 Okay, well, well, July that you have our second speaker of this morning Maggie Miller from Stanford who will talk to us about non isotopic cyphered surfaces. Thanks. And this is joint with Kyle Hayden, Sungwon Kim, Junghwan Park, and Isaac Simberg. I think Jung is here. Is he here? Yeah, okay. So I've given this talk a few times in the last few weeks and Tom Roff already saw it once. I'm going to do it different this time. So I'm going to talk about Kevanophilmology, so it's still boring for you, but confusing for everybody else now, so it's like, you know, skip the easy part. Okay, so let me start off by by motivating the problem that I'm going to talk about. So here's the the background motivating principle motivation, which is that cyphered surfaces can be interesting, which is pretty non-intuitive for me as a four-dimensional topologist, which is nicer than how I said it two weeks ago, but Jen told me that was controversial. So when I say that cyphered surfaces can be interesting, what? I don't know, it's being recorded. Okay, so when I say that cyphered surfaces can be interesting, I mean that for a fixed knot we can find distinct cyphered surfaces in S3 that are the same as surfaces, but different as embeddings. So I could have a knot k, here is my second favorite knot, the unknot, and it could bound two surfaces, let's say S0, which is genus one, and S1, which is also genus one, so let me draw another genus one surface. Okay, great, which are not isotopic to each other, and in this particular case it's very easy to prove that these two surfaces aren't isotopic to each other, because what you could do is compute the fundamental groups of their two complements, and we'll see that, well, Pi1 of S3 minus S0 is just a free group, I guess on two generators, whereas Pi1 of S3 minus this other surface S1, well you see that it looks like a figure eight, so in fact this whole surface is just a figure eight thickened up with a little band attached to it, so this is the figure eight group, that looks confusing, free product Z, figure eight. Okay, so these aren't the same group, which means these surfaces can't be isotopic to each other, I don't even have to say raw boundary, but these surfaces also aren't very interesting, and I think that's not very controversial, because they can easily be simplified, because both of these surfaces are compressible, by which I mean I could find a small disc properly embedded into the complement of the surface in either the S0 complement or the S1 complement, and use that disc to cut the surface or compress it to make it a genus zero, and if I do that, then these two genus zero surfaces' discs will be isotopic raw boundary that follows from the Schoenflees theorem in this dimension, or you could just like see it in this one, I don't know. So these surfaces are different, but somehow they're like barely different, because they would be the same, if you did a very simple operation. So the first legitimately interesting examples, I think are due to Alfred in the 1970s. So I want to, am I gonna write over this line? No, probably not. So Alfred in the 1970s constructed the following pair of cypher surfaces, now I have to use a slightly more interesting knot, it's still not my favorite knot, I don't care about this knot. So let me draw a non-trivial knot, I'll draw like a clasp, and then I'll sort of tie something in it that's shaped like a trefoil. I was really nervous during Robert's talk about how many pictures he had, and I was like, I'm using chalk, I can't compete with that, so I'm really pushing myself. So here's my non-trivial knot, and I'm gonna construct two different cypher surfaces for this knot that will not be isotopic to each other, but they also won't be compressible, okay? So I'm going to temporarily add two more boundary components to this picture. That's two small unknots that link these two strands, I hope you can see the colors, it looks kind of okay on zoom, pink or blue. And I'm gonna start filling in a surface starting at this pink like hole, coloring in pink, and have my surface come through the blue hole, like this, and then flip around to the blue and tie a band shaped like this trefoil, and eventually come through the pink hole and end at the blue circle. Okay, so this was a planar surface whose boundary was the white knot that I drew along with the pink and blue circle. To get rid of these two boundary components, I'm going to attach a tube whose boundary is the pink circle and the blue circle. But I want to get two surfaces, so I'm gonna consider two different possible choices of tube. So one of those choices is going to look like, well, here's my pink circle and my blue circle. I'm just gonna take a tube that starts at the blue and just like completely ignores the trefoil, like sort of draw it faintly here. Sort of like a bubble that like swallows the whole trefoil, so it's called a swallow tube. Let me erase this because that's gonna make my picture a little round. Okay, but something like this. I could choose a different tube, still with these same two boundary circles, but now I'm gonna take a tube that's shaped like the trefoil. Okay, so it's called the follow tube because it follows the path of the band that I drew. Okay, and so it's a little bit less obvious in this diagram than in the first one. Not that this should have been obvious if you didn't know that this was true, but again, these two surfaces have compliments with different fundamental