zariski closure of the image. That's also a, this is a zariski dense representation. And now I just look at the inverse limit overall rho that a zariski dense of the u rho. And this is going to be equal to gamma unipotent completion. And in, in this case here if you take a, I'll do this in more detail later on. But, so this first start is infinite dimensional. Here, the free, the Lie algebra in this case of pi 1 unipotent is, is isomorphic to a freely algebra on two generators, x0 and x1. And then you have to complete it. So you take Lie power series. And this thing sits inside, say q power series in non-commuting, these are non-commuting indeterminates. So we've got power series. And all right. So, so Delin and Gontroff says that the Lie algebra of pi unipotent is a pro-object of, say, mixtape motives over q. In fact, but in fact it will in general not be, well it won't, unless we choose the base point carefully, which is what I want to discuss now, it won't be a mixtape motive over z. So, so for example, not a mixtape motive over z. So, and because I want to say that why is this the case, if we look at p1 over z, and we take some x here, so we map this down to spec z. And now you've got some point here, x. If you look at the picture here, here's spec z, and you've got 0, 1 and infinity. Every such finite base point will intersect, you know, be 0, 1 or infinity at some, at some places. So if you think about it, if you try to compute the pi1 here, say this prime here, the base point has become 1, it's run off the edge of the variety. So pi1 with that base point doesn't make sense. So what we're forced to do here, or what Delin was forced to do, is introduce tangential base points. And let me just be very brief here. So if I take, if I take a Riemann surface, you can do this in higher dimensions but it's easiest for curves. So if I take some Riemann surface, and say I've removed some points here, and suppose I want to, so I can use a tangent vector, say, at one of the missing points. So my x is going to be some x bar minus some finite set. This is, say finite. And I'm going to take s in s, and say v inside the tangent space at s of x bar, and v not equal to 0. So here's my v right here. What do I mean by pi1 with this tangent vector? I, I take paths that leave here with this velocity vector, travel around. So between 0 and 1, the path never leaves x. And then it can return here with negative that tangent vector. You take those guys, mod homotopy. That's definition one. A way I prefer, in fact I think it's conceptually nice is, or, so you, what you can do is this. Let's take x hat to be equal to the real oriented blow up. So, alright. The real oriented blow up of, of x at, what do I call it, s. So that means here's the point s here. I, I replace that with all the different directions I can leave this point. Alright, so when I blow up, this becomes a circle. So I've, I've now got a bordered Riemann surface here. So, so the boundary of x hat is equal to s1, which is equal to the tangent space of s at x bar minus 0 mod r plus cross. And now, so when I do that, the picture up here I would have replaced this by a circle. Everything else is the same. Okay, still got my missing points. And now this tangent vector determines a point here, which I'll call, say, brackets v. And you can define the, the fundamental group with this tangent vector as base points as being pi 1. So pi 1 of x with base point v, you can define to be pi 1 of this blow up, this oriented blow up at s of x bar minus s minus s comma v. Right, and that's clearly the same as the usual topological fundamental group. And while I'm here, let me mention that sometimes we're going to have a variation of hot structure over such a curve and we want to be able to talk about the fiber over this base point, it's going to be a limit mixed hot structure. So, so let's, so here to what you can do is, so we have p1 and we want to find a base point that is non-zero mod p for every p. So we're going to look at p1 minus 0, 1 and infinity. And there's a coordinate here that is 0 at 0, 1 at 1 and infinity at infinity. And let's let that coordinate be z. And we're going to take d dz. And we're going to take this guy here. It'll be an element, say, of the tangent space at 0 of p1. And that's non-zero, visibly non-zero at every prime. And if we take the theorem of Deline and Gontra office that pi 1 unipotent of p1 minus 0, 1 and infinity, d dz, this is also commonly written as 0, 1, d dz is a pro-object mixed-take motives of the z. And so, so there's some remarks in order. So this is an overview at the moment. So remarks are the hodge periods, 1 is that the periods of the mixed-hodge structure, so on the hodge realization. So this will be, this is an easy kind of limit mixed-hodge structure here. So on the hodge realization are multi-zeta values. So zeta of n1, nr, this is something like the sum 1 over k1 to the n1, k to the nr. And I think I need nr bigger than 1 and 0 less than k1. So these are the standard multi-zetas, perhaps written backwards. Yeah, let me finish this remark and then we can take a break. And then the second remark I want to make is that, yeah, let me just stop here. Right, so I was making remarks. There's a few important remarks I want to make. So the second one is, well we know that mtm, I'm just going to use this for mtmz, mixed-take motives over z. So this is, this is tenakian. That means it's the category of