 So today I'm going to talk about not flow homology and concordance. So today we're focusing on knots in S3. So here's a Hager diagram. What B-manifold does this describe? S3. Yeah, this describes S3. And move over, it's like a standard sort of Hager splitting for S3 in that sort of, well, this alpha torus. It is just sort of the solid torus sort of standardly embedded in S3. And we'll isotope the beta curve slightly, but that is sort of also just sort of the standard torus that kind of goes out towards infinity and comes back. Great. OK, so this is describing S3. And so now the claim is I can describe a knot in S3 by just adding one extra bit of data to this diagram. So if I just put a second phase point, and I'll choose to put it here, the claim is that this data, so a surface of genus G, G alpha curves, G beta curves, and a pair of base points. Well, the claim is that this is going to describe a knot in the S3 manifold. And so how does this describe a knot? Well, the rule is let's connect Z to W in, so I'll do it in the surface, but away from the beta curves. And then I'll push it slightly into the beta handle body. And then let's connect W to Z, but we'll do the same thing but with respect to the alphas. So in sigma minus alpha, and then push slightly into the alpha handle body. So let's do this in this picture. So I want to connect Z to W missing the beta curve. OK, so let's see. We can go down this way. And then if I go around the back, and then if I come around this way, I've followed my instructions. I want to connect W to Z missing the alpha curve. Now, we have to get the crossings, right? Because the crossings data is important and determined on the knot. And so well, the alpha handle body is on the inside. So here, you should think about this. The first curve I drew is slightly outside of the surface. And now what I'm drawing is slightly on the inside. So this here is going to be under here. And then if you stare at that knot, so if you stare at that knot, it looks something like this. So since today, the question was about admissibility. Basically, could I put the points anywhere? And so today, we're working with knots in S3. And admissibility only comes into play when you have B1 non-zero. So for S3, and in fact, for any rational homologous here, you don't have to worry about admissibility. Other questions? All right. So now I was in the process of drawing the knot. I think it looks something like this. If I haven't messed anything up, it's meant to be a truffle oil that looks like a truffle oil. It's alternating. It has two crossings. Questions? Yes. So the question was about uniqueness. So it's a unique way to connect them. And the answer is yes. So another way to think about this. So let's think about, say, on the inside. So on the inside, well, it's this torus, and I had this condition of missing the alpha curves. Basically means that on the inside, I'm missing those disks. And on the inside, well, if I'm missing the disks, what's left is going to be a three ball. And so basically, I'm connecting them to be an unknotted arc in the three ball. And that's unique up to isotope. And that's happening on either side of the surface. So yes, this does uniquely specify a knot in earthly manifold. And also, any knot in any three manifold admits a double-pointed Hager diagram. It's possible that maybe you need a very large genus Hager splitting for your three manifold. So for example, some knots in S3, you need more than just a genus one Hager diagram or nor is it described in this way. But every knot admits some such diagram. So the question was, can you relate the minimum genus of needed for such a description to the genus of or not? As far as I know, they're independent. So for example, any torus knot admits a genus one double-pointed Hager diagram, but torus knots have arbitrarily large genus. But one, you could just define a measure of complexity of a knot. Well, what's the minimum genus you need for a double-pointed Hager diagram? And that's some interesting measure of complexity of a knot. We played the same game that we played before. Great, so we're going to build a chain complex out of this. Now I'm going to tell you about the knot floor complex. And you basically do the same thing. So it's going to be a chain complex. Well, you sort of do the same thing in the sense that the generators are still going to be intersection points between this alpha torus and this beta torus and SIMG. But now, so remember, before we were working over the polynomial of an effer during u and the variable u counted how many times we crossed our base point w. Well, now we have two base points. So one way to deal with that is to just have two variables, one that counts the base point w and one that counts the base point c. We're effer during uv. And then the differential is going to look exactly the same except for this extra bit about the second base point and the second variable. Usual. So counting points in some zero dimensional moduli space. And now u counts how many times across the w base point. v counts how many times across the z base point. So the point is you can