groups. So they're certainly not isotopic to each other. But these are also genus one surfaces. And this boundary knot K is not the unknot. You could compute like just literally anything about it, basically, and determine that it's not the unknot. So these are certainly minimum genus, which means that they're also incompressible. Okay, so cypher surfaces can be interesting, I guess, and people write a lot of papers about, you know, constructing different kinds of non isotopic surfaces. They don't all differ by this like tube construction. There's things involving satellites, from Morisugi Sum, there's ways to get infinite families of non isotopic surfaces. But then you have to question like, is this interesting? So I am more interested in four dimensional topology. So from my perspective, this setting is a little bit unnatural. The problem is that, you know, my surfaces have boundary, but they're embedded in S3, which is not as interesting as S4. And it's not even a proper embedding because they have boundary and S3 doesn't have boundary. And there's like sort of no way to fix that unless I remember that S3 is actually like best viewed as the boundary of B4. So here's my picture of S3. I draw it as like one dimension less than we see because it is one dimension less than the preferred dimension, which is four. So this is S3. And now it's very easy to see that it's the boundary of the four ball. It's sort of this lower half space here. And up to now I've been drawing surfaces that all look like, well, I have a knot that's in the boundary and I have a surface that's also in the boundary. And I guess I have another surface that's also in the boundary. And now it's clear what to do to make these embeddings proper. I'll just push the interiors of both of these surfaces a little bit into the four ball. So I'll just keep this here and sort of push this down. So let's make the embeddings proper. It takes me so long to write. Okay, a lot of our interesting behavior when we do that. So let's look at these two surfaces, which were like borderline interesting before. And we push their interiors into B4 and suddenly they actually become isotopic even real boundary. And it's very easy to see in this particular case because what is really the difference between these two surfaces? It's this crossing. We're really unable to guess, but if I were allowed to just do this by isotope, sort of pull this part of the tube in front. Well, now this looks like a trivial knot instead of the figure eight. And so I could sort of like, oh, not everything and make it look like this, just by isotope. Of course, in S3 you can't do that. Crossing changes are not isotope. But if we had a fourth dimension of freedom, then I could pull this tube that looks like it's in front and pull it into the fourth dimension first and then backwards and then back to where it started. And I would achieve the crossing change. OK, so these surfaces are not isotopic in any sense in S3, but they are smoothly isotopic in before even real boundary. OK, these two surfaces seem more intimidating than the first ones, but actually it's the exact same behavior when we push their interiors into the four ball. Close to the boundary, they both look like this, exactly the same. And then a little bit further into the interior of the four ball, we see one of these two tubes. But again, the difference between these two tubes is just a crossing of this trefoil. And now that I have a fourth dimension of freedom, I could change the crossing and make the tubes look the same. So these kind of seem scarier, but it's not any harder to argue than a fact these two surfaces are also smoothly isotopic real boundary once you push them into the four ball. So not isotopic in S3, but they are isotopic in B4. And so as I mentioned, not every construction of non isotopic cyphret surfaces looks like these examples where the difference is a tube. But yet in almost every example or construction that I'm aware of, there's a fundamental reason that the surfaces have to become isotopic smoothly, even real boundary. Once you push them into the four ball, I say almost because, well, okay, never mind. So here's the motivating question. This is in sort of in text in a paper of Livingston in 1982. He was studying cyphret surfaces of the unlink and showing that any two surfaces that were the same as surfaces became isotopic once you push them into the four ball. And he also showed that that was true for a certain family of surfaces with non trivial boundary constructed in the 70s. And he asked, like, is this a general principle? So must, oh, I put the lid on my chop, which means I can't make the more chop from out. Okay, must my two surfaces, let's just call them S0 and S1, genus G cyphret surfaces with the same boundary be isotopic. Once you push them into B4. And he said, you know, probably not presumably not, and we just don't have the right example. And then I guess at this point, well, I mean, what invariant would you use? It's hard to guess in advance, like how you would distinguish the surfaces, I think. So the theorem that we proved, which Robert actually implied in his talk at the end or said a particular statement from the paper. So what are our initials? H K M P S, this is easy for me because Kyle and I wrote another paper that was also a comps, but with like two Ks was like a comps. So I've just been giving the same thing for a year. So a theorem 2022. No, great. Okay. So I'll draw