All right, thanks very much, Jen. And thanks for being back, especially at this early hour. Really appreciate your being here. So today, we have to do a little bit of work. So last time was mainly motivational. Today, I have to tell you the main construction in this whole area. So we have to get through quite a bit of material.  So I hope to get through all of it today. I put everything on slides just because there's quite a few definitions that I want things to be as clear as possible. And also my handwriting, apparently, was quite small last time. So I hope that by making slides, I could remedy this. So I put everything on slides. I chose the smallest font that Beamer would let me. And here is the result. OK, so let's start with a brief recap. Last time, after this letter of Zaghe that we read and we got all into the mood of the theme of this conference, we started discussing quadratic forms ABC. So those are polynomials homogeneous of degree 2 with integer coefficients. And we studied the action of SL2Z that exists on these quadratic forms. Now, this is a photograph of Hendrik Lenzstrah looking quite unimpressed. Now, why is he unimpressed? Last time, I made a definition. And I let FD be the set of primitive forms of discriminant D defined to be B squared minus 4AC. And of course, it's my constitutional right to make any definition I like. But a little bit later, I made a claim which is not quite true unless I put in that additional condition there that A needs to be bigger than 0 if D is less than 0. So let me do that now and make up for that. Oh, hey. Sorry. And the statement was the following. We saw that if we looked at the set FD up to SL2Z equivalence, we got a finite set. Not only did we get a finite set, we could put it in bijection with a finite abelian group so it inherits also a group structure. And the explicit bijection was given by this particular map. Now, this is the collection of the different SL2Z orbits. And we'll be delving into one specific SL2Z orbit. And so we'll be interested in what's happening inside of this orbit and what kind of structures we can find there. And people have found a lot of amazing structure there. For the purposes of these lectures, we'll only need very basic concepts. And a notable one is the notion of reducedness. And so we call, we try and identify distinguished representatives for these SL2Z equivalence classes inside of an SL2Z equivalence class. And we take these reduced forms, which when D is negative, is defined by this pair of inequalities where B is non-negative, if either inequality is an equality. In terms of the first root we defined last time attached to a quadratic form, this condition is equivalent to saying that the first root is in the standard fundamental domain for SL2Z in the upper half plane. Now, more interestingly, and that's really what the case that we'll be looking at, when D is bigger than 0, we will define a quadratic form to be nearly reduced if A and C have opposite signs. If A times C is less than 0. And we'll say it's reduced if, in addition to that condition, also B is greater than A plus C in absolute value. This is a little bit different from the condition of reducedness that you usually find in the literature. So Gauss and many other people like him would have defined reduced forms by the following pair of inequalities. So that square root of D minus twice the absolute value of A should be less than B should be less than the square root of D. This condition here looks, well, I think, a little bit simpler. You don't really see it in the literature. I don't really know why, but they're entirely equivalent. And this is a problem that you find in the lecture notes. Yes, absolutely. But it won't necessarily be in the same equivalence class. So you can apply the matrix S, which changes A and C, but it also negates the sign of B, and that violates the second condition. So that's why it's not quite symmetric. Yeah, that's a great question. So in addition, I also threw this thing at you without explaining it too much. But I saw in the exercise sessions that a lot of you were getting deeper into this topograph. So I wanted to quickly mention it again. So the topograph, maybe it's not for you. You might be perfectly happy with the algebraic approach. It's a way to conveniently visualize the entire SL2Z orbit of a single quadratic form. And so this topograph is a tree, and the quadratic forms in a SL2Z orbit, they correspond there in one-to-one correspondence with the oriented edges of this tree. So for instance, if I take the middle, the left picture, the oriented edge going up, right in the middle, I can read off the coefficients of the quadratic form. A is on my left, that's one. C is on my right, that's two. And B is going to be the number in front of me, which is three, minus the sum of the numbers next to me. So that's three minus one plus two, which is zero. That's the middle coefficient of this. And so on the right, we have an indefinite form. And an indefinite form, of course, represents both positive and negative numbers. So there's a boundary between the regions where it's positive and it's negative, and that's what Conway calls the river. And so the nearly reduced forms are precisely the forms that correspond to oriented edges on this river. You can think also about what reducedness translates to in this visual picture of the topograph. I'll give you a hint. It has something to do with the side rivers branching out either to the left or to the right. And when they switch sides, you'll have precisely one reduced form at that edge. OK. So today, we're going to start talking about co-cycles. And I'd like to begin by making a couple of definitions. So these definitions are quite formal, and they look a little bit strange, maybe, the first time you see them. But bear with me. There is a payoff, which I'll get to next time. And hopefully, I can give you a little glimpse at the end of today's lecture if I get that far. So we'll consider the additive group of rational functions. So CZZ will be the name of the variable that I'll consider. And I'll make that into a module over GL2Q. I'll use left modules throughout today's talk. If you prefer right modules, you just have to make your own adjustments. A matrix ABCD will act on a rational function by the weight to action. So that's the determinant times this factor minus CZ plus A raised to the power minus 2. And then F, where the argument is translated by a linear fractional transformation by the adjoint of the matrix that we started with. So this is the usual action that you see also for modular forms, except in most textbooks, traditionally, this is done with the right action. So everything is the same up to an adjoint. Now, what we'll want to consider is the rational co-cycles. So it's a group of rational co-cycles, by which I will mean elements of Z1, so one co-cycles, for the group SL2Z, valued in CZ. So that's this additive group of rational functions on P1C. And if you are unfamiliar with this notation, let's give a very concrete characterization of such elements. Such elements will be cross-tomomorphism. So they will be maps from SL2Z to CZ. Not just any maps. These maps have to satisfy this identity for all gamma1 and gamma2 in SL2Z. It should be true that if I take phi of the product, it is phi of gamma1 plus gamma1 acting in this way that I've just defined on phi of gamma2. So these will be the objects that I'll want to consider. Of course, I can make any definition I like. And since this is not something you may have encountered before, let me give you an example. Now, let's start with an example that's not terribly interesting, but it's still very instructive to work through. Especially maybe if you want to implement all of the knowledge that you've gained throughout this workshop, these things are very nice to implement also on a computer and to play around with a little bit to get a feeling for how they work. Or by hand, whatever you prefer. So here's a toy example. The definition of this toy example depends on an auxiliary choice of a cusp. So a cusp for me will be an element of P1Q. So it's a pair of rational numbers well-defined up to multiplication, so up to a scalar. And here's the definition of this co-cycle. So I'll call it P sub C. I didn't really know what a good name for it would be. The sub C indicates the dependence on this cusp C. And it's a map that sends a matrix gamma to the difference of two rational functions. So I have my matrix gamma here and I send that to the difference of these two rational functions where that rational function is defined in this way. So you see it's a rational function that takes these coordinates, R and S, and it puts S in the numerator and SZ minus R in the denominator. Of course, R and S were well-defined up to a scalar, but the definition of LC is independent of that scalar. Okay, bizarre definition, but it turns out that this is a co-cycle. There is also this additional choice of a cusp that went into the definition, but it turns out, and this is also very easy to show, that it only changes the result up to a co-boundary, yes? So