Pt. 5 – Khovanov homology and surfaces | Robert Lipshitz, University of Oregon | IAS/PCMI

My pleasure. So, thank you for coming back. Those of you who have come back. Um, Paul complained that I started lasagna modules without him. So, here it is again for Paul's benefit. The skein lasagna module of a four manifold X, that's what's this thing is, and a link L in the boundary of X. So, this is boundary X. Is um the free abelian group. I should have written angle braces, but okay. The free abelian group generated by all fillings, which are surfaces connecting the link on the boundary to some links in ball in the boundaries of balls inside. And labelings of the links in the ball in the balls inside by elements of Khovanov homology of those links, cuz these are in S3, so that makes sense. Um, that one isn't, so it doesn't. Mod Okay, somebody maybe Michael pointed out I missed this yesterday. Isotopy, no, somebody else. I knew I don't remember who. Um, multilinearity in the labels and a relation for engulfing balls with bigger balls if you uh, and mapping forward the Khovanov classes. So, um >> [clears throat] >> as I jumped up and down saying yesterday, this is a ridiculously large um, abelian group modulo a um, very confusing equivalence relation, not because it's complicated, but because it's not involves all the Khovanov cobordism maps and so it's not clear what it's saying. So, there was a significant amount of skepticism in the community for some years about um whether this would actually be useful for anything. And then there was some work of Manolescu and Natanzon some follow-up work that made it look a little more computable that maybe I'll say something about at the end, and there's a little bit on the notes about um so that reduced the skepticism slightly. And then there was a amazing beautiful paper by Wren and Willis, which I'm going to sort of read to you for the next part of this lecture. Um that shows that this actually does see some exotic four-dimensional phenomena. So, um next um sort of aspects of the Wren I should write bigger Wren Willis paper um from 2024 five four um >> [snorts] >> which is referenced in the notes. I suppose I could I could check when the paper was from that. 2024, good job me. Um The paper is beautifully written. So, um I encourage you to look at it. I'll try to do a little bit to help make that even a little bit smoother in in a little bit, but um you don't really need me to read it to you. It's a very nicely written paper. Okay, so um I'm going to pick out one of the theorems that they prove. They actually give two proofs um using the skein lasagna module. So, um the knot traces um X minus one of the mirror of five two and um the minus one trace of the pretzel P three minus three minus eight um are homeomorphic. But, not diffeomorphic. Um There's a nice concrete pair. So, not trace. I think this is probably came up in other lectures uh this summer, but X M of K means you take the four ball and you attach a N framed two handle along K. Probably I'm required to draw a picture. Before K, and then you attach a two handle along K. Um, give it some thickness. Okay. Um, this theorem is not original to the Wren and Willis paper, that's why I didn't write an attribution. They say it was proved by Akbulut in 1991 and cite two papers for the proof, which um it always makes one a little bit nervous when when you say you see a citation to two papers for a result instead of one. I didn't actually find it in the papers, but I'm sure it's there in some slightly disguised form. Um, then there's a simpler proof due to Akbulut and Matveyev from '97. I love preparing these that I realized that you guys weren't born yet then mostly, so um some of you were. I think Paul was. Um, so uh here's the proof by Akbulut and Matveyev. Um, which Wren and Willis very nicely Matveyev very nicely explained in their paper also '97. Um, so first thing is that these manifolds M5 2 and X so the boundary of this is homeomorphic to the boundary of the one from the pretzel P3 minus 3 minus 8. I'll come [clears throat] back to why that's relevant and I mean obviously this wouldn't be very I mean that's obviously needed for the homeomorphic but I'll come back to that in a sec. Um this one is Stein. Um and there's an a junction inequality for Stein surfaces proved using Seiberg-Witten um inequality implies that a minimal genus representative for a generator of H2 >> [snorts] >> is one. It's a torus. Um oh this XN of K of course this has homology Z in dimension 2 and Z in dimension 0 just by cellular homology. Um but by contrast um P3 minus 3 uh minus 8 is slice. So here um this uh generator of of H2 is represented by a sphere. >> Um the fact that they're homeomorphic follows from Friedman's work if you have two simply connected homology spheres with the same I mean sorry, two simply connected four manifolds with the same boundary and that boundary is a homology sphere then um the homeomorphism type is the determined by the intersection form. Um and again, they very nicely cite the relevant paper of Steve Boyer which explains some more general case and has that statement in the introduction. Okay. Um So this is the old proof. I'm going to kill off the that board. Any questions about the recall of the definition before I erase it? Excellent. So um uh so Ran Wolfe is give two proofs. They both involve developing