you an example. I guess I'm going to put it here. Okay. So I'm going to draw two cyphret surfaces whose boundary is the same not. Again, it's kind of like a mess to have both of the surfaces completely in one picture, but this time I'll just draw two copies of the knot. So I'm going to start off by drawing a torus link. So something like this. So come around and connect everything. Okay. I hope you were impressed by that. And then I'm going to draw like another torus link because I'm going to be constructing two knots. So the same link again. Okay. And then connect everything up. Oh, no, oh, no, no, no, it's okay. I almost got too cocky, but it's fine. It's fine. Okay. So I'm going to start constructing two surfaces. Right now the boundary is a link, but let's like ignore that little problem. I mean, maybe you don't even think that's the problem. I just don't like that. Okay. So I've drawn here a two component torus link. Both components are left-handed trefoils. No particular reason. I just think that looks nicer. And because it's a torus link with two components, that means I have two knots that are like parallel to each other. They lie on a torus where they cut the torus into two annuli. So I'm going to draw those two annuli. So here's one of them. I have to use yellow for the first one or else I'll be confused later. So I'm just going to color in a strip between these two parallel strands. And this annulus is going to be shaped like a left-handed trefoil. Okay. So I hope you can see this trefoil sitting in the picture. And to get the other annulus, well, I could just rotate this picture pi over three, but not would look the same, but the annulus would look different because somehow it's like every other piece is filled in. So this one obviously has to be blue. So see here this wasn't filled in before. And I'll draw again an annulus that is shaped like a left-handed trefoil. Okay. Great. So right now these two surfaces actually are isotopic. If I don't say real boundary, that's obvious because I just sort of rotate pi over three and I really don't want to worry about boundary. So the way I'm going to fix that and at the same time fix that their boundary is the link is by attaching a band to connect the two boundary components. So I'll attach a band here. So probably this in yellow. And I originally I had an annulus and I attached a band. So this is now a genus one surface. So I'll call this Sigma zero. And then I need to attach, well, at least something with the same boundary here because I want to still have the same knot. So I'll just attach like literally the same band. So something like this. So I'll draw a blue band here. Okay. And so this is Sigma one and it's also genus one. And actually it's very easy to argue that these two surfaces are not isotopic in S3. I don't want to say real boundary or anything. Because even though this looks like a complicated picture with a lot of crossings and stuff, it is just an annulus with a band attached to it. And that annulus was shaped like a left handed truffle. That means this whole surface could deformation retract onto this left handed truffle with a band attached along an arc like this. Okay. On the other hand, well, the surface is also like a left handed truffle annulus with a band attached. But it looks kind of different. If I draw this left handed truffle, it's sort of rotated a bit. And this band, see how the ends are all on like one petal of the truffle? So it looks more like this. Is that showing up? Yeah, I think that's okay. Okay. So actually it's just the exact same thing as before. If you've ever seen Tonal number, then you might just recognize like why this picture looks good. But very easy to compute that pi one of F3 minus sigma zero is free. And pi one of S3 minus sigma one is the trefoil group free product Z, which is not free. Okay. So there's certainly not isotopic in S3, but that doesn't tell us anything because we don't care about S3 right now. We care about B4. And these guys had different fundamental groups in their compliments and they were isotopic. So the theorem that I'm not going to prove is that sigma zero and sigma one are not topologically isotopic in B4. I will tell you what the proof is and I'll just tell you that it's very easy. And if you felt like it, you could do it maybe. So the proof is that the two-fold branched covers, so the double covers of B4 branched along the surfaces are not homeomorphic. That's a pretty good obstruction. They are distinguished by their intersection form on H2. So if they were topologically isotopic, then what I'm saying, not homeomorphic, this is all like just topological topology. So that would mean that the double branched covers would be homeomorphic, but they're not. Okay. So here's a much less impressive theorem and I think I'm going to move back to the whole other side of the board, even though maybe that's weird. Well, I think I'm going to work my way back over there by the end. What time is it? Oh yeah, definitely. Okay. So here's a much more difficult and much less impressive theorem, which is theorem sigma 0 and sigma 1 are not smoothly isotopic rail boundary. Okay. So you see the first problem here, like we don't care about this because they're not topologically isotopic. So what's the point of that? But let's prove this. Okay. So in particular, let's focus on this word smooth and then I'll justify, I'll justify myself later. Okay. So I'm going to prove that these two surfaces