analogously define a subspace of SIMG using the base point z instead of the base point w. And that's what this is counting. So this is what the differential counts. So let's look at an example. So let's look at that example, which I'll redraw over here without the knot. So we'll have three generators for our chain complex. So we'll call this a, b. Boundary of a. Yeah, great. I saw Jacob holding up a zero. We did an example like this on the first day, where we just didn't have the base point here. So it's actually going to be the exact same calculation, but now the one that crosses the space point now is to get counted with a v. So if you remember, well, right. So from b, there's this disk here. So that's going to kick out a ua. And then this disk here. And so now it's crossing z. So z gets counted with a v. So that's going to kick out a vc. And the boundary of c is zero. Great. OK. And so now, first time of the week, I'm going to say something substantive about gradings. So for knots in S3, there's actually a way to compute absolute gradings. OK. So this is a bi-graded chain complex. We have the u-grading. We have the v-grading. And I guess we also have the Alexander-grading, which is just 1 half the u-grading minus the v-grading. What happens if we set v equal to 1? So v is the variable that was counting how many times across the z base point. So if you set v equal to 1, that's saying, well, you can still cross the base point, but we're not counting it any differently than if it weren't there. So setting v equal to 1 is effectively ignoring the z base point. So this is basically saying, ignore z. Well, z was just what was, if we ignore z, right, then this is just a pointed Hager diagram for S3. So in particular, if we set v equal to 0, we better get out something whose homology is giving us hf minus of S3. And in particular, I declared for you the absolute graded on hf minus of S3. I declared that the element 1 in there was in grading 0. So if you set v equal to 1, this should give you something that's homotopy equivalent to cf minus of S3. So if I set v equal to 1, then you can check that. So this is going to give us f rejoin u. And OK, so I guess it's an exercise for you to check that if you set v equal to 0, the homology is going to be generated by a. Oh, I mean, the generators of this chain complex are these points a, b, and c. So I'm saying that a, a is equal to 1 in here. Oh, sorry, sorry, sorry, yeah. You said v equal to 1. Yes. Yeah. You said v equal to 1, wait. And then you exercise it for you to check that if you set v equal to 1, well, you do indeed get the correct homology and it's generated by a. So that's going to, so that was, we said v equal to 1, so you threw away v and we kept u. So that's going to tell us the u grading of a is 0. What else can we use? Well, this is a graded chain complex, right? So when we set v equal to 1, then this is telling us that, well, the boundary of b is ua plus c. So this tells us, since this is graded, this tells us that ua and c have to be in the same grading. And remember, we declared that the u grading of u is minus 2. So what is the u grading of c going to be? Wait, so I saw some, it's going to something like 2. Yeah, it's going to be minus 2, right? Because if this isn't grading 0, well, that means that this is in grading minus 2. So that means that c is in grading minus 2. And our boundary lowers grading by 1. So that tells us that b is going to be in grading minus 1, because it has to lower the grading by 1 to get something that's in grading minus 2, so it has to be in minus 1. Also, instead of setting v equal to 1, you could set u equal to 1. If you set u equal to 1, well, it's using a different base point, but it's still giving you back s3, and we still know that the element 1 in hf minus of s3 should be in grading 0. So then it's an exercise, compute v gradings, analogous later how we did here, but just instead of setting v equal to 1, set u equal to 1. I'll tell you the answer so you can check your work. And then, well, you can also compute the Alexander gradings, so this is going to be 1, 0, minus 1, okay. And so this lines up with the gradings that I put up on the board on Monday, so that's how you can determine those. Great, and then, well, right, if you set u and v equal to 0, this is what gives you hfk hat, and this is the invariant that categorifies the Alexander polynomial. Great, so the question was, which variant of novel homology is cfk minus, right? So this is sort of this full chain complex over effort drawing uv, and then if you set v equal to 0 and take homology, that gives you hfk minus. Was that another question? Yeah, so the question was, why is it hard to get absolute gradings and for not an S3, hopefully I convinced you it's not too hard, but you just do this process, you set v equal to 1, compute homology, find the generator for hfk minus of S3, put that in grading 0, and then what I implicitly claimed about any justification is that the differentials can help you determine the gradings of everything else. You could have the issue that maybe not all of your generators are connected by