a certain amount of machinery. I'll tell you what the machinery is in a little bit. So neither is really easy, but um the easier one is you okay, imitate the argument over there. And the point is we need to replace this statement which in the background use Seiberg-Witten theory for by something that just uses Khovanov homology. Um So, but uh use a version of the s invariant um to uh get the genus bound for um X - 1 of the mirror of 52. Um harder, they build on that a a little bit to show that uh in fact this Skein lasagna invariant we defined itself of these two manifolds is not are not isomorphic or different. So, whatever it was, 3 - 3 and - 8. So, in fact, the Skein lasagna invariant itself distinguishes them. You don't have to use the s invariant. They give some more applications. So, this one was in the literature. They give some other examples that were not in the literature and that maybe would be hard to get from gauge theory or at least that people haven't written down a gauge theory proof of so far. So, uh I think they had a back-and-forth with uh uh Anna Bova Mukerjee about, you know, we can prove this. Oh, I can prove that with Seiberg-Witten theory, too. Well, how about this? Until eventually they um at least for now won. Uh Great. Questions. So, the plan is I will at least well, tell you the main results going into the easier proof. Then I have an and maybe I'll say something and maybe I'll even tell you about the the results going to the harder one, but I won't prove any of them. And then I have an IOU about the naturality that we needed in order to define the skein lasagna module. So, I think I'll go back and pay that off at the end cuz maybe there's some nice ideas there that I can explain in finite time. Um Okay. I should have said at the beginning, I was absolutely among the extreme skeptics of the skein lasagna stuff. And you know, this is pretty nice and convincing that one should pay attention. But, this is not my area. So, um So, you guys should have a fun time asking me questions I can't answer. Let me say before I um outline their their results, a few words about the S-invariant. So, um back to Lee homology and the S-invariant. Because the framework they work in is a little different from the framework I set up for you. And so, if you do go and read their paper, maybe it's helpful if I translate a little bit. So, um my way I mean, nothing is due to me, but my way in these lectures, we had this Khovanov complex C K H T of K. This was over um Z adjoint T. And um we argued that KHT of K is um Z adjoint T um plus Z adjoint T plus some torsion. So, this is T torsion. And these happen in some grading and those gradings happen to be S plus one and S minus one. In particular, this is a bigraded uh mod- module over the graded ring Z adjoint T with a grading of T is four. Um The original Rasmussen approach, uh this is completely false. Shall we try to make it not completely false? Q. Q. Okay. Now, it's less completely false. It's true for any field of characteristic not equal to two. So, you could do it over Z3 if you prefer it. Um Uh the original um Lee Rasmussen phrasing Um they set t equals 1. So, um KHT equals 1 of KQ um is a finite dimensional Q vector space. Um I just said something with grading 4 to 1, so I lost that grading. Okay, so lost the um quantum grading. Um I guess you keep around a Z 4 grading, which in this setting nobody ever talks about, but the Ren and Willis paper talks about a Z4 grading, and so it's this Z4 grading. Um Okay. Uh but and here um KHT equals 1 of KQ is isomorphic to Q squared. Um for any knot K. Uh follows from that on the left in the universal coefficient theorem. Um >> So, can you can you explain a little bit more about the the Z4 grading? Like, is that the quantum grading and for some reason now just like this Z4 cyclic and so on, why? >> The question was what the hell am I saying here? And the answer is, yes, that's the quantum grading. I set something in grading 4 to be equal to 1, so what was left was a Z mod 4 grading. Yeah. Um >> [snorts] >> Okay. Uh C K H T equals 1 uh is still a filtered chain complex by the quantum grading quantum There's no way to say this that doesn't sound idiotic. Quantum grading, which is not a grading, it's a filtration. Um >> [clears throat] >> And so you ask where in this quantum filtration do those two does that Q squared live? So um Uh there are lots of ways to say that. So um I should choose one. So this gives me a spectral sequence from um the Khovanov homology uh So this is the T equals 0 of K Q to the Khovanov homology with T equals 1 of K Q. This is the associated graded. This is the um the homology associated graded. This is the homology of the E infinity page. This is Q squared. And um then each of these Qs appears at some quantum grading on the E infinity page. The E infinity page still is a bigraded thing. And so you look at the So the S invariant is the quantum grading where um Well, it's the average quantum grading where the Q squared exists is at the infinity page. Okay. And you can say that more concretely, but apparently I didn't. Um Okay. A lot of the literature does things in this, you know, filtered chain complex framework. Um I think chain complexes over polynomial ring are less confusing than filtered chain complexes, but um reasonable people can disagree. Anyway, the Ren-Willis paper is in this