 Thank you for the introduction, thank you for the invitation, thanks Tom for existing. Great, so I'm going to tell you that I'm not in number and cabling and this work is joint with Ty Libman and John Long Park. So here we have the truff oil, here we have the truffel. It differs from the truffle by a single crossing change and if you notice that this is in fact the unknot, great so today I want to talk about the unknoted number. Great so the unknoted number of k, well it's the minimum number of crossing changes maybe to unknot k, so one thing that makes a unknoted number particularly difficult to study is that it cannot always be realized in a minimum crossing diagram for your knot, right? So in 1984 Steve Blileau came up with this example of a 10 crossing knot where it has, so in its minimum crossing diagram, well there's no two crossing that you can change to a knot or knot but there's some 14 crossing diagram for the knot where two crossing changes do suffice to a knot or knot. What are some of my favorite knots? Well the torus knots are one of my favorite bosses of knots so the truffel is the 2-3 torus knot, more generally the p-q torus knot is the knot that sits on a torus that winds p times longitudinally and q times linearly and without loss of generality we'll always take p to be strictly bigger than 1, p is 1 then it's the unknot and by doing some symmetry properties you can always assume p is positive. All right, so what do we know about the unknoted number of torus knots? Tom, what do we know about the unknoted number of torus knots? Okay, well Tom doesn't remember but in 1993 he and Peter proved how they resolved the Milner conjecture so they proved that the unknoted number of tpq is p minus 1 times p minus 1 all over 2. So the unknoted number is one measure of complexity or not, what's another measure of complexity or not? Well the four ball genus is another measure of complexity so the four ball genus of a knot k is the minimum genus of a surface s where s is smooth, properly embedded, oriented surface and before his boundary is k sitting in the boundary of your four ball. Great, and so it's an exercise to show that the four ball genus is a lower bound for the unknoted number, basically you can use each crossing change to sort of sweep out a genus, but each crossing change sort of gives you one genus. Great, and so in fact what Peter and Tom actually proved, they actually proved that this is also the same as the four ball genus, the pq torus knot. They proved something more general about algebraic knots and sort of that goes by the name of the, it's called the local tom conjecture. Okay, and so maybe let's keep track of these facts that we're collecting. Great, so we know that the unknoted number of tpq and that's the same as the four ball genus. Any questions so far? Let's talk about some operations on knots. So let's talk about connected sum, and so one question you might ask is well how does a knotting number behave under connected sum? And so well the conjecture is that it's additive, that the knotting number of the connect sum of two knots is the sum of their unknoted numbers. Okay, so well one direction is clear, right, but certainly this direction is clear, and the other direction is open. Great, so what's sort of the state of the art about the unknoted number connected sum? Well the state of the art is due to Charlemagne from a while ago. Okay, so he proved that the unknoted number of a non-trivial connected sum is at least two. But for example, as far as I know it's still open, the unknoted number of the left-handed truffle connects some t25. It's either it's either two or three, but we don't seem to know. And then I guess maybe you'd also wonder well what's the four ball genus of a connected sum? So in some cases the four ball genus is additive, so for example the four ball genus of the connect sum of the right-handed truffle of itself, well that's it's additive, but in general this is sort of wide open. So for example the four ball genus of figure eight knot is one whereas the four ball genus of the figure eight connect sum itself. Well this is slice, right, so this is zero. What's another operation we can do a knot? So we can take the satellite of a knot. So take the satellite, we have a pattern knot p in the solid torus. Great, so maybe you have some pattern p, solid torus, and you have some companion knot. And then you basically tie your solid torus into the knot or more precisely remove the tubular neighborhood of this knot and glue in the solid torus where the longitude of your solid torus goes to a sacred frame longitude of your companion. And so we'll be, one particular number associated to the pattern that will be important to us is the winding number, the algebraic winding number. That plays an important role. So for example, it plays an important role in the formula