 Okay, I'll start. So today's lecture is about Mathemic Periods, and it will be entirely mathematical. So classically, the way to think about periods are coefficients in the Duran isomorphism. So let me explain what that means. We have two functors, two cohomology theories, which are called Betty or Duram, B or DR, which go from the category of smooth algebraic varieties over Q, if you like, to the category of finite dimensional vector spaces over Q. Okay, so let's explain what these are. So Betty am cohomology is also known as singular cohomology, as we learn about in algebraic topology courses. So the i-Betty cohomology of x will be defined to be the i-Singular cohomology group of the complex points of x. So the complex points of x is a smooth manifold, so we can take its cohomology with coefficients in Q. This is singular cohomology. So it's constructed out of singular co-chains. And then we have algebraic Duram cohomology, which was defined by Kotendijk. So the general definition is that the i-Duram cohomology group is defined to be the hypercohomology of x with coefficients in the chief of regular differential forms on x, this is Kotendijk. In all the cases we'll look at today, x will be affine, in which case this simplifies enormously. And the Duram cohomology is the naive thing. So it's closed forms, modulo exact forms, except instead of working with smooth C infinity forms, we work with algebraic forms. And the miracle is that it actually gives the correct answer. So it's the cohomology of the following complex. So D is the usual differential of differential forms. So omegadot xq here denotes the complex of global regular algebraic differential forms, which are defined over Q. I'll give an example in a minute. So we have two different co-multi-theories, and they can be compared by integration. So this is the comparison map. So it's comp bdr, which goes from the i-Duram co-homology group of x tensored with C, because periods are complex numbers in general. So this is an isomorphism. And what is the map? The map is integration. So we take a differential form omega, and to every homology class, sigma, we assign the integral of omega over sigma. So how do you think of, how to make sense of this? Well, sigma here is an element of the dual of the Betiko homology. And because we have Q coefficients, this is hom hibeti xq, and this is just the usual singular homology of our space, a singular homology. So chain of integration. So a period is a certain type of number obtained by integrating an algebraic differential form over some appropriate chain of integration. So here's an example. So I like to call this the left shet's motive. So x here will be the projective line over Q minus 0 infinity. So it's the punctured Riemann sphere in two points. And so its complex points is just C star, C minus the origin. And so omega 0, xq is just the regular functions on x, O of x, and the regular functions on this are just polynomials in x and 1 over x. And then omega 1, xq are just one form, so it's the same thing times dx. So what is this complex? It's 0, O of x, d omega 1, x. So this is Q of x, 1 over x, Q of x, 1 over x, dx. So what is H naught? H naught are things in the kernel of this map. So polynomials with zero differential and they're just the constants. And what is H1? It's the co-kernel, so it's the elements, the one forms which cannot be written as d of something. And as we all know, the only element in this ring that is not an exact differential is dx over x. So H1 diram, so this is a diram here, is isomorphic to Q dx over x. As we learn in high school that dx over x has a primitive, which is called the logarithm and that is not an algebraic function. Okay, so now for homology. So the punctured complex plane is homotopy equivalent to a circle. And so that means that the betti homology is again isomorphic to Q. It's connected and H1 of x is mapped by the class of gamma and naught where gamma naught is some path winding around the origin. So let's put infinity in as well. Okay, so what is the period? There's only one period. So of course, as predicted, as expected, the diram and the betti spaces are both one dimensional. They need to be isomorphic after tantering with C, so they better have the same dimension. So the period of H1 is obtained by integrating in this isomorphism. You integrate over gamma naught, the differential form dx over x. And this is equivalent to Cauchy's theorem, states that it's 2 pi i, 2 pi pi. Okay, so we need something more complicated, slightly more complicated, which is relative cohomology. So it's the cohomology of something modulo, something else if you like. Yeah, so what is? I mean, I understand what you wrote, but what exactly is period here? Period, so you have a period is where you take an element in diram, which is a q vector space. And then you take an element which is in the dual of betti, which is a q vector space. And this comparison map will enable you to pair them together, but only over C. So it will produce a complex number. If you like, write a basis of this vector space, write a basis of that vector space, and write this linear map as a matrix in that basis. Now, a period is an element of the matrix that expresses this linear map. So linear map can be written as a matrix, and it has complex coefficients. But does it have to be arbitrary? Yeah, yeah, absolutely, I didn't say how to choose a basis, it's not, there's no canonical basis. Change of basis is rational, because they're rational vector spaces, exactly. So here I could take three times omega and 16 times gamma naught. So I'll end up multiplying, the periods, the ring of periods will be two i pi times rational numbers, if I change the basis. We'll see some more interesting examples in a minute. So now I'll take two spaces, z contained in x, I'll be slightly dishonest. Let's assume that they're smooth for now, but in the examples, z will not be smooth, it will be a simple normal crossing divisor in x. Sorry for disturbing this one. But again, there are periods to be a matrix, right? There'll be a period matrix and an element in this matrix. If you look at the vector space spanned by the elements of the matrix, we'll form a ring, that's a ring of periods of that situation. So any such element can be written as linear combinations of integrals. So relative homology is something that sits in a long exact sequence. So here this applies to both Betty and Durham homology. So I don't want to spend a long time defining this. I'll just say that the elements in relative homology are not represented by chains anymore, but represented by chains in x, c, singular chains, whose boundaries are contained in z. So in the previous example, the chain of integration, gamma naught, had no boundary. But we need things with boundary and that's what forces us to look at relative homology. And for Durham, it's defined in a similar way. One way to think about the relative Durham form is a differential form on x, whose restriction to z vanishes, for example. It's a bit more complicated than that. But that will be enough for all the examples we need in this entire course. So again, there's a comparison isomorphism similar to before, which goes from relative homology tensored with c to Betty homology tensored with c in much the same way. And we're going to look at a very special case. So suppose that the dimension of x is n and so dimension of z, let's say, is n minus 1. Then any algebraic, let's take x, r, fine. So any algebraic n form omega, because it's an n form, then automatically its restriction to z will vanish. And furthermore, it will automatically be closed. So in fact, we can just take any form of top degree and you can check that the differential form omega defines a