differentials, and then in that case, there's a formula to compute the gradings. So you can figure out the gradings of one element, and then the difference between the gradings of any two other elements, you just look, so you look for some disc connecting them, and then you look at the, that's the index of that disc, and then minus nw of v. And so this formula is gonna let you compute the gradings of any other generators. V-manifold invariant, this invariant behaves nicely under orientation reversal. So let minus k denote, I'll just denote the reverse of the mirror image of k. Okay, so reverse means change the spin orientation, and mirror means change any over-crossing to an under-crossing and vice versa. And then while the not-flow complex of minus k is the dual of the not-flow complex of k, so you just look at homomorphisms over a ground ring from your chain complex to the ground ring, and then it also satisfies the current formula, so the not-flow complex of a connected sum of two knots is a chain homotopy equivalent to the tensor product of the chain complexes of the respective plane. So one of my favorite things to do is to use this invariant to study not-concordance. So now I wanna tell you what not-concordance is, so any questions before I move on? Yeah, if you don't do that, you get a chain comp, so the question was to get hfk minus, we said v equal to zero and take homology, what if you don't do that? Yeah, then you get a chain complex over the polynomial of an effigyron UV whose chain homotopy type is an invariant of the isotopy class if you're not. That's like, it's like if you don't set v equal to zero, you have what presumably is a stronger invariant because it has sort of more algebraic structure. Effigyron UV is not a PID, so it's like finitely generated modules over that or like not classified. So let me tell you about concordance. In K1, in S3, concordant, which I'll denote with a twiddle, I'm denoting it like this because this is gonna turn out to be an equivalent solution. So if they co-bound a smoothly embedded annulus in S3 cross i. So the cartoon picture is that, okay, so maybe here is S3 cross one, here is down here is S3 cross zero, and then up here you have one down here, and then you have some annulus. Like okay, so to see that this is an equivalence relation, well to see that it is reflexive, you can just take the product, annulus, to see that it is symmetric, well you just turn this thing upside down, and to see that it's transitive, well I'll just take two of these things and stack them on top, and then you care if you parameterize it, so it's S3 cross i. Certainly if two not so isotopic, they're concordant because you can just sort of take the isotopy in the i direction. Okay, so let me give you maybe an interesting example. So let's take this knot, and so I'm gonna think of the i direction of like time, so maybe this is what we're gonna see in S3 cross one, and then I'll move down in the i direction, and we'll see what we see at each different cross section for different values of t, which is the i direction. Okay, so first let's just do an isotopy. And so if I just do an isotopy, well that's still just gonna be sweeping out a cylinder in S3 cross i, so I start it off. So here's a sort of like what the, what this is sort of me telling me abstract of what the surface looks like. So I've just done an isotopy, so the surface just sort of looks like a cylinder so far. And now, well let's bring these two pieces together. Right, so topologically this looks like a wedge of two circles. So here, well what I'll see here, something like this. Okay, and now next, well let's have these like split apart but in the opposite direction that they came together in. Great, so now if you look at this, now a two component link, there's one component here, there's another component here. And so in terms of the surface, well now we see this. And now if you stare at this link that I drew very small at the bottom of the board, well this is actually the two component unlink. Right, so this is the two component unlink. So we can cap off one of the bounded components of the disk. Wait, so top a lot, so what I've drawn for you, right, this is home, Diffie and Mort would say if it's everything's smooth, well this is gonna be an annulus, right. So what I've just drawn for you is concordance from the square knot to the unknot. It doesn't seem, no, you could have smoothed this out so that, yeah, so it just looks like, yeah. So that's the definition of concordance. K is concordant to the unknot. And an exercise for you is to check that K is slice if and only if K bounds a smooth disk in the four ball. So what do I mean by that? So by that, I mean think of S3 as the boundary of B4. The boundary is S3, okay, I'll draw the knot K as an S0 and then K is gonna bound some disk here. You'll keep this. So we have this equivalence relation. Okay, so there's a natural binary operation on knots, the connected sum. If you look at knots in S3, another operation of connected sum. It says it was set with a abelian binary operation. Is this a group? But just look at knots in S3 up to isotope, yeah. So