 Our next speaker of the morning is Juanita Caizón-Pinzon Caizado, who will speak about satellite operations that are not homomorphisms. Is this working? Okay. Okay. Thank you, Paola, for the introduction, and thank you to the organizers for inviting me. It is a huge honor to be a speaker in Tom's birthday conference. Tom's work has been incredibly influential in my career. Unfortunately, the topic that I chose for today is not one in which I use a lot of his work. So my apologies, Tom. Thank you. So for the people in the Internet, he forgives me. So satellites that are not homomorphisms. So this is joint work with Ty Leedman and Allison Miller. And for details, you can already find the preprint in the archive. Okay. So, you know, I think that the first thing that I am going to do is define satellites and then what do I mean by homomorphisms? And especially for the grad students, you know, this talk, I want you to make the most out of this talk. And so if there's a concept that I talk about that I don't define or that you would like some clarification, this is my invitation for you to just stop me and ask. Okay. So questions are welcome. And there's no question that I am not going to try to answer. Try. So what's a satellite? I think we already saw it a little bit on gen stock, but let me do it again. Let's say I have a knot. Well, this is a funny picture of the knot. But the thing is that here what I want is I want to embed it in a solid torus, in an unknotted solid torus in S3. And then I'm going to take any knot, say the trefoil. The trefoil does have a tubular neighborhood around it. And the tubular neighborhood is homeomorphic to a solid torus. So the satellite operation, so this is going to be the pattern. This is the companion. What it produces is a third knot in which I basically, you know, like I cut this open along this meridional curve. I tied along the knotted companion. And whatever image of my pattern I get in these knotted solid torus, that's the satellite knot. We need some twists here because I want this operation to be untwisted, which means that I want meridian to go to meridian, obviously, but I also want longitude to go to longitude. There's another, so let me call this or a meridian of the unknotted solid torus eta. And the winding number of my satellite operation is the linking number between the pattern, let me call it P, and the meridional, not the meridional disk, the boundary of a meridional disk. This is the definition. So if I have a pattern and its winding number, this is going to give me a lot of information about this function. So here already I am sort of hinting at the homomorphism, but I have to somehow give you a group from the knots. And so the second notion that I need is the notion of concordance. So I have that two knots are concordant. There exists an annulus, smoothly embedded in a cylinder over S3, in such a way that if I restrict to the ends, then I get the knots that I started with. So they co-bound an annulus for dimensionally. So the theorem is this is Fox and Wheeler. If I take the set of all knots in S3 and I'm out by concordance, then this is going to be, well, let me call this quotient space C. But what they tell me is that, well, one, concordance is in fact an equivalence relation. And then this quotient is an abelian group with connected sum as the operation. What do I mean by connected sum? So I say I have again my trefoil, why not? And then I can have any knot, let me be lazy and then just hide it there. The connected sum is remove a little arc from each one of the knots and then identify the ends. Then so just some observations. Class of the un-knot, let me say if K is concordant to the un-knot, this is equivalent to saying that K bounds a slice disk or a disk. And so these are called slice knots. And inverses. So say I have any knot and I want to take the inverse. So what I get is that this is the same as the class as the mirror. What's the mirror? So for the trefoil, here's my right-handed trefoil. And the inverse I'm going to get by changing all of the crossing. Whatever strand was going above now goes below and by servers. And it's in this setting that I have my group. So here I have just an operation that goes from knots to knots. But the thing is that, well, basically a tubular neighborhood of this annulus in this four manifold is basically the same as taking a solid torus cross i. So I can actually do this operation at each level in the concordance. And so what I have is that my pattern induces an operation in concordance. So I want to obstruct these maps from being homomorphisms in concordance. Where does this come from? So this is a conjecture of Matt Heddon. Well, I heard it first from Matt Heddon officially in 2016. But you might correct me if you know more about this. This is when I heard about it, is that if we have a satellite operation, then it is going to be a homomorphism. P is the trivial homomorphism. P is the identity. And P reverses orientation. Or is the map that reverses the orientation of the knots. Okay, I'm going to obstruct these satellite being homomorphisms. So there's one obvious one which I am just going to take for granted, which is I need this knot, meaning when I remove the solid torus, when I forget about this curve eta, I want this knot to be a slice knot. So I'm just going to assume that. And then the next one is I need for any knot. I need the connected sum of its image under P because they match of its inverse under P to B slice. And this is what I am going to use. So this is