 of this co-pairing and identity to construct a co-multiplication. Is it okay? I mean of course you can, as there are some in equivalent definition of, but you usually need kind of three structures to relate the other structure. I mean one might try to define Rabinus algebra by using three other kind of structures. So now we want to see kind of, so the 3d to qt's. So what we want to take is we can start them up from what is called the easy, from a given modular tensor category, which is more like C. We want to construct 3d to qt z, C. So yeah, so I don't have much time to go into details of the MTC structure on a category, but so this requires probably a separate course. So let me very briefly mention what this MTC, what kind of basic MTC data. And so first there is a following statement. So an MTC can be understood as a, there is a ribbon category with, so is a semi-simple ribbon category with finite number of simple objects and a non-degenerate S matrix. So now of course I need to decode what this statement means. So what is a ribbon category? So again I'm not gonna go into very detailed definition. So a ribbon category is an abelian and more it's a linear C. So roughly speaking linear C means that all homes are vector spaces over C. So abelian means there is always a well-defined kernel and co-kernel in the category. So it's an abelian, monoidal, and moreover there is a notion of direct sum and kernel and co-kernel. Monoidal this means that there is a tensor product, which now it's important that it's not symmetrical. And such some, with additional structure, which, so let me kind of write it in words and then I will explain what this means, which allows to define invariance of framed links via the braid representations. So let me, so I will explain what this, so this is actually the reason of the name ribbon because as I mentioned you can say frame links as ribbons, linked ribbons. And so to explain this additional structure, it's useful to introduce a diagrammatic approach where which essentially can be related to do some brain representation, so braid representation of link. So we have the following diagrammatic rules. So first of all, if I have a part of the diagram which looks like this, we'll have a bunch of lines here. So all lines will be colored, so they are colored by objects of my category C. Here are some objects. And so I can introduce, so I can descend lines as a kind of parts, as a strands of some framed links. And A, I can then send as a whole picture kind of embedded into R3. And I can then send this thing as some sort of rectangle, also called coupon sometimes. And so for this picture I will, so the A means, so if I have something here, there's something here, it can then show the element of home from tensor product of objects associated to strands below to tensor product of objects, which color these strands above, okay? So if I have just nothing here, nothing here, so I start with some object V and just draw a line, this means that there is, this corresponds to the identity morphism from V to V. And now the other rule is that, so one part of the treatment structure is that every object has a dual, has a well-defined dual, and so if I draw something like this, this is equivalent to drawing the line in the opposite direction and changing color, changing this object, coloring the strand to its dual. And so another part of the ribbon structure is that there is a well-defined, there is a choice of the morph from V tensor V to identity object, so this is the identity respect to monoidal structure and it's represented by this diagram. Again, so if you have no strands here, so no strands means we assign the identity object. Okay, and there also a choice of the homomorphism from identity object to V tensor, V star tensor V, which corresponds to the following diagram, to this diagram, so we start with V tensor W and we end up with W tensor V, so this should be a map from V tensor V to, V tensor W to W tensor V and this is denoted by R V double. And it should satisfy, should satisfy a braiding relation, particularly if I draw a picture like this. I can see the certain braid where I start with U, V and W and I can, so this corresponds to a certain combination, so a certain composition of morphisms R and identity here, so here R U, U V, and here I apply identity, so here I apply identity on the table and here I apply R V W, sorry, R U, W and here I apply R V, W tensor identity, this should be the same as a morphism corresponding to a different diagram. So I'm not gonna write explicitly, but this gives a certain, so this also provides a certain, this diagram according to the rules provides a certain morphism from U V, U V W to a tensor product of W V, and U and they should be the same, so this gives a certain equation on R matrix, which is usually called Young-Bakster equation, or sometimes R calls braiding matrix, so this is essentially, so if you, but note that there is no relation which tells us that this should be the same as this, so if you construct, using this data, we construct invariance of framed links, not unframed links, and any questions? So I mean, how do we, so using this structure, we can construct invariance of framed links so by considering that, so if I have a framed link, I can start reading it in some direction, and so, and apply, so links color it by objects. So how do I do this? So if I take a framed link and I start reading it using this rules, I obtain some morphism from identity object to identity object, and this is where, so in simple cases, this is just as a morphic to C, and we just get a complex value to our hand. Okay, so this was the review of the ribbon structure on the category, now we want to, so the model tensor category structure is, we want to assume some additional conditions. So as I mentioned, the first, so we first require that the category is semi-simple, and so any object for any, this means for any object V, in, there is, we can decompose it into a direct sum, maybe there's some multiplicities of simple objects, so this is a space of, this is a set which is a finite, finite, this is another assumption of the model tensor category, finite set of simple objects, which we can see them up to isomorphism, okay, and then there is, one can define what is called unnormalized S matrix, so it's