are not smoothly isotopic rail boundary and I'm really going to work smoothly and use something that obstructs specifically smooth topology. And there's not like that many options. There are several papers now about obstructing smooth isotope of surfaces in the four ball. Robert talked about that at the end of his talk. You could use like the, the not for our cabortism maps or the Kovanov cabortism maps or I guess that's it. So let's use the Kovanov cabortism maps. So proof is that sigma 0 and sigma 1 induce different maps on Kovanov homology. Okay. So I thought it would be fun to show you how to like actually compute something and prove this. So this is going to be, I mean, if you know about Kovanov homology, then this is going to be very basic, but I think it'll be fun to like see an example. And if you don't know about Kovanov homology, this is going to seem like a series of random steps, but then you'd know how to compute the maps, even if you don't know what they are. So that's pretty good. So let me erase this and explain myself. So the point here, a surface co-board is in this case, a surface sigma, which is a co-board is from a not K. Does anybody know what our not is? Like what's the boundary of sigma I? That's a very familiar not. So it's the negative six frame negative whitehead double of the left-handed trefoil. Okay. So we have a surface co-board is and it goes from K. So that's shorthand for this case that to the empty not. But really, I could just take any surface co-board is in between like two links. Then we get and induced co-board is a map co-board is a map of K H of sigma, which goes from the k-phomology of K. So the k-phomology of the empty not, which we'll see what that is. Okay. And I'm going to remind you, if you haven't seen it before, I'll tell you, well, sort of what this object is, and then some examples of what this map is that are enough that you could compute it in some cases. Okay. So first of all, what we need to know is that the chain complex for co-phomology is generated. I'm going to work over Z mod two by labeled smoothings. This means smoothings of my not. Okay. So I guess maybe I should not use K if something else is already named K. So what do I mean by this? I mean, while I take in particular actually a diagram of the not, I take a fixed diagram of my favorite not. This is not my favorite not. So here's, here's D and at each one of the crossings, I'm going to make a choice of how to get rid of that crossing by either replacing it with like two vertical arcs or replacing it with two horizontal arcs. And there's like a way of naming these like zero or one. That doesn't matter. So I'll get something that looks like, I don't know this free. So that's smoothing. And then for every circle that I get, once I do all this smoothing, I need to choose on that circle to label it with either a one or an X. Okay. So I could label this like, so this is a perfectly good labeled smoothing and it shows up in the chain complex. And then the rule for when a labeled smoothing gives me a cycle, a cycle, if and only if for each zero smoothing, it actually does matter. But for each of one of the types of smoothings, replacing it with one would merge two X circles. So we're not actually going to really worry about this rule right now, but I do want to just convey that even though I'm talking about the chain complex, it's very simple to decide if any element represents a cycle, which would be in Kevanophomology. So that's not really a problem here. I want to focus on Kevanophomology, but I know exactly which things are in Kevanophomology. So that's fine. Okay. So let me give you an example of a co-ord is a map. So I said, Alfred, that was right. That was right. It was. Sometimes I say, Trotter. Okay. So example, here's the best example. The Kevanoph map of the disk, which would go from the Kevanophomology of the unknot to the Kevanophomology of the empty set. Well, first of all, I said we were going to see what the Kevanophomology of the empty link is. And actually from this definition, now we can. So I've pre-prepared a diagram of the empty link. Okay. So to get the Kevanophomology chain complex, we will consider all possible labeled smoothings. As you can see, there are no crossings, so there's no choices to be made for smoothings. And you don't get any circles, so there's no choices to be made for labeling. So there's exactly one labeled smoothing. It's this one. It's generated over ZMA2. So we get ZMA2, and just, it's like obviously a cycle. So the homology is ZMA2. So this is ZMA2, and it's, yeah, okay. And, well, there's a clear diagram of the unknot. That's our favorite diagram. I think that's not controversial. And there's no crossings in that diagram. So we don't have to worry about smoothings, but there is one circle, and we could label it one or X. So to tell you what this map is, I'll tell you that, well, when you take the label X, you go to one in ZMA2, and when you take the label one, you go to zero in ZMA2, which does make you question the labeling conventions. So that's okay. Okay, so that is the example. And I'll tell you that probably my favorite thing about Kavana homology, well, first of all, I care about surfaces. So Kavana kabordas and meps are great, but the thing that I really like about them is that you can compute them. That's like a backhanded disk at not for homology. And there's lots of references for this. So the sort of canonical ones, I think, are Barnaton