class in h relative homology. And now we can take any sigma, a chain contained, a topological singular chain whose boundary delta of sigma is contained in the subspace. And so that then defines a relative homology class. So an element in the dual function on relative cohomology and the period of this pair of data, the differential form and the cycle and the chain is the integral of omega along sigma. So this relative business is just to take account of the fact that we need chains of integration which have non-trivial boundaries. So example, same as before, x is p1 minus 0 infinity and z is just a pair of points, 1 and alpha, where alpha is a rational number, and let's say bigger than 1. Then from this long exact sequence, it's a very straightforward calculation to check that the Duram cohomology is two-dimensional now. So I'll spare you the details. There is the differential form we had before and the class of dx. So dx viewed as a differential form just on x, not relative to z, is of course exact. But as a relative form, it is not exact. So likewise, what's the Betty homology? Relative Betty homology again has now is two-dimensional. So there's the class of gamma 0 and there'll be gamma 1. So if we have 0 here and infinity out here and 1, then gamma 0 is a path all around 0 and gamma 1 will just be the straight path from 1 to alpha. So these are indeed relative homology classes. The boundary of gamma 0 is trivial so it's certainly contained in the points 1 and alpha. But the boundary of gamma 1 is clearly contained in the 2.1 and alpha, which is in z. So gamma 1 indeed defines a homology class and these form of basis. So now the period matrix, which I hope answers the previous question, we take this basis, we could choose another one if we wanted to, and we compute all the integrals. So we integrate dx over gamma 0 and gamma 1 and dx over x over gamma 1 likewise. And at this point to make the answer cleaner, actually I'm going to rescale this to be, I'm going to divide by alpha minus 1 to get a nicer answer. I'm allowed to do that. And so the integral of dx along this path is, the integral is just alpha minus 1, but I've divided by alpha minus 1 precisely to make this number 1. The integral of dx around a closed loop by Stokes formula is just 0 because the loop has no boundary. Then the integral of dx over x from 1 to alpha is the logarithm of alpha minus the logarithm of 1, which is the logarithm of alpha. And finally the integral of dx over x around gamma 0 where we already did that. 2i pi. So the next example I'm going to try to interpret zeta of 2 as a period in this language. The example is due to Gonsrov and Manin and it's a very, very nice example because it illustrates a lot of the phenomena that are going to come up again and again in the context of Feynman integrals. So I'll try to spend some time on this example because it's crucial to understand it in some detail. By the way, can we understand quickly so pi is not the period on the i pi? Yes, pi is a period as well. You can cook up an example. All algebraic numbers are periods. So you can multiply by i, periods form a ring. So if you write i as a period, you multiply 2i pi by i and you get 2 pi minus 2 pi. So here's an example. I may have to cut the example in two if I run out of time. So we want to interpret the following numbers a period. So zeta 2 is the integral. So to check that this is zeta 2, I think for first this is very trivial, you just do an expand 1 over 1 minus t1 as a geometric series and you just integrate, you just do the integral, it's very easy. So you'll find sum over 1 and squared. In fact, this integral goes all the way back to Leibniz. So this is the prototype for all the integrals we'll see in all the Feynman integrals. Okay, so what we want to do is to interpret this geometrically. So the differential form, what we want to do is we want to view this as a pairing between a Duram cohomology class, a differential form, and some relative cycle. So we want to take x as a first approximation to be affine 2 space and we need to remove the singularities of the differential form, otherwise it won't be defined on that space. So we have t2 equals 0 and t1 equals 1. And the domain of integration is going to be... the domain of integration is going to be... okay, so not yet. First of all, the differential form, omega is dt1 over 1 minus t1 dt2 over t2, which is indeed a regular global differential form on this space x. So that looks good. The domain of integration, sigma, is the domain is given by the set of points t1 t2 where this inequality holds. So it's this region here and whereas it's boundary contained in, the three white lines, which is t1 equals 0, union t1 equals t2, union t2 equals 1, intersected with x. And this is where the problem lies because the problem is the following is that when we remove the red lines from this space, we remove two extra points, which are actually in the boundary of the domain of integration. So the problem is then that sigma does not define a relative homology class and this will be the case in all the interesting examples. You always find this phenomenon why every time you have a period that's arithmetically sort of interesting, this type of phenomenon happens and it happens in abundance in physics. And the problem is that the points 0, 0 and 1, 1 are both in the closure of the boundary of the domain of integration but the points 0, 0 and 1, 1 are not in fact in the space x. So in some sense when you remove the red lines, you throw out the baby with the bath water and you lose your chain of integration because you've just deleted two of the points in the corner. So what is the solution? So the solution is to, mathematicians call the blow-up and in physics the relevant words would be I guess HEP sectors sort of does the same job in quantum field theory. So you want to blow up the points 0, 0 and 1, 1. What is the name of this? In the case of the Feynman diagram, HEP sectors, HEP, HE class HEPs. So next time I'll do the compactification in great detail in the case of Feynman integrals and it should be very familiar to people who know the old physics literature. So blow-up 0, 0, 1, 1, what does that mean? It means that you replace each of these points with a copy of the projective line. Sorry, that should be in... Yeah, that's okay, it's white. So this is the new space. So we get a new space that I'll call A for now in which we've added, replaced the two points by P1s. This is called the blow-up. Oh, let me label these devices once and for all. R1, R2, let's call this B1, B2, B3 and these are called exceptional devices, epsilon 1 and epsilon 2. Is this a real blow-up or a complex blow-up? This is a complex blow-up, but I'm drawing the real point. A circle of your head or of the two opposite points of this? No, it's the complex blow-up. But of course I can only draw a real picture. But how do I draw the real... I would have identified the opposite if it had been... Okay, so the inverse... We just think of this as a change of variables on the integral. So the inverse image of sigma now is the shaded pentagon with boundary contained in now five devices, not three. So B is the union of B1, B2, B3, epsilon 2. And when you pull back the differential form up to A, you have to check, and I'll do one of the cases in a minute, that it has no poles. So what can happen is it can acquire poles along the exceptional devices, but it doesn't. We have to check that. And this is in omega 2 of A minus R1 and R2. And so what this means is that this integral... So I should say that these are now simple normal crossing devices. And so what we get is that we now write the integral z to 2 by change of variables as pi inverse of sigma, pi upper star omega. So