 So, let me start by reminding you what we did last time, right? So, well, on Monday, we defined Hager diagrams, and then on Tuesday, we defined a chain complex from a Hager diagram. Sort of the most robust version of this chain complex was CF minus, which was freely generated over the polynomial ring Fijoren U by intersection points between T alpha and T beta. And then the differential counts holomorphic disks from X to Y. And then we count how many times they hit the base point with our variable U. Oh, yeah, yeah, this is a compact zero-dimensional manifold. Yeah, so just counting the points. And we're doing everything over the field of two elements, so we're counting my two, so don't need to worry about signs. And then we defined this simpler chain complex CF hat where we required our disks to miss the base point. And now, if you look carefully at these two definitions, you'll notice that you could actually... CF hat can be obtained from CF minus by setting the variable U equal to zero. And I guess taking the convention that zero to the zero is one. Right, so if you set U equal to zero, that says that any time NW of phi is non-zero, well, it's not gonna count. So if we set U equal to zero, it's saying don't count disks across the W base point. And that's exactly what the differential here counts. And certainly, in terms of generators, if we set U equal to zero, then you're just freely generated over the field F. Right, so that's how these two chain complexes are related. And so what I wanna do today is from a doubly-pointed Hager diagram, which is gonna describe a knot. I wanted to find a chain complex associated to that. Great, so as we saw on Monday, well, you can describe a knot via doubly-pointed Hager diagram. So the Hager diagram itself describes S3 and the two base points describe the knot. The two base points sit in the Hager surface and then you connect W to Z in the alpha handle body missing the alpha disks and you connect Z to W in the beta handle body missing the beta disks. Great, so we have this Hager diagram for a knot and then our goal is to build a chain complex which we'll call CFK, whose chain homotopy type, so in particular, whose tomology is an invariant of K. Right, and on Monday I stated to use any two doubly-pointed Hager diagrams, so to say I'm not related by a sequence of doubly-pointed Hager moves, so to show this is an invariant and just show that the chain homotopy type is unchanged under doubly-pointed Hager moves. So just like the three manifolds invariant, this comes in different flavors and so the simplest flavor is the hat flavor, so that's a HFK hat. So this is a bi-graded vector space. So we'll write it as a direct sum of its bi-graded pieces for sort of, I guess, historic reasons that bi-gradings are often referred to by M and F and then we'll put it in like this. Great. And it's the simplest version that categorifies the Alexander polynomial, so we have the following theorem due to Alge-Volstavo and independently, Jake Rasmussen, which says that the Alexander polynomial is the graded-oil characteristic of not-floor-homology. So let me give you an example. Is this too wet? Great, so for example, the not-floor-homology of the luck hander trip-oil. So I'll plot this for you on a pair of axes where the horizontal axis will be the S-grading and the vertical axis will be the M-grading. So this is three-dimensional, looks like this. And so the graded-oil characteristic, we'll just, in each S-grading, take the oil characteristic and then give that a power of T to whatever S is. So here, this is gonna be a T to the minus one since S is minus one. This is gonna be a minus one times T to the zero plus T. And that's indeed the Alexander polynomial of the luck hander trip-oil. This is gonna be something called the Alexander-grading, which I'll say what it is when I actually define the chain complexes. So the short answer is that, right, so we saw on Tuesday that we sort of got this one grading from one of the base points and the other grading's gonna come from the other base point, essentially. So it does the not-floor-homology categorify the Alexander polynomial, but it also strengthens certain properties of the Alexander polynomial. So recall, oh, so first, so let's write our Alexander polynomial in the following form. So, great, so, as you saw last week, the Alexander polynomial is symmetric, so you can write it in the symmetrized form like this. So would this be the symmetrized Alexander polynomial? On the Alexander polynomial, it gives a lower bound than the genus of k. So the way I've written things, the genus of k is gonna be less than or equal, sorry, greater than or equal to the maximum s, such that As is non-zero. Great, and so not-floor-homology improves this in the following sense. So, not-floor-homology detects genus in the sense that the genus of k now is gonna be equal to the maximum s, such that h of k hat and Alexander gradient s is non-zero. So for example, so for the truffle we'll receive that, well, s equals one is the largest Alexander gradient, or this group is non-zero, so the genus of the truffle is one, which I guess we already knew. So also recall that the Alexander polynomial obstructs fibrousness in the sense that if an odd is fibrous, the Alexander polynomial is monic. So recall, if k is fibrous, then, well, I guess in the way I've written the Alexander polynomial, it implies that A sub g of k is equal to plus or minus one. And then the updated version, which is due to genie and knee, is that non-floor-homology detects fibrousness. So k is fibrous, if and only if the non-floor-homology in the top Alexander gradient is one-dimensional. So in particular, you can see that for the truffle over there in the top Alexander gradient, the non-floor-homology is indeed one-dimensional, so the truffle oil is fibrous. So that's a corollary of these two results, but there's only two genus one-fibrous knots. So in particular, it follows some of these two theorems. The non-floor-homology detects the truffle oil