this co-cycle doesn't really depend in an essential way on C, only up to a co-boundary it does. So we understand very well the ambiguity in making this choice. Now, why is it a co-cycle? I can claim this. This is also one of the exercises. There's lots of exercises on these examples in the notes. I'll give you three ways to check that this is a co-cycle. So the first is by a direct calculation. This is your preferred method. There is no shame in that. Just work out the identity, verify that it checks out through a calculation. The most direct and seems very satisfying approach to showing that this is a co-cycle, but we'd like a little bit more because we'll identify also other co-cycles where these calculations become so horrible that we probably don't want to do them anymore. Now, some of you in the audience will know that this is a co-cycle without doing this calculation, and they'll know they'll prove it by memory because they'll remember when they took a course on modular forms and they defined Eisenstein series, that when you define Eisenstein series in the usual way and you specialize to weight two, it's that you don't have absolute convergence anymore. And when you have a lack of absolute convergence, the transformation law doesn't let you rearrange things in the usual way. And you find that E2, the Eisenstein series of weight two, is invariant under the action of gamma up to this error term, which is precisely the co-cycle that I've defined for you. So more precisely, so E2Z, if you slash that, and here I'm using the right action that you usually find in the literature of weight two, by gamma minus one, if this was an honest modular form, you would get zero, but you don't get zero. You get, in fact, the co-cycle P infinity gamma inverse, where infinity is a particular choice for the cusp, so one comma zero, and the gamma inverse is there because I have a right action on the left, so I need the right action on the right, but I had a left action in the definition, so that's why the inverse is there. That was a lot of lefts and right, so I hope that wasn't too confusing. From this formal identity, it follows entirely, formally, that this is a co-cycle without doing any laborious calculation. Then there is a final way to check that it's a co-cycle, and this is really the way that we'll prefer, but I have to make a small definition first. So if M is any SL2Z module, a modular symbol, usually people in the literature have slightly less stringent conditions than the ones I'll put, so they might call this an invariant modular symbol, it's going to be a map that takes a pair of cusps and sends it to an element in this module M. But it's not just any map, it has to satisfy some conditions, and I apologize for the notation if some people don't like this, but it's entirely traditional to use these set brackets for the arguments of a modular symbol. I don't know why, but this is how the literature works.  So M of r, s, a pair of cusps r and s, is going to be minus M, s, r. So it's anti-symmetric in the arguments S and r, the cusps that I take. Moreover, it's additive in the arguments, and that's what this condition here is a very important condition. And here at the bottom says, if I go from r to t, if I take the pair r and t, or I take the pair r and s and the pair s and t, and I add them up, I get the same answer, the same element of my module M. Then the condition on the right, that's the one that indicates the invariance, that says that this modular symbol interacts well with the action of sl2z that I have both on M, because that's a sl2z module, and on the cusps, because there we already have an sl2z action, and I want this to be equivariant in precisely this way that's on the slide there. Very strange. You may have encountered many modular symbols in your life. Now, what's convenient about these modular symbols is that if you have such a map, you have such a modular symbol that satisfies these axioms, and you take any cusps c, then the map that sends gamma to M of c, gamma c, is a one-coh cycle. This is very easy to check. So if we could only identify this PC as a special case of this, if we could construct some modular symbol that does precisely this, and that is obviously additive and anti-symmetric and gamma invariant, you wouldn't have anything to prove at all. So I'll leave that on the slide, and it's an exercise to make this a reality and to check that this is a co-cycle in this way. Whichever method you prefer, but the third one's the best. Okay. Now, besides this toy co-cycle that we had, which came from a Wait 2 Eisenstein series, we won't have much use for this co-cycle. We'll actually be interested in other examples that live inside of this group of rational co-cycles. And they were constructed for the first time in the late 70s by Marvin Knopp. Now, the definition's a little bit involved. I want to take some time to walk you through it. And the first thing we're going to do is just like before, we're going to choose a cusp. So c is going to be an auxiliary element in p1q. And in addition to that, there's an additional piece of input, and that's going to be a hyperbolic quadratic form. So I'll take a discriminant that is not a square, and I'll take f to be the element of the set fd that I defined. So it's a quadratic form of discriminant d in particular, and that will be part of my input for the co-cycle. Okay. So what is the co-cycle? Here it is. So let's take some time to parse this definition. So I'll denote it by kn sub cf. The sub denotes, so it reflects the dependencies here, and the kn, I guess, refers to Knopp. It sends a matrix gamma to a sum where the sum is indexed by all of the elements in the sl2z orbit of f. That's the tilde notation I had also last time. So q in the sl2z orbit of f is another way of phrasing that subscript of the sum. So a priori, it's an infinite sum. It's an infinite orbit. But what am I going to sum over? I'm going to sum over a rational function whose numerator I have to explain entirely to you. So let's put a pin in that and return to that in a second. And the denominator, actually, we can already appreciate what that means. So z is our variable, and rq is the first root of the quadratic form q that we're summing over. So I'll remind you the first root last time of a quadratic form abc, we're defined by minus b plus square root of d divided by 2a. Now the only mystery is in this numerator. What is this complicated sign c comma q of gamma? Here's what it is. The numerator of this expression is going to be, as the name suggests, a sign. So it's going to be minus one, one, or zero, depending on the following conditions. So if I take my quadratic form and I evaluate it at the cusp c, which was my auxiliary input, sorry, and that's less than zero, and that, in turn, is less than q of gamma acting on c, then I return minus one. If I get that they have opposite signs, but the inequalities are in the other direction, then I get plus one. And if neither of these conditions are satisfied, I just return zero. So that's what this sign function is. Now it's a lot to parse. And of course, everything is entirely algebraic, but there's also maybe a more visual way of thinking about this. And it's to have this picture of intersecting geodesics in mind. So what you can do is when you have your quadratic form q, it has two roots. It has a first and second root, which we denoted by rq and rprimeq, respectively. If I draw those on p1 of r, and I have the upper half plane sitting above it, I get a geodesic from the second root to the first root, which I've denoted here, I've rendered here in red. So it's oriented from rprimeq to rq. That's a geodesic. It's a semicircle. I can do the same with the cusp c and make the geodesic to gamma of c, the gamma translate of c. And that gives me a second geodesic. And I'm going to choose the usual right-hand orientation on the upper half plane. And I'm just going to compute the intersection number of these two geodesics.  And if it's pointing up, then I get a one. And if it's pointing down, I get a minus one. And if they don't intersect, I return zero. So it's an intersection of two geodesics defined in this way. Okay, now what a bizarre definition, right?  