 sigma and sigma prime. Then I can do a similar thing. I can stabilize the ambient form manifold. So the statement is if I take sufficiently many connected sums with S2 cross 2, these exotic services will eventually become smoothly isotopic. And this operation where we change the ambient form manifold is called an external stabilization. Alternatively, I could modify the services themselves in some way. So here the statement is, again, if I've got some exotic surfaces, the analog of stabilization here is I'm going to be attaching some tubes to the surface. And if I attach enough tubes, again, these two surfaces will eventually become smoothly isotopic. And the conjecture in all of these cases is that a single internal or external stabilization should be enough to dissolve the exotic behavior. Since it's a little bit of loom mating, let me sketch the proof of this theorem of bicarin synixians. The idea is, okay, sigma and sigma prime, they're homologous. So they co-bound some safer hypersurface embedded in X, which I've sort of drawn cartoonishly here. And I can think of this as a co-bordism between sigma and sigma prime. And I can give this a handled decomposition. So why is it three manifold? So it's going to be built out of some one handles and some two handles. And let sigma tilde be the top boundary of the one handle co-bordism and sort of the bottom boundary of the two handle co-bordism. I can describe sigma tilde in two ways. It's the result of doing S0 surgery on sigma, just sigma. In other words, it's sigma with a bunch of tubes attached. Dualy, it's also the bottom boundary of Y2, which means it's also S0 surgery on sigma prime. So it's sigma prime with a bunch of tubes attached. But these were two different descriptions of the same surface. So this is the claim, actually, that if you attach tubes to sigma and sigma prime, eventually they'll become smooth like X topic. But this co-bordism is sort of found abstract. You have no idea how many one handles are in this handle decomposition. You have no idea how many tubes are really necessary to dissolve the exotic behavior. Okay. So and again, in a closed case, I believe that every known example of closed exotic surfaces in a four manifold, a single stabilization external or internal is enough to dissolve the exotic behavior. But the sort of positive evidence is countered by negative evidence in the relative case. So last year there's this paper of Lyndon Mukherjee where they found a link in S3, which bounds an exotic pair of linked disks in a punctured K3 surface. And they show that this link actually remains exotic after a single external stabilization. So you can take a connected sum with S2 class S2 and this link pair will remain exotic. And our main result is that there's an analogous thing for this internal stabilization. So we show that for any positive integer P, there's some not J sub P in S3, which bounds a pair of exotic discs, DP and DP prime in the four ball. And moreover, these discs remain exotic after P minus one internal stabilizations. So attaching P minus one tubes. And just some terminology here. We say that the stabilization distance between a pair of exotic discs is the minimal number of internal state stabilization is required to make the two smoothly isotopic. Excuse me, sorry. So rephrasing, our theorem just says that the stabilization between these two discs we construct is exactly P. So the main tool we use to prove this result is not fluorohemology, which we've heard about a fair amount this week, so I won't say too much. Recall the structure is given a knot, there's some associated effo joint U module, HFK minus and if I have a co-partism between two knots, then there's an associated map between the not fluorochemologists. And since these maps are invariance of the surfaces up to ice-to-be-roll boundary, if you want to distinguish two surfaces, it's enough to distinguish that they're induced maps. Not fluorochemology is really well suited to studying stabilizations because a stabilization has a very simple effect on the induced maps. So if sigma tilde is the result of an internal stabilization of sigma, then you can write down explicitly what the induced map is in terms of the map induced by sigma. F sigma prime is just U times F sigma. And this means that not fluorochemology immediately gives you a method for finding a lower bound for the stabilization distance. If you've got a pair of surfaces, sigma and sigma prime and you want to show that they're still going to be distinct after you stabilize n times, what you can try and do is show that the two maps are going to be distinct after you multiply it by U to the n. Give me a moment here. So maybe the first obstacle is that there are not a ton of known exotic disks floating around. One example is due to Kyle Hayden. So he has this knot, which we'll call J, that bounds a pair of disks, which are topologically isotopic but not smoothly isotopic, no boundary. On the left, there's sort of an immersed picture on the right. There's a handle decomposition and the bands here are the saddles in these two disks. So maybe the first thing to try is, well, one, can you compute the maps induced by these disks and two, after you multiply by U, will the maps be the same? It turns out that you can compute the maps induced by these disks and moreover, they're distinct. So not fluorochemology can tell the difference between these two exotic disks. So the next thing you want to do is say, well, if I multiply by U, are they still different? And unfortunately, the answer is no. The maps end up being equal. And this follows some from this general fact, that if I have a ribbon knot K, so a knot that bounds a ribbon disk in the four ball, then I can consider its fusion number, which is the minimal number of bands that appear in a ribbon disk per K. And then it turns out if I have any two ribbon disks with boundary K, the maps always are going to be equivalent after I multiply by U to the fusion number. So if I go back to these examples here, these disks both have a single saddle, which means their fusion number is one, which means a single power of U is always going to be enough to make the two maps equivalent. But that's actually fine, because you can show explicitly that if you attach a tube in the right spot, these two disks actually become smoothly exotopic. So the surfaces are the same. So the maps better be the same as well. So the punchline is, if you want to use Hfk to find a pair of disks with a large stabilization distance, you need to work with a knot that's got a large fusion number. Otherwise, the single power of U will always sort of make the two maps equivalent. And the way we'll find such candidates is by using a satellite construction, which I'll remind you what that is. So here's how satellite works. You pick