for the Alexander polynomial of a satellite, which is the Alexander polynomial of your companion evaluated at t to the w. So this is sort of like, I think of this as like stretching your Alexander polynomial by a factor of w times the Alexander polynomial of p of the knot. Studying the knot in number of a satellite. So if you notice by the single crossing in your companion, well that becomes w squared crossings when you satellite. And so maybe a naive guess that the knot in number of your satellite is maybe at least w squared times the knot in number of your companion. Great, so that's maybe a guess. Well, the answer to that is nope. Wait, so let's look at the two one cable of the trough oil. One, two, okay. And so I'll leave it as an exercise for you to check if you're not a number of this is at most three. So there's three crossings you could change to get the unknot. And as far as I know, it's also open what the unknot in number of this knot is. There's bounds, I'd say it's at least two. So maybe by the end of the week someone can tell me what the unknot in number of this is. That would be great. I would be very happy. Maybe that was a bad guess. Maybe the new guess is that the unknot in number of p of k. Maybe it's at least the winding number times the unknot in number of k. All right, and so what's the state of the art in unknot in number of satellites? There's a result of Charlemagne and Thompson from 1988 that the unknot in number of p of k is at least two when the winding number is down zero and k is non-trivial. And then maybe a big open question is, well, the four ball genus of a satellite in terms of p and k. I don't have a good answer for that. So maybe study in general satellites is too difficult. So what if we focus on cables? So what if we focus on satellites where the pattern knot is a torus knot sitting in the solid torus in sort of a standard way? Think about, go back to our chart. Okay, maybe let's talk about the four ball genus of cables first. Maybe that's a difficult question, but I guess maybe I'll make a conjecture. Maybe we'll conjecture that the four ball genus of a cable is zero. Possibly exactly when the four ball genus of k is zero and q is plus or minus one. That seems like a reasonable guess. I guess there's some recent sort of evidence and maybe why you might believe this. So recently Dai Kang, Malik Park and Stoffigan proved that the 2-1 cable of the figure eight knot is not slice. Strong evidence for why you might believe this conjecture. But anyway, so what I want to talk about today is right here. I want to talk about the unnotting number of cables, which is joint with Kai Libman and Zhang Baik. Right, so we'll let kpq, I'll be the pq of kable, pq kable of k, where p denotes the winding number. And then if k is not the unnot, the unnotting number of the pq kable of k is at least three questions. Like most of my theorems, the truth relies on nozzle homology, which the graduate students are still here from last week. I've learned about it. And right, so that's due to Ajvap and Zabo and then independently, Jake Basky-San. And it relies on a cabling formula for inverse curves. So that cabling formula is due to Kanselman and Watson. And it relies on an unnotting bound coming from not-for-homology that's due to Akvamalashahi and Iman Ektifari. So before I talk about the proof of this theorem, I want to talk about some correlaries of the proof. So notice, in general, the unnotting number of a knot doesn't have to equal the four-ball genus of a knot, but for torus knot it does. And so one question you might ask yourself is what about iterated torus knots? So iterated torus knots are cables of cables of cables of torus knots. And so the answer to that is no. So this is the corollary of the proof of that theorem, which is that there exists a family iterated torus knots, K i, such that the gap between the unnotting number and the four-ball genus goes arbitrarily, goes arbitrarily large. And so I can even tell you what that family is. So K i takes the two negative five torus knot and then takes the two five cable of that and then takes the i comma one cable of that. And then it turns out that the four-ball genus of these knots is two i. And the unnotting number is at least four i minus one. And so maybe for the torus knot experts in the audience, maybe you look at this family, maybe something about this family that won't surprise you is that we have a negative parameter in here. So for example, for positive rate closures, the unnotting number always equals the four-ball genus. And for sufficiently large iterated torus knots, so those are torus knots where we're sort of the, sort of the cube parameters are all sufficiently large relative to the cube parameters in some society sense. Those, for that family of knots, the four-ball genus also always agrees with the unnotting