filtered chain complex instead of the chain complex over Q adjoin T framework. Okay. So, the new contribution the new technical contributions some of the new technical contributions of the Ren-Willis paper, um they define a skein module using this um uh what I called A T equals 1. So, this is the Lee deformation. The thing that gave this complex. Um They call that thing S. Where do they put their Lee superscript? S Lee. >> [snorts] >> Well, I mean, congratulations to them. You can also define this. That's not very interesting. You take exactly the definition I wrote at the beginning and you replace Khovanov homology by the Lee homology everywhere. So, I've thought I've given you this definition. Um >> [clears throat] >> Uh So, great. Um S Lee of XL um still has a quantum filtration. It has lost the quantum grading, but it still has a quantum filtration, just like the homology here had a quantum filtration. Um Okay. It's badly behaved. Um Some classes are in filtration level minus infinity. Um that doesn't happen here. That's because of this enormous um abelian group and equivalence relation that we had. Um great. This Lee homology wasn't interesting. So, if I forget about the filtration, I just had Q squared for the Lee homology. And the same is true here. So, um the Lee invariant of XL um is uh what I want to say. Um has no information if you forget the quantum filtration. For example, um the Lee invariant of X empty set um is the Q vector space generated by H2 of X cross H2 of X. >> [snorts] >> um So, this is a vector space with one basis element for each element of H2 of X times H2 of X. Okay. um Fine. um If I give you an element alpha plus {comma} alpha minus in here, it's alpha grading. uh is um alpha plus plus alpha minus. This is a some version of my size two like Q squared before except it's bigger because there's more homology to to mess with it. um Okay, but using the quantum grading given a class um alpha in H2 of X that get um an invariant S of X alpha. So, this is like our concordance invariant S. An integer except it can be minus infinity because as I said the quantum filtration is badly behaved. Okay, sorry. This was not inspiring. Let me tell you the main um the main result or a main result why this is interesting is it's a genus bound. So, um the genus of alpha, that is the minimal genus of any surface representing alpha, is at least S of X alpha plus um alpha squared over two. Uh That means that alpha dot alpha, let me write alpha dot alpha. So, this looks a little bit like the adjunction inequalities from from uh Seiberg-Witten or Heegaard-Floer. If S is minus infinity, this doesn't tell you anything, but if S is not minus infinity, sometimes it does. Um And so, um to prove the theorem, you just need to compute or if you're lucky, you just need to compute and then where this number S of X alpha. >> [snorts] >> A lot of work goes into that, but um 1.14.2. Here's part So, theorem numbers are to their paper. Um for positive knots then uh for n less than or equal to zero um s of xn of k zero is zero. Um but s of xn of k one is um s of k minus n. You had a related uh sort of philosophically related problem on the homework about the Milner conjecture and computing the ordinary Rasmussen s invariant for positive knots. And maybe if you do that problem, you sort of imagine that you might be able to to say something in this setting also. Um Okay. For instance um m of five two is a positive knot. It would be really nice if I drew it for you on the board and then you could see, but I'm not going to. Um and um exercise the s invariant of m five two is two. So um what does that give me? The S-invariant of um X minus one of M52 uh {comma} one is um the S-invariant of K minus one three. >> [snorts] >> This is a really exciting part where we see if I've got the formulas right. So, the genus of the homology class one, this is in H2 of X minus one of M52 is bigger than or equal to 1/2 times three plus the self-intersection of that class is minus one, which is one. And that was the piece of that was the piece of the um Atiyah-Singer argument that I needed to prove for you. So, this So, in that sense, I'm done proving that the that these are distinguished. Let's take a vote. Would you like me to tell you some additional results which give you that the skein lasagna modules like how you leverage this is the skein lasagna modules themselves are different without proof or prove the uh nat- rest of the naturality, do some of the proof of the rest of naturality that you need in order to get the skein lasagna module to work? No? Okay, we can just quit. Who votes for option one? Who votes for option two? Okay. Option one has it for now. You're going to you know, well I would say you can change your vote at any time, but that might be disruptive. >> [snorts] >> Okay. Um this theorem has an additional part then, so wonderful. You've all just proved that you're not me. I hate watching people assert theorems without explaining the proofs, but um but if I were a great a better mathematician, I would enjoy that, too. So, um continuing this theorem, and um the skein lasagna module in homological grading zero and quantum grading Q of this map this not trace for alpha equals one, okay, tensor Q, um I wonder what that's supposed to be, is um Q or zero. Uh Q if Q is um S of Xn of K one um or S of Xn of K one minus two, and