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.  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.  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.  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.


Transcribed. Ladies and gentlemen, the railroad hour. And here comes our star-studded show frame. Tonight, the Association of American Railroads presents the beloved and tuneful operetta, Countess Maritza, starring Gordon MacRae and his guest from the Metropolitan Opera, Blanche Theebaum. Our choir is under the direction of Norman Luboff, and the music is prepared and conducted by Carmen Dragon.  And now, here is our star, Gordon MacRae. Hello, Miller, and good evening, ladies and gentlemen. Tonight, the leading mezzo-soprano of the Metropolitan Opera, lovely Blanche Theebaum is the Countess Maritza, and I am Countess Silo, and we shall be singing of the gypsy life. I ask you, Baron, what more could a man want in a morning like this, a spirited horse? I must say, Countess Silo, you're in rather high spirits for a man without a penny in his pocket. Why not, Baron? I had a good time spending the money. One of the finest estates in Hungary is squandered, and what do you have to show for it? The memories of a very good time. You have a very carefree attitude toward life. Baron, that's my philosophy exactly. Gone by like a slave was I, and a woman's smile then was all worthwhile, was a siren song I heard all put my heart in her smile. My friends' smiles seem to sympathize laugh while they're gleaming. Countess Silo, I don't trust you at all when it comes to women. I don't think I'll recommend you to the Countess Maritza after all. Please, Baron, I need work. I have to support my sister, Lisa. Well, I know that Countess Maritza needs a novice here, but your sister and Maritza are friends. What if Lisa finds out? Well, tell her that I'm carrying out a masquerade. Tell her that I'm secretly in love with Maritza. Maritza finds out who you are will both be in trouble. She thinks every impoverished nobleman who pays any attention to her is out to marry her for her money. Well, here are her gates. We turn in here. This gypsy wanker. Yes, they come every year at this time and camp on the estate. Ha-ha! Oh, our gypsy's a song! Albert sings where the gypsy music rings. People, thank you. You are welcome. This spring is always. Set up your camp and make yourselves at home. Oh, bless you, Countess Maritza! Countess Maritza! Oh, Baron, forgive me. I was so happy to see the gypsies, I didn't notice you or your friend. Countess, this is Bella Torek, who used to be overseer for Countess Hilo. I'm at your service, Countess Maritza. An overseer? But he looks like... I can recommend both his diligence and his honesty. Bella Torek, you come then with high recommendations, indeed, from a gentleman I greatly respect. If you want the position, it is yours. Thank you, Countess Maritza. I do want it very much. Oh, it's you, Torek. Forgive me, Countess. I didn't realize anyone was in the garden. I was sitting here listening to the gypsy fiddler. Sometimes I wonder if there's gypsy blood in my veins. I have such a longing to follow them. Oh, there's gypsy blood in all of us, Countess. We all love the open road, the open sky, the road that curves ahead into adventure. Yes, and the wild and intoxicating music of Roman age. Have you ever been in love? No, Torek. But there's never been anyone in love with me. You were young and lovely. Hasn't any man ever told you then? Oh, yes.  If you go through life mistrusting everything and everyone, you'll never find happiness. This has developed into a rather personal discussion, hasn't it? It was the spell of the gypsy fiddler's countess. I quite forgot for the moment that you are a countess, and I am merely your overseer. Big pardon, Countess Marica, but the Countess Lisa has arrived. Lisa? I'll be right in, Charles. Yes, your lady. Torek, are you in the habit of referring to the sister of your former employer as Lisa? I apologize, Countess. It was a slip of the tongue. Kindly watch yourself in the future. You may go to your quarters. Greetings, gypsies. You seem unhappy, overseer. Come, forget your troubles. Come join the gypsies around the gypsy fire. Yes. The fires are the carefree and the happy. Nobody here ever has to pretend. Oh, I wish I could spend the rest of my days with you. They used to call me countess, and I had sweethearts followed, sunny shore. I could have my old friendships I'd not care. I'd wonder free as air. Comrades, I would say gypsies dance, gypsies play while... Marica. Marica, you forget yourself again, Torek. Yes. Yes, I do, Countess. But I'm intoxicated with gypsy music and gypsy dreams and thoughts of you. Of me? Of you. Here by this gypsy fire, we are equal. Really, Torek? For these few moments, Countess, I can tell you that there is no girl you're equal the entire length of the Romani trail. Act of Countess Marica in