and the figure eight. And the odd knot, since the odd knot is the only knot of genus one, of genus zero. So h of k hat detects the odd knot, the truffle oil, and the figure eight. All right. So now that I've advertised some of the properties of non-floor-homology, let's dive into the definition. And it can just happen between the left hand and the right-handed. Yeah, the gradings tell you that. Oh, that's right, it's only well-defined up to plus or minus the power of t and the gradings in non-floor-homology are chosen, so you're always gonna recover the symmetrized Alexander polynomial. Yeah, it's the gradient convention. So first, let me just introduce a polynomial ring. So let's polynomial ring in two variables, u and v. This is gonna be bi-graded. So we'll suggestively call one of the gradings the u-grading and one the v-grading. The gradient of the variable u will be minus two, zero. So negative two in the u-grading is zero and the v-grading in the gradient of v will be zero negative two. And then, right, so it's bi-graded and at times it'll be useful to consider a linear combination of these gradings, which we'll denote by a, which will be the Alexander-grading and this is gonna be one-half the u-grading minus the v-grading. Oh, and for those of you in the audience who are already familiar with novel homology, the treatment I'm gonna give here is sort of notationally different than the original, than sort of the definition in odds-wasn't-zopper's paper, but it's equivalent. So we have a doubly-pointed Hager diagram for our knot. So we'll have a chain-complex CSK and then I'll denote our ground ring as a subscript. So this is going to be really generated over this ring I just described for you by what you're now used to, so by intersection points between t alpha and t beta. So now I just need to tell you what the differential is. Maybe you can already guess. Right, so I have this differential. All right, so, well, let's just do exactly what we did on Tuesday, but except now that we have two base points, well, we have two variables, so we'll count u, we'll count, u will count how many times you cross the w-base point and v will count how many times you cross the z-base point. So as before, we'll sum over other generators y, phi and pi two from x to y, mu of phi equals one, the number of points in this zero-dimensional modular space and then u to the nw of phi v to the nz of phi times y, and then, well, Ajvat and Zabo prove that this is a differential, so that d squared is to zero, the proof is essentially the same as in the three manifold case and then they also prove that the chain homotopy type of this chain complex is an invariant of k. So yes, this is proved by both Ajvat and Zabo and independently Jake. So d squared is to zero and the chain homotopy type, in particular the homology, as an invariant of k. Make a few remarks about this chain complex, right? So if you remember, on Tuesday, we defined a relative grading and so now we'll actually have a relative by grading. So we have a relative, we have relative gradings. Great, and so let me tell you how to define this relative grading. Well, the u-grading is gonna be exactly what we saw on Tuesday, so the u-grading, the relative u-grading between x and y is just gonna be mu of phi minus two and w of phi. All right, and now with the v-grading, just do the same thing, but instead of for the w-base point, do it for the z-base point. So the v-grading, the relative v-grading between x and y is again gonna be mu of phi and now minus two and z of phi. And as before, you can check the differential lowers the u-grading by one and it lowers the v-grading by one. And so remember, we define this Alexander-grading, one-half u-grading minus the v-grading. So the differential preserves the Alexander-grading. Okay, so I've defined a relative-grading. Well, it'd be nice to list this to an absolute-grading and so we can, in fact, do that. So remember, on Tuesday, we computed hf minus of s3 and we saw that it was just a ground ring, hf would join u and then we declared that one was in grading zero. That was how we normalized our gradings. And in fact, we can use that convention to normalize these gradings. So notice, so what happens when we set v equal to one? Right, so first of all, if we set v equal to one, we have to trash the v-grading because with multiplication by one, but it can't change the gradient, but multiplication by v lowers the v-grading by two, so set v equal to one and forget the v-grading. Okay, so, well if v is equal to one, it's like now, well, it's like forgetting the z-base point, right, because oh well, no matter if v is equal to one, this doesn't matter, this can be whatever you want and it's not gonna change anything. So the point is that setting v equal to one is like forgetting the z-base point. So if you've forgotten the z-base point, well then what we have is a singly-pointed haggar diagram for S3. So if we take the homology, well also if we set v equal to one, well now we're just working over the ring, F would join u, so well if you think about this, now this is exactly just gonna give us C of minus of S3, so the homology is gonna be H of minus of S3, which is F of join u. So this gives C of minus of S3. So the homology of the complex, so we set v equal to one, well this is just gonna be isomorphic to F of join u, which is the H of minus of S3, so this is gonna be in grade in zero. So you declare this homology to be in grade in zero and that pins down the u-grading. So this determines the absolute u-grading. Now if you want to determine the absolute v-grading, well just do the same thing but reverse the roles of u and v. So similarly, the homology of the chain complex when you set u equal to one, this is gonna be isomorphic, okay. So if you set u equal to one, now we're working over the ring, F of join v. So the homology is gonna be F of join v and then declare this to be in grade in zero and that pins down the absolute v-grading. Let's compute an example. So let's consider the following