number. The four-ball genus agrees with the unnotting number for positive rate closures and hence also sufficiently positive iterated torus knots. Well, what's another quarrel I want to tell you about? First, let me remind you of some things. So recall that j is concordant to k if the four-ball genus of k connects some minus j is zero. So here minus j denotes the reverse of the mirror image of j. I also want to talk about algebraic knot. A knot k is algebraic if k is the link of an isolated singularity of a complex curve. And this is equivalent to k being a sufficiently large iterated torus knot. So k of this form where qi plus 1 is greater than pi qi pi plus 1. And so, well it's the consequence of the of the local tom conjecture. Well, this implies that algebraic knots are in certain senses the simplest knots in the concordance class. So what do I mean by that? Well, this implies that algebraic knots minimize both the sacred genus, the unnotting number, in their concordance class. So in other words, if you have a knot that's concordant to an algebraic knot, well, its genus is at least that of the algebraic knot that it's concordant to. And it's a knot of numbers at least that of the algebraic knot that it's concordant to. And so, well, you might ask, well, are there other notions of complexity to which algebraic knots are sort of the simplest knots in their concordance class? Okay, the unalgebraic knot or an iterated torus knot, basically any iterated torus knot with one maybe small exception. So here we just, the first thing can't be the torus, it can't be the chuck oil. So this can't be the chuck oil. But besides this, you can have negative numbers here. So that's why this is more general than just being an algebraic knot. Okay, and well, the statement is then that k minimizes both the braid index and the bridge index in this concordance class. So in other words, in other words, if j is concordant to k, then the braid index of j is at least that of k, and the bridge index of j is at least that of k. And this relies on work of U-Haas, Maggie Miller, and Rengi, where they have a bridge index bound coming from novel homology. P2 minus 3 is also not okay. Thanks. This, right, so we expect that it should be true for all iterated torus knots, but sort of they're proofalized on novel homology, and sort of you need the novel homology of the starting knot to sort of be complicated enough, and it is for all torus knots except for either chuck oil. That's sort of what goes on here. Great. I'm impressed by your level of attention. Okay. Well, then I'm honored to be the first speaker. Great. Other questions? Great. So let's talk about, let's talk about the proof to a knot k. We can get it's not for homology, so it comes in a few different flavors. So we have HFK hat, which is, this is the invariant that categorifies the polynomial. Great. And so, well, the flavor of knot flow homology that we'll be using is the minus flavor. So, okay, so the graduate students who are here heard me talk about this last week. This is a finitely generated rated module over the PID F join U, where the degree of U is minus 2. Wait. And then right, so a finitely generated graded module over a PID always can be written as a direct sum of three parts plus torsion parts. And then it's a result of Peter and Zoltan that this is a non S3 was exactly one free part. And since since U has degree minus two, well, the only homogeneous degraded polynomials are monomials. So the torsion pieces look exactly like this. You can define a measure of complexity associated to this called the order of k, which I'll just declare this to be the biggest n i. If you're worried that that depends on your choice of splitting, well, you could just look at the torsion sub module and look for the smallest n that annihilates the torsion sub module and that'll be the same as this. Knot flow homology detects genus, so that's a result of Peter and Zoltan. And I'll phrase that in a particular way that sort of in terms of h of k minus. So the way I'll phrase that is that equal zero exactly when k is bad not. And so well, this this torsion order is exactly what features in all of Shahi and after parties are not in bound. So the car is that the not a number of K is at least the torsion order of K. And this is the same invariant that also features in the U Haas-Miller-Zemke bridge index bound. So for the proof of that corollary, it's also useful to have the following theorem that the bridge index of K is bounded below by the torsion order plus one. Okay, and so what's what's the main idea behind the proof of our theorem? Great. So the main idea is that well, if K is not well by Aschbach and Zauberl, that implies that the order is at least one. And then the cabling formula of Hanselman and Watson helps us that the order of the PQ cable of K is at least P. And so you should think