in 2002, wrote a paper about how to compute these kabordas and meps by like subdividing the surface into a sequence of very small standard pieces, at least in some cases. And then Elliot maybe did a few more cases in 2009. And then my favorite reference is the Hayden Sunberg paper from last year, 2021, which has some very nice tables of certain small surfaces and the corresponding maps really on the Kavana chain complex, okay, surfaces with like a fixed diagram. So let me tell you a few basic ones that we're going to use to compute the induced maps from these two surfaces on at least one cycle, because that's all we need to distinguish the surfaces is produce one cycle where we get one for one surface and zero for the other. Okay, so here's a few, here's a few surfaces. When I say small, I mean like really small. Here's a surface, which starting at the top, I'm going to have like a fixed diagram. So a diagram that looks like it has this little kink in it, and as I go down, I'm going to do a righto maestro one move that gets rid of the crossing, and this traces out a surface. The surface looks like this, does that help? Of course, you know it's isotopic to a product, but it's a fixed diagram. And so this induces a map on the chain complex. I'm only going to worry about circles that are all labeled X because I'm very lazy. And so then there's only two different labeled states, smoothings that I'm going to worry about because I could smooth this way or I could smooth this way. So this is the rule. This one maps to like the corresponding state downstairs, and this looks confusing because it's like a circle, but this is a zero. I don't really know if this is zero. So this disappears. So this is very easy to deal with whenever you have sort of a righto maestro one move, simplifying your surface. We're going to see that in a second. We can also deal very easily with righto maestro two moves, at least in the simplest case. So let's take the trace of doing a righto maestro two move, by which I mean like sort of this surface. And I'm going to say all X, I'm just going to give you the non-trivial ones. Well, if I smooth sort of the obvious way where like I ignore the righto maestro two, no surprise. This maps to the standard thing down here. And then something weird happens, which I'll explain in a second. But specifically, if I smooth like kind of the other symmetric way, and these two strands happen to be in the same circle, then this also maps to this. And all the other all X labelings go to zero. Okay. And now we have like the big guy. So what's what's the most useful small surface? It's this one. So this is a picture of a saddle. At the top, we have like two strands sort of vertical with some perspective. And at the bottom, we have two strands horizontal, and I've drawn an index one point, which I flattened out. Okay, because that's kind of the best I can draw, I guess. So, well, there's no issue with like smoothing up here. Luckily, I haven't drawn any crossings. And again, I'm taking all X labelings. But there's something weird here, which I'm going to motivate, which is that if I'm splitting one circle into two, then that's okay. Like this, this smoothing goes to this smoothing. But if I'm trying to merge two distinct circles into one, that is just the zero map. Okay. So let me give you some circular logic for why these are all reasonable. So why is this reasonable? Here's a fact. If your surface is smoothly compressible, I already told you about compressible meant. So that's great. But now I mean compressible in B4 or S3 cross I, then its induced map is just the zero map. I think this is a cool feature. So this is circular logic. So I'm telling you that if you know this is true, then these two rules make sense because really a tube looks like two bands, sort of the bottom and the top of the tube, but they're each like different of these types. Like you'll have one that looks like this, the one that looks like this. Okay. I won't draw that or explain that. If you've seen this before, you either know that and that's obvious or you don't know that and that's hard to visualize. But that's true. And so if you have a tube, you'll get at least one instance of this map and therefore the map is zero. So that's circular because really like you know these are the maps from Kovanov and then you prove this and it's like, ah, what a great feature. But if you don't know Kovanov homology, then you should like this and remember this and then believe that this makes sense. And then the same thing here. Why did I want these to be in one circle? Well, it kind of looks like that one. That makes sense. Okay. Okay, fine. Yeah. This merge the two, but this isn't merging. This is, um, what do you mean? Well, it's, it's, there's no critical point. It's, oh, don't throw me off. Okay. Please save any mistakes until after the talk. Okay. Um, so great. Uh, this is enough information that we can actually compute the induced map of both of these surfaces on one particular, uh, labeled smoothing. And I thought that I was going to have more space and now I'm like, where am I going to put it? Uh, this, oh, we don't care about the topological isotope theorem. This is not redundant. It's actually much better, but we don't want to think about that. Okay. Um, so let me draw a particular, uh, labeled smoothing of the diagram of the knot that I've drawn here. Um, I don't want to sort of draw it over one of the pictures because I think that'll