 Okay, let us start So just so in the in at the end of the last lecture, so I introduce the explicit formula For WRT environment, which essentially defines it. So I was a bit in a rush So there were some questions. It was a bit the formal was a bit messy. There were some questions So I rewrite it and so just what is the idea so we take the color of John's polynomial of This frame linked and so you've seen the uncolored John's like It's got a John's polynomial of unframed links in jCasmos and lectures, and I told you how how it's How it depends on the framing so dependence on the framing is very is very simple and explicit you just Have some extra factors and so what do you do? You take this a bunch of so you calculate a certain sum of all colors so for each link component you assign a color and each color runs in the sum from one to k minus one and so you add some weights and then you you introduce some denominator which normalizes it in a certain way So one part of the numerator is essentially do the same thing but you replace Your link by a bunch of unnot this framing plus minus one and their number is B plus the number of positive eigenvalues of linking matrix and B minus number of unnot this frame minus one and then you add also normalization corresponding to the number of zero eigenvalues of Of the linking matrix, so let me make a A couple of remarks. So the first remark is that so normalization here so the standard confirmation in the when people talk about double-edged invariant, which is denoted by tau is such that on a sphere It's equal to one. So this was part of the exercise and But there are also another standard another natural choice of normalization and which I will denote by that s u2k and So it's essentially the same thing But you multiply it by The following factor and So it's this it has a property which was also the result as a direct result of the one of the exercise is that It's normalized such that now it's normalized such that on a s2 times this one is equal to one and this is a normalization is monochrome from Tqft point of view and As you will see later in the lecture the double-edged invariant Can actually be understood as a evaluation of a certain Tqft on a close to a manifold and so So People you usually use this normalization so in this normalization it behaves well under the connected some operation on the connected some It's just a product of the Dumbarty environment of the corresponding manifolds But for this it's it just it will it will also be a product, but with extra factor So the remark number two is that this can be generalized so now so we define it as a function from positive integer to see but this can be generalized to the function from rational numbers to see and What what you do is so if you take a rational number and represent it as a ratio of r and k Where so that r and k are coprime The result you will be given by the same for essentially the same formula, but you instead Replace so here you do replacement like this, but now you just want to replace q by e to z to apply i or Okay, sorry to be precise. It's a it's a mark from integers Modular from rational numbers modular That of course Okay, and it's actually can be generalized to kind of for to what is called the WG environment to see to any Modular terms of category, which I will talk a bit later Okay any questions so another remark is that it was motivated by more physical And not and not so mathematically rigorous approach by written and So in this approach The double or tenor and so now I will be mostly talking about the environment, which is normalized like this, which is different by this Factor simple factor. So kind of the physical definition of this of the value of the environment on a close to the range of three manifold is given by a certain pass integral so this is This part is mathematically defined in principle, but I will comment on this later so over a certain space and Here I integrate exponential of 2 pi i k minus 2 Transimons functional of a so let me elaborate a bit on different indigrants here So transform so a is a connection one form of offer SU 2 principle bundle over M3 So any SU 2 bundle so as you to as you to bundles any as you bundle where m3 is trivial So it's always So that's why we can all we can globally choose a connection one form to describe the connection on this on the principle bundle and So then the Transimons functional is defined as integral over M3 of our trace of a da plus two-third a Cube so the connection one form I understand So since I can I can globally trivialize my mind or just in terms of the element of the easy one form while it in SU 2 the algebra and The space which I integrate our is So is a space of all Connection one forms on M3 Modular gauge equivalence So the gauge equivalence is so I can identify a is another a which is Related to each to it by the following explicit formula where G is a function from m3 to SU 2 group and So this the transimons functional is well defined so it's It's a well-defined function of from a SU 2 to To real numbers Modular integers and so this is because it's so it's invariant under all gauge transformations This can be connected to a trivial one, but if you take a gauge transformation Which is not connected cannot be connected to a trivial to a territorial map to identity as in its can change by an integral number and But the exponential of the sink is of course since K is integer here The exponential is is in wire is a well-defined function on the space to the complex numbers Okay, so again, this is You'll define but I actually Kind of can make so in the next lecture we talk how can