doubly pointed Hager diagram. Conveniently it will be genus one, so the one-fold symmetric product of our Hager surface will just be our Hager surface. And so T alpha is just the alpha circle, T beta is just the beta circle, so the intersection points are just these three points. All right, and so now we'll compute the differential and so now the only difference from Tuesday is that we'll count how many times you cross w with the variable u and you count how many times you cross z with the variable v. So the boundary of a will be have this disk right here. So that's gonna give us a u b plus u. The boundary of b is zero and the boundary of c is v b because there's a disk that crosses the z base point. All right, so now if we want to determine the absolute u-grading, so if we set v equal to one, we'll now our differential looks like this. And so the homology, well now what's in the kernel, b is in the kernel, as is a plus u c. And now the image is b. Great, so the homology is generated by a plus u c, so that means that the u-grading of a plus u c should be zero. So let's fill in the gradings. Great, so this has to be a gradient homogeneous element, so that means that both a and u c are in gradient zero, so the u-grading of a is zero. So the u-grading of u c is zero, u lowers gradient by two, so that means that c itself is in gradient two. And then what else you know that the differential lowers the gradient by one, so that means that the u-grading of v b should be one, v does not affect the u-gradings, that means the u-grading of b is one. So a similar calculation, reversing the roles of u and v, and you can obtain the following. And then I'll also put in the Alexander gradient, so this is gonna be minus one, zero, and one. Great, okay, so we've computed the differential, we've computed the gradings of everything. I guess we can compute the homology now, if we'd like. So the homology, great, so what's in the kernel? b is in the kernel, as is v a plus u c. And, well, what's in the image? Both u b and v b are in the image. So the homology is isomorphic, so we're gonna get a copy of, a ground ring generated by v a plus u c, and then plus a copy of f, generated by b. And then if you wanna know the gradings of these things, well b is in gradient one, one, and then this is gonna be in gradient zero, zero. So this gradient here is the u gradient, comma the v gradient. Any questions so far? Terrific. All right, so this is the most robust version, but just like for three manifolds, we have the most robust version, hf minus, and then while we saw he said u equals to zero, we got hf hat. So similarly here, well now if we want, we can set u and v equal to zero, say, so there's some other flavors. Great, so the bi-graded reference space that I advertised before, hf k hat, that's just the homology of, well take this chain complex and set u and v both equal to zero. And so if you think in terms of the differential, what is the differential counting? Well if we set both u and v equal to zero, well that's saying we have to miss both the w-base point and the z-base point. Great, and then this is a bi-graded vector space and the gradient conventions, so the gradient that I referred to is m that comes from the u-grading and the gradient that I referred to as s is the Alexander-grading. So let's continue this example and compute hf k hat of the left-handed trough oil. Great, so there's differential that I just started to erase. So now if we want to compute, so if you want to set u and v both equal to zero, well now that means that the boundary of a is gonna be zero and the boundary of c is also gonna be zero. So in particular, it tells us that hf k hat of the left-handed trough oil, well, I sum up to three copies of f generated by a, b, and c. And so if we're gonna view the bi-grading on this chain complex as the u-grading and the Alexander-grading, let's tell us the gradient here. So in particular, maybe if I wanted to plot this on a pair of axes representing my bi-grading. Great, so a has some, this is the u-grading and this is the Alexander-grading. So a has u-grading zero and Alexander-grading minus one. b has u-grading one and Alexander-grading zero. And c has u-grading two and Alexander-grading one. So this matches up with the example that I put up at the beginning of today. And so you can see while the graded oil characteristic is t inverse minus one plus t. Is it easy? You don't just, yeah, you shift the gradient. Is it easy to see? It can be seen, it's maybe not completely obvious. Actually, next we're actually gonna talk about what happens under mirroring and orientation reversal. But yeah, there's basically just some gradient shift that happens. Other questions? Great, yeah, so, right, so natural question to ask is, well, how does this invariant behave under some familiar operations that we'd like to do or not? Maybe first, let's start with connected sum. So there's a Purness type formula. Before this, oh, before I do this, I wanna talk about one more favor. Right, so we saw that we had sort of this full knot floor complex that was over the ring after join UV. We saw that, well, we could set u and v equal to zero, and then we got sort of the simplest version, hfk hat. But you could also, you can sort of, since the chain homotopy type of this effigent UV module is an invariant, you can sort of do any like these sensible algebraic thing, and that's still gonna be an invariant. So here, the algebraic thing we did was set u and v both equal to zero. But if you want to, you could like set only one of them equal to zero. So that's a relatively common thing to do. So if you set v equal to zero, then that gives you the invariant that's called hfk minus. So this is the homology of v equal to zero. So if you were in the learning seminar this morning, right, Irving talked about hfk minus, which is the effigent UV module. And then, you know, there's other variants. So for example, in certain settings, it's convenient to set maybe the product uv equal to zero, and then you get.