be really hard to look at. Cool. Okay. I am going to draw it over this line. Yeah, I think it's fine. I think it's fine. Oh, it's not fine. Okay. This is like this top part here. Like, see, this is like sort of the top of this band. Okay. Um, okay. So I'll just claim that this is like a labeled smoothing. It kind of looks like the two knots. That's fine. Um, and I'm going to label them all X because that's all I told you how to compute. So if I have any one, so we won't know what to do. Okay. And let's, let's look at these two surfaces, how they behave with respect to this particular smoothing. So I'll draw parts of the surfaces in the picture so that you can see. So let's, let's start with Sigma zero and maybe I'll note that, well, Sigma zero includes a strip. So when I push the interior of Sigma zero into the four ball, it's going to look like I could start with the boundary. And then maybe I see this little strip first. We'll keep that a little bit higher than everything else. So it's like a band in the surface. Um, so if I look at how it hits this picture, that band sits here. Um, so when I'm computing the effect of the induced map of Sigma zero on the state, the first step is to say, well, this looks like, um, I have an index one point. That's what that band is. And it's, uh, it's merging two circles that are both labeled X. Um, so the effect of the map is zero. So just immediately with very little work, I can tell that, um, let's name the state. Fine. This is a cycle in Cavanaugh homology that, uh, the induced map from Sigma zero on this particular cycle is zero. Okay. What about Sigma one? Well, I'll just tell you, like, we can't find a band that looks like that. Um, so, so what do I do? Well, Sigma one, it's genus one. When it, when I push it into the four ball, what will happen is attached to the boundary. I'm going to see two bands that sit in the surface, maybe something like, uh, I don't know, this, okay. So I'll draw those in this picture. You see that neither of them merge two circles. And once these two bands appear, the rest of the surface is just a disc. Um, so I know this is kind of bad, but it's like blackboard, so I think nobody will really mind if I just sort of delete these for a second and call this D. Oh, well that's confusing. D. Okay. And I really want to think of this disc as, as not just a disc floating around, but a disc drawn in a particular way. Okay. And, um, I'm going to, um, I'm going to the effect of like doing both of these two band maps is to split one X labeled circle into two, which is fine. I just split it and I get a new state for this diagram. So I would like actually do the splitting. Okay. Well, so now, um, maybe we can justify why I chose the rest of the state, um, or like why I chose this particular state. This is carefully orchestrated so that, well, first of all, I notice that this disc is really nice, like this drawing of it is really nice. And I'm not just like complimenting myself. I mean that, um, I can unknot it by what you make it small and round and standard by only performing exactly these two moves, um, right in my stir twos that just pull two pieces of the disc from being over under each other to being apart and don't involve anything else on the surface. Um, and right in my stir one moves that just sort of like untwist a little piece of the disc. Um, so I don't mean that just as a move on the knot. I mean, that is a move on the disc. Okay. Um, so I could do right in my stir two moves here and then a one here to get rid of this whole part and then keep going around and just doing much of two and one moves. So it's really good that I never have to do our three moves, which are much harder to deal with. And it's nice that it's a very simple form of right in my stir one and two. Okay. And in particular, this state has been chosen so that every time I do a writer my stir one or two move, it looks like these non trivial models that I've drawn here. Um, so you can see like this right in my stir one move, uh, here that I would do to start simplifying my disc. It looks like the resolution that gave me the non trivial map. Okay. So, um, when I start simplifying my disc, I can change my state by just getting rid of the circle and I'll still have a bunch of circles that are all labeled X. Um, and similarly, I guess, uh, well, if I start doing right in my stir two moves, this R2, uh, looks like this local model. Um, so I'm just going to like sort of keep the same state and everything is just labeled X. And again, like I start doing another right in my stir two move and I'll be able to check that it will look like this local model. And this always happens. So at the end, what I get, I get that K H of sigma one evaluated on this same cycle. I'll be able to simplify it and get that it's K H of just the disc without any crossings, uh, evaluated on a cycle with all labeled X's, uh, which I carefully made sure to tell you was one and not zero. Okay. So I'll get one. Um, okay. And so we conclude the theorem. Um, well, this means these maps are well defined up to smooth isotope real boundary really like smooth if you morphism real boundary, um, which means that these two services can't be smoothly isotopic. Okay. Um, awesome. So, uh, without too much work, we can get rid of this real boundary thing by just, um, investigating the symmetry group of the knot. Uh, I did tell you it was a minus six frame negative white head double of the left-handed trefoil. So it