 So today I'd like to talk about the Humphley-PT polynomial and its categorification. But before I do, I'd like to continue a little bit with a big picture that we were talking about at the end of the last lecture, and say a few words about the general notion of quantum invariance. OK, and so probably I should mention Witton's name here. So the setting is that I have, let's say, G, a Lie algebra, and a tangle, say t and scripty t and m. And I color the strands of t by elements, the space of representations of G. OK? So maybe just to draw a picture, here's a tangle where I have three colors, say A, B, and C. And now I see I've forgotten something very important. So I want this to be an oriented tangle. So far when we were thinking about the Jones polynomial, the orientation wasn't particularly important. But now all of a sudden it is. OK, so maybe I'll orient the tangle this way in this picture. And then I'll color the ends of the tangle, t, let's say it this way, of a lambda colored strand, lambda if the orientation is sort of right moving at that end. So for example, here I'm moving to the right. I'd color this point by A. I'd color this point by B. Here I'm moving both of them though. I'd color both of these by C. And by the dual representation, lambda dual, if I'm left moving. So for example, this would be labeled by an A dual. And this would be labeled by a B dual. And then so let's say nu bar is the sequence of labels, my set of left endpoints xn. And similarly I'll say mu bar over here. So I'll write this tangle as being t going from x nu bar to x mu bar. And then what we have is to the pair x nu bar x mu bar, I get a vector space, or I guess r module. Again, r is this ring z adjoin q plus or minus 1. And let's call that v nu bar mu bar. I have a composition. And you should think of this as being completely analogous to the vector spaces that we wrote down for the Jones polynomial. OK, in the last two lectures. So I have a composition map that goes from, say, v nu bar mu bar to v tensor over r v mu bar lambda bar to v nu bar lambda bar. And if I have a tangle t from x nu bar to x mu bar, I get a bracket, bracket of t in the nu bar mu bar. And this behaves well with respect to composition in the same way that we saw with the Jones polynomial. So the bracket of t, t prime is the bracket of t times the bracket of t prime, where I use this multiplication here to define this. OK, so this is sort of the general setup for quantum invariance of knots and tangles. So the question you have to ask yourself is, what are these vector spaces? And how do I evaluate this bracket? OK, and I won't give you a general answer to that question. But let me just remark that the dimension of v lambda bar, oops, let me write lambda bar mu bar here. This is the same as the dimension of the space of homomorphisms between the representation, the tensor product of all the lambdas, to the tensor product of all the mus. And so what is the Jones polynomial? So Jones polynomial and the Kauffman bracket correspond to taking little g as SL2, simplest, interestingly algebra. And I color all strands with the same representation v, which is c2, this is the vector representation of SL2. And then, so in particular, I see that v is isomorphic to v dual. That's why the Jones polynomial doesn't really care about the orientation on these strands. And also, if I take v tensor v, that's wedge 2 of v plus sim 2 of v. These are both irreducible. So this says that the dimension of v 2 2, so that's sort of things where I have two endpoints on either end. Colored with v is the dimension of the space of hams from this to itself, which is just two dimensional, because space of hams from one representation to itself is one dimensional, and from one representation to another one is zero dimensional. So this is 2. And so for example, that's y. This is why the Jones polynomial satisfies a skein relation. So for example, the tangles that look like, let me not draw, but this tangle, this tangle, and this tangle, they all give me their brackets live in a two dimensional space. So there must be a linear relation between them. And then so the other thing that I should say is that if L is a link, so L's a co-bordism from the sort of empty boundary to the empty boundary. Here in this tensor product, I have no representations. So this v empty empty is always one dimensional. So the bracket of L I could think about as being a polynomial, which I'll label pL lambda. That's the coloring of L, pG lambda. And that's a polynomial in z adjoining q plus or minus 1. So in fact, this theory of quantum invariance gives us a whole pile of polynomial invariance of knots. Any time I have a Lie algebra and I pick a bunch of representations and color the strands, I get some not polynomial. And of course, that's a lot of information. And a very sensible question is to ask whether there's any relation between all of these different polynomials. And we'll talk about that in the next lecture and in this one a little bit