representations of some proalgebraic group. And I'll give a brief explanation of what this looks like. It, so we'll have pi1, so it's the category of representations of some group. This is a proalgebraic group. Pi, I'll just write it as pi1 of mtm. And I'll be vague about the fiber functor, but it's, the common one is the diram fiber functor. But, so what do you want to think about is this also, you could look at the category of, say, split mixed-take motives over z. Right, so what would these be? These, what do I mean by split? It's a direct sum of, say, qm's to some power nm, and you would, you would sum over m. So there's no extension data, just a semi-simple thing. So what would the fundamental group of this category be? It's just the category, pi1 of this category is just equal to gm. Right, you just need to know the weight of a mixed-take motive. So qm, so how does gm act on qm? It just acts by the mth power of the standard character of the defining character. I don't know how to say that. Right, so gm, say, if q sits inside q cross and you just take the mth power of that. And that representation tells you what qm is. And so this is a subcategory. So if you think about representations, things go in the opposite way, so we're going to get a map to gm. And so if we had a representation of this that factored through this, it'd be completely split. And the kernel, again, you can think about this, is, I like to call it k. I'm sure Francis has a different notation. So this guy here is pro-unipodent. And so, and it tells us, it's telling us about it governs extensions. So let's just do a little calculation here. So let's try to compute what it is. I mean, I could just write it down, but it's a good exercise because we need it in the elliptic case. So we know that x in mixtape motives, xj of, say, q by qn. So yeah, we'll compute this. And so this is something I think that was done long ago by the likes of Dylene and Baylinson. They thought about what this should be. So, but this should be, so x in the category of representations of a pro-algebra group was given by group homology. You can define it for pro-algebra groups. And we're going to take pi1 of mtm, say, with coefficients in qn. And now, if you look at the spectral sequence for this group extension, you'll actually see that this is h0 of gm, hj of k tends to qn. And why is that? Because gm is reductive, right? So every representation splits. So there's no higher hj. You only get an h0. This is because gm is reductive. And now, now you look at this here. And strictly speaking, I should be dealing with you. So this is, this is going to be h. Now, if you have a pro-unipotent group, it's, it's, homology is the same as that of its Lie algebra. And so, now we're going to tensor this with qn. And now we're going to take invariance, right? That's what h0 means. So, right. And so what we know is, we know, what do we know over here? We know this is equal to 0 if j is bigger than 1, right? So this implies, this implies that hj of k equals 0 j bigger than 1, right? Because, and that implies, I can give you references for this. This is all very easy stuff. This implies that k is free. I should say strictly speaking, it's the completion of a free Lie algebra, right? But it's, and you can walk here and, because of the exactness of g, the exactness of the graded, weight graded quotients, you can walk here with everything. And then you actually work with a graded Lie algebra here. So the first thing we know is k is free. And then to know what k is, we need to calculate the generators. And so, so k, so this is, this is Lie defined to be Lie of k, is topologically generated by, well it'll be the duals of, it's going to be, okay, but this, how do we know what this is here? Well, whoops sorry, yeah. I want to take gm and dual. If you, anyway, so basically we want to pick up the generators because they, they're telling us about the one extensions. And we know for z, we know that x1 of say q by q of n in mixtape motives over z, remember this is given by the k theory. This is equal to q for n odd and bigger than one. And it's equal to zero for n even. And if you think about this for a second, what this tells you, this implies that, that k is isomorphic to, though not naturally, you look at this statement here. So, sorry. But okay, we're over, okay, so what happens with n equals one? It's zero n equals one because I, I guess I did it for a field before, maybe I should have done it for the integers. It's because k1 of z is equal to, the units in z is plus or minus one. When I tense it with q, I get nothing. So, this is isomorphic, though not naturally, to a free Lie algebra, the completion of a free Lie algebra with one, one generator for every odd integer greater than or equal to three. And so, this is quite a famous Lie algebra through the work of various people, including Francis. And I should also say this extension up here is split. There's various reasons. I mean, basically, Levy's theorem implies that this is splitting here. But also, this comes from grw dot, or there's, there's various ways to construct a splitting here. And so that means this, this Lie algebra has a gm action that, that sort of splits the weight filtration here. We can think of this as a graded, the completion of a graded Lie algebra. And yeah, and how does gm act? So, to get the semi-direct product, how does gm act? So, gm acts on z to n plus one by the two n plus first