And the first time you see it, you're like, whoa, what's this? But then the fungus, it grows and you kind of learn to appreciate the fungus and you get used to it. And ultimately, you love the fungus. And so this is, this co-cycle, it takes some time to parse. But I think the best way to engage with it is to really verify all of the properties that I'm about to put on this slide. And this is also in the exercises. And try and do this in the optimal way. That really gives you some insight as to why this definition, why on earth you would make this definition. So the first thing you would have to check is that it's well-defined, right? I'm trying to map to the rational functions and I've written an infinite sum there so that sum had better be secretly a finite sum. Meaning that the sine condition is only non-zero for all but finitely many terms in this sum. Okay, that's the first thing you need to check, which is in the exercises. Once you've checked that and you know that the map makes sense, you have to verify that this is a co-cycle. And I've given you three methods to do that so you can choose your favorite method to try and verify this. Either you do this by a direct calculation, as you see sometimes done in the literature, but you might have quite a bit more gray hairs at the end of this workshop if you choose this method. But you know, it's your right to do that if you choose to do that. The second method, constructing this modular integral, so finding the analog of the E2, that's probably the hardest. You should save that one for last. In fact, Yukima, Mogul and Toth have an annals paper where they construct this form so it's a pretty difficult problem. The last one, however, so the method using modular symbols is I think by far the best method to check that this is a co-cycle because it gives you some insight as to why on earth this definition works. Why on earth this miraculous expression is indeed a co-cycle. And it comes from the additivity of intersection numbers of geodesics. So the crucial thing to check is that when you define the modular symbol, it's additive. And if you have three geodesics between R and S, S and T, and then R and T, and you intersect with some other geodesics, then the intersection numbers are going to add up between the three. So if you intersect one, then you have to intersect another with the opposite sign, so they have to cancel out. And that's kind of the idea. I don't want to give away the exercise, but that's really the idea to check and it gives you some insight as to why this definition works and what makes it tick. Once you've checked that it's a co-cycle, all the other properties are easy to check. So like before, the auxiliary input of this cusp doesn't really matter much. So in fact, if I choose a different cusp, I will only have changed the answer up to a co-boundary. So another way of saying that is that the co-homology class is completely well-defined and independent of the cusp scene. Now finally, since we know that the choice of cusp doesn't really matter all that much, a natural choice to take is the cusp infinity. Yeah, why not? This seems to be the most obvious choice. If we do that, we get a co-cycle, KnoP F. And a co-cycle, it's a lot of data because for every matrix, it's a rational function and it satisfies this co-cycle condition. Now SL2Z has two very convenient generators, which I think we put also on the board last time. I had some big chalk, which here it is. So SL2Z is generated by the matrix S, which is minus one, one, and then zero is on the diagonal. And then you also have the matrix T, which is a translation matrix, one, one, one like this. And if I know what the co-cycle does on these generators, I can figure out through these identities what it is on any matrix. So it's enough for me to tell you what the co-cycle does on these generators. If I choose the cusp infinity, then I see that gamma, so c and gamma c are the same for the matrix T. The translation doesn't change the cusp infinity. So this intersection number is always zero. So the matrix T just gets sent to zero. That's what parabolic means. Parabolic here, I put an orange, that means the same as that condition there that the co-cycle vanishes on T. By the way, whenever I choose the cusp infinity, I'm going to drop the subscript c from the notation to make it a little bit lighter. So when you don't see the c in the notation, that means I've chosen the cusp infinity. That's my canonical choice. Now on the matrix S, however, this is not going to be trivial, otherwise this wouldn't be a very interesting example. On the matrix S, if we translate these conditions, we see that we go from infinity to S of infinity, which is zero. So that's a geodesic that the black geodesic is just going down. It's like a vertical line. And then this condition of intersecting is entirely the same as saying that the roots have opposite signs. And one of them is positive and the other is negative. That determines the sign of this thing. So if you plug everything into this definition for the matrix S and the choice of cusp infinity, you get precisely this, where you see that the sum now is replaced by Q inside of the set of nearly reduced forms. So these distinguished elements that we found in the SL2Z orbit, that's what will index over. That's a finite set, by the way. So here we see this is indeed finite, which was something you had to check for every gamma, but for this particular gamma being S, this is indeed finite. I've defined it here on the slide again. So the sigma F is a set of nearly reduced forms. It's A, B, C in the orbit of F, such that A and C have opposite signs. Okay, so this is the rational function that we get sent to. And of course, last time I gave you a reduction algorithm, and I know that some of you have been computing some of these nearly reduced forms in a given orbit, you can compute this. So in any given example, you now have the power to find out what this rational function is. The only thing that's not clear to you yet is why on earth you would do this. So that's what I have to deliver on, and next time will be the big payoff of all of these definitions. Okay, all right. Any questions about this before I move on? Yeah, that's a great question. I mean, I never met Knopp, so I can't be sure, but my impression is that his paper was entirely curiosity-based. So his starting point was the theorem of Eichler-Schmura, which gives a way of thinking of modular forms in terms of their period polynomials. So elements also in these cohomology groups, but with values in spaces of polynomials with a very similar transformation law but where the weight has the opposite sign. And so what he did is he said, we look at these cohomology groups to study modular forms. That's what Eichler-Schmura tells us to do. And if I naively change the weight to a negative one, I don't have any modular forms anymore, but I can still look at these co-cycles. And of course, polynomials doesn't make sense anymore with a negative weight. He then had to change the rational functions and he wrote down lots of these examples. So as far as I can tell, that was the main motivation, but I can't be sure because I never knew Knopp. Now, let me give you a very quick, let's see how I'm doing on time. Okay, a very quick example. So when the discriminant D is five, there's a unique SL2Z orbit of quadratic forms. F being one comma one comma minus one is an element in there. And the set of nearly reduced forms, you can very quickly work out, you don't have to program anything. Just from the conditions, you see that everything has to have absolute value one. And so you have a finite number of possibilities and this is all of them. So there's four elements in this set of nearly reduced forms. And so the Knopp co-cycle, okay. Okay, this is true up to square root of five. I forgot that the right hand side should be multiplied by square root of five. I apologize for that. I'll correct that on the slides. Anyway, it doesn't really matter. It's a scalar. The essential part of the rational function is there on the right hand side. It's a very nice rational function. And it happens to be the value of this co-cycle at S. If you make this computation, of course you wanna check that you did this properly. You know, this really works. And of course, the fact that it's the value of a parabolic co-cycle at S puts some really strong conditions on this particular rational function. So you can think about this as also an exercise. It translates because of course the S and T have various relations existing between them. And so if you go back to the presentation of SL2Z in terms of SLT with the relations that they satisfy, that translates into the following transformation laws for this particular rational function. So they're really strong. That's what I'm trying to say. You have to satisfy really both of these transformation laws. So the first is, if you hit it with the left action by one plus S, so by one you add the action of S, you get zero. The same you do when you hit it with the matrix ST, which is of order six in SL2Z. And you do just with the power to the zero power, the first power and the second power, when you add them all up, you also get zero. So these are very strong conditions that these things must satisfy. And if you check those, for instance, once you have computed your example, you're very confident that you got it right. In fact, any rational function satisfying these two identities is the special value of a co-cycle at S, of a parabolic co-cycle, I should say, at S. OK, so that's an example. I just wanted to briefly mention, so you can forget everything on this slide, just as for a little bit of extra motivation, because I know that these definitions look a little bit and so what Jordan was asking, I want to kind of maybe continue along that thread because there's another motivation for considering such co-cycles, which came much later. And that's in fact where we learned about the existence of these co-cycles. It