what we obstruct. And just in case, sometimes I will refer to this as a pseudo-homomorphism. Those who do satisfy this condition, I will call pseudo-homomorphisms. Great. So let me show you as an example that this cable is not a homomorphism. So how am I going to do it? Well, actually, the obstruction is really going to come from obstructing three manifolds from bounding simple four manifolds. So what I am going to do is instead of looking at these knots, I'm going to look at some Q-fold covers of S3 branch along these knots. Why am I allowed to do that? So there's a theorem by Kasun and Gordon that tells me that if I have a knot that is slice and then I have a number Q that is a prime power, the Q-fold cover of S3 branch along my knot is going to be the boundary of some rational homology for a ball. So if I show that one cover is not the boundary for rational homology for a ball, then I am showing that a knot is not slice. That's what I'm going to do. How am I going to do it? There's this wonderful invariant by Peter. Am I going to get it? Did I get it right? So they define for, let's say that I have why rational homology three sphere, meaning a closed oriented three manifold that has the same rational homology as the three sphere similar for the four ball, then why is the boundary of a rational homology four ball? If this happens, then there's an invariant D with some spincy structure that is going to be zero. Or if you want, if I take, this is maybe the way I'm going to use it. I'm just going to forget about the spincy structures and the way I'm going to be able to do that is instead I'm just going to define the max of y to be the maximum of all of these D invariants for any spincy structure. And if you don't know what a spincy structure is, really they're going to be featured as, because they are required as an input for these correction terms. Can you really see here? Okay. Oh, okay. So this is in one-to-one correspondence with two times the second homology of my rational homology three sphere, which, you know, because it is a rational homology sphere, this is going to be a finite abelian group, so I can in fact take the maximum. And so what I'm saying is, or the way we're going to use this is that if it bounds a rational homology four-ball, then this is going to be bigger than or equal to zero. Okay? This is the way we're going to use it. So if I show that this D max is negative, then I'm showing that these three manifold does not bound a rational homology four-ball and therefore my nut is not going to be sliced. That's the way I'm going to argue. For this particular cable, so this is actually called the two-one cable of a nut. Remember, the winding number is two in this case. It's the linking number between eta and my pattern. And so I'm going to choose q to be two because two divides two. So really I want q to divide double. In this particular case, so let me draw a picture of this rational homology sphere or which homology sphere. So to understand the two-fold cover of the two-one cable of any nut, one way is that given by Abel and Kirby. So notice that here I have a merbius bound and this is going to be true all cables. You see here this is a merbius bound for all two-one cables. And so, again, for any cable, if I am lazy and I draw it like this, if I take the central circle, what Abel and Kirby tell me is that then the two-fold cover is the result of cutting this circle in half and then rotating it to the bottom and changing orientation. So these are here, means reverse orientation. And then I'm going to put a framing number here which is going to be twice or the number of half-twists that I find in my band. So in this case, I have plus one. So with this description, what I have is that this is plus one surgery on K connected some KR. Great, this is wonderful. Because we understand surgeries, especially when it comes to Dean variants, I think that Peter and Sultan, even when they defined the Dean variants, they gave us the following computation. If I have, and by the way, in this case, this is not just a rational homology sphere, but an integer homology sphere. So I really only have one spincy structure. So I don't even have to take the max, but whatever. So for the two-fold cover of the two-one cable of the two K plus one torus knot, this is going to be minus two K if K is bigger than zero or zero if K is less than or equal to zero. And so what do I have again? So I want to obstruct this knot from being sliced. I want to show that this is not sliced. What I am going to do is I'm going to do it in terms of the two-fold cover. But turns out the two-fold covers behave very well with respect to connected sum. And so do the Dean variants. So here, the Dean variant here is going to be the sum is that this is going to be negative two K plus zero. So if I choose K is bigger than or equal to one, then the knot is not sliced. So it's enough to have one. In this particular case, we're giving you infinitely many. It's an overkill. One would have been enough. So the trifle is just enough. But life is not that good and not every cover is going to have this very simple form. It's just not going to happen. For example, I think that if you do the three-one cable, which means you put one more strand and one more crossing between the new strand and the next one, you're not going to get surgery. So what do we do? Well, there's a second description of these three manifold, which must actually generalize to all branch covers, cyclic branch covers over satellites. So notice that