power of the defining character. So, so that tells us how gm acts here. So it tells us what this fundamental group is. It's a semi-direct product of gm in this group. I've just told you how the group acts on the Lie algebra. I mean, that's, it's, yes, it's q, it's q of one. And if, if you ask me to put a gm to give you the weight, I would do what I do with Mikoto and put a minus two there. Sorry? Yeah, this is the two n plus first power of the standard representation of gm. Okay, so the next remark, you know, I should say here, so we now understand pi one of mtm is isomorphic to gm, semi-direct product. We can think of it like this, z three, z five. So, by this I just mean the pro-unipotent group with this, this is Lie algebra. And so, I think it's the third remark is that, so we know that pi one minus zero one and infinity, so if we look at pi one unipotent of this guy here with the base point, alright, ddz, is a pro-object. And so, more accurately, I should say that the coordinate ring of this is an in-object of mixtape motives over z. And so, therefore, it's a representation of this group, so pi one of mtm acts on it. And now, you could say how rich is this as a mixtape mode of how representative is. It could be really boring. It could be completely split. No interesting extensions or it could have some. So, there's the theorem of Francis Brown. I won't ask you to stand. This action is faithful. So, the importance of this result is it's telling you that you see everything you want to see. You can construct all of these mixtape motives over z inside this unipotent fundamental group. And we'd like to have something like that in the elliptic case. We don't, not yet. But that's sort of what, that's the direction I want to go. And so, I should say a corollary. And so, this was an open problem. It says the periods of every object of mtmz are multisatas, right? Because this is basically saying that the subquosions of this guy here generate this category. And so, their period, yeah. Sorry? The periods of mtmz are multidatas along with all the... Oh, thank you. And 2 pi i. So, there's another point I want to make here. So, I think of this stuff in terms of modulite spaces of curves. Let me fix some notation. So, so let's, let's, let's just let m, and I'll say more about this in the elliptic case in the next lecture. But let's let mgn plus r be the modulite stack of... So, so here we are going to require the 2g minus 2 plus r plus n. So, r and n are in z greater than or equal to 0. And I put an arrow over this. This is the modulite stack of, and I'll just draw a picture. I take a smooth projective curve of g minus g. I'll have n points. So, I'll have something like x1 down to xn. And then at r are the points. So, here I'll have tangent vectors, nonzero tangent vectors, v1 up to vr. So, so it's, it's a curve plus n points plus r tangent vectors. Whoops. Maybe I'll be inconsistent about how I do this. And, and every vj, vj is not equal to 0. And they lie, you know, the points where they're anchored are distinct and distinct from all of these guys. So, again we're only going to need this with a few small numbers, but I thought I'd throw this in. I, I spent a chunk of my life thinking about these guys. So, these guys are defined over z. And so the way I like to think of this is that the genus, so m, t, m, z is sort of, it's closely related to sort of the mz, n plus r story. And to, to bring this home a little bit, p1 minus 0, 1 and infinity is equal to m04. So, the next, so basically, yeah, it's, it's genus 0. I mean there are other things that somehow aren't genus 0, but, and so the next step is the elliptic. And I claim to understand all of these higher genus ones, and I may say something about this at the very end, you only need to know genus 0 and 1. It goes back to an old result of John Harrah. It's been elaborated on by various people. So let's, let me start talking about the elliptic case. So, there's what I'll call classical and I'm not sure who's worked on this. Gontrof certainly worked on it. And so you'd like to look at motives that come from an elliptic curve. And the key thing is you start out with an elliptic curve, and to it you would associate a motive say h1 of e, and maybe the sense of Grotendick that would represent, say, the first homology. So let me talk about what this will be. So let's just say, let's suppose that we've got that e is defined over q and, and it has say equation. We can always write it in the form y squared equals 4x cubed minus ux minus v. So uv are in q. And we have d, which I'll denote as, is defined to be, say, u cubed minus 27v squared, which is up to a factor of 4, the discriminant of this polynomial. We want this to be not equal to 0. So, so let's, let's talk about what, I'll call it h1 of e. What is it? And I'll talk about just the various realizations. So, Betty is of course, it's just going to be h1 of e analytics, say q. Let's talk about, I'll still call it q to Ram, or to Ram. And so it's, I'll call it, let's just call this h. I'll call this, or I'll call it h Betty, h to Ram. So you can write the homology. The Abelian differential here is dx over y is an Abelian differential. And you can also span the homology with a differential of the second kind, which is x dx over y. So this is of the second kind. It still has a pole, but all the residues are zero. So it gives you a well-defined homology class. And even though everything here