just a moment. You know, you probably live a long way from the factories that manufacture the things that make your home such a pleasant and comfortable place. That's why without the railroads, your home would be pretty empty. But the railroads part in furnishing your home involves more than just seeing that everything you need and use gets to your hometown. Those things must arrive safely and in good condition. To begin with, goods must be packed properly and in the right kind of container. Every container must be tightly closed and marked correctly. Now, that part of the job is the responsibility of the shippers. Then, all containers must be carefully loaded and firmly braced in freight cars so they will not be damaged in movement. Finally, the railroads must see to it that the freight cars are carefully switched and smoothly handled all along the line that leads from factory to your hometown. And because the prevention of loss and damage depends on teamwork, the 25,000 members of the shippers advisory boards and America's railroads are once more joining together in the observance of April as Perfect Shipping Month. This country-wide observance of Perfect Shipping Month is just one of the ways in which shippers and railroads cooperate with each other in striving for a safe and prompt handling, shipping, and delivery of America's freight. And today that job is more important than ever, not only in April, but also the year round with every month a Perfect Shipping Month. For as America's production lines and America's railroads step up their twin effort to meet the needs of national defense, at the same time that they keep our civilian economy rolling along, the nation can ill afford lost or damaged goods. That's why these days both railroad men and shippers are exerting extra care to see that America's productive output is fully protected from source to final destination. Now here is the second act of Countess Maritza starring Gordon MacRae and his guest Blanche Fibon. You sent from me, Countess Maritza? Yes, Torek. I, um... I wanted to talk to you about your work. Has my work been unsatisfactory? No. As a matter of fact, your work has been most satisfactory. But I'm afraid I'm going to have to discharge you. Why? Please don't feel that it's because you're inefficient. It's because I am inefficient. I just can't seem to think of you in the terms one should think about one's overseer. Why, Countess Maritza? I shall miss you very much. I refuse to go. I love you. What? How can I leave? I love you, Maritza. Oh, Torek. Oh, Maritza. Maritza, my darling. Oh, Torek, my dear. It's Lisa. Come in. Oh, Tessilo, it's all right now, isn't it? She loves you, I'm sure of it. Lisa. Lisa? Tessilo. Then you must be Lisa's brother, Tessilo. Count Tessilo. It was most ingenious of you to come here and pose as an overseer. I was lonely and much too sentimental, and you played on it very cleverly. Maritza, listen to me. Oh, I congratulate you on your performance, Mr. Torek. I love you, Maritza. I know the exact state of your finances, Count Tessilo. Your coffers are as empty as the words you would like to use to convince me now. Maritza, I wouldn't dream of trying to convince you of anything.  Well, I almost loved you, and I'm sorry for you now.  Oh, the skies were blue. Calm, Lisa. Yes, Tessilo. Goodbye, Maritza. Stop, my brother. I loved him, Lisa, when I thought he was Torek the Overseer. But when I found out who he was... Do you think he wants you for your money? There are a dozen women of fortune he could marry. He loves you, Maritza. He truly loves you. But if you let him go now, you'll never see him again. I know him well enough for that. Where is Tessilo? He's waiting for me at the Gypsy Camp. He doesn't know that I'm speaking to you. At the Gypsy Camp? Oh, I'll go there at once. It's a terrible mistake. Please, forgive me. You don't think I'm a fortune hunter any longer? I love you. That's all I know about you, and that's all I want to know. I knew that when you walked out of the door. If you left me, I should be lonely all the rest of my life. Maritza. This is like a golden dream. Lovely Countess Maritza will be back in just a moment. And meanwhile, thanks to Lorraine Tuttle, William Johnston, and our entire company. Countess Maritza with music by Emory Kalman and booked by Harry B. Smith was adapted for the railroad hour by Gene Holloway. The railroad hour is brought to you each week at this time by the American Railroads. Marvin? One of the most important factors responsible for saving the lives of our fighting men wounded in Korea is the availability of blood and blood plasma. American servicemen are recovering from their wounds at greater speed and in larger percentages than in World War II. Now, the reserve of plasma retained from World War II and built up in the years of peace is virtually exhausted. That's why a real emergency exists today and why all adults are urged to call their Red Cross chapter or local blood donor center and make an appointment as soon and as often as possible. Thank you, Marvin. And now here again is Blanche Theebaugh. Thank you, Gordon. It was a genuine thrill to join you on the show train.  