here, what I am doing to get my pK, I did was I consider the complement of the tubular neighborhood of my three manifold, and then I replaced this tubular neighborhood with these solitors. So let me call these solitors the complement of this unnotted editor. And so in this case, what I have is that p of K is really going to be in this part. It does leave in this part. So here, let's say H is that identification between these unnotted solitors and these notted solitors. And I have that I can then represent p of K in a much simpler way. I'm just going to redraw this. What I am saying is if I replace this component by the not exterior, let me call this p of K in such a way that the meridian of K goes to the longitude of eta, and the longitude of K goes to the meridian of eta, then in this case, what I see here what used to be just a knot knot, if I replace by a non-trivial knot, then this is precisely my p of K. And I want to branch over that. But to be honest, I only really know how to branch other than Aguilu Kirby that says if you have a cyphered surface, and then they give you an algorithm to construct a surgery presentation for the branch cover, the only other thing that I can do is I can just branch along the knot, because that's just a matter of rotation. Well, I am super lucky that this knot is symmetric, which means that I can just basically change the colors. And now this is my eta, and this is my p of K. Again, I am replacing the two-learned neighborhood of eta by the not exterior. And now I can actually take the two-fold cover. So I have my crossing here, my crossing here. This is my axis. So again, I am going to be gluing a knot exterior to each one of them. That didn't change. But now I have to tell you what the gluing looks like here. And for that, well, this part is going to remain the same. This is fine. That checks. But I have to be careful about this identification. I have to know where the longitude goes. And so a longitude here, it's very easy to see. But a longitude here, well, now I have right. So I have to compensate for that right. And so in this case, I would have to draw a curve that looks like this. And so I need a curve like that in both components, which really is the minus one one curve on each component. So what I am saying here is that in this case, in the two-fold cover, the meridian of my knot is going to go to the minus one one curve, each lift. So now notice that here I am completely describing the branch cover of the satellite by telling you, by giving you two pieces of information, basically, the lifts of this eta curve and the curve, the lifts of this lambda eta, the lifts of those to the cover, which is not the same as the longitude of each one of these eta one and eta two. Great. Okay. So why am I bothering describing all of these and taking a long time to do that? Because in the end, well, one, as I had sort of mentioned, a similar composition is true for all branch covers. So if I have my satellite operation, I have my winding number. So from now on, I'm going to think of my satellite operation in terms, or my pattern, really, as a two-component link. Instead of defining the precise embedding of p into an unnoted solid torus, what I can do is I can give you a two-component link such that one of them is unnoted. So here I have my winding number, and then I want q prime power such that q divides w. And so in this case, I know that the q folk over any satellite is going to be composed of two things. I take the branch cover over the slice notch that I started with, except I remove the tubular neighborhood of all the lifts. And because I'm choosing q to divide the winding number, I am actually going to get q lifts. So in this case, I have two. I'm going to glue in q copies of my node exterior in such a way that each one of these meridians is going to go to the lift, the height lift of the longitude of my error. And so that's just basically the same as what I am doing here. So how is this going to work? Why am I talking about this particular example? Well, because in the end, our theorem, which I'm about to write down, is going to follow from trying to massage any one of these branch covers to that. In what sense? Well, first let me write the theorem, and then I'll tell you how to do it. So let me say the same, well, I have my pattern, I have my winding number, and I have a prime power that divides the winding number. If I have the following two conditions. One, if I look at the linking number in the q-fold cover of p of u, any pair of lifts, I want this to be non-negative and not all zero. I want each one of them to be a null homologous knot. Satellite operation is not a homomorphism. Okay, so a few comments. This linking number, this is a rational linking number. Possibly, you know, I have a rational homology sphere and so if I have a knot, it's not necessarily going to be null homologous, and if it is not, then the linking number between two non-null homologous knots is going to be a rational number. Just as a word of caution, I don't mean the linking form. The linking form collapses all of the integers to the same thing, and I do want to make a difference between, say, negative one and four. So it's not the linking form. It is the actual linking number that is well-defined up to isotopic. Okay, in this case, because I am assuming that they're null homologous, these are actually integers. Questions so far? Any comments? 