is holomorphic, this is the, this is f1. So this is the, this is the Abelian differential. And now, so if you take h Betty, tensed with c, it's isomorphic to h to Ram, tensed with c, and this gives you a hodge structure. Here's the q-struct, the underlying rational vector space, and this gives you the hodge filtration. With f dot, this is, this is, so this is hodge. This is a hodge structure. So you can write it as, so h01 is not necessarily spanned by that guy there. Right, so let's look at the eladic. And we have that the Galois group, say, of q bar over q acts on, you can look at, say, the L to the n torsion and this guy here. And so this guy will be isomorphic or essentially equal to, I guess, the L to the n torsion of the complex points of the curve. And that's just going to be h1 of the complex analytic curve with z mod L to the n coefficients. So let's, so pi1, a tile of e over q bar, 0, it's just going to be the inverse limit of these guys here. So it's, it's equal to the inverse limit of these guys here. So it's equal to the inverse limit of these guys here, h1. So this is stuff that follows from fancy machinery, but here it's pretty basic. And so this is just isomorphic to h1e. Alright, so, and so, right, so the cohomology, if we want first cohomology, h1 of e, say, a tile, so with ql coefficients is going to be isomorphic to, harm of this guy here, into ql of minus 1. Okay, sorry? Sorry. Hold it. What? Ah, yeah, sorry. This will be the pro-l part. So maybe I should just put tensor with zl. Yeah. So this is the pro-l completion. This is abelian so we can split it as, alright. And so we have, and then all of this gives us a Galois reaction here, gq. So that's this guy here. And so what should we mean by mixed elliptic motive? So, say, iterated extensions of symmetric powers. So we're going to take, say, Snh, a tape twist of symmetric powers. And by this I mean the nth symmetric power of h, tensed with ql of r. And now, so the category of mixed elliptic motives, maybe e, should depend heavily on, depend heavily on what the curve is. For example, if it has complex multiplication, you will get a lot more algebraic cycles, for example. And so I don't know much about this, but what I want to do is look at the universal case. So what I would like to do is just now give you an overview. I mean, it's reasonably complete. If I try to dot every i and cross every t, it'll take a long time. It won't be clear what's going on. So, and I should say here, most of this is joint with Makoto Matsumoto, joint with some joint with some other people and some's joint with myself. So what's the idea? So I want these, so these are elliptic motives defined, this is a vague statement, for all elliptic curves. So what should that mean? So what we're going to do here is, so more precisely, they're motives associated to the universal elliptic curve. So, and I'll make a comment here. So k-theory, as I understand it, k-theory is an absolute theory. You can't look at, you know, schemes over something or other. It's an absolute theory. And so I think the theory of motives, maybe there's some version, relative version, but I think most of it's just absolute. Is that correct? That's what I, so, but this is a relative theory. We want to look at this guy over this. And so, what we're going to do, so let's give this a name. So we're going, what's the analog of the h over there? We're going to set h. So I use this black bolt, bolt for local systems. It's equal to R1, F lower star of q. And again, to be absolutely precise, I have to explain what this means in all these different categories. So it has to be suitably interpreted. So you don't take any level of structures, you don't take any level of structures, you agree with the stack? No, I mean, so if you go back to the case of mixed-tate motives, I mean, the ones we understand are the ones where we're just working over z, right? You know, the analog, I think, of a level might be to work with rings of integers in some other number field, right? So the one that you can define it for having levels and you can define it over bigger rings, but it's going to be harder. And this is like the first case, at least I think it should be attacked. We'll see. I mean, curious to hear your reaction. I mean, it certainly would work with levels, but we'll see. I mean, the general setup will. Okay, so what is this? This is the local system over say M1, 1 whose fiber over the moduli point of an elliptic curve is simply, it's just going to be H1 of Eq, right? But again, interpret, so the approach I'm going to take here is that this is all, I'm going to look at compatible realizations. And so then, although it's going to turn out that we actually get actual motives, but that's through some strange fluke. Okay, this always, okay, I go backward and forward on this. If I wanted H lower one, I'll just take H1, right? So it, somehow my brain functions better if I put H in weight one. So this is, this guy here is a, it's a local system of weight one in all senses of the word. So for example, if you think of it as a variation of hard structure, it's a variation of weight one. All right, so let's talk about the realizations of H. So we understand it. So and again, I'll do, we need to do more detail here because we need to do computations. Yeah, in some sense, so let me say a word about, what's M1, 1 analytic? You can view it many ways, but the way that's most convenient here is