this seems like total nitpicking, but it's very important because it now expresses this number as a period of a certain relative homology group. So the pi upper star omega has singularities in R, so minus, and then we take relative to B. So what that means is that as before, this differential form means that pi star omega is in the dirame cohomology of this relative thing, and pi inverse sigma, which is this pentagon, is indeed now a relative homology class. So z to 2 is the pairing between the two. Okay, so that's the crucial example. So I'll now explain how to do the blow-ups using explicit coordinates because next week there'll be lots of blow-ups. I want this example to be quite familiar. So let's just blow up the origin. So in the neighborhood of the origin, we change coordinates and put x equals t1 over t2 and y equals t2. So what that does is in the t1, so we're looking at, in the t1, t2 plane, we have this sort of sector near the bottom left-hand corner, and it's going to correspond to the xy plane. It's going to correspond to this picture. So x equals 1. This is t1 equals t2. This is the piece of the exceptional divisor, and this is t1 equals 0. And we see that what was previously a point here corresponds to an entire line in the new coordinates. And so let's just check that the differential form is compute what happens to the differential form. So we started off with, so the epsilon1 in these coordinates is given by y equals 0. And the differential form omega is dt1 over 1 minus t1 d2 over t2. And if you do the change of variables, you get dx dy over 1 minus xy. It's a simple exercise. And we notice that in the new coordinates, there is indeed no singularity at y equals 0. It has no pole, y equals 0. So this seems very pedantic, but it's crucial because all the qualitative information about the numbers e to 2, about the period, is contained in this infinitesimal data of what's happening in the corners. And it's that data that will govern this. It's sort of Galois conjugates. And so we have to study this in great detail in the physics situation. So perhaps I'll give the general coordinates, a full set of coordinates on this blown up space because it's recently, it's something I did a long time ago, but it's recently become of, it's had several different applications in parts of physics in Super Yang-Mills theory, for example. And I think it may be of interest to physics. So let me do this example and explain how to write down coordinates on the whole space. And the general case I may postpone to the end if I have time. So let's write down the coordinates on A. So we can write down u, 1, 3. We can do these two blow-ups simultaneously by choosing coordinates in a nice way. And in fact this picture has a symmetry group acting on it. And these coordinates are chosen in such a way that they're completely symmetric with respect to that symmetry. It's a dihedral symmetry of order 10. A dihedral group of order 10. Okay, so we have, these coordinates will do the job of blowing up the origin, which was what we did before with these, just looking at these two coordinates. But the other ones will also blow up the corner 1, 1 for you. What do you have in u, u3, 5? Yes. Is it t2 minus t1 and t1 minus t2? No, it's absolutely. Thank you very much. I've miscopied. Yeah, because that would cancel. Thank you. Okay, I need to wake up. So we get on, so the space a, so these equations satisfy, these coordinates satisfy certain equations. And if I've done it right, we get equations like this, and we get five equations obtained by cyclically permuting the indices. So incidentally these are, this is a cluster variety in this particular case, but it's not in general. And we set a to be the variety defined by the solutions, I mean the points of this variety are solutions to these equations. So it's the spec of z, ui, j, modular of these equations. So this space I call m0 of 5 delta. It's a partial compactification of the moduli space of Riemann spheres with five marked points. And on it there are devices, di, j, defined by the vanishing of these special coordinates, ui, j equals 0. And they form a pentagon. So these coordinates are chosen in such a way that they vanish exactly then one-to-one correspondence with this, the components of the boundary, and they vanish precisely on each. And so the co-module we're looking at is h2 m0 of 5 delta relative to d, where d is the union of the di, j. So this generalizes in a very straightforward way and enables you, so you can play the same game and express all the multiple zeta values so zeta n1 up to nR, which I defined last time. So these can be written as integrals over simplices. So n is the sum of the arguments, also known as the weight, and it can be written as an iterated integral dtn where the epsilons are the sequence 1, 0. So 0 to the power of something is a notation to mean a sequence of so many zeros and followed by 1, then another batch of zeros, 1, 0, and R minus 1. So if you do the similar calculation... You neglect the order, like d14 is the same as d41. I neglect the order, yeah, absolutely. So ui, j, yeah, I've been a little bit sloppy. So ui, j equals uji, and so di, j equals dji. So the multiple zeta values can be written as periods in this strict chronological sense. So the upshot is that mz, v's, which were examples of periods we studied last time a little bit. So they are periods of certain spaces. So this, let's say the zeta n1 to nr are periods of h, n, m0, n plus 3, delta, some space which can be defined in a very easy way. I may do it if I have time at the end relative to some special divisor inside. In other words, they're pairing, but that means you have a differential form, a Duramco homology class in here. You have a Betty relative cycle given by blowing up the simplex in some way, and by pairing them together you get multiple zeta values. But now the d is singular, so... d is singular, yeah, but it's smooth normal crossing. So that's fine. When I define the Duramco homology, you have some... you have some simplicial complex, and you take the hypercomology of the associated triple complex. So it's not a problem to... it is singular, but it's not a problem to define the Duramco homology. Yeah, I might... I'll gloss over that. Okay, so... Just so I understand, the figures you define here are more special than just saying everything that you write seems to be quite maximum. Yeah, so, I mean, so you can show by general themes in mathematics that, I think here on Akastem, that if you write a general integral of a differential form that's algebraic over blah, blah, blah, then you can choose a nice compactification like this. But it won't be canonical. It will depend on some choices. And the point of what... we won't see this yet, but later on, the point is that in the physics case, we can do it canonically. We can do it explicitly and canonically. Because if you do it a different way, then you might find you get... the homology that you get may be full of junk. And this will mean that you expect to see Galois conjugates of numbers that actually don't occur. So you want to do this in a minimal, cleanest way because the data of this geometry is going to tell you about your period. Okay, so... what am I going to do next? So the first stage was... as Thibault asked, we started off with a period which was just an integral some integral that we're interested in. And we want to interpret it as coefficients in the pairing in the Dirac isomorphism of some cohomology, some vector space, some vector spaces which were Betty and Dirac cohomology of something. So far we've... the cohomology is just a vector space. And the next stage is to explain that there's more structure going on here. In fact, there are