comes from this recent work of Duke, Imamoglu and Toth on linking numbers of modular geodesics. It's a beautiful paper, and what's happening in that paper is the following. So you can take SL2R, and it's a nice, it's a three-fold, you view as a three-fold and you quotient out by the arithmetic subgroup SL2Z. There's a very classical result that says that this three-fold is diffeomorphic to a three-sphere with a truffle not removed. And so in this three-sphere with a truffle not removed, there's a very nice flow on this three-fold and it's called, it's a so-called diagonal geodesic flow. Now, I'm not gonna use it in the lectures, this is entirely just to show you that these co-cycles are good for something, before I tell you that they're good for something else as well, but initially this is where we learned about this. Whenever you have a hyperbolic matrix in SL2Z, you can associate a canonical knot inside of this three-fold by following this diagonal geodesic flow. Doesn't really matter so much what it is. I've put the definition on the slide, so you start with your matrix, gamma, you diagonalize it with some matrix G, that's what this thing here on the right is, so you diagonalize it with this matrix G, and then you put this matrix G here and you consider the coset of SL2Z of G times e to the t, e to the minus t. Now you see that when t is zero, and t increases all the way to the logarithm of the first eigenvalue of this matrix, you can flip them around and you see that you end up where you started, so you get a periodic orbit this way, as you get a closed loop inside of this three-fold. And so if you see this picture, a very natural question to ask is, what's the linking number between this hyperbolic knot and the trefoil knot? And this is a question that was raised and answered by Etienne Guisse in his ICM address, I think in 2002, I wrote it, maybe 2006, I'm sorry, I can't remember exactly, I think, yeah, maybe it was 2006. He raised and answered this question, a very, very nice talk that he gave, and he showed actually that the linking number is encoded in the logarithm of this toy co-cycle P that we started off with, this innocuous-looking thing that came from the lack of invariance of the Eisenstein series of weight two. So the logarithm of that keeps track of this, you can squeeze out the linking number in a very concrete way that I won't tell you about now. Now, a very natural follow-up question, which is a question that Duque Mamoglo and Toth asked, is forget about the trefoil. If I take a pair of hyperbolic matrices, I get a pair of knots in this three-fold, and I can ask what their linking number with each other is. And the answer to that question, what the linking number with each other is, seems to be encoded in precisely the same way as in the work of Gis, but using this much more involved Knopp co-cycle. So this is one other instance of where this Knopp co-cycle was used for something quite productive, and that's where we learned also about its existence. So I just wanted to mention that in both cases, the procedure for getting from the co-cycle to the linking number involves some integration procedure, and that's precisely what we'll do next also. Now, the integration procedure that we consider is quite different from the one that they consider. We have a multiplicative integration. They have an additive one, but nonetheless, there are many similarities that happen between these two works. Okay, so let me tell you back to, after this motivational slide, we had our additive co-cycles. We have two interesting examples. Well, in fact, we have infinitely many interesting examples. We had the toy co-cycle, and then we had all the Knopp co-cycles for all of the hyperbolic quadratic forms that we could take. What we'll do now is we'll consider also the multiplicative group of non-zero rational functions on P1. These are elements in the function field of P1 of C that are not zero, and they are a left GL2Q module for the weight zero action. So here's the action. So it's just acting by the adjoint on linear fractional transformations on the argument of the function F. Now, the connection between the additive and the multiplicative co-cycle is furnished by the following map. It's what's traditionally called the logarithmic derivative map, and it takes one of these multiplicative functions and it maps it to an additive function by sending a function F to its derivative divided by the function itself. Now, this is a morphism. Multiplication turns into addition, and it also respects the action of GL2Q that I defined on both sides, the weight zero on multiplicative, weight two on additive co-cycles. Now, the kernel of this map is precisely the constant functions C star, and note something quite funny is that both of the examples that we wrote down are valued in the image of this logarithmic derivative map, meaning that the functions we wrote down where we send gamma to, they all have a very specific shape. They're always of the form of elementary factors that have a simple pole, and so therefore they naturally arise in the image of this. In fact, I can write down how they're in the image of this completely explicitly. So if I map gamma, so this function L, if I map that to Z minus gamma infinity to be my cusp, again with the usual conventions of this L function that I had last time, and I can do the same with the Knopko cycle where this sum that I had is now changed to a product where the numerator in these elementary fractions, which was always plus or minus one, has now become the exponent of these factors. And just entirely formally, I haven't done any mathematics here, if I apply the D log to this product, I get precisely the sum that defines for me the Knopko cycle. So it's entirely formal to see that the expressions that I wrote down allow for such a lift for the logarithmic derivative map. As a consequence, these particular maps that I've written down here are actually co-cycles, but they're only co-cycles modulo scalars, because that's the way the D log map works. It has a kernel, which is scalars, and so if I write down this formal map, I know that it's going to satisfy the co-cycle conditions up to some scalar. Okay? Can we do better? That's the question. Can we get rid of this scalar ambiguity? I know that co-cycles up to scalars, can I make them actual co-cycles, valued in Cs at star?  That's a natural question to ask. The answer is yes, and the reason is that the group SL2Z is not so rich, comologically speaking. So here's a little lemma. I have to explain what this lemma means. So let's start on the right-hand side. These are co-cycles, so forget the 12 for the second. These are co-cycles, the subscript denotes parabolic co-cycles. So those are co-cycles that vanish on the group of translations inside of SL2Z. They're co-cycles for SL2Z, valued in Cs at star over Cs star. So those are the objects that we have. We would like to lift them to actual co-cycles, valued in Cs at star. Now of course, you have a natural map from Cs at star to Cs at star modulo scalars, just by taking the co-cycle modulo scalars. And all of the elements, all of the co-cycles that become parabolic modulo scalars, I'll denote with this subscript F on the left-hand side. So the existence of the map is really the definition of this subscript F. So it's all of the co-cycles that become parabolic modulo scalars. So if I do that, if I take that projection map, so I forget the scalars, I take everything projectively, that's actually an isomorphism, or at least it's an isomorphism if I remember to raise everything to the 12th power. So if I multiply everything by 12. So this lemma really is telling us there's no real difference. If I have a parabolic co-cycle, it actually lifts uniquely after I take the 12th power to an actual co-cycle with values in Cz. So I'll leave the proof as an exercise. I'll give you the idea though. So the idea is to start with, oops, with a short exact sequence, which I think no one will object to, and passing to the associated long exact sequence in co-homology, so group co-homology for the group SL2Z. And that shows you that the group that we actually want to be a part of maps to the group where we found some classes defined by the co-cycles that we wrote down there. So this is modulo scalars. And this maps in turn to an H2 of SL2Z with values in C star. That's not good. Did that make a huge racket? No, okay. Can you still hear me? Okay, great. Sorry about that. Now there's a kernel also to this map. And this comes from the group H1 SL2Z C star. So to prove this lemma, the first thing you're going to do is to show that this group is zero. And this group is torsion. So it's cyclic of order 12. And you can use the usual presentation of SL2Z to prove this. So this is the first part of this exercise. So that shows actually that once you multiply by 12, you get an isomorphism between these co-amology groups. Now the lemma is stated in terms of co-cycles and not just their co-amology classes. So you need an additional ingredient at the end. And that is that you see, when I have a parabolic co-cycle and I look