of this here as it's somehow somehow under a cabling things get stretched by a factor of P. And that's analogous to the formula for the Alexander polynomial, the satellite, where you evaluate the Alexander polynomial at T to the P where it sort of stretches Alexander polynomial curves. So these immersed curves are a reformulation of the bordered invariance of Lipschitz-Aldschbach and Thurston, where to a three-manifolders boundary, they associate, say, a type D structure over some complicated algebra. But for three-manifolds of torus boundary, this has been reformulated in terms of immersed curves in a punctured torus. Great. So basically they take the not flow complex, or maybe the experts in the audience, they work over the VIN, F-adjoint, UV, mod, UV. Great. And then, well, to that they associate some immersed curve in a punctured torus. Great. And then, so this curve has lots of nice properties. So for example, if you want to glue together two-manifolds of torus boundary to get a closed three-manifold, but you just sort of intersect these curves in a certain way and then do Lagrangian intersection homology on the result. But what will be important for us is the following cabling formula, which says that if you want to take the cable of K, will you take the immersed curve gamma that's associated to K? And, well, if you want to take the PQ cable, what do you do? You take P copies of gamma each stretched by a factor of P, and then you shift them by a factor of Q, and then you connect them up in some certain way and then push them. So maybe that's best illustrated by an example. Okay. So I said that these curves live in T2 minus a point. Well, we'll draw them in the universal cover of T2, and then the point will just lift to sort of give us a lattice upstairs. Right. So being very unassociated to the figure eight knot looks like this. So sort of this is lattice here, and then this is sort of just sort of the fundamental domain of the lift of the curve. And so what's the relationship between this immersed curve and the knot flow complex? Okay. Well, if you look, if you look at these like, okay, I'll call them like right arcs. So you look at this, you see this arc here? It sort of is on the right of some of these points. And it sort of, it sort of arcs around a single, a single one of these marks points. Great. And this, such arcs are in one-to-one correspondence with torsion summands in Hfk minus. So this arc here, since it sort of is length one, this is, this is sort of in one-to-one correspondence with a effigy and u mod u to the one summand in Hfk minus. And so, well then, then this cabling formula of Hanselman Watson, so it tells us we stretch out, it tells us we stretch out, and then we, we smush everything. So, right, so this is the immersed curve for the figure eight. And then what I'm going to draw here will be the immersed curve for the two one cable of the figure eight, which looks something like this. Okay. And so, so what did we do, right? We, we stretched the picture, and then we had these two copies of it. And then from this picture to get from here to here, we just compressed everything horizontally. And so the key thing to observe, right, well we had one of these arcs here. Now, now let's sort of follow through the image of this arc in this picture. And now, now you see it here. And so, well it got stretched by a factor of p. And so now, well it used to just sort of go across one point. Now it goes across p points, where here p is two, two one cable of the figure eight. Great. And so, well this is going to imply that we have, in this case, an effigyuanu mod u squared sum in an hfk minus. And then using a la shahi and eptakaris, a knotting bound, well that's going to tell us that a knotting number is at least two. That's everything I wanted to say, so thanks for your attention. Oh, what's next? It'd be great if we could just look at all satellites. Sort of the Hanselman Watson formula is sort of very special to cabling. Maybe it could be adapted to what are called one one cable, one one satellite knot. So those are ones where the pattern in which the Hager diagram that's genus one. Oh, well the, a knotting number of a whitehead double is always one. So it's a sort of non-zero winding number is the interesting case to look at. Great. So the question, great. Yeah, the question was, well, there's a knotting bound coming from Kovanov homology. That's sort of the analog of this and it's the same true for the a la shahi eptakari knotting bound where I guess rational and not a number means instead of just changing a single crossing you replace a rational tango with some other rational tango. Yeah. Don't know if there's anyone in the audience. Great. Robert says that he believes that, so great. Yeah, so I guess, I guess, yeah, the rational and not a number of cables. Yeah, I'll tell you, PXPose.