filtrations on these vector spaces and they carry what's called a mixed-hot structure. And this is what will enable us to speak, to think about the weights of periods and periods of different types. And then the next step up is to view these vector spaces not just as mixed-hot structures but as representations of a certain group. So what I want to do now I'll start it before the break is to explain to you what a mixed-hot structure is. It's just some linear algebra structure on the cohomology given by some filtrations. And then after that I want to explain something about tenacian categories and how we can then promote this to representations of a group. And then the point is that the group this group will be a version of this Galois group I mentioned in the first lecture and it will move the data of this differential form in this cycle around and will enable us to define Galois conjugates of of the period. So that's the plan. That's where we're headed. No, there's no ring at this stage. The ring will come in here in the tenacian category. You're allowed to take tensor products. Yes, yes. That will come in the final step. Okay, so I'll now give a summary on a little bit of hot theory. I apologize to mathematicians who have seen this a hundred times, but I think it's good to give all the definitions. So recap on mixed hot structures. So a pure Q-hot structure of weight K is actually incidentally I should say that this we don't need the mixed hot structures to define we can go to the last step from here to here without the mixed hot structures, but the reason I include it is because it's very useful to speak about weights and there's a lot of applications in physics for the notion of weight and so I think it's important to emphasize it. So Q-hot structure of weight K is a finite dimensional vector space V over Q with a decomposition of VC which is V tends to QC into types VPQ where P plus Q is equal to K satisfying VPQ equals V QP bar. VC you can say that VC has a decreasing filtration such that FP VC direct sum F K minus P plus 1 bar VC is isomorphic to the whole space. So the example to bear in mind here is differential forms with P DZs and QDZ bars. That's the prototype. So to pass between these two definitions then from the second definition to the first you set VPQ C is FP VC intersect FQ bar VC and in the other direction to pass from the first definition to the second you set FP VC equals direct sum R greater than or equal to P P R K minus R So this is something like differential forms with at least P DZs at least P holomorphic components. So example Q minus N is the unique is the hot structure the following hot structure of weight to N So the vector space V in this case is just Q, it's one-dimensional and the hot filtration FI VC is VC if I is greater than or equal to N sorry less than or equal to N of course and it's zero if I is strictly bigger than N So this means so Q means it's the underlying vector space is Q it's one-dimensional and the minus N is a notation it's a take twist it's a notation to tell you that you're to keep track of the weights so it's not just a vector space it's a vector space with a weight so it occurs in weight 2N so this notation I don't know where it originates from but it's very very standard so this is the hot structure of type N comma N and so in fact VC is isomorphic to VC N comma N in this case You have not said what you could with type oh yeah thank you so the type the type are the so HPQ is the dimension of VPQC and the Hodges numbers are these dimensions so type NN means that the only non-validating Hodges number HPQ is zero unless P equals Q equals N that's what it means to be of type NN thank you and just before the break so of course a classical theorem that if you take a smooth projective variety over Q let's say then the cohomology HKQ has a pure hot structure of weight K and perhaps I'll stop there and then continue after the break okay so now a mixed hot structure so the what sorry what is this here from the definition I think if I come to any vector space I have a piece of trivial science hot structure it's a canonical and functorial so morphisms of varieties induce morphisms of hot structures you should take a nature like okay now yeah a natural these words are so abused and yeah otherwise you could put the sort of trivial hot structure on everything it wouldn't be very interesting no no of course of course a mixed hard structure is a finite dimensional vector space V over Q again so there's a Q mixed hot structure with one an increasing filtration W dot V on V which is called the weight filtration and a decreasing filtration F dot VC on VC again called the hot filtration such that the associated weight graded pieces of V so this is WK over WK minus one V equipped with the filtrate the hard filtration on its complexification defines a pure hot structure of weight K okay and the key with theorem is due to Deline and that's in particular the sorts of groups who are interested in the relative homology the singular relative homology has a canonical and functorial mixed hot structure then the next key point which is very important but it will become clear in a minute why it's important that the category of mixed hot structures is an Abelian category so you have kernels and co kernels and you can also take tensor products tensor products with mixed hot structures and duels so why that's important will become clear later on a definition that's important and very relevant for physics is it a mixed hot structure is mixed Tate if graded VK is 0 if K is odd and graded weight KV is a direct sum of Tate hot structures of weight K if K equals 2R and all the examples we saw were of this type so why is this important it's because so the philosophy is that the periods of co-homology which are mixed Tate should be very special and they are things like multi positive values and poly logarithms are all of this type so is it the initial mixed Tate or you assume that you know what it is this is a definition okay so that's it for mixed hot theory okay so now so we've promoted the co-homology to a mixed hot structure by adding these filtrations and then now we want to put a group action in so for this I need a reminder on Tanakian categories I'm not going to give the full details because it's very tedious and so the idea of a Tanakian category is that it formalizes the the categories which occurs representations of groups so as you know if you have representations of a group you can take direct sums you can take tensor products of representations you can take duels and so the axioms of a Tanakian category formalize what the properties satisfied by categories of representations of groups and the game is the following that you think of a group or an algebraic matrix group and you give me the category of its representations and then I will then reconstruct the group and give you back your group so that's that's exactly what's going on yes yes so here's the setup so you have to be a bit patient with all these axioms and I'm not going to give the full glory but think in your mind of the representations of a group so k is a field of characteristic zero and we start off with t a k-linear category oh I might want to make a billion straight off so k-linear means that the homomorphisms are k vector spaces and the composition of morphisms is k by linear then t is what is called a neutral Tanakian category if if it also has okay so I said you could take tensor products of representations so the first thing is a tensor product tensor t cross t to t which technically is a bi-linear bifunctor it has a unit so think of this as a trivial representation and some extra data given by natural isomorphisms of the form a tensor b is isomorphic to