Oh, no. Tell me, Gordon, where are you traveling next week? Well, we're traveling into a wonderful world created by the one and only Jerome Kern. We're all going to get smoke in our eyes, Blanche, for Roberta. And Nadine Conner is our guest star. Oh, sounds wonderful. I'll be listening. Good night, Gordon. Good night, Blanche. You were wonderful. And come ride the show train with us again real soon. Oh, the board? Well, sir, it looks as though we're ready to pull out and so until next week, and Roberta, this is Gordon Macrae saying goodbye. Countess Maritza was presented by arrangement with Century Library, Incorporated of New York. Gordon Macrae will soon be seen starring in Warner Brothers About Face. Our choir is under the direction of Norman Luboff and our music is prepared and conducted by Carmen Dragon. This is Marvin Miller saying goodbye until next week for the American Railroad. Now keep tuned to your Monday Night of Music on NBC. Transcribe there's more great music with the telephone hour next on NBC.


An important relationship between the tangent and quadrature problems was discovered by Isaac Barrow. Barrow took an unusual viewpoint. Researchers had an obligation to publish their work. And so Barrow's geometrical lectures appeared in 1670. Now when Barrow identified it as an obligation, he meant it was an obligation, something he had to do, whether or not he really wanted to do it. And so in the introduction, Barrow notes that they are school lectures, read by duty, and sometimes spoken to hastily that I might end my design talk within the hour. You should not expect to hear anything accurately perform or beautifully digested. In his geometrical lectures, Barrow linked the quadrature and tangency problem and created calculus as we know it. Let's take a look at Barrow's work. So Barrow begins with two curves D, C, and S, R, where if O, B is equal to M, Q, the ordinate R, Q is equal to the area O, B, C. In modern terms, we might call MSR the curve of area because the ordinate of the curve gives the area under O, D, C. Now again in modern terms, let RT be a line with slope M equal to CB, and consider any other point S on MSR, where P, Q is equal to H. Now remember O, B, and M, Q are equal, so we'll mark off AB equal to H, and we'll fill out some parallel lines, we'll draw SL parallel to the axis MT, and let RT intersect SL at some point K, and we'll also draw AD perpendicular to our axis. Now since K is on the line RT, which has slope equal to M, which again is equal to CB, we have RL to KL as M, and so RL is M times KL. But note that RL is the difference between the ordnance PS and QR, but remember that our curves are such that PS is the area O, AD, and QR is the area O, B, C. So RL is the difference in the areas OAB and ODC, it's this area, and so that means RL by itself is equal to the area A, B, C, D. But since ODC is increasing, this area must be less than MH, that's the rectangle with width H and height M, and so we have MH must be greater than M times KL, which says that H has to be greater than KL. And consequently this point K has to fall somewhere between L directly below the point R and S, another point on the curve, which means that this line RT has to fall below the curve on the left of R. And by essentially the same argument, we can show that RT must also fall below the curve to the right of R, which means that RT is entirely below the curve, and so consequently RT is the tangent line to the curve. And so we might summarize Barrow's result as follows, if ODC and MSR are two curves where the ordnance RQ are everywhere equal to the area O, B, C, then the line RT with slope equal to CB will be tangent to MSR. Or if we introduce some very modern notation, let F of T be some increasing function, let capital F of X be the area between 0 and X under this curve, then the derivative of the area function is equal to the function values. And this is one form of the fundamental theorem of calculus. Barrow then goes on to claim the following. Let PR be a curve where if OA equals PQ, then the square of QR is twice the area of OAB. If we take QT equal to AB, then RT is perpendicular to the curve. Again, in very modernized notation, let the square of G of X be twice the definite integral of 0 to X of F of T dt, then the normal to the curve Y equals G of X at X equal to A has slope negative G of A over F of A. The proceeding proof is a rigorous proof based on accepted Euclidean geometry. However, Barrow was aware of a new approach to solving such problems, infinitesimal quantities. Barrow questioned the value of the methods until a former student and later colleague convinced Barrow of their usefulness. The student was Isaac Newton.