too. And the other thing that I wanted to say, so at the end of the last lecture, we talked about the sort of general categorification project. So what we want to do here is we want to upgrade this r module to some category. And given a tangle, we want to see an object in that category. And a co-bordism between tangles gives us amorphism. Usually, this category is chain complexes over some simpler category. And the morphism is a chain map. This situation is now quite well understood in the situation where G is SLN. And these representations are all exterior powers of the standard representation. And in contrast, almost nothing is understood about the case where if you chose any other Lie algebra other than SLN. OK. All right, so now, where does the Humphley-Petit polynomial fit in all of this? Let's be concrete now instead of, all right. So what is this? So this assigns to a oriented link L inside of S3, oops, R3, let's say, P of L. And it satisfies a skein relation. I should see where P of L lives. So this is in Z adjoin A plus or minus 1, Q plus or minus 1. And I have to invert Q minus Q inverse as well. And it satisfies a skein relation that looks like this. So if I take AP of a knot with a negative crossing minus A inverse P of the same thing where I change the crossing to positive, I get Q minus Q inverse P of the resolve thing. So just let's see what that means in an example. So for example, if I take P of this diagram, oops, AP minus A inverse P of this diagram, that should be Q minus Q inverse P of this diagram of the unlink. Now both of these diagrams here are the unknot. So that says that P of the unlink is A minus A inverse over Q minus Q inverse P of the unknot. And now we have a sort of normalization question, the same way that we do with a Jones polynomial. So we'll say this sort of normalized polynomial satisfies P of the unknot as 1. And unnormalized satisfies, let's say, P twiddle of the unknot is A inverse minus, oops, A minus A inverse over Q minus Q inverse. So now you'll notice that the skein relation encompasses both the Alexander and the Jones skein relations. So if I take P of k and I set A equal to 1, and this is actually just the Alexander polynomial of Q squared, and if I take P of k and I set A equal to Q squared, this is the Jones polynomial, the way we've defined it. In general, if I take P of k and I set A equal Q to the n, this turns out as the SLN polynomial in this sense, colored with a vector representation by cn everywhere. And let's just see that that makes sense. So again, if I worked with SLN instead of SL2 and I colored everything with a vector representation, then it would still be the case that this dimension, if I tensor the vector representation with itself, I still get this decomposition. So this space associated to this tangle with four n's is still two-dimensional. So I still expect a linear relation. But I can no longer, this is now not such a good object, because there's a distinction between v and v-dual. But for example, there's a linear relation between this, this, and this. And in particular, let's notice that PSLN of the unnot, and let's take the unreduced polynomial, this is Q to the n minus Q to the minus n over Q minus Q inverse. And this object shows up everywhere in the theory of quantum invariance. It's called quantum n. So now I'd like to talk a little bit about how this Humphley polynomial is actually defined, how you would go about proving that there is an invariant like this. And to do that, I have to talk a little bit about braids in the braid group that we saw in the second lecture. So let's recall that we have the braid group, BRN. So this was generated by sigma 1 up through sigma n minus 1 with the braid relations. So I have far commutativity sigma i sigma j is sigma j sigma i, where the distance from i to j is bigger than 1. And the third right of Meister moves, sigma 1 sigma i sigma i plus 1 sigma i is sigma i plus 1 sigma i sigma i plus 1. Now, if I give you sigma and BRN, I can form what's called its closure sigma bar, which is a link, it's an oriented link. And what does that look like? Well, maybe here's my braid sigma, and it's got n strands. Maybe n is 3. And to form its closure, I just take sigma and I join those strands up to each other like this. So let's notice that if I take the closure of sigma tau, what does that look like? So that looks like I take sigma and then tau, and then I close it up. But that's the same thing. I can take this tau and just slide it all the way along here. Here I get tau and sigma. So that's the same as tau sigma bar. So let me now tell you two classical theorems about braids. The first theorem is due to Alexander. And it says that so any oriented link L and R3 can be written as the closure of some braid. So sigma is in some BRN. OK? So it's not that every link is the closure of a three-strand braid, but if you give me a link, then I can find some n such that it's a closure of an n-strand braid. So the minimum of such n is what's called the braid