13 minutes to tell you how to prove this theorem. Proof of the theorem, basically, so let me do the sketch in case I don't finish. So I need three steps. Step one. I start with a three-manifold that is not a homologous sphere. So I am going to do null homologous surgeries or integer surgeries along null homologous knots. Let me say that sigma is of p of u to get acobordism that takes me from this sigma, which is my link at a one through at a q, two s3 with another knot, sorry, with another link, feel a q-component link, and it's that the linking information is the same, including the framings. So whatever I have on one side, I want to have on the other side. And then because, so remember here that I said, for any companion, really the defining piece, you know, like this is constant to all of them. All of them have this piece. So if I have sort of like the idea is if I have a statement about this side, then by just extending it to this other side, I have a statement about all of the covers. And then because I have acobordism, then I am going to get a bound. So then get a bound. Like I said, what I want is to get to this point, to this link over here. Well, I don't know what these linking numbers are in general, but here I do know that the linking matrix, this knot is going to be negative one, one, one negative one. Right, by framing I mean, you know, like there's a way of keeping track of which curve the meridian goes to by just having an integer. So I want to somehow get to that point, except, well, if I start with, if Q is five, then how am I going to go back to these two only two by two? Well, I'm going to allow this matrix to be a little bit bigger. And then here I'm going to have just a bunch of zeros. And I'm going to have some unlinks, or Q minus two unnoted unlink components. So here I do change the crossing. Yeah, surgery alone, unnoted, negative one frame unnot. So this gives me a second co-board design. Now from this L to a new L prime, such that now the linking number of this new one is precisely what I have over there. So it's going to be one for one particular pair, say for the first two components, negative one if I mean framing, and then zero for everything else. So basically I am just changing it so that I get this linking matrix. And so now I have the linking matrix, but I really want to get not just the linking number, but the precise link. And so the third step involves two class per surgeries to transform S3 with this L prime into now the actual, let me call this the HQ. So the two first components are hopefully with negative one framing, and then all of the others are zero framed unnotes that are unlinked. Okay, good. So this is the sketch. Now let me give you a little bit more information. Questions so far? Anything? Okay. Yes. Yes. So, okay. So maybe I'm just going to start third to first because Dani asked a question about the third one. So class per surgery. A class per surgery on a three manifold is the result of replacing this genus three handled body. Let me draw it by. So where are the handles? Let me cheat because I don't want to get it wrong. So, wait, so you just have the handles joining these, the same end points in such a way that they alternate and same thing here. So I have these with these. I have these front and I have these. So there's, I wish I could show you a better picture. This is the same or equivalently, you replace the Borromean links by you do the surgery along this Borromean links. Or equivalently, if pi is the element of the mapping class group of the surface of genus three, such that I have a three handled body union, another three handled body along phi, this gives me a Hagar decomposition for these three torus. Then the class per surgery or Borromean surgery is the result of, you know, so if I start with any three manifold and I have any handled body genus three, I have it with F with my another genus three handled body, then the result is just replacing this by the composition of this new one. So let me just say that I just, I just change it by, in whichever way this element, this self-deferred morphism of the surface tells me. And what's important is that, well, maybe I'm not, maybe it's not clear by this, but if you want more details, I can tell you. Because I only have two more minutes and this is not where I wanted to go. So this is an element of the Torelli group of the genus three surface, meaning the induced map on homology is the identity. And so it really doesn't change the homology at that point. And what we have, so let me say, so part of our results, so I always do, in each step I do topological operation that has a four dimensional counterpart and I use that four dimensional counterpart to get a bound in the D invariance. And so in here, what we have is that if I have an element of the Torelli group of really any surface of genus G and maybe here I can just call it M sub phi instead, if M is any three manifold and M psi is the result, if I have a genus G Hager splitting, then I am just going to get a bound on the D invariant by a constant that it depends on the expression of the factorization of phi in terms of the generators of the Torelli group which has bounding pair maps and dain twists. Okay, so basically what happens in the end is by step one I get that the D invariant of sigma will be max minus the D invariant actually and this is the important part. I have, again, I have these operations that give me a precise way of changing my link and so because I have it with the link then I can say that this is true for any companion. It's not just a statement about the first one of them all. And then on the other side, well, what