at its co-amology class, I lose some information. So but if I have a co-amology class that has a parabolic representative, it must have a unique such representative. And that's because there are no co-boundaries that are parabolic. Because co-boundaries that are parabolic, they would correspond, they would come from actual functions, rational functions whose divisor is invariant on the translation. And no such things exist besides constants. And therefore the co-boundaries must all be trivial. So that's how you prove this lemma and hopefully that gets you going to actually solve this exercise, okay? Okay. So with this lemma in hand, I can take my two examples that I have and I can lift them multiplicatively, at least after I take them to the 12th power. So I take these co-cycles that I have there, modulo scalars, I raise them to the 12th power, I know I get a unique lift and those unique lifts, I'm going to call P upper times and KnoP F upper times. So they're elements, they're co-cycles for SL2Z valued in CZ star and they're parabolic modulo scalars, even though they're not necessarily parabolic themselves. And neither of them are in fact. Okay, no, that's a great question. Of course you don't really need to take the 12th power. I'm doing it here to get the uniqueness statement, but in practice actually we don't do this. We're happy to live with this ambiguity of 12th roots of unity in practice because in practice later on, we'll try and recognize algebraic numbers that are absolutely massive. So we want to keep them as small as possible in height and therefore we often don't take this 12th power. But for theoretical purposes let me just take it here to make the lift really unique. So everything is really now canonical with this 12th power. Otherwise, all these choices of lifts, I mean the lifts still exist because the H2 vanishes, but the choices of lifts, it's like a principal homogeneous space for this H1 of SL2ZC which is cyclic of order 12. So you can only change it really through homomorphisms from SL2Z to C of which there's a finite number. So if you're happy to live with that ambiguity, you don't have to take the 12th power. So that's a great, great common. Okay, we're almost there guys. So we started with our additive co-cycles. We lifted them to multiplicative co-cycles. And now what are we going to do with co-cycles? I mean we started off this whole story with the J function and the arguments of the J function at CM points. And if you have a function, it's clear how to talk about values. But now I'm taking co-cycles. What's the value of a co-cycle? A co-cycle is a collection of functions. So what do I do to produce an actual number out of a co-cycle? And this is what we do. So we're going to define a mechanism. It's a very simple mechanism. It looks a little bit bizarre, but it's really simple to evaluate a multiplicative co-cycle at a hyperbolic form of non-square discriminant D that is positive. So let's call that form G. So G is going to be an FD with D greater than zero non-square. And a nice feature that such matrices, that such forms rather have is that they're stabilizers in SL2Z or actually infinite cyclic modular torsion. So there's always plus or minus one. And then you also have this free part. And the free part, there's a canonical generator for it, which people usually call the automorph of G. So the automorph of G is defined by this identity here. So gamma of G, which has entries that depend on the coefficients. I didn't write this. I should have written A, B, and C are the coefficients of G. I forgot to write this. A, B, and C are the coefficients of G. And then there's also this T and U. And T and U are going to be solutions to the Pell equation. Pell equation T squared minus D U squared equals four. And in fact, for any solution to the Pell equation that you take, that matrix gamma G is in the stabilizer. And if I take the minimal solution, let me take the minimal positive solution to that Pell equation, I get actually a canonical generator for this stabilizer. And that's what people usually call the automorph. So solving Pell's equation is entirely equivalent to finding the stabilizer of this form G in SL2Z. And in fact, it's also equivalent to applying the reduction algorithm that we did last time. So you'll quickly find that when you compute this, I mean, you can try and solve the Pell's equation or you can work with these cycles of nearly reduced forms. So once you have a reduced form, you can keep applying the reduction algorithm to get more and more reduced forms. But there's finally many, so eventually you look back onto yourself and the matrices that you used in the process to make this transformation will then be precisely this automorph. And if you up to a power plus or minus one, I guess, but if you do it the right way, you get it on the notes. And so that algorithm, of course, in turn also is again equivalent to finding the continued fraction expansion of the first root of G. So all of these algorithms are essentially all equivalent and you have the tools now to find these things in practice, okay? Now, why am I telling you this? I'm trying to evaluate a co-cycle. Oh, yes, question. Sorry? And non-splitjust means the discriminant is not a square. No, so it doesn't factor as two linear forms. Okay, now, back to our co-cycle. We take the co-cycle phi, which is a multiplicative co-cycle, any multiplicative co-cycle. The value of this co-cycle at the form G, it's a co-cycle, so it eats matrices and it spits out functions, yeah? The matrix that will feed it is precisely this automorph that we just computed. So once you feed it the automorph, it spits out a rational function and that rational function, I'm going to evaluate at the first root of G. Outcomes a number, yes? If you consider infinity a number because of course you're taking a rational co-cycle so it could happen that you evaluate at a pole in which case you get infinity. So you get a number or infinity, so a nice element of P1 of C out of this construction. Now, this looks a little bit bizarre and you have to kind of take my word for it that this is a bona fide operation. One clue that you get that this definition might be a good definition is that in fact this value that I've just defined for you is independent of the choice of G in its SL2Z equivalence class, yeah? So if I change it to a different G that's equivalent to it, that doesn't affect the value. That's already kind of nice. It has a good canonical feature that doesn't depend on this particular representative G that you chose and that's in fact also why I chose to put this notation here with the square bracket, you see, to evoke the sense that it's only the equivalence class of G really in the Picard group that matters for this value. Okay. So for the examples that we have, and this is in the exercises, you can try and work out what are these values of these rational co-cycles that we wrote down. And the answer is that for the toy co-cycle what you get out is the 12th power of the fundamental unit in the order of discriminant D. Of course, the fundamental unit is only well-defined up to science and then taking it inverse, but it's precisely the one that corresponds to that choice of the Pell's equation that I made in the automorph. The very canonical element inside of that order. Okay, so something vaguely arithmetic, not terribly interesting yet. For the co-cycle you can do the same thing and here I'm not really sure what to say about this number other than that it's a number in the field that you get by joining the square roots of D1 and D2. So D1 is the discriminant of the form F that I use to define the co-cycle and D2 is the discriminant of the form G that I use to evaluate that co-cycle. If these discriminants for instance are co-prime, which is the generic case that we'll be interested in, this is a biquadratic field and what I've done here is I've defined a canonical element of this biquadratic field that depends only on the class of F, the SL2Z class of F and the SL2Z class of G. That's about all I can tell you about this. I'm not really sure what else to make of this element in this limited setup of rational functions. That's something to think about I guess. But anyway, that's where it lives. So in summary, nothing too terribly shocking has happened. We've made a bunch of definitions and I just wanna break it down before we move on. There were three steps. The first step is we constructed an SL2Z co-cycle valued in the additive group C of Z. Now interesting here is in quotation marks because that hasn't been apparent yet. We've just made this definition. It looked like a vaguely interesting, miraculous looking co-cycle but why it's interesting remains to be seen. The second is to take that additive co-cycle, lift it to a multiplicative co-cycle which we could do in a unique way because SL2Z had such limited cohomology. The final thing that we do is after we take the multiplicative lift, we evaluate it at a second quadratic form or a form G with discriminant D and the discriminant here has to be positive and out comes a number. So a co-cycle