defined, so let me, so also let me mention the following statement, so there is a following statement that, if we consider our homes from a simple object to itself, then it's isomorphic, so all homes, all possible homes, they are proportional just to identity, and this is the result of the Shor-Liemma and the homes from different symbol objects, the trivial, so in particular, so if we, so the S matrix unnormalized, so it's denoted by tilde is defined as follows, so I can see there's this diagram, so for this diagram, according to the rules above, I should associate the map, the home from identity object to itself, which is from this, so identity object is also a simple object, so it's just a complex number, and so this is, so I can send this, the elements, so I color the, so I can see the Hopflink with components colored by simple objects, the i and vj, and the result is a complex number, and this, I say this is a matrix element of a matrix S tilde, and the condition of the model category is it should be, this matrix, it should be non-degenerate, in particular, I should be able to inverse it, questions? Okay, so it's useful to introduce some other things, notation, so if I consider just a node colored by vi, so this also gives me a homomorphism from one, from identity object to identity, and again this is just a complex number by the same argument as before, and these numbers are called quantum dimensions of simple objects, and there's also a notion of T matrix, which is a diagonal matrix with components defined as follows, so if I consider a diagram like this, so this should be, this should provide me a home from vi to vi, and again, from the short lemma, it follows that it should be just identity times sum of number, and this number I will denote by ti, so essentially this means that I can always say, if I have this piece colored by vi, I can replace it by this piece with factor ti, so another, so it's also useful to introduce the following quantities, so d e would be a sum of square root of sum of squares of quantum dimensions where i runs over the set of simple objects, and also kind of normalize this matrix without a tilde, which is given by just rescaling of this guy by d, okay. So now we are ready to construct a TKFT associated from this data, so let me take sigma g to be a Riemann surface of genus g, and so let me understand this Riemann surface as follows. So let me consider a rectangle and a bunch of oriented strands attached to this rectangle, and then my Riemann surface of genus, so the number of rectangles will be g, and my Riemann surface I understand as a boundary of a handled body, which is obtained by sickening this picture, okay, and so first I need to say what is the, so what is this, the TKFT function assigns to a Riemann surface of genus g, so the statement is that the value on sigma g is can be understood as a space of homes from identity object to tensor product of g copies of the following sum, where I take a sum of simple objects of tensor product of vi, tensor vi star, so for example, if for a sphere, this will be just a home from one to one, which is just c, and the other simple example is for torus, genus one, it's a space of homes from one to the sum star, the sum star, but this is of course, is just a direct sum of homes from vi to itself, because here we have a dual, the maps from identity to vi and the vi dual is the same as the homes from vi to vi, and so each of these guys is just as a morphic to c from the showing. So we have, in particular, we have the dimension of the vector space associated to T2 is just the number of simple objects in my category, okay? And now I need to define, so if I have a, so let's m3 take m3 to be a bordism from sigma g1 to sigma g2 to a pair of Riemann surfaces is general g1 and g2, so this means I need to give you, this can be as a map from home from one to, so let me, don't repeat this, so this is the parenthesis means the expansions, this expression turns the product of g1 copies to homes from one to the product of g2 copies and so I can, so just, so I have to end soon, but let me just briefly finish what I wanted to say, and so this can be understood as, so if I take a, so this map can be understood, it can be completely determined if I take a, so I have to take a, some element in the first space and b to be element in the dual to the second space but making dual is equivalent to just swapping these two guys around and then I, so then what I need is, so for any a and b I want to calculate the value of that c of m3 related on a and then pair it with an element of the dual space to this one and this can be, so this is realized as follows, sorry, I'm out of time but let me just finish this. So I, do I have like three minutes? So let me, so what I want to do, I want to replace this, so I have some bordism and say to between Riemann surfaces, g2 and g1, so what I want to make this, I want to construct a closed manifold and I want to do this by gluing in the handle bodies which appeared before and those handle bodies which can be understood as sickening of these rectangles, rectangles which I wrote here. So imagine gluing in to copy, like this thing for g1 and g2. So this gives me some, this can be, this gives me some three, the closed, closed three manifold with coupons embedded, by coupons I mean these rectangles with attached strands and this can be realized can be realized by surgery which I write here. So the surgery, so the same approach, we do the same approach as before but the surgery looks as follows. So instead of just a link, I have a link with some coupons and of course we can have this link and the components of the link can also link with some strands are connecting the coupons and so I consider this picture and so I assign the value a and b here and this gives me a map from identity object to identity object which is the complex number and if I color all my strands by some simple objects and the result, so this expression is given by the following. So I do a sum over all colors for strands and so this is essentially the same formula as for WRT environment and I have a product of a link components and for each link component I write d color and I do this factor d color divided by d two. So let's say a is a link component, color of a. Okay, I stop here, sorry. Questions?