b tensor a a tensor b tensor c is isomorphic to a tensor b tensor c you know 1 tensor a and so on and so forth and I spare you the details and they satisfy a bunch of axioms which you can imagine it also has a duality so this is a a contra variant functor from t to the opposite category which we denote by m goes to m dual that satisfies all sorts of compatibilities with the previous data and the most important thing which I'll put here is what is called a fiber functor so this is an exact faithful tensor functor omega from the category to vector spaces over the field so in the case of the representation of a group you have to forget the group action so you have a vector space with a group acting on it you just forget the group action and you get a vector space the first time you see this it looks a bit silly but it's very important and so such that so being a tensor functor means that omega a tensor b isomorphic to omega a tensor omega b isomorphic to the vector space of dimension 1k and so on and so forth and there's a whole list of axioms and compatibilities between these axioms that I don't want to write yes, does everybody agree with the definition of the tanakin category because at the beginning it's because it's a neutral tanakin category so this is I should have said this is due to Saavedra and if you're paranoid you can check that it follows that N1 is isomorphic decay but it follows from the axioms so that was the missing thing that works when it fails if you try to consider a non-neutral tanakin category in this case it's the definition of Saavedra and it works fine that's all we need okay so the main theorem is that from there in fact any category like this is the category of representations of a group so what it means is that there is an affine group scheme that I call I'm going to call G omega that's technically defined to be the tensor automorphisms of the fiber function of omega, I'll explain what that means and the map from T to to representations so this is the category of finite of finite dimensional representations of G and the map is just M goes to omega M and such that, sorry, such that such that this map is an equivalence equivalence of categories so the upshot is whenever you have any category satisfying the axioms that would lead you to think that it might possibly be the category of representations of a group then indeed there is a group such that it is the category of representations of that group finite dimensional and here also omega is for me, VEC is finite dimensional for me, VEC is finite dimensional always yeah, thank you okay, so what is phi so what is this group so let me describe its points so you can extend this to infinite dimensional representations by taking inductive limits there's no problem yeah, I believe so so if phi is oh well, yeah, no there's no problem yeah, I believe so so if phi is oh well, yeah, no if your representations are are limits of finite dimensional representations then yeah so if phi is ought is a a point in this group so R here will be some a commutative unitary k-algebra then how do we in fact this group is just the biggest possible group it's got to act on these vector spaces so it's going to be the biggest possible group which acts on all these vector spaces simultaneously and preserves all the structures so phi is given by an element phi m in gl omega m subscript r, this means tensor with r for all objects such that so if we have if we have two objects m and n in t and a morphism from m to n then the following diagram commutes so this is phi m and phi n, I really should put an r in here but I won't bother so for every morphism between objects m and n in t we get a commutative diagram and then there's the data of the tensor product so we need so for any two objects so I should put little r's everywhere sorry this is phi tensor phi n and this is phi m tensor n so the tensor product of so in other words this collection of matrices respects the tensor product so this is not such a helpful way to think about the Tanaka group for the Tanaka group is something slightly mysterious because it how do you specify an element of the group like this in the art points of the group well you have to specify an automorphism for every vector space for every possible object in your category it's an infinite amount of data so what's a much better way to think about the Tanaka group is it's dual, it's functions so the better thing to do is to construct the functions on the group by something called matrix coefficients so the functions on a group is from the Hopf algebra and so we want to write down Hopf algebra from from a Tanaka category so definition a matrix coefficient or sometimes called a framed object of a Tanaka category t is a triple consisting of an object in t and v in omega m and f in omega m dual so let we're going to think of this is very so in the back of your mind you should think of the cohomology of something and differential form and domain of integration so this triple defines this is the data which allows you to define an integral and hence a period and this as we'll see in a minute should be is very close to this idea so we should keep this in mind so let p omega omega denote the k vector space spanned by the symbols m vf modulo the following relations which are going to be analogous to relations which you know to occur for integrals so the first relation is linearity and that's that if I take m lambda 1 plus lambda 2 v2 f this is equivalent to lambda 1 m1 v1 f plus lambda 2 mv2 f so to pursue this analogy with integrals we know that integrals are certainly linear in in the arguments in the differential forms and also in terms of the domains of integration so the other so of course by symmetry we have another linearity relation so this is for all lambda 1 k and then we need another relation which corresponds to changes of variables on integrals so to for every morphism m to m prime alpha let's say there is an equivalence mvf is equivalent to m prime v prime f prime if v prime equals omega of alpha v and f equals omega alpha transpose f prime so we have omega m m prime we have a map omega alpha and the transpose goes in the opposite direction on the duals so denote the equivalence class by square brackets m vf and if I want to be put an take I'll put a little omega, omega in the superscript which reminds me that v is in omega of m and f is in omega dual of m but I will sometimes drop this out of laziness so that defines a we have a vector space of symbols and relations there's a product on this which is m vf multiplied by m prime v prime f prime and that's m tensor m prime v tensor v prime f tensor f prime that is well defined and there's a co-product so it's a map from p omega omega p omega omega tensor p omega omega and the formula which will be absolutely crucial for what happens next is simply the sum m e dual tensor m e i f so here the sum is over e i basis of omega m and so e i dual is the dual basis and if you don't like the choice of basis well it's clear that this doesn't depend on the choice of basis but of course the element e i chech tensor e i is the representative of the identity in home omega m omega m which is omega m chech okay so we have some abstract algebra of symbols modular relations and from that we define the Tanaka group in fact the definition of the Tanaka group so this is the p omega omega is the affine ring of this Tanaka group in other words what this means is for all our commutative unitary k algebra take k itself if you prefer then the group the r value points of the group is given by the homomorphisms from homomorphisms of rings from this ring to the ring r and the group law on g is is encapsulated by this coproduct but the key remark is that it is not