zero otherwise. That is in the process of proving this theorem, you also get some information about the what the homology looks like. This again has an analog for just ordinary Khovanov homology for positive knots. You can see not only what the S-invariant is, but also that the Khovanov homology, not the deformed one, the ordinary Khovanov homology is non-trivial over there in that grading. Um Okay. Uh here's proposition 1.13. Um Let K and K' be um concordant knots. Also works for links, but okay. Um then Swee of Xn of K is isomorphic to the Lee Rosansky module of Xn of K' respecting the gradings and filtration. Um and the S-invariants agree. That follows from this. Okay. So Let's see if we have enough. This one I claim this is the ordinary skein Rosansky module, not the deformation. Um >> [snorts] >> So, the point is like if you have a positive knot, the S-invariant you're over at the extreme end of the Khovanov complex. And in particular, you can also see that these classes live in Khovanov homology themselves. Yeah, that's a good question. Is that genus bound always an integer? Um I don't know. Probably not. I don't know. I mean, the alpha {dot} alpha is probably if you Well, maybe I missed a hypothesis. Unless I missed a hypothesis, no. But but maybe I missed a hypothesis. It's a good question. More questions. Good. S is always even. No, it's not. It wasn't that example. Nice try, Robert. Which is good because N isn't always even, either. >> [snorts] >> The parity I guess if the boundary is a homology sphere, then the two probably have the same parity by what we were saying before. So, you probably got an integer there. If you're not in a If the boundary is not a homology sphere, then um we better check whether they prove this result in that case. It's great I'm being recorded. Okay, let's see if we can prove the second half of the detection theorem. So, um So, we have M uh 5 2 is positive. So, um this additional part of the theorem applies and we get that uh Okay, and we had uh S of X N of M 5 2 um one was three. So, that gives me that S 0 Q of X minus one Sorry, X minus one of M 5 2 one is um Q if Q is one or three and zero otherwise. Um On the other hand, uh S of X minus one of uh I don't think I need this P3 minus three minus eight uh one is S of X minus one of the unknot um in grading one. Um and their Lasagna modules agree. >> [snorts] >> Okay. Um But now, applying this theorem to the unknot um or using earlier Manolescu and Natale computation we get that the skein lasagna module for the unknot The unknot is positive, see. Um Okay. Uh We get the Lee lasagna module. Okay, it's all right. I'm I misled you. We get the Lee lasagna module um for the unknot is Q in um gradings 0 plus or minus 1. And then the relation between the skein the Lee and ordinary S um In ordinary Khovanov homology, there's a spectral sequence. Here, there's not a spectral sequence, but there's some remnants of the spectral sequence tell you that um S uh 0 Q of X minus 1 of P3 minus 3 minus 8 is um contains a Q contains a Q in gradings 0 plus or minus 1. I mean, the Lee lasagna module did because of this concordance invariance property, and so the ordinary one also does. Morally because of the spectral sequence, but that isn't quite right. There isn't but um you still get a bit of a rank inequality. So, it wasn't this Q that caused the non-isomorphic, it was the Q in grading minus one, the zero else that caused it to be not isomorphic. Okay. Do you feel enlightened? Questions that I won't be able to answer? It's very nice that they did this extra part because this checking that the skein lasagna module is actually different. Because if they hadn't done this, um old insert your epithets here people like me, um would have said, "Okay, but this is really just using genus bound information. We still don't know whether the skein lasagna module itself has any smooth information in it." So, they refuted that claim. It's good having young people working in the field. Um especially very uh clever ones. Okay. Let me say a few words about the part that you didn't want to hear because we still have 15 minutes left. I I've hidden a lot. I didn't prove these theorems, right? And the proofs involve some fairly um intimidating computations of Khovanov homologies of cables of knots. Um and a lot of cleverness. So, there's a lot of work in the paper. I just haven't done any of the work. Okay, so we had key to the um to this lasagna um is given a cobordism goes so let's say given uh four manifold um X diffeomorphic to B4 and um submanifolds um B sub I inside X let's call it B B sub I inside B uh diffeomorphic to B4 um and um links L inside the boundary B LI inside the boundary of BI um >> [snorts] >> We're doing great. um and a cobordism sigma from these LIs to um L so sigma sits inside B minus B1 through BK you get a map of Khovanov homologies. It's untied. I I proved that yet. Um What I proved mostly proved was given a cobordism given L uh zero L1 in R3, the usual R3 not just something homeomorphic to R3 and um sigma from L0 to L1 inside 0 1 cross R3 get a map of Khovanov homologies. Um So, maybe I'll say what the key ideas are in reducing this needed lemma, the key the key lemma of lasagna to the thing that we already proved. Questions? Irritation? Anger? The question is am I going to tell you the sweeper ad move and I am. Um but let's get