it's the quotient of the upper half plane by the upper, so this is the upper half plane. And it's really, I'll write that for overfold quotient. So it's overfold quotient. And what that means in practical terms is you're just working on the upper half plane, but SL2Z equivariantly. It's sort of what everybody does in the theory of modular forms. For a base point, I can choose anything. But in the upper half plane here, we have, you know, we can take a strip high enough up. You know, we're going to have some fundamental domain here. It looks like this, the action of SL2Z. We can, say, take the stuff of imaginary part sufficiently large. And then, when we take the quotient by SL2Z, it's just taking the quotient of this by the unipotent subgroup. We'll take that here. And so I'm going to set D star as basically the quotient of this strip up here, S. It's just 1Z01, the quotient of this strip, sort of like m tau bigger than about 1, I guess. And this is, and so I'll take Q, this is all very standard, e to the 2 pi i tau. And this disc here, D star, it maps into m11 analytic. I want to take this, this actual disc. I mean, you can, you can take, right. Every point in here has stabilizer minus the identity, so I could do something else. But I want a genuine punctured disc. And now, so what can I use for a base point? I can use the base point, DDQ, for the base point. So my base point is going to be equal to DDQ, and it naturally, if I look at the imaginary axis here, I can think of it as corresponding to this arc here, you know, the imaginary axis. And so I'm going to include this, if you like, into the upper half plane as the imaginary axis. I mean, maybe I'm being a little too fussy here, but, you know, I'm usually fussy about base points. But anyway, so you can also use the upper half plane as the base point. What I'm saying, here's a map between these base points. So with this convention, what I'm trying to say here is that if I look at pi 1 of m11 analytic, DDQ, this is isomorphic to pi 1 of m11 analytic, and we have the projection here from H into m11 analytic. Maybe I'll call this, I don't know, I was going to, P is maybe bad, but I'll call it P. And this is just SL2Z. And much to the annoyance of algebraics, I use the topological convention that I multiply paths in the topologist's order. Right, so pi 1, with these choices, you can work it all out as naturally isomorphic to SL2Z. So a local system, so 1 is Betty. So a local system is really given by a representation of pi 1, and H, Betty, just corresponds to the defining representation of SL2Z. It's just a two dimensional representation. I'll be more precise about this later on. So I'm going to skip Kudoram. I actually worked it out, but it's actually very interesting. It's more or less done in the work of Levin and Rassanay, although they left quite a chunk out, actually. And I will write it down later on, but I'll skip that. Let me just do Hodge. So what you can do is you can take H, Betty, you know, tensor it with O of M11, analytic. And so this will be, I'll call this H. So this is a vector bundle. So H is a locally constant sheaf. This is a vector bundle, and it's got a canonical flat connection. And this is a, and I'll write this down next time, but this is a polarized variation of Hodge structure. This is the most canonical one of all. It's what's called an SL2 orbit. It's locally homogeneous. And I will describe this in complete detail. It's completely elementary. And it's also a good way if you don't know the theory of, so polarized just refers to the fact that it has an inner product. So that's actually a flat inner product, say, from H over E, tensor H over E into, say, Q of minus one, right? It's just, this is just the cut product. And polarization just means, basically it means that this current, it's an inner product that satisfies the Riemann bilinear relations. I mean, we won't need to lean on that at all, but it's good to know it's there. But in the general theory, you need a polarization. So this is a least sheaf. And it's, let me just describe it like this. So this fiber, say, E, say, where E is defined over Q is pi one of E over Q bar, say, zero, tensed with QL of minus one, just what we wrote down before. Like I said, I'll do the Q Durand theory later on. I should say more about this base point. For lots of reasons, it's convenient to work with the base point I wrote down, DDQ, because, right. So let's say more about the base point. So the first thing is, is that since every, so we need a base point that's non-zero at every prime, right? So if we took some, you might think, okay, let's take some E defined over Z or over Q whatever in M11, say Q. And now, but there'll be some prime at which this has bad reduction. So, and at that prime, we can no longer use this as a base point because it's no longer in our space. So we're forced to use bad reduction somewhere, have to use, and so there's only, you know, various ways to draw this marginalized space. You can draw it as this castle picture out here, where this is basically where infinity goes and say i and rho, rho being a cubed root of unity. And our Q disk is out here. So this is really the Q disk out here. If we choose DDQ, that works because the Tate curve, so the Tate curve, so you've got E, say E Tate over, say, Z power series in Q, this, if you reduce it mod P everywhere, it's still smooth. And so, and you know, the analog of the tangent vector