the group the key remark is that it is the functions on the group which is the more fundamental object not the group itself if I want to specify an element of the group I have to know how to map every matrix coefficient into r the point is that when we do this Feynman game and look at integrals we only work with a finite subspace of this space and we can do computations if you work with a group you have some infinite data that goes into defining an element of the group ok so now a variant and then we'll be near the end of the technicalities so a different situation is when we have two fibre functors omega and omega prime to find a dimensional vector spaces of a k and then I can define p omega omega prime to be the linear span of symbols mvf exactly as before but this time v is in omega of m for the first fibre functor and f is in the dual of the image of m under the second fibre functor and we take modulo exactly the same relations as before so linearity and equivalence is defined in an identical way and now we don't get a co-product we get a co-action we have delta which is given by the same formula I think I'll I'll write it out again do you mean omega prime omega prime in the second one? no omega omega omega prime to omega omega omega so I will write it out sorry sometimes I'm single so omega abstract category t to express representation of a group yes but then you say so category abstract category you need to define the group yes I mean I have feelings that omega is somehow unique up to some so why do you need omega omega prime which are different? in what sense are you different? to give me more flexibility and the reason I'm going to give omega prime to be different because one will be diram and the other will be betty functor and I cannot define a period out of two differential forms if you give me a differential form and another differential form sometimes I can but in general I cannot define a number but if you give me a differential form which is in diram and you give me a cycle, a homology cycle then I can integrate and I get a number so when we think about periods it's crucial that we have two different fiber functors and so that will come in in a moment yeah so the formula here let me put sigma to differ from the previous case is now some m so if I'm going to be pedantic I should put omega omega prime m v e i chech omega omega tensor m e i sigma so you see that the sigma because here so it must be an omega prime up here and it's as before and so how to think about this this is a co-action to O by definition to the functions on the group and so this is equivalent to an action of g omega on p omega omega prime so we'll think of p omega omega prime as some sort of ring of periods and purely formally we get a group that's going to act on it so let me do that now ok so long-last periodic periods so now I'm going to take t, there are many choices for t but the bare minimum that one needs actually it's not the strict bare minimum that I need will be the following category of triples vb vdr and c where vb is a finite dimensional vector space over q so here k equals q and it has an increasing filtration weight filtration w called the weight filtration we have another q vector space called v diran with also an increasing filtration called the weight and a decreasing filtration called the hodge filtration both defined over q this time and c is a comparison isomorphism c is an isomorphism from v diran tends to c to v betti tends to c such that it preserves the weight filtration and such that the betti vector space v equipped with the weight filtration and the image of the hodge filtration on its complexification is a mixed hodge structure so those are the objects in the category there are triples of pairs of vector spaces which are isomorphic after tensioning with c and the morphisms are the morphisms of these objects which respect this data so this category oh sorry I want to call this category h let's call this category h for hodge so h is a tanakian category so the proof of this fact is due to the delin and it has two fibre functors omega betti or diran which go from this category to vector spaces over q and to vector space we associate either vb or dr depending on which fibre functor we are looking at so at long last we can define an abstract ring of periods so pmh is going to be defined to be p omega diran omega betti in h so I like to write sort of as a mnemonic m for the pair omega diran omega betti and I write dr for the pair omega diran omega diran so we have a ring so we call this I don't know what to call it maybe the ring of h periods and we get a group so we get g diran or betti we get two groups in fact we can take spec of p omega omega where the dot equals diran or betti so we get two groups acting on this ring in fact I never go I prefer to think about diran so actually let's let's scrap that so we'll just have one group and so what do we have as advertised in the first lecture so the elements of this ring of this ring of periods are equivalence classes m v sigma m where m is a triple consisting of two vector spaces and a comparison isomorphism v is in m diran and sigma in m betti dual so we have an abstract ring it has a period homomorphism to complex numbers so what it does is it takes a triple and it maps to so what can you do here we have v is in m diran and then we have the data tells us that after tensing with c we can send that into v betti so c v is in m v tends to c and sigma is in we can view it as in the dual so then we can pair these two things together we have an element in a vector space and an element in its dual and we denote the pairing by this and that's a complex number so what have we gained is that we have an inverted comma as a Galois action not much to do with Galois for now but it's how we want to think of it so we have this affine group scheme g diran which acts on this ring of h periods so we have some types of numbers so there would be analogous action if I looked at the betti group but I'm not going to what else do we have we have a weight filtration so we can talk about the weight of an element of this ring so the elements of weight less than n is an element of the vector space spanned by these equivalence classes where v the differential form if you like is of weight at most n so we have an abstract ring with lots of properties and of course some of these are not in fact you can write that this map will actually be surjective you can write any complex number as a period of a random mix-hold structure it's not very interesting so the remark is that this ring is too big because not all objects in h come from geometry so they're not periods to start off with but we will only the point is, the key point is that we will only we shall only consider examples of the form or elements which actually come from geometry and are hence periods in the sense which I described earlier so the objects will be equivalence classes of h elements in h of this form and with the canonical comparison which is given by integration and so we get periods of the form so this comes from geometry and indeed defines a period when you run your ring you have of course relations and the relations hold in non-geometric case so when you take on the geometric generator are your relations still the same? I will come to that so the word that I'm not saying yet is a motive and so the way people normally set up such a try to think about periods is by defining some category of motives we don't have a good definition of a tenacian category of motives but here's the key point that any reasonable definition of mixed motives so that means you're only considering cohomology of algebraic varieties with some relations given by certain sort of standard