Lokal language 2 Stück pase ǎb nā ër gĒ bhaḍhā coop applicig lāt. Lokal language 2 ǎs a dope tool. Lokal language 2 ǎs a dope-ered commencer PlayStation ǎs .  Doug oh yertدي Windows ǎn. Transliteration is a quote that is jot epa liqhami we wassapa anthuman mapping kora hai phonetic artistic beauty. Udha hong sorabhe jitirami Latinakhor orthak. Imraje akhor at baik te baik koru. He baik chyo laksha bhakha thokha ekhi utsaronon akhor loe puri borto loe baik transliteration orthak. Versus transliteration. And we look at my original imrajean baik koto hepe what is your name.  Rent in India located in India means the South Indonesia Transglation  Kadak would now use subscriptions towick the world                          Loka


On Wednesday, we found out that the U.S. big banks trading desks posted their strongest results in years during the first three months of 2020, even with the coronavirus stressing everyone out. Meanwhile, the U.S. retail sales collapsed as coronavirus kept consumers at home. The French retail sales also plunged 24 percent in March. The Bank of Canada lowered its benchmark overnight interest rate to 0.25 percent to address the economic consequences of the new coronavirus pandemic. Welcome to the Tick-Mill Update, I'm Kana Danielle, the founder of the Investeva movement. Make sure to subscribe to the Tick-Mill YouTube channel and support us by liking and sharing this video with your forex trading friends. Thursday, we'll be looking at Bank of England's Liabilities and Quoted Conditions Surveys, the U.S. retail initial jobless claims, and China's GDP. Today, I'm looking at the Swiss Yen pair, the two safe havens that have been basically rain trading ever since the coronavirus became a thing beginning of 2020. The pair has been ranging between 111 and 113. The Ichimoku Cloud has turned in consolidation mode as a result. Currently, the pair is approaching the lower band of the range at 111. So there's a chance that we could see another bullish move up to 113 starting next couple of weeks. Now do you think the pair will continue to range in the next couple of months? Head over to the comments section and let me know. Of course, trading in the financial markets involves a risk of loss and you should only trade the money that you can afford to lose. If you like this video, give it a thumbs up and subscribe to the Tick-Mill YouTube channel. I'll get back to you with more updates tomorrow.