will be to pull it back to say Q bar Q 1 over N. You know, you pull this back here. So this is going to be our base point DDQ. And this guy is non-zero at all primes. So this gives a base point, which is non-zero at all primes. And so the only two possible base points we can use are plus and minus this guy here. And it's also very natural when you're working in the upper half plane. Like I explained over here, it's more or less topologically using the upper half plane, the projection from the upper half plane to be your base point, and distinguishing the imaginary axis. So I guess it's a few minutes to lunch. So maybe I won't, let's see what's a sensible place to stop. Yeah, I'm going to make one point here and then stop because everybody wants to eat. So this is the point that's most confusing. So that's why I'll give it to you right now so you can be confused for a week as this. And this is why I have two different notions for weight filtration. So if you have, first of all, this is just some scribble. If you have a variety, say a curve, and you've got a variation of hot structure over it. So here it is here. It's a local system. It won't have a fiber over here. And by method, I'll review all this because I expect most people here don't know it. But you have this local system. So for example, this might be our punctured Q-disc, and this is our local system H. And if you look at, so the first thing you can do is you can take H tensor O and you get the flat vector bundle, this guy here. It has a natural extension to the entire disc, which is called Deline's canonical extension. And you get a fiber of that canonical extension over here. And then to every tangent vector here, you get a mixed hot structure on this. Even if you started out with the pure hot structure, you get a mixed hot structure on the fiber of the canonical extension over the origin of the disc. And so what happened, and this is a very elementary calculation I'll do next time and I'll completely explain it, is what's the limit mixed hot structure. So the limit, and I should say also the mixed hot structure depends on the actual tangent vector. If you use different tangent vectors, give different results. So the limit mixed hot structure on H over, even though this was pure of weight one, over dDQ, I'm going to denote it by H. So this is, I'm going to denote that fiber by H dDQ actually turns out to be Q plus Q of minus one. And you go, this is weird because this was pure, the variation H is pure of weight, this is pure of weight one. So what I want to do is this. I want to define, well, I should say we have weight zero of H equal to zero and weight one of H equal to H. That's because H was a variation of weight one. Because H one of an elliptic curve is of weight one. As H one of E is of weight one in every theory here, every interpretation. But here we're getting something mixed. So we have to give this weight filtration a different name. So this has got weight two, this has got weight zero. It should not be average of these as one. And so we're going to get a weight filtration here. So we also have a weight filtration, which is called the monodromy weight filtration, which is going to be M on H. And on this guy, on this limit mixed hard structure, which is going to be M zero contains M minus one, which is equal to zero. This is going to be equal to M one. And this is contained in M two equals H. And so you've got two nonzero graded quotients. This one here and this one here. And this is Q of minus one and this is Q of zero. And so the similar thing applies in Galois theory. If you use this, if you look at the elliptic, the Tate curve, where did I put it? If you look at the Tate curve over this guy, it's H one splits as Q plus Q of minus one. Similarly in the etal picture. And I'll make my final point. I guess I got one minute and 40 seconds according to my watch. Whoops. This is a suitable place to stop. So the, so, so H ddQ is equal to Q plus Q of minus one in all realizations. But because it's a direct sum, this is the, or these I should say, these are the realizations of Q plus Q of minus one, which is a mixed Tate motive. And so this is going to be the starting point. And this is why I was denoting the weight filtration here by M dot. Because so in a word, universal mixed elliptic motives are going to be extensions of Tate twists of symmetric powers of this variation, whose limit mixed HOD structure is a mixed Tate motive. Is the realization of a mixed Tate motive. And this forces this things to be genuine motives. And makes the story very rigid. And then you can ask, well, perhaps there are no interesting examples. And what I'll explain next time is there are interesting examples. If you look at an elliptic curve and a nonzero tangent vector, you can take the unipotent fundamental group of E minus zero with this is, so you've got an elliptic curve. You've removed the identity. So it's got a free fundamental group. And you've got this vector here. And this guy here. And you look at the variation of these. Maybe I should put the Lie algebra here. So this gives you a local system actually over M1 vector one. And this is a universal mixed elliptic motive. And it's fiber. It's limit mixed HOD structure, which is the one associated to the Tate curve up there is a genuine mixed Tate motive. And so I'll try to explain all of that. So we'll stop here. Don't have any questions? I should have asked before. Okay.