can be evaluated at such forms but nowhere else. So this evaluation procedure is very unique to these real quadratic singularities. So these G's of this particular form. So those were the three steps that we went through. Okay, now we've constructed an element of a bico-dratic field with this knob co-cycle. It's not so clear what this element is good for or if it's even good for anything but the motto and that's what we'll see now is that we will see that very specific periodic limits of these values actually converge to tremendously interesting algebraic numbers that exhibit a lot of similarities with the differences of singular moduli studied by Gross and Zagier. So this I'll try and convince you of next time. I first have to tell you how these periodic limits work but this is the kind of motto, the rough version of what I'm about to tell you. We'll identify very specific periodic limits of these numbers that converge to interesting looking limits. Now the idea is this is the setup we've been operating on. We've been looking at the action of SL2Z on CZ star. It was a left action that we defined by weight zero and outcome these objects. What we'll do now is we'll enrich this situation considerably by replacing this setup with a much richer group and a much richer module. The group that I'll choose is a group that was considered by Ihara in the 60s and of course by many other people but there's a good reason for singling him out in this work because a lot of the inspiration came from there. It's the group SL2Z1 over P. Still has a kind of a global flavor. I haven't completed anything. I've just inverted P in the arguments of my matrices and I'm going to make that act not on rational functions on P1 but on meromorphic functions on the piatic upper half plane. That's what I'm going to do. So I have to tell you what that means. So M star will be the multiplicative group of non-zero meromorphic functions on the piatic upper half plane HP. Now if you haven't seen the piatic upper half plane before let me just give you a quick idea of how it works and in fact this pedestrian definition that I'll give you now will serve you for all practical purposes. If you want to know more, it's a beautiful object about which you can say a lot. I think the best, if you're looking for a way to get into this topic of the piatic upper half plane which is good for many things, the best way in is probably these Arizona winter school notes from 2007 by Das Gupta and Titlebaum and if you're interested actually the travel library at the PCMI has a copy in the office so you can have a look there for the full story. So I'll only give you a very half baked kind of version of it and I'll say that the piatic upper half plane it's defined as a limit of HP less than or equal to N. So I have to define now for you what this HP less than or equal to N is. And HP less than or equal to N is defined by these conditions. So it's pairs of elements in CP in P1 so up to a scalar and I normalize it so that they're both integral in OCP and at least one of them is a unit. That's what primitive means here on the left. So Z1, Z2 primitive. Now I'll not take all of these pairs I'll take all the ones that satisfy an additional condition and it should be true for any CUSP R, S that I take which I normalize to be integral and co-prime these RNS so they're co-prime integers that define a CUSP in P1 of Q. And I want that inequality there to be satisfied for all CUSPs. So this is the piatic absolute value of SZ1 plus RZ2 has to be at least P to the minus N where N is of course the thing I'm gonna let vary that's the thing that will go off to infinity. So a more geometric way of thinking about this is that I'm taking P1 of CP and I'm taking away open piatic discs around all of the rational numbers in P1 of Q and the radius of these discs is gonna get smaller and smaller. So at step zero I'm gonna take away P plus one discs around P1C with radius one and then at the next step it's gonna be P times as many with radius one over P those and so there's more and more discs of smaller and smaller radius that I keep taking away to get to the actual piatic upper half plane. Now it's a rigid analytic space that's really the category that it lives in and this covering HP less than or equal to N is an admissible covering by affinoids. If you don't know what that means it's not so bad you can think of it just on the level of points the CP points of the piatic upper half plane are just the CP points of P1 where I take away the QP points of P1. And you can think by analogy and this explains also the name of this space if you take P1 of C, oh that's in the shadow, thank you, all right. No, no that's fine, it's very quick. So if instead of this non-archimedean field I would take C, so if you think of P1 of C minus P1 of R which would be the archimedean analog of that what you would get is sort of the union of the upper and lower half planes and so usually in the theory except when you forget to say that A is greater than zero and D less than zero so the thing I forgot to say last time that's the same as forgetting to discard one of the two half planes. So usually we work just with the upper half plane which contains the first roots of the positive definite quadratic forms and then the lower half plane contains the first roots of the negative definite root. So this is kind of why in the archimedean case you get the union of the upper and lower half planes. In the non-archimedean case when you take P1 Cp and you take away P1 Qp it's still connected so the upper and lower half planes are kind of blended together in some interesting way that you don't see in this non-archimedean setup. So maybe the upper half plane is a slightly bad name but that is what people use. So that's what I'll use also. Now here's a note. It contains points of Cp that are not in Qp. So they're not Qp rational. In particular it contains the first roots of the quadratic forms with positive discriminant I should have said. Well actually it doesn't matter I shouldn't have said that. So just quadratic forms with I guess non-squared discriminant even that I don't have to say. Just quadratic forms of discriminant D that satisfy the condition that this particular chronicle symbol is minus one. So that P is not a square in the so that P is inert let's say in the order associated to D. If you have this condition then you know that the first root is not gonna line Qp so it's gonna lie in this piatic upper half plane. Of course all of this is implicit with fixed choice of embeddings of Q bar both into C and into Cp which are fixed once and for all at the beginning without telling you. So it contains those roots and that's an important observation for later on. Now if you're more visually inclined you can think of this piatic upper half plane as a tubular neighborhood of again a tree not the topograph this time but Brouhat-Titts tree. The Brouhat-Titts tree is a P plus one regular tree whose vertices classify Homozi classes of lattices and Qp squared and there's a natural map from the upper half plane to this tree which gives you a convenient way of visualizing the piatic upper half plane. Again you can read more in this Arizona winter school volume about all of these details but it's a convenient way of visualizing it and so you see on this picture here I guess P is two in this picture I stole this picture from Mark Mazdeo it's a very nice picture. P is equal to two I guess in this picture and what's happening is that this is kind of HP less than or equal to two I guess in this picture. So you've taken a bunch of disks at the ends there and what remains is kind of this tubular neighborhood of the Brouhat-Titts tree so it's one step of this process and you can keep repeating it like a fractal which comes down to taking this limit and then the limit you get HP. Absolutely. So HP is the rigid analytic space let's say so it's a geometric object and HP of Cp is the Cp points of that space. Yes it's in fact defined over Qp but for us it'll suffice to just view it as a rigid analytic space over Cp it doesn't really matter. Yes precisely, absolutely. So you can ask for HP of Qp and that's empty. There's no Qp points in this space. There are however lots of Qp squared points where Qp squared is the unramified quadratic extension let's say of Qp and so in every extension we find lots of points but in Qp itself the set of points is empty. Okay. So this M, the meromorphic functions on this space again is something very concrete. What it'll mean for us is it's uniform limits of rational functions on this affinoid covering. So that means that if I restrict a meromorphic function to any of these affinoids it's going to be uniform limit of rational functions on that affinoid. And this should be true for any N. So that's the definition of it, of meromorphic functions. Okay so essentially you know what's coming now and this is what I'll end the lecture with. I've already announced all of these steps and we went through them in great detail on the special very concrete baby case of rational functions. Now it's the same story on steroids but with rational functions replaced by meromorphic functions on the Piadeck-Kaprahavt plane. So it's the same steps except everything is slightly richer. So the group now is richer, it's SL2Z1 over P. Again we take a cusp C and in fact let me just take infinity because why not? It doesn't matter all that much. I'm going to choose a quadratic form F with positive discriminant that is not a square mod P or at least the Kroniker symbol over P is minus one. If I have this the first step was to consider interesting additive co-cycles. So co-cycles valued in the weight two module of meromorphic functions defined in exactly the same way as before with the weight two action. So that additive co-cycle is going to be inspired by this Knopp co-cycle and this time it's going to be sum over the gamma orbit of F. That's what the subscript here is saying. So I'm letting gamma act on these forms.  