one try to make sense of this integral So the idea is that so I mean suppose you have some so the problem is that It's not here how to define a measure on this infinite dimensional space But so usually I mean the point if you if you calculate some integrals Like of this of this for example of this form some complex integral I mean some integral over over over real line. So you actually don't like calculate don't calculate it by by the definition using measure so you do some counter deformation and you You do some sort of easy you you you collated you collated using the singularity structure of this all this function so essentially one can try to define it by just using the Kind of using the topology property of the space which is can be which can be mathematically well defined Yes, well, I think you can get any any integer Yeah, any other question? Okay Good so and Well, but the point is that from this motivation is that it's actually is a what is called partition function so is a partition for what is called partition function of Tqft so this is So this is how you usually write a partition function of quantum fields here You do some you calculate a pass integral over fields and the fact that the the here the action Come the exponential of the action so action is the functional of the field the fact that the the action doesn't depend on the metric means that this series topological and but the The the Tqft is a quantum field theory kind of is this it's usually described some and value is evolution in the time So you can see the manifolds particular direction and we want to understand the time direction So we also should be able to define this The Kind of what is what do we associate not to close manifold by some Bardism and So the Bardism can be through the time evolution And this can be so If one understand Tqft this way this can be formalized So the definition of Tqft sub already kind of appeared in the lectures So cheaper and mentioned and also in the lectures by Søren it was mentioned. So let me Kind of remind you what exactly what I mean will mean I Remind you the definition and also fix kind of notation what by What will I mean by Tqft? so an n-dimensional Tqft Form is a symmetric anoidal function From the So earlier I will use the word Bardism instead of co-Bardism. So I mean there's subtle difference, but in this case I mean, there's some difference when one for example studies the generalized homology or co-homology here associated this but Here there's not actually much difference to the category So I will I will in my in my right hand side the The category will be a very particular category. So it will be the function will be valid in the category factor spaces over complex numbers and So in principle, of course one can consider any field here But at some point it actually will be important that this filter for us as this filter is algebraic closed And on the left hand side So let me clarify what I will mean. So what do I mean here by the category of? n-dimensional Bardism so objects and of course, I will be interested in the case n equals three objects are So I can I can assume smoothness, but it's not it's not in this three-dimension. It's not really important as smooths and Manifolds so n minus one Manifolds oriented. So for me, this would be always a rented and closed for now and And morphisms so morphism from Let's say M and 1 2 So, okay, let me let me denote it by homes homes from m1 to m2 are Also, so oriented So again, I can assume if smoothness, but it's I cannot just consider them as topological manifold oriented and Manifolds M such that There is explicit there's explicit choice of so in the in the political case there is choice of homomorphism just a choice of homomorphism and In the smooth case, I need to choose a different morphism together with some neighborhood such that homomorphism from So this should be homomorphism preserving Orientation to and one bar where bar means reverse of orientation to in the joint union to the joint union of Yes. Yes, the choice of homomorphism is extradite Yes, and so more I can see that with this modular I homomorphisms preserving Boundaries, so I can see the home I can see model homomorphism which preserves this And so So what is he what is the symmetric monoidal? so the magnetic monoidal function means that the function which respects symmetric monoidal structure on this Categories so So I'm not gonna go so Roughly very briefly the symmetric Monoidal structure so on so on that so let's first Consider the right hand side on vector. So there is a notion of of course tensor product So monoidal structure means there is a notion of a standard product and here This is it will be a standard tensor product of vector spaces over C and Moreover, there is a canonical. There is a choice of isomorphism to a tensor product reverse and there also Identity Object and from generally denoted by one. So in this case this object is just my field C and So it's object. It's also it's a it's a cubed with So the tensor product with this is equipped with canonical isomorphism to the V itself and swapped around so there is data which consists of this isomorphisms and So on the left hand side So if I take two objects the the tensor product in the in the sense of the monoidal monoidal structure is This joint union and of course we have a natural canonical isomorphism to the joint union swapped I'm sorry. There is there a standard homomorphism from this to this and So the identity object is Empty space Well, I'm saying three dimensions and lower it it doesn't really well the Yes, well, you have well you of course Yeah, I mean the statement