actually has a lot of symmetry from the white head double part and the trefoil part. Um, but we know explicitly what the symmetry is. You could just check. I just checked. It was easy. Um, okay. So they're not smoothly isotopic. Um, the problem is that I originally told you that they're not topologically isotopic, which is obviously better. And this, I have to say it was harder than the not topologically isotopic thing. So what was the point of that? Well, the point of that, I lost my cloth. Uh, the point of that, well, I told you that they weren't topologically isotopic via an intersection form argument. That's algebra. Um, on the other hand, they're not smoothly isotopic by the Kavana Kubordism maps, which I'm just going to say is diagrammatic, uh, a diagrammatic argument. Um, so to get an interesting statement, we should do something to the theorem premise that is like good diagrammatically and bad algebraically. Um, so here's a theorem. Uh, the whitehead doubles. I'll say what that means because their surfaces of sigma zero and sigma one, these two particular surfaces, people on zoom can't see them, but you know, um, whitehead zero, whitehead, uh, sigma one are also not smoothly isotopic. Um, so I'll just draw a schematic of what I mean by whitehead double of a surface because usually we say whitehead double of a not. Um, well, let's, okay, let's say that this is, uh, one of my surfaces, sigma, um, its boundary is this not K. Um, this is probably how like, well, I'm going to construct a surface where I'm going to start off with two parallel copies of sigma. Um, motivated by how we construct a whitehead double, I guess, of a not. So here's K, uh, they each have genus. Um, and then I'm going to connect these two disjoint surfaces by a band, um, in, in the boundary. I can push the interior of the band a little bit in. Okay. So I hope this is clear enough. Um, so in particular, the boundary of whitehead of, if, if, if boundary of sigma is equal to K, then the boundary of the whitehead double of sigma is the whitehead double of K. Uh, so that's good. That would be awful if that weren't right. Um, we've also doubled the genus of the surface. So originally sigma was genus one, and now the whitehead double is genus two. Okay. Um, well, why did we do this? Um, I said that this is like really bad from an algebraic viewpoint or really good depending on like, I don't know your personality. Um, this has, uh, non, this has trivial Alexander polynomial. So it's, it's a theorem, uh, of Conway and Powell, uh, from 2020 that because, uh, the Alexander polynomial of whitehead K is one, uh, these two surfaces, I mean, they weren't thinking about these two surfaces in particular. I just, this just follows from a theorem, a whitehead sigma zero and whitehead sigma one are topologically isotopic. Well, boundary, um, so they prove that, that disks with the same boundary, um, with, with group Z and trivia Alexander polynomial are always topologically isotopic for positive genus. It's more complicated, um, but for cypher surfaces, it's easy. So, um, these two surfaces are topologically isotopic, but nevertheless, they're not smoothly isotopic. So that means we have an exotic pair, uh, which I guess, I don't know if you're interested in four dimensional topology is kind of like the goal, right? So that's good. Um, so cypher surfaces can actually be like very interesting. Okay. Uh, so in my last two minutes, I'll just tell you an open question. And I guess I'll mention that in fact, we have lots more examples. We have infinitely many pairs of examples or, or bigger finite families. Um, they can be higher genus that can be non-orientable. We prove a general statement about strongly quasi-positive surfaces. Um, and in particular, these, these two surfaces, well, the whitehead doubles aren't isotopic to each other. Um, and we produce explicit states or cycles in the Kavanaugh homology where the maps are different, but they both surfaces induce non-trivial maps. But we also have examples where like one of the surfaces is a trivial map and one's not, which is weird. Um, okay. But here's the question that's open, which I have no idea how to do. So question. Um, or problem. I don't know. Uh, can you find, uh, an infinite family? Uh, let's just say sigma i for i and z of ciphered surfaces, uh, same genus, genus-g ciphered surfaces, which, uh, first of all, you could just say like our pairwise, our pairwise, uh, not topologically isotopic. Um, I told you that the obstruction for us was the intersection form and the two full branch covers. Algebra makes me like really nervous. I just don't know if you can have an infinite family. Uh, that seems like weird to me. Like that seems really hard. Uh, anyway, uh, I don't know if you can make an infinite family that are topologically isotopic, but I also don't know if you can make an infinite family that are top isotopic, uh, but not smooth isotopic. Uh, and this is all in B4. Um, the short of two problems, or I guess really one fundamental problem here with, with either of these is that your, your first step will be coming up with this family sigma i and then proving that they aren't isotopic to each other. Right. Um, but I, I really, I really mean it when I say that, um, almost every construction that I'm aware of, uh, of surfaces in S3 that, that are not isotopic to each other are fundamentally isotopic in B4. Like just obviously isotopic in B4 shouldn't say isotopic. Some people proved it. So