But it doesn't really matter. So you can take this sum, it's still of course an infinite orbit, it contains the index of this set contains the SL2Z orbit but it contains even much more. What I'll do is I'll make this rational function in precisely the same way as before except the index set has changed here. And it turns out that when I do this I get a meromorphic function. Now it's very important to note that this sum here unlike what happened before is an infinite sum. That's very important. We have to be allowed to converge to something interesting, because we need infinite sums. The sum here is infinite, the reason that it converges is because the divisor, so the set of poles that I've introduced here, it's infinite but it's discreet in HP. So it kind of converges to the boundary and at every finite stage if I restrict to any aphanoid with a fixed N I get finitely many poles. So this thing makes sense only as a meromorphic function no longer as a rational function. The second step was to define multiplicative lifts which nothing new is happening. It's exactly the same as before. Formally I write down this infinite product where I just multiplicatively integrate everything entirely formally. There's one little subtlety that I'm glossing over here. In order to make this product converge, this converges already, but this product you have to be a little bit careful. So you have to, the factors that I put there you have to put a precise, well you have to put some constant in, a damping constant to make this infinite product converge to a meromorphic function. The reason I'm glossing over this is because it's entirely irrelevant what constants I put in because the only thing I'm interested in is the co-cycle modulo scalars. So I'm only assuring convergence and then the scaler I don't really care about so that's why I gloss over this. So that's the multiplicative lift of this thing. So again we wanna play the same game, we wanna lift it now to a co-cycle, valued in M star, not just modulo scalar ambiguity. Alas, you can never do that. So the group SL2Z was homologically quite poor but the group gamma is homologically much richer and the H2 that I have there on the left is decidedly non-zero. It is something you can appreciate, it's essentially isomorphic up to torsion to H1 of gamma zero of P with values in Cp star. So a group that is very closely related to modular forms. But the lifting obstruction is never going to be zero and that's life. So you might think, okay, game over then. I guess we're done, how do we define a value? I have to lift it multiplicatively first but the thing is it doesn't really matter. If you want to evaluate it at a form G that satisfy these usual conditions then we're just going to restrict this co-cycle to SL2Z where we know we can lift and we can lift uniquely after taking a 12 power and then evaluate at the automorph of this form gamma G which is in SL2Z, the standard copy of SL2Z inside of SL2Z one over P. So we want to evaluate it at some element which happens to lie in the standard copy of SL2Z inside of this group and so there's no lifting obstruction at all and that's precisely what we'll do. So we'll define this and I'll just finish with this. I'll say that after all of these definitions we have this quantity, you have a naive kind of algorithm for computing this using only what I've told you but that algorithm would be exponential in the required precision for computing this quantity. So in our paper we find an algorithm that is actually polynomial time in the required precision and I just want to send you off with one example and I'll tell you that next time we'll explore this further and further and further. So historically this was the first time we ever computed an example. This is the first one I could find in my records where we did this whole procedure with all these bizarre co-cycles, out comes a number and we recognize this number as 24 square root of minus one minus seven over 25. So an algebraic number has appeared, hallelujah. So next time we're going to explore them much further and so already I hope that this gives you some ideas to why all of these bizarre definitions make sense and we'll really dig into them and get their fine tune arithmetic in the next lecture next time. So thanks very much.


Hi everyone, Joe for JaspysCaseBreaks.com. We did it. We filled up that 2020 Topps Transcendent Tennis Hall of Fame Collection Break, Random Hit Break number one, a 34 spotter. Big thanks to all of these folks for getting into it. Thanks to these guys for buying their spots straight up. Remember, you get times 10 of those spots in the break credit promo which is happening after this. Big thanks to everyone who, and congrats to people who won spots in all those different breaks. So there's UFC, B.S.'s Bowman Sapphire, there's Immaculate, the basketball mixer, Salopax, what else do we do? The multi-year mixer, black gold football, Gold Rush Baseball, the tri-star break, Leaf jerseys, Prism Draft, so on and so forth. Thanks to the winners here and thank you for getting in that way as well. So here's what's going on. I'm going to open up the case that's right here and I'm going to keep all the stuff on there and try not to, I'm going to do my best not to reveal the cut auto and the one-of-one too early. I'm going to pull everything out and then I'm going to bring over the base card set, which I'm not going to open up, but I'm going to pull out the base card set and just kind of show it to you. I'm going to try to hide these. I'll show you this and then I'm going to pull out the sleeve of framed autographs and rip that open and as we're doing that, Nick Jaspie is watching as well and he's going to be using the Auto List tab and he'll be typing in all the autographs here. That way I can just quickly grab it, put it into the randomizer, then I'll randomize your names, then I'll randomize everything we pulled out, then we'll put it all on the hit list right here and we'll see what you get matched up with. Sound good? Thank you very much. I'm a big tennis fan, so I'm pretty excited about this, so let's see what we got. We got the knit cam expanded a little bit, cut away from myself, safety first, sponge thingy here, there's the case right here, a sponge on all sides, nothing else in here, empty box, just more sponge. We've got the invite right on top, so let's slice this open here. That's pretty cool, right? It's got a magnetic flap right here, gold top, more sponge. Oh, they covered them, nice, thanks guys. So the cut auto will be in the top spongy area and then right where my thumb is wiggling right here, that's where that one of one is. This is the base set, right, I guess I could bring it over here now. This is the bright high, multiple cameras here. So this is the base set which someone is gonna get the whole thing of and we're gonna keep it shrink-wrapped as well. There is the gold-framed auto set right here which we're gonna pull open right now. Let's take a look at this. Now obviously because of the times that we are living in now, there is a possibility that this event could be delayed. I don't think they're gonna cancel it, they'll eventually do it, but there's a chance at that. Let's take a look at this. So you're awake and open this up delicately. I'll just keep that sealed and I thought we were gonna be able to look at the invite, but all right well I'm sure it's nice. I'm sure there's all sorts of fun details there. Someone's gonna win all of this. Let's keep that right there. Now the autos. All right good luck everybody. First autograph, is that back or should I go the front ways? First autograph is gonna be, it's gonna be Andre Agassi. Nice, 21 out of 25. You know what, I'll top load all these a little bit later. I didn't bring enough top loaders. Let's just get through this. Next is, are they in alphabetical orders as AJ? There's Tracy Austin. Nice, 3 out of 5. So the Agassi was out of 25. The Tracy Austin's out of 5. The next one, so someone's gonna get a train with us out there. There's Boris Becker, 20 out of 25. Everyone's familiar with the transcendent sort of design by now. Nice gold frame. Nothing loose here. Good construction. Boris Becker to 25. That's pretty nice. Next is Bjorn Borg. 15 out of 25. If you're a tennis fan, these guys like Becker, Bjorg, Agassi. Pretty strong right there. Good photography as well. And supply, think about it. Supply and demand, ladies and gentlemen. There's just really not a lot of high-end tennis stuff out there. Next up is Michael Chang. Nice, 16 out of 25. I guess it's alphabetical.  