is that Yes, I any for any homo there is a unique The unique smooth structure if I take my manifold there's a unique smooth structure Yeah, up to dimensions Yes, yes here. Yes, this is why it's called cement Okay. Yeah, the statement that well, this kind of is inverse Just if you apply the another I mean it's inverse to itself In a sense that if you so okay, so as a like before proceeding to three dimensions. So as a warm-up And consider two dimensional to give teeth. So this was briefly mentioned by chip ring and And So there is statement that if I take a Frobenius Algebra, so now here I can see the algebras over C This me this produces me a 2d 50 and And this map is actually one-to-one correspondence and so So how do we construct this? So if I take so if I take a Frobenius Algebra V so as a So now I have a statement how do I construct this response so I want to produce some function that V and So first statement is that that V of s1 is equal to V as a Just as a vector space and now, what is the what is the additional structure on the on this V and how does it? correspond to How does it correspond to the properties of this function so first of all there is a so what is the data of? Frobenius Algebra so first of all there is a Multiplication so it's a map from V tensor V to V and This map will correspond to the following Bardisson and second Part of the data is a unit is a map from C to V Which correspond to the following Bardisson and the third part is well, let me write it and It's in particular one which means I mean the unit map means that one is maps to the unit with respect to multiplication and and Three is a core unit or like we can call it trace And I will do it by trace map. It's a map from V to C and It will correspond to this cap So this is enough for you can feel this is all the whole data which you need, but you require some properties So it's not arbitrary Collection of mu Counit epsilon and this trace map. So first conditions Is Associativity so you want So for example if I can see there well, let me let me draw it now the Bardisson seen the vertical rotation So if I can see there Uh Then the product of a triple of V three copies of V. I can first Apply multiplication to the first pair To get some element in V tensor V and then I can apply The multiplication here to get some element in V or I can go the other way around I can apply identity times mu and Then apply mu here and So this diagram should be commutative and this is reflected in the in the in the fact that So how do I how do I realize these maps the composition of this map as a composition of Bardisson's it's Goes like this so the identity The identity morphism the identity morphism in the Bardisson correctly is just Is a Bardisson which is given by the product of the manifold with It is an interval and this is of course homeomorphic to the following to the other composition of Bardisson's and the second property which the second condition which we need is that Which so the it's required in the Frobenius algebra the The Following composition trace with mu which is a map from V And there we to see so there's some billionaire firing on V is non-degenerate And this follows from the fact that we can explicitly find its inverse so so this composition corresponds to the following composition of the pair of pants and the cup and But of course if we can find it inverse By compositing with this banded cylinder and then identity here and This is the same as identity up to Up to homeomorphism. So indeed This should be this part. So it should be non-degenerate So we see that there is a kind of the structure of two-dimensional to kft is encoded by some algebraic structure and So we want something similar in three dimensions Any questions? Yes by twisting well, I can consider the topological Kft as a kind of as a supersymmetric kft with just superchargers actually so Kind of in this sense The statement is becomes true. I Mean I can always consider to kft as kind of Three way like twisted sympathetic kft in some three in some were in some two logical sense Just all superchargers suck trivially like all states are already bps states Well, I don't know maybe you mean something more some some well you At least in this sense is the question The other answer is yes Yes, but it can be it can be Obtained using this data this data is enough because you can so if you want to construct co-multiplication so let's see you So what do you do you you can start with some sort of multiplication and Then So first you can So you can you can apply so first you can you can define this thing and you can apply a cup kind of and you can express co-multiplication Using sorry not no just don't you can express co-multiplication using I'm multiple using multiplication on this and this kind of Kind of copiering and the copiering can be understood as inverse so the copiering is fixed by the condition that it's it's It's inverse to pairing which is defined by the by this composition So you can you can unambiguously determine co-multiplication using this unit co-unit So they'll be in the indexes there are some problem there are some problem which kind of about some simple example Yes, but co-multiplication I can Can constrain requiring that this in its inverse. It's an adverse of multiplication Sorry, what do you mean? No, it's okay. So first let me so first I can I construct the pairing right this pairing by by composition of multiplication and and the trace or co-unit Then I construct this then I can construct A copiering which is a map from C to V tensor V by by by the condition that this is Identity and then I use a composition of