Welcome back, we are here at Perkota Live 2014, Santa Clara Convention Center, Santa Clara, California, the heart of Silicon Valley. I'm Jeff Frick, you're in theCUBE. We take theCUBE out to all the great events. We get the smartest people in the room. We ask them the questions you'd like to ask them, extract the signal from the noise and really bring you the story if you weren't able to get here. So I'm joined here in theCUBE by Alex Yurchenko, co-founder of Codorship. Welcome to theCUBE. Thank you, Jeff. So you've got a really interesting product name. I want to jump into Galera Cluster. And I was struck by the name because a close friend of mine, Portuguese buddy from Brazil, used to say, we are Galeta. We are Galeta. We are a kinship, you know, a group of guys, a group of five or six guys. I'm just curious if that has any play into the name of your product. Well, it's definitely a Roman word and it must have some connection to folks, to people. But initially where we started off with was that Galera is a Roman word for what's in English is called a galley, a ship with multiple oresmen and maybe multiple even rows of ores, you know, Privyamas, Privyamas and stuff like that. And why we choose this name is that first of all, oresmen must work very synchronously in order to not interlock their ores. And secondly, and Galera is a synchronous replication solution. And secondly, the more oresmen you add to Galera, the faster it goes. And we were aiming for scale out and the more nodes you add to the cluster, the faster it goes. And ironically, or maybe not ironically, anyway, this synchronous operation is exactly the thing that allows us to scale up pretty much like in Galera ship. So synchronous and able to scale. Just add more rowers to the boat. Yes. As long as they're in synchronized fashion they'll make the boat go faster. Yes, exactly. All right, well that's a good picture. So let's talk about the technology. Where is this being used? What exactly have you built on the technology side that's different, that wasn't there before? And give us a little bit of background. Right, I've been waiting to say this. Here we are, we're breaking it live on the queue. Yeah. For many Galera users and just people who are familiar, have heard about our technology, it's just a replication solution for Maya scale. Not that sexy. Well, it's not that sexy, but in reality, Galera is a library, is a generic library which implements an efficient way to serialize access to a global resource in a high latency environment. Okay. To put it in more concrete terms, global resource may be a row in a table in your database. High latency environment is pretty much anything that is not in FiniBand. Anything where global locks are too expensive to obtain. And that's pretty much the whole world. If you want to have concurrent global access, write access to your data, Galera is pretty much the most efficient way to achieve that I know. So it's about knocking down latency in what you call, what you define as the high latency world. Yes. So that's interesting, because all we ever hear about is less latency, less latency, really driven by the mobile first. You get the Google developers at any of the shows that they're at and they're talking about, just shaving, shaving, shaving, bits of time here, there and everywhere. So what are some of the, I guess, legacy environments that are high latency where you've got an opportunity to go in there and trim? The applications and the customer use cases. Well, just this Monday, J-Pipes gave a very compelling use case for Galera in OpenStack, where you want Keystone, Keystone is a part of OpenStack infrastructure, where you want Keystone data to be known globally. In all your data centers, you want to distribute and you want to have full access to this data. So Galera is the perfect tool for this job. And basically I think it's the only solution that I know that allows you to distribute Keystone data across your data centers. And the Keystone data in the OpenStack is really the key piece again, taking the great naming convention. It is the stone that holds the arcs together. It's what makes it all hold together. Awesome. So then you play in the OpenStack then obviously as well. Well, many OpenStack installations have adopted Galera. Okay, great. So talk a little bit about the company. Where are you in terms of the company, how big are you guys? Where are you based? And talk a little bit about having kind of an open source component in your company. But then how do you make money? So obviously you got to have some non-open source stuff or services. Give us a little background on the company. Well, perhaps I should start a little bit with some history. Sure. That our company is called Codorship. And this is because the founders, me and Sepe Yacola and Temo Olaka, we met 10 years ago in the company that was then called Emic Networks. And now it's called Continence. All right. And due to certain events, we left that company and then we decided that we wanted to start our own company. Of course. That's what we do here in the Valley, right? And we kind of, we're thinking that we, you see, we started this work like 10 years ago. Everybody has experience in implementing database replication solutions in Emic Networks. We started with the M cluster. Then it was Uni cluster. Then