But remember, there's 51 autographs in the auto list, but only 30 autographs. Who loves me? I'm assuming Jesus loves me. There's Michael Chang, one of the very speedy guy on the court. We've got Jimmy Connors, 23 out of 25. He's got a nice auto. Jimmy Connors, 23 out of 25. And there is Cliff Drysdale, 14 out of 25. Then we have Stefan Edberg, 17 out of 25. Stefan Edberg. Then we've got Chris Everett, 5 out of 25. Nice Chris Everett. I'm pretty sure she does a lot of commentary, TV commentary for the ESPN broadcasts. Then we've got Gigi Fernandez, 12 out of 25. Then we've got Steffi Groff, 6 out of 25. Is Steffi, Steffi Groff still married to? Steffi Groff is pretty incredible. Yeah, Steffi Groff and Audrey Agassiz still, still married, still residing in Summerlin, which is in Vegas. Nice. They have a couple children. They're not pushing them to be tennis players, although genetically they would be pretty well inclined. Next up is Justine Hine, or Justine Hennan, I think. 24 out of 25. I kind of remember her. Then we've got 2 out of 25. It's like a hockey name right here. Yevegny Kofelnikov. Got it. 2 out of 25. Then we've got Yan Codes, 7 out of 25. Correct. Not all autos are included. Jack, there are 50 autographs, like 50-51 autographs on the checklist, but only 30 autographs in the sleeve here. We've got Rod Laver. Nice. 9 out of 25. Nice Rod Laver. He's got a stadium arena named after him in Australia for the Australian Open. I think he's still pretty active too. I think they still kind of get him to do some public appearances and stuff. 15 out of 25. Amelie Merezmo. Then we've got, nice, Martina Navratilova. 10 out of 25. That's pretty strong. Next up we've got Manuel Orantes. 11 out of 25. Spaniard. Nice. Manuel Orantes. I don't know where they do this or where they don't. I went to Spain and I feel like they're just arbitrarily throwing the list.  It's like, when do you do it? I don't know. We just know. All right. 21 out of 25. We've got Charlie Passarell. Roger N. wondering how long it took them to construct each case. I'd love to see a behind-the-scenes video on them putting together a set like this. We've got 8 out of 25. We've got Dennis Ralston. 12 out of 25. Tony Roche. And there should be two one-of-ones in this sleeve right here. And it's Marat Safin. One out of one. He's a pretty strong tennis player. Nice one-of-one. So a nice train whistle hit there. Should be one more in here somewhere. Nice. There's Pistol Pete Sampras. 3 out of 25. And a one-of-one. Pam Shriver. Nice. 19 out of 25. Stan Smith. That's right. From Pasadena. Not too far from us here in Hermosa Beach. And a 2 out of 5. Margaret Smith Court. Yes. The base set has all of the cards. Correct. So you can tell that it's a little bit shorter than the autograph set. Well, I gotta find out who the one-of-one goes to, J-Dog, before I do the train whistle. 24 out of 25. We've got Michael Sitch. Few more to go. 8 out of 25. Helena Sokova. And we've got a 1 out of 5. 1 out of 5. That's Virginia Wade. And the last one. 11 out of 25. Mark Woodford. Nice. Alright. So let's stack that up. I'll top load all of these a little bit later. Now let's do the one-of-one. And then we'll do the cut. So the one-of-ones that bottom card right there where my thumb is kind of wiggling at. Good luck, everybody. Someone's going to get this. One-of-one right here. It's going to be Manuel Lorantes, the Spaniard. Nice. That's pretty cool. International Tennis Hall of Fame Class 2012 from Granada, Spain. Turned pro in 1968. And a Grand Slam title under his belt. Nice. And that 1952 design as well. This is really, what a good idea. Tennis was a great idea. Alright. Now the cut autograph. Or the cut, the cut auto, yeah. One-of-one. Cut auto. There it is. It's Dorothy Round. Dorothy Round. I must admit, I do not know who Dorothy Round is. Let's take a look. Well, we're going to learn together. She is, she was born in 1909, passed away in 1982. And she was an English right-handed tennis player, member of the International Hall of Fame in 1986, ranked number one overall in 1934. Wow. Won the Australian Open in 1935, won two Wimbledon's 1934 and 1937. Wow. That's pre-World War II. That's crazy. And semifinal in the U.S. Open, semifinal doubles, French Open finals, U.S. Open doubles, mixed doubles, third round in the French Open, and won it three times, won mixed doubles Grand Slam three times, 34, 35, and 36. That's pretty crazy. British tennis player active in the 20s until the late 1950, achieved her major success in the 30s. There you go. After her wedding in 1937, she played on her married name, Mrs. D. L. Little. During the Second World War, she played in North America, became a professional coach in Canada and the United States. Post-war, she played in British regional tournaments, coached and wrote on tennis for newspapers. So there you go. Dorothy Round, old school, old school tennis player right there. And there you have it, ladies and gentlemen. There she is. She's got a pretty big Wikipedia page. There you go. Nice. Now, big thanks to Nick, who typed in all 30 of the autos right here. So let's copy these names, put them into this list right here. So there's 34. There's the, once again, the base set, the one of one, the cut signature, the VIP invite, and then the autographs. Good luck to everyone here. Let's roll it and randomize each list four and a one five times. Your names first. One, two, three, four, and five. After five times, we got Karen down to Dan. Let's go to the hit list tab right here. Karen down to Dan. After five, four and a one. Good luck, everybody. Four and a one five times for the hits. One, two, three, four, and a one, right? And a one. Fifth and final time. Good luck, everybody. Thank you for getting in. After five, we've got the Yan Auto down to the VIP invite and the one of ones down there too. All right. So Karen with the Yan Codes Auto, Ian with Bjorn Borg. Nice. Boris Becker for Derek Molanson. Marco, you got the Helena Sokova. Pete Samper is going to Joe P. That's out of 25. Evan, you get the complete base set right here. Sealed. We kept it sealed for you. You get that whole set. Joe Andreessen with the Andre Agassi. Dan Tyson with the Michael Chang. You also have the Margaret Smith Court out of five autographs. All aboard the Big Hit Express. A.J. with the Gevny Kafehlnikov, autographed to 25. Evan with the Dennis Rawlston. Michael Jones with the Mark Woodford. Dom, you got the Chris Everett to 25. Elliott with the Manuel Orantes. We got the Brian Peeples with the Gigi Fernandez. Steffi Groff going to Arthur. Anthony with Martina Navratilova. Jimmy Connors for Joe Clanton. Derek Molanson with the Pam Shriver one of one. All aboard the Big Hit Express. Jack B with the Michael Stitch 25. Dusty, you got the Amelie autograph. And Jack with the Marat Safin, autographed one of one. All aboard the Big Hit Express. SKS with the Stefan Edberg. Scott with the Charlie Passerelle autograph. Aaron with the Tracy Austin out of five auto. All aboard the Big Hit Express. Aaron Pettit with the Tony Roche to 25. And Curtis, there you go. You got the Cut Auto. Old Dorothy Round. Tennis legend. Nice. Curtis. All aboard. Derek, you got the Rod Laver autographed 25. Nice. Curtis, you also have the Stan Smith auto to 25. Ed with the Cliff Drysdale auto. Steve Locke with the Justine Hanine autographed 25. Evan with the Virginia Wade autograph. And the one of one. All aboard the Big Hit Express.  You have the VIP event invitation. Are you going to go? Let us know if you go and let us know who ends up there. That's pretty awesome. And there you have it ladies and gentlemen. Let's take another quick look at it. There's the invite right there. We kept it sealed. I think Dan was thinking about going if he said he gets the invite. There's Dorothy Round. Nice cut autograph. Old school tennis legend. There's the bass set right here. There's the Manuel Orantes. One of one. And Dan, you are going to go? Nice. Let us know who ends up there and what kind of swag you end up with. And there's a quick look and we'll top load all these two, but there's a quick look at all of the cool autographs that were pulled. Folks, I think we're going to have one more case of this. So keep an eye on that on jaspyscasebreaks.com. Well, unfortunately, Joe, I don't think that's going to happen. Considering the licensing. Nice. There you go. Hey, everyone, thank you very much for getting into it. Is there anything else in here? Nope. Empty. That was the tennis. That's pretty exciting. It's kind of a different look at a different look, a different thing to add to your collection. So thanks for watching everyone. Thanks for joining all these breaks and making this happen. Joe for jaspyscasebreaks.com. I'll see you next time for the next one. Bye-bye.