exact sequences between cohomology so let's suppose that there existed such a tenacian category of mixed motives over Q so any one of the things you would like when you have such a category is you would like to have a functor mixed motives over Q to this category H so a motive should have a better realisation and a drama realisation that's one of the bare minimum requirements and it should have this should be an example of an object in MMQ so if you have and also it's expected so we get a map we get a linear map from the motivic periods so this would be motivic periods in the true sense of the word where you had a category of motives and it would map into this slightly more naive version this ring of H periods and one of the things that you would like about the category of motives is that this functor should be fully faithful and that would imply that this is injective so that would answer your question that there should be no new relations here that you can't actually lift to geometry and the key point of this which is maybe not obvious the first time you think about this is that if you have mixed motives all these constructions that we're going to play with in this category H we can immediately lift them to MM what that means is that we're going to have we can define the analogous ring of periods in this category and it's going to have a ring structure and it's going to have a co-action in exactly the same way but the point is that this diagram where I work in this elementary naive category H is going to give me exactly the same answer so if you play this same Tanakin machine applied to the true category of motives for your co-action you will get exactly the same formula as the one I wrote down before so this diagram commutes and if you have a period here that comes from geometry and if you could lift it to this Tanakin category of motives then when we compute the true motivic Galois action up here it will give exactly the same answer as if we calculate it in this category H so what could happen is that if you give me a nice Tanakin category of motives everything we do here we can immediately lift to this and do it on the level of actual motives but if it turns out that your category of motives does not have an injective map from here to here then it's simply the wrong thing to do and what remains is this game which works so examples the examples where you know this are mixtape motives over where OS is S units in a ring of integers in OK where K is a number field that's essentially the unique case where we have the correct definition of mixed motives it's not clear, you can construct some toy categories of motives but we will gain absolutely nothing by doing that so by this reasoning it's perfectly good to work here so in some sense we completely circumvent all these conjectures on motives so what do you mean exactly by geometry here so here I will only so a motivic period will be will be the so let's put it here a motivic period for me will be an element in this ring so in my head I want to think of it up here but we don't have such a category so I'll view its image down here of the form something like this so it's H this will cover all the examples that we need and some differential form and some Betty cycle so I had smooth scheme over Q smooth quasi-projective scheme over Q and Z will be a simple normal crossing divisor and X all defined over Q that will be enough so the subtlety here is that if you give me two different ways of writing a given number as a period of cohomology then I have a problem, I have to show that they define the same period but that's an inevitable problem because you can write different integral representations for a given number and you don't necessarily know how to prove the same this category is much more flexible it's much easier to prove relations here than it is up here up here you tie your hands you say we're only allowed to use certain types of algebraic operations and it's very restrictive so for all those reasons we work in this category H and it gives the right answer, it will give the correct answer we see that the period is already here that we can multiply yes here so here we take a triple equivalence class we multiply two together that's the formula for the product and the period of this times the period of this equals the period of that so a long last examples so because of classes of notation normally I think of these the following examples in a true dynamic and category of motives but because in this case it injects into PMH I'm certainly allowed to identify elements with their image so it's a tiny abuse of notation it's not very drastic so the first example that we saw was 2 pi I left shits motive so this was H1 P1 minus 0 infinity with a differential form dx over x and what I called gamma 0 which is a loop around 0 so this what is this H1 it's the take motive Q of minus 1 it's in weight 2 and it's of type 1 1 and so the period of LM was given by the integral of dx over x on gamma 0 and it's 2 pi I so LM is the motivic version so it lifts the number 2 pi I and sorry N on the right hand side N it's okay so the Galois action well we have G Duran is some huge group we don't know what the group is but how could it act on this thing well this is a one-dimensional vector space so the only way it could act is by a linear map on a one-dimensional vector space is just multiplication there's nothing else it could be so how does an element act on this number in this motivic period G LM is going to be some rational number times LM let me look at Q points so I think I gave this formula these groups of Q points are always a risk-ident in groups of essentially extension of by a pro-unipotent group no it's an extension of a pro-reductive group by a pro-unipotent group yes as a pro-reductive quotient so here we see that because G gives a map so we think of LM as a map from GDR to GM it's a homomorphism so there we've computed the action of the Galois group on 2 pi i and the other example I gave was the logarithm so let's do the logarithm so what does it mean that means you're working yeah, so yeah exactly so another way to say it is that so let's take omega d'Ram so the definition of the Tanica group is that we take the we take a fibre functor which is in this case the d'Ram of what I wrote by abusive notation as this one I mean really the triple of betting d'Ram so this is just omega 1 d'Ram minus 0 infinity which as a hot structure is Q of minus 1 but as a vector space it's just the vector space Q and so we get so for element G here we have omega G is a map from Q to Q so what it does is this differential form as you say of so omega d'Ram of G of dx over x what it does is it scales the differential form by a number and therefore it scales the integral by a rational number because the period is non-trivial you could also use weight arguments but it's because if it was trivial this would be the trivial representation and by the Tanacian equivalence we have omega maps the Tanacian category to vector Q is an equivalence so if this was the trivial representation it would be the zero object and therefore it would be the zero object as a mixed hot structure but it can't be because it has a period that's non-trivial well no it can't be because it has weight too that's what I mean to say okay so I'll try and finish very soon so the next example is the logarithm so the motivic logarithm with the same alpha I had before so alpha was in Q alpha bigger than one so it's the triple I just write this h1 and then it was the differential form was dx over x and then the path was gamma one gamma one if you remember was the path from one to alpha and so this thing here is called the cum emotive k alpha and in fact it sits in an exact