there. Let me think I'm going to go the notes right in in I'll just tell you something about it. The notes right in in straightforward order, let's go in reverse order instead. So, let's let's start with this and try to reduce to simpler and simpler cases until we get to that. Okay, see what what the ingredients we need are. Um so, reduction one it's enough to enough to um consider a um single BI. So, I have this ball abstract ball B and I have inside that abstract ball B some B1 B2 B3. Um choose a graph a tree connecting the different BIs and drill out a neighborhood of it. So, choose a tree connecting the BI and disjoint from sigma and drill out. Uh I can certainly do this. Then I have a you know, then I just have a single ball inside. But I made some choices. So, um we have some things to check that I'll leave as IOU's for a few minutes. So, must check. Um one is obvious. What if I choose a different tree? And two is that uh we could have sigma isotopic to Sigma prime um in B minus the B1 through BK but where the isotopy goes through the tree, but where the isotopy intersects the tree. At least the dimension is right for that to be a problem. Okay. >> [snorts] >> These are not unrelated problems, obviously. If I can just choose an arbitrarily different tree, I can just move it out of the way at each point. Okay. Now um I have a cobordism let's say L0 to L1 in a four-manifold W um diffeomorphic to 01 cross S3. Because I have a ball minus a smaller ball that's diffeomorphic to 01 cross S3. So um and I only have a hope of writing down Khovanov maps in um 01 cross S3. So obviously what I'm going to do is so um I'm going to fix um fee from W to 01 cross S3. I didn't name my cobordism, so call it sigma. And consider fee of sigma. That's now a cobordism in 01 cross S3, the standard thing where I can write down not diagrams, so I can do a sequence of elementary cobordisms, etc. Um but what if I change fee? We don't really know what the So, change fee. That is, there's an action of the diffeomorphism group of 01 cross S3. Um and that acts on my set of choices of fee freely and transitively. So, um I think this group is not known. I mean, the pi zero of this group is not known. So, um that's an active topic that maybe was discussed elsewhere in the in the lecture series. So, that's a little upsetting. I need to show that any different fees give me the same map. And the last thing is uh finally, um I had functoriality in 01 cross R3 and I want for sigma inside 01 cross S3. Okay, cuz that's where I'm ending up here. Okay. Wonderful. Um let's unwind these in reverse order now that we understand all the things we need to check. It's going to boil down to one lemma that I want prove and then a little bit of a little bit of cleverness. So um last one first, 01 cross R3 versus S3. So um if sigma is inside 01 cross S3, then a small perturbation doesn't intersect. I mean, generically it doesn't intersect 01 cross infinity. Because this is a two-dimensional thing and a one-dimensional thing inside a four-dimensional thing. Okay. Um but if sigma t is an isotopy of surfaces in 01 cross S3, then generically it does intersect um 01 cross infinity um finitely many times. many times. Okay, that's totally electoral. um 4 minutes Claim spelled out a bit on the in the notes. uh So, if sigma t So, sigma t passing through um 0 1 cross infinity transversely once um corresponds to the following move. You have some knot diagram. And I need more space. So, here's sigma zero. I have some knot diagram. And I have a piece of the knot that um goes So, this turns into uh, try again. This by going over the top. That is you pull this strand around over the top like this. That's one of the hemispheres around infinity. Or, um, you go around the bottom. That's you pull it around the back. That's the other hemisphere around infinity. So, the knot is going around one hemisphere at infinity or the other hemisphere at infinity on the two sides of this of this passing through infinity. Okay, so this is called so, replacing going over the top by around the bottom one by the other by the other is, um, MW W's sweep around move. And you have to check um, by bizarre but correct argument that it doesn't change the Kauffman map. Time is nipping at our heels. Um, >> [snorts] >> this also so, okay, so you have to check that. I'm not I mean, in the remaining 15 seconds I'm not going to do that. Um, and anyway, the proof is I can't can't improve on the proof. Um, that deals with the changing graph problem and the passing through the graph problem for my first reduction. And maybe in the remaining -5 seconds um there was this diffeomorphic surfaces give you the same map uh question. And the solution to that is, well, you can arrange the diffeomorphism to be the identity near infinity and it's the identity on the boundary by construction. And so what you do is you push the surface so that all the interesting you do an isotopy of your surface into all the interesting surfaces out near infinity where the diffeomorphism is the identity. And then you apply the diffeomorphism and then you bring your surface back. So you can see that you get that you actually get um isotopic surfaces when you apply this diffeomorphism of S3 * R1. So with it might be interesting diffeomorphism of R1 * S3, but they're not really seen by surfaces. Okay. I will stop there. Thanks for listening.