a knot in S3, and you choose some pattern not P, which lives in the solid torus. So here's sort of a cartoony picture of this. I've got this knot that lives in the solid torus. What I do then is I take my knot K that's living in S3, I cut out some neighborhood of that knot, and then I glue in this new knot that I have in the solid torus. When I do that, I get what's called the satellite of P with pattern K. And as you can see, well, you know, this satellite knot is sort of clearly related to K, but it's got some more complexity due to the complexity of the pattern. So this pattern that we've shown is called the 3-1 cable pattern. So cables are when I pick the pattern knots to be the torus knots. And here 3-1 means this knot wraps around the longitudinal direction three times and once around in the wraps around once in the meridional direction. And the really powerful thing here is actually there's a four-dimensional extension of the satelliteing business. If I have the concordance between two knots, which I'll call C, and recall the concordance is an annulus embedded in S3 cross I, such that one boundary is K and the other boundary is K prime. This induces the concordance between the satellites of those two knots in almost the same way. I take some pattern knot in the solid torus and I cross this with I. Then what I can do is I can just cut out a neighborhood of the whole concordance in S3 cross I, and I can glue these two things together. And just by construction, this new thing is going to be a concordance between the satellite of my first knot to the satellite of the second. So back in our situation here, we had these two disks that both bound this knot J that Kyle Hayden found. If I puncture those two disks, that will give me a pair of concordances, which I'll call C and C prime. If I take this pattern to be this cabling thing, this P1 cable, then I get concordances between the P1 cable, the unknot, and the P1 cable of this knot J. But if we go back here, if I wrap around P times in the longitudinal direction and then once in the meridional direction, this knot here is actually still the unknot, which means I really have a concordance from the unknot to the cable of J, which means I can just cap off this concordance and this will actually give me two cable disks, which now bound, which have bound to the cable of J. Even better, it turns out this knot J has fusion one, fusion number one as we saw, it had a single band, but after you cable it, the fusion number increases. So the P1 cable actually has fusion number P, which is what we wanted. So it also turns out that these cable disks that I've shown here on the right are still topologically isotopic and this just follows from the fact that I had a topological isotope from D to D prime and the existence of topological normal bundles tells me that this immediately gives me a topological isotopic between the cable disks. Like I said, these disks have now sort of lots of saddles necessarily, so there's some hope that if I multiply by U lots of times, the maps will remain distinct. So the last thing to do is figure out if you can compute the maps induced by these cable disks by bootstrapping some knowledge about the maps induced by the original disks and that's what we do. And it turns out the right tool to do this is bordered Haggard-Fleur homology. So let me briefly tell you what this is. So bordered Haggard-Fleur is a package of invariants for three manifolds with parameterized boundary. So to a surface, there's an associated differential graded algebra, which we call A of F. And then if I have a three manifold with boundary, there are two invariants. One is a left differential graded module, which we call CFD hat, and the other is a right A infinity module called CFA hat. And the really powerful thing about bordered Haggard-Fleur homology is that if I have two three manifolds with common boundary, I can glue them together. And if I want to know what C have hat of this glue three manifold is, that's homotopy equivalent of this derived tensor product of CFA hat. The first thing tensor CFD hat of the second. And there's a version of this pairing theorem for Knott-Fleur homology as well. If I just think of Knott-K is living in one of these three manifolds with boundary, this box tensor product recovers the Knott-Fleur homology of K living in this glue three manifold. And this is sort of really well suited to studying satellite operations, because that's exactly how we define these satellite knots. I had this pattern knot that was living inside the solid torus, and I glued it to the complement of my knot in S3. And bordered Fleur then tells you exactly how to compute H of K minus of this satellite knot. So then we also describe these cable disks as sort of luing together two co-bordisms with corners. And the next theorem, which I won't read all of, I'll just try and give you the idea. Excuse me. Given a concordance between K and K prime, we saw that that induced some cable concordance between the satellite of K and the satellite of K prime. And the fact is that you can compute the maps induced by this satellite concordance by passing through bordered Fleur homology. And sort of exactly the way you want. We built this satellite concordance by first cutting out a neighborhood of the concordance and then gluing in a solid torus cross pi. And you would really want the map induced by that concordance to be like the identity on the solid torus tensor product with some map induced by the complement of the concordance. And that's what we show. So here is basically how the rest of the proof goes. You can use the fact that, okay, you know what FD and FD prime are, and you can sort of bootstrap that knowledge to figure out what the maps associated to the complements of these two disks should be. And then that previous theorem tells you that if you want to compute the maps induced by these cable disks, then you can compute these as some tensor product with these maps associated to the complements of these disks. Once you do that, you use just a direct computation that shows if you multiply these two maps by U to the P minus one, the maps are still distinct. But a single additional power of U, the maps become one. So again, that shows you that the stabilization distance is at least P, which is already maybe great. And it's not so hard to show that stabilization distance is exactly P by just constructing an explicit isotope between the P fold stabilizations of DP and DP prime. And that was our main result. And I think I'm roughly in time. So let me stop there. Thank you so much for listening. Yeah. So the knot you started with has fusion number greater than one. It's actually, I think it's open that the N1 cable has fusion number N times P. I think Jen would probably know the answer to that question.