I mean, it shouldn't say obvious. Some people proved it. Um, so the exception is that these two services are based on a paper of Leon 1974. Um, a little bit different because his whole thing was making hyperbolic examples. And like these are twisted whitehead doubles of trefoil. So like super not hyperbolic. So probably wouldn't like this. Um, but it's basically the same thing. Um, but you know, he, he uses this symmetry, uh, of this tourist link to get a pair of examples. Um, and I don't see any way to adapt this to get an infinite number of examples all at the same time. Um, so the, the, the fundamental problem here is just how would you even produce your like infinite family, let alone show that they're different. But I don't know. I think this is like a really interesting problem to think about. Um, so now it's 46 and I can stop. That's right. So in fact, that's, that's how we get some other weird examples. Um, if you take like, uh, well, these aren't minimum genus, but they also have non-trivial induced map on Kavanaugh homology. But on the other hand, you could take the genus one surface and out of tube, and that has zero map of Kavanaugh homology. So it clearly can't, like it's compressible. It can't be smoothly isotopic to either of these. Uh, we do some things to, uh, we make them minimal genus by, um, I think Robert actually showed a picture. We instead of just actually stop instead of just whitehead doubling, we can then do like a band-sum with a trough oil surface. Um, we just check minimum genus with snappy ears. If you keep whitehead doubling, do they stay non-isotopic? Um, I'm, it's a diagrammatic argument, like, uh, uh, let me just think for a second. Like what, what we do is produce an explicit cycle here and then basically whitehead double the cycle with the one that we use. We make sense of what that means. Um, my guess is that if you keep whitehead doubling, you can keep repeating the argument for that particular cycle. But I didn't check and I hope is no, uh, uh, I don't know. I would ask Isaac to guess before I tried probably. I think probably. Oh, so he asked, do we have genus one examples that are exotic and, uh, no. Um, yeah. All of our genus one examples have non-trivial Alexander polynomial and I just don't know how to, well, I think, I think all of the ones that I can think of just aren't topologically isotopic and I definitely don't know how to deal with this without the Conway Powell theorem. So that's a good problem too. Are they cycles? Um, you can just check, like all of the zero smoothings in this diagram are, uh, it's, it's like they're all. Oh, actually, no, I don't have to say that. I think, um, every crossing, uh, involves two different circles in this state and everything is labeled X. So you're not, it's just, uh, 8 20, obviously. Yeah. Yeah. It's fiber and it's ribbon and it's a pretzel and it's better than everything. Oh, show that the two, uh, covers are not dif, oh, that's, so he, he asked, uh, for, for Zoom, he asked, um, the topological isotopy obstruction went through the double covers are not homeomorphic. So could you have shown that the double covers are not dif-humorphic using Hagar floor and like, I, I've never tried to compute that before. So I don't, maybe. Yeah. I mean, it seems really reasonable, but I'm, I would ask Irving. That's what I do. He's not here, though. Oh, he's here. Oh, like, yeah, I definitely don't know how to do that. Sorry. Yeah. So this is, um, uh, so, so, yeah, he said, uh, if, if you, like Kavanagh, the Kavanagh chain complex is so big, but somehow we've given away of, of just looking at like one specific part of it and using that to distinguish the surfaces. And really, I think this is what maybe all of Isaac's papers are about. It's kind of like Isaac's thing. And they're very good at it. And it's also an, it's Kyle's thing, too. They have a paper with this like table. And, um, this is how they distinguished, uh, the discs that, uh, Robert mentioned in the previous talk, uh, by producing an explicit cycle. And somehow the cycle, it's natural when you look at the picture, um, this cycle somehow looks a lot like Sigma one. Um, we sort of have these like regions that look like the overlapping parts. And whenever they actually overlap, we get a circle. And there's some sort of like general rule, um, for how to get a, uh, a cycle, or at least for like a strongly positive surface where you know the map will be in on zero. Um, and then it, we're just lucky that it doesn't look anything like this surface. Uh, no, actually, so I knew about these surfaces with, um, I was talking to someone and someone was talking to Jung. And we sort of like knew these surfaces existed. And then I gave a talk at, uh, a conference at ICERM. And I said, you know, like they're probably distinguished by their Kovanov or not formats, but I don't think anyone can compute the not formats. And I personally don't know how to compute the Kovanov maps. And within like three minutes of the talk ending, Isaac Somburg emailed me was like, what are the surfaces? And like I sent Isaac the surfaces and I was like, you know, maybe I'll hear back from them like eventually. And then like two days later, uh, they sent this picture. And I was like, and then, and then, uh, uh, Kyle had this like cool thing about strongly fuzzy, positive surfaces. And then we realized that it all didn't matter and they weren't even topologically isotopic. And that was like a whole thing.