we worked with Continence on Tungsten and kind of, we felt that we have accumulated enough experience that we know how this should be done. And we wanted to implement it. So what was the whole, what was the whole that existed that gave you guys the thought that you could go out and do it a better way? What couldn't be done that you can do now? I think... Or was it just a marginal improvement in the process? No, it wasn't a marginal improvement. It's, our idea was to, you see, we saw how complex replication solutions are. We saw the deficiencies and we were thinking that we now know how to, you know, pretty much make a qualitatively new solution which replication that just works. Unfortunately with replication, you know, replication is never simple. But I believe that we have achieved our goal, say, we have only two mandatory configuration variables in our cluster. Two? Yes, so to set up the cluster, you need to configure only two variables in MySQL configuration file. Wow, so talk a little bit about hyperscale and, you know, big scale, big data. Obviously, you know, there's a lot in Hadoop and there's a lot in the NoSQL databases but a big topic of the conversation here at the show is, you know, making MySQL better for large volumes of data. Here you're doing replication that would only seem to make your job more complicated, more difficult as the front end of that pipe is getting bigger, bigger, and faster, and faster. What does it mean for you guys? What's the opportunity? How's it impacting your customers? Well, to be honest, we haven't yet tapped into that Hadoop pool, but we're starting to get some ideas how we can improve, say, Hadoop, for example. Hadoop has a name node and currently there is no kind of good solution for high availability for that node. It's pretty much single point of failure and there are some workarounds for that but, say, what we can do with Galera is make a multi-master cluster out of Hadoop's name node. Well, these are kind of plans or some ideas that we're having but it's all... Still pretty new. Yeah, depending on how much resources we can gain, it all requires some work. So, even though Galera kind of is not exactly targeted for big data, even there we can... There is an application for that. Sure, because it's coming, right? Big data is coming, it's everywhere. Big data is coming and what's, like, one year seemed impossible. Right now we have already clear ideas how to do it, sharding and stuff. Okay, so why are you here? What's important about Percona Live that brought you here to the show? Users and partners, we're here to talk to people. Okay, and how's it been going? What's the vibe on the floor, Howard? What are the feedback you get from customers on kind of the state of the union and some of the partner community? It's been very wonderful for us but, you know, it's kind of... It's been very wonderful for us for probably the last half year or so. People are getting aware of the technology. They are trying it, they are building expertise. They are using it in solutions, migrating from proprietary solutions to Galera. So it's not just this conference, it's the overall process what is happening. In our business. Sure, and will you be at OpenStack Summit? Yes, we will be. Maybe not me personally, maybe, but we will have a boost there as well. Because we'll be there, the queue will be there in Atlanta this year, closer to Finland than in San Francisco, maybe, because it was important last year. So that's good. So what's the next big challenge to overcome? What's your next big kind of technical or business objective you're trying to achieve? Well, there are certain limitations that Galera has and that we resolve to solve by the end of the year. But these are kind of usability problems and what people have long asked of us, for example, support for arbitrary large transactions and stuff like that. But I don't see it as an next big thing, it's just, you know, finishing the product. I think the next big thing will be sharding. We are working on design of a sharded Galera cluster that's pretty much addressing this big data challenge and we are trying to address it in the most optimal way. And it's, this very thing is actually, the challenges, what kind of keeps us going. Yeah, it's very exciting, yes. I can't wait to start working on that. But you know, there are users, emails, like mailing lists, negotiations. I mean, when I have really very little time to actually code. That happens when you grow business, right? You have a business instead of writing the code. So is that, is the sharding project, is that a three month project from now? Six month project, year long project? It will start probably in a year. Start in a year, okay. Yes, I think, well, we will probably have some alpha. Okay, so we're here next year, Percona Live 2015, we get you on theCUBE. We very much hope to have something to show. Okay, good. Well, good luck. This sounds like an honorable objective and certainly a fun one to keep you working and keep the team going. So again, thanks Alex for coming on from co-founder of Codership, the Galera cluster. Best of luck to you and the team. Thanks for coming on theCUBE. We will be right back from Percona Live 2014. We won't be at 2015 for a year with our next guest after this short break. We're watching theCUBE from Percona Live.