sequence of mixed by these pure hot structures it's a mixed hot structure and the period of log alpha was by definition the integral of dx over x along gamma and that was log of alpha so this is the motivic version of the logarithm let me give the gamma action so how does the how does the tanika group act well it the tanika group acts on the diram realization of this thing in such a way that it preserves the image of this sub thing and it maps to the quotient and gives the action on Q of minus one which you've already computed so the tanika group will act here in a trivial way and here by this character lambda g so g of log alpha is going to be let me just put alpha as a prime just for psychological reasons then this is going to be log g of p plus I think I called it new p last time in Q so here g is in so that's how the Galois group acts on log of a prime number I beg your pardon oh yeah new p of g sorry yeah new p uh put in brackets new p of g yeah so so this I did all this I gave examples last time but we didn't really understand where they came from but so now um so I called this representation rho last time g diram Q let's say to g l2 and g maps to well last time I think I wrote it the wrong way around but I'll do the same so it gives a representation of this group on g l2 and so why g l2 because it acts on the diram realisation of this and so g l2 we think of as g l2 is g l omega diram of h blah which is equal to g l of h1 diram p1 minus 0 infinity relative to 1 alpha and as I explained earlier this was a 2 dimensional vector space spanned by the 2 differential forms so that's how we get a 2 dimensional representation and that means that the worst thing that can happen to a logarithm when we hit it with the is that it's going to pick up a period of this integral which is a rational number sorry here you mean that you see the teacher the marital product of 2 primes and UP will get the sum yes because there's a functional there's a relation here which is the functional equation of the logarithm that of course a log log a b equals log m a plus log m b so if you like for every prime number I get a new degree of freedom I get a new map from g to rational numbers but to understand that what do you need to do to have another algebraic geometry which includes the two structures you see that they need to be compatible in a way or how do you mean I don't understand the question probably you have this motif with alpha and you have at some moment the motif corresponding to p to q and to p times q exactly so you have to do a little bit of you have to do a little bit of geometry it's not it's not very difficult it's a nice exercise and you have to check so you have to look at what's the definition of this equivalence it means that to show that to so you've got to write log a b as a period and log this period of something else and you've got to construct a morphism of mixed hot structures such that the differential forms match up and the integration domains match up so in practice that means you have to write this integral and the sum of integrals and find an algebraic change of variables which sends one onto the other this kind of thing that has to be done in practice is always done in the literature oh this of course not no yeah I think so well I don't know there are gaps of course but this is there are many ways to do this but this is the point that in general it's not the case so here's an example where I've been dishonest so I'll confess to my dishonesty and I wanted to find the motivic multiple z to value just for psychological reason it's good to see the definition or a definition I should say so I define this space in the case this modulate space in the case n equals 2 for z to 2 so so this is the motivic mzv so I say this because I gave all these formally last time for the Galois action so here n r is greater than or equal to 2 and n is the weight is the sum of the indices omega is the form you get the relevant form on this modulate space that was obtained from the integral representation I gave earlier for the multiple z to values and x is the analog the higher dimensional analog of what I called pi inverse sigma in the example when n equals 2 so this gives you a motivic multiple z to value but you have to be on the period n r equals z to n 1 up to n r and but then here so here's a caveat so there's another definition of motivic multiple z to values so the usual definition involves a different geometry involves the motivic fundamental group of p1 minus 3 points so we have two different definitions of a motivic multiple z to value and we expect them to be the same but actually it's not known expect this to give the same motivic period that means element of this ring so this is viewed in p mh so this is an example where we have two different ways of compactifying the space doing these blow-ups to make it all nice actually they're extremely close to each other so I have no doubt whatsoever that this is doable but you have to do something else and for example all the combinatorial relations that are in the literature for motivic z to values are being checked by some people to be motivic there are two senses in these two senses oh I'm not even in no there are relations which are not known to be motivic in either sense I can't think of an example but yeah there are some not everything is known to be motivic when you define such a character like you see you know that they are all independent like you break the traditional literature or you remain some um so that's the thing so um um what can I say so that's essentially a question about independence of these motivic periods can we verify that there are no relations so here that the relations between the news come from a relation so how can we check that such a relation cannot occur and and that's quite easy to do in practice in these situations because we have weight we have all these structures we have weight filtration, we have this group action so it's not very difficult to show independence so here's an example maybe to illustrate, example z to motivic 5 special case of this um because of um so in this case in the take case the weight is in fact a grading so we can speak of the weight this is of weight 5 as an mzv mzv weight but um this actually weights 10 in the hotesteretic weight so it corresponds to something of type um 5 5 and then you take the number 1 in this ring which is of weight 0 and so because this is of weight 5 and this is of weight 0 we deduce automatically that um they're linearly independent so if alpha z to 5 plus beta equals 0 then that implies alpha beta equals 0 whereas for actual numbers it is still an unsolved problem it is not known whether z to 5 is irrational so in this setting its transcendence questions are very easy because you have all this structure so in practice if you want to prove that two numbers are linearly independent in this ring um you can do it quite easily I think there's no way to go back to actual numbers no but in some sense no but in some sense it's not a it's a good question because um because you need to understand the relations in this ring before you can even attempt to understand the relations as actual numbers because if there was a relation in this ring you'd apply the period you get a relation in the actual numbers so before you can even think about transcendence questions you need to know what you're supposed to prove which numbers are independent and which numbers are not independent so you first have to understand independence in this ring first and so I think that it's necessary to understand this before one can do transcendence questions ok so that was the technical bit done next week we'll be entirely graph polynomials and more physicsy