index. Question? Oh, good question. Yes. So notice that there is, in fact, a canonical orientation on any braid, which is I point the arrows out like this. All the time. Excellent point. OK. So now the second theorem is due to Markov. It says, say that I have two braids that close up to give the same thing. So say if I give you sigma 0 in BRN, sigma 1 in BRM, and they have the same closures. So sigma 0 bar is sigma 1 bar. Then sigma 0 and sigma 1 are related by a sequence of Markov moves. OK? So this is analogous to the theorem of Reitemeister. It says that if I give you two diagrams of the same link, they're related by Reitemeister moves. OK, and there are two Markov moves. One is I can take sigma and exchange it for, let's say, tau sigma tau inverse. OK? And you see by this property over here, I could move the tau inverse around and have it cancel the tau. So this clearly doesn't change the knot type or the link type. And then the other thing that I can do is I can take, so sigma is in BRN, I can send it to sigma sigma n. That's now in BRN plus 1. OK, so what does that look like? It looks like I take my braid sigma. That's an n-strand braid. Maybe n is 3. And now I add in one more strand. Oops, I should put plus or minus 1 here. And I put in a crossing like this. And now you see if I close this up, I can just undo this bit by the first Reitemeister move. So this Markov move is really just the first Reitemeister move. OK? And that should sort of make sense, right? Because the braid relation sigma i times sigma i inverse is the identity. That's the second Reitemeister move. And this relation here is the third. So the only one that's sort of not covered is this one here. All right, so now let's talk about the Jones polynomial from the perspective of braids. So remember, we had this algebra Tln that naturally came up when we were thinking about the Jones polynomial. So this was generated by B1 up through Bn minus 1 with some relations. I still have far commutativity. And I have Bi squared. Now I can write this as quantum 2, since we know what quantum 2 is, times Bi. And then I had BiBj Bi is Bi if the distance from i to j is 1. And what Jones tells us is that there's a homomorphism. Let's call it psi 2 from braids on n strands to Tln that takes sigma i to q minus Bi and sigma i inverse to q inverse minus Bi. We should multiply these together and check that this really is a homomorphism. And so if I want to evaluate the bracket of sigma, this is really none other than psi 2 of sigma. So this says, let me draw that in pictures over here. It says that the bracket of sigma i, that's this minus q times this, but this is Bi. And this is 1 in the temporally-leaved algebra. So in particular, that means that if I take the bracket of sigma bar, this is now the Jones polynomial, this is the same thing as taking psi 2 of sigma, closing it. This is now some crossingless planar tangle that I've closed up and counting the number of components. This is q plus q inverse to the number of components psi 2 of sigma bar. Yeah, what do I mean by that? So this thing here is a sum of a whole bunch of terms in the temporally-leaved algebra. This is an algebra, so I can add things up. I close up each of those individual diagrams, count the number of components, and then extend linearly. So now, the thing that I want to point out is that the bracket of B, B prime bar by the same kind of picture that I had here that I just erased is the same thing as the bracket of B prime B bar. So in other words, the map that sends Tln to R, so B goes to the bracket of B bar, is a trace. It satisfies the defining property of a trace map that I can switch these across. And in fact, this is the way that Jones discovered the Jones polynomial, as he was thinking about traces and representation theory on von Neumann algebras. So it wasn't discovered by any of these easy diagramatics that I've described to you, but rather, it came from thinking about representation theory. OK, so now what about the Humphley-PT polynomial? So the definition of the Humphley-PT polynomial that I'll give is very similar. But instead of the temporally-leaved algebra, it uses an algebra called the Heck Algebra, which is a sort of universal object in mathematics. It shows up all over the place. Let's give a definition. So the An Heck Algebra is, this is an algebra over our ring R. And it has a presentation that looks like this. I managed to erase the braid relations. So it's generated by T1 up through Tn minus 1. And again, I have far commutativity. Ti Tj is Tj Ti for i minus j bigger than 1. And I have the right of Meister-3 relation. Ti Ti plus 1 Ti is Ti plus 1 Ti Ti plus 1. And then I have a quadratic relation that says that Ti squared is q minus q inverse Ti plus 1. So let's notice that we have a homomorphism psi from braids on n strands to the Heck Algebra on n strands, actually to, let's say, the group ring of the symmetric group Sn. So remember, Sn has a presentation. Let me call these two braid relations star.