do I do? I take the complement of my link L and then I take that with the union of Q copies of the nut exterior which is exactly, so I just carry the same gluing that I had and I get bounds in each step officially out of time and I didn't finish the proof but, you know, we get one bound here then I get another bound here and then I get another bound here and all of these are independent. This is the key. They don't depend on the companion so I get to choose and so then what I have is that then I have the difference between this D invariant and the D invariant of my beautiful spaces that I had here that is bounded by some constant so I can just choose this D invariant to be large enough or negative enough to compensate for the existence of the constant and then I can just choose my K so that I get something negative and then therefore I have non-slice this of the connected sum of P of K with P of minus K. Okay, that's it. Sorry. Questions for Juanita? Yes, yes. So we start with a factorization of psi. Okay, so let's say that first, so first let's do it for dame twists. Well, a dame twist plus or minus dame twist, let's say. That's nothing other than surgery. So I already have surgery and then if I have integer surgery, I have the trace co-boardism, the co-boardism that is induced by attaching the two-handle and then I get a bound from that. So that's sort of for free. Then what happens if I have a bounding pair map? Well, in that case, the proof is similar in the sense that what you need to do is just in your four-manifold, you want to find a closed surface that sort of contains all of the second homology and you want to remove it so that the complement is a semi-definite four-manifold because then you have bounds for what happens with semi-definite four-manifolds with bounding. And basically, that's what we do. We just construct the correct surface. If you have a bounding pair map, then you start with... So bounding pair map means... So say I have something like this and you have alpha plus, alpha minus. This is a bounding pair map. So I can construct a surface by taking, say, this side and I take a cypherd surface for these two and then that's the one that I excise and when I excise that, I get a lot of control and that's what allows me to get this. So when I get a tubular neighborhood of this closed surface, I am introducing a three-manifold into the co-boardism or into the new four-manifold which is the circle bundle of the only number N over a genus, whatever surface and then you just have to be careful with the spincy structures over there and they bounce on the D invariance for those manifolds, but those have been computed. Yeah, so there's a theorem. I mean, it's just not clear. A lot of people did work to tell us that if you have an element in the Torelli group of a closed surface, then you're going to be able to factor it in terms of dame twists along non-separating curves. Sorry, separating curves. Yes, the opposite. Separating curves and bounding pairs. That's the theorem. So then from there we can go. Any more questions? Marco. For this? Yes. This one, there are formulas. You know, there's a paper by Cha and Ko that tells you if you give me a presentation of the rational homology sphere as surgery on a link, integer surgery on a link, and I'm going to be able to compute the linking number of that. So that's a formula. But you do have to produce some sort of surgery description. But you can do it using a curved curve. So that one is very computable. In fact, I kind of want... I think Snappy should be able to do these computations, but I don't know enough about Snappy too. I don't know what to tell the computer, so if somebody wants to teach me I'll be so incredibly grateful and happy. And then the second one. Yeah, the second one is a little harder. But then in all honesty, we can replace this one by either we have even winding number or the order of each one of these curves in first homology by W. And in those cases, you also get that it's not a homomorphism. So there's a little bit of a problem. So do I need all of these linking numbers to be non-negative? Is it enough to have a lower bound? I do absolutely need them to be non-negative because here when I am changing the linking numbers, if I want to get to these, I have to start with non-negative. Because what I'm doing here, this trick just tells me you can go down. You can go down a little bit and a little bit until you get to these. So if I start in negative, I'm not going to be able to get there. Yeah, that's a good point. No, absolutely not. So Tom's question is, at first it seems in my proof that all of these co-boardisms are definite co-boardisms and in fact it is not. So this whole project started because I wanted to use to turn Simon's bounds to get this obstruction. But to use turn Simon's or instanton's, you really need definite manifolds and definite for manifolds and in our three steps, the only one that has a definite for manifold is this one. So it's just not enough. That's why we had to go to the invariance because the invariance, you get a lot from semi-definite. So from this co-boardism and from this co-boardism, I can produce a semi-definite for manifold that is going to give me the bound in the D invariant. So, I mean, if you know how to control turn Simon's using semi-definite co-boardisms, I mean, or if anybody knows, I would just love. I would love to understand how instantons behave under non-negative definite formats. Any more questions? If not, then let's thank Juanita again.