if you look at knots in S3 under isotope, under connected sum, it fails to be a group. One way to see that is that genus is additive under connected sum and the only knot with genus zero is the unknot. So if you have two non-trivial knots, i.e. knots whose genus is strictly positive, well, it won't have an inverse. But the fix is, okay, well if we mod out by concordance, this is gonna be a group. So this is called the concordance group. Great, so somehow up to concordance everything has an inverse. So an exercise is to show that the inverse of k is appropriately enough what I've been calling minus k. So remember minus k, well wait this is an abelian group so it's a natural way to denote the inverse and wait, minus k is the reverse of the mirror image. For example, yeah all of my knots are oriented. Oh great, so Paul asked, does the proof of the exercise go like this? And the answer is yes. And so if this doesn't make sense to you, then your exercise is to make sense of this. Now what's our goal? Well, our goal is gonna be to use non-floor homology to study concordance. The chain complex, so. And so, well just like in the three-manifold case where we said nice co-borderisms induce nice maps on the Hager-Floor homology of your three-manifold, well the co-borderism between knots, so even if it has genus, we'll do some map on the non-floor homology, but concordances which are particularly nice co-borderisms between knots are gonna induce particularly nice maps on the non-floor complex. Co-borderisms. So the question is, how do you see the inverse is unique? And so, well I guess you can check, it's unique up to concordance, right? So just say okay, I taught intro to Proust last year, so how do you show uniqueness? Well assume you have something else and then check that it actually has to be concordant to minus k. Oh, the question, the statement was if you show it's a group then the inverse is unique. Yes, that probably also works. Great, all right. So co-borderisms between knots induce this chain complex and, oh I wanna say maps, so I mean module maps, so they respect the UNV action. And concordances, IE genus zero co-borderisms in S3 cross I induce especially nice maps. And so now I guess I should tell you what I mean by especially nice. Due to Ian's empty. Okay, so it's concordant to k1. Then there exists i-grade in preserving effa-join UV, effa-variant chain map, the knot flow complex of k0 to the knot flow complex of N. Moreover, this is an isomorphism if we invert UNV. We get this chain map from the knot flow complex of k0 to the knot flow complex of k1 and this is an isomorphism on, okay, so I'll just write UV inverse for HFK, right? So this is, okay, so take the homology of, well, take your chain complex and invert UNV, so tensor with effa-join U, U inverse V, V inverse. So this should look a little familiar to you from the statement about three manifolds last lecture, right? So you get a chain map that induces an isomorphism once you've inverted all your variables in sight. Oh, yes, I just wrote complete nonsense. Other comments, questions, corrections, great. Oh, maybe one thing to observe, right? So since concordance is symmetric, right? So remark, well, if k1 is, k0 is concordant to k1, then k1 is concordant to k0. So it also exists g going the other way, one to k0 inducing isomorphism on this localized homology. So if you have a concordance, you get maps in both ways with this property. Yeah, oh, yeah, what is HFK without a definition yet? I'm just defining this thing to just be this. Yeah, there's a definition here. Yeah, the question was, if you compose concordances, the maps compose and the answer is yes. So the question was, we have f1 way of way of g, the other way, can you say anything about the compositions? Well, just in general, you can say just whatever is implied by the statements on the board. If you know things like one of them is a ribbon concordance, if you know what that is, then you can say stronger statements. In general, not inverses of one another. So for example, the unknot is concordant, I drew a concordance between the unknot and the square knot earlier. I didn't tell you this, but the unknot, the chain complex just looks like the ground ring with trivial differential and not the homology detects the unknot. So the unknot is in fact the only knot whose chain complex just consists of the ground ring, but there's concordances between the unknot to other more complicated knots. In particular, in general, this thing is not gonna be an isomorphism just on the chain complex itself. This is gonna be an isomorphism on this localized thing. And this strong statement is very useful, but as I mentioned earlier, yeah, are not so oriented, and so the surface is better be oriented so that the oriented boundary matches up with the knot. F would join UV, it's not a PID, so maybe that's not great. So maybe it's hard to work with. Okay, so remember mark F would join UV, not the PID, but if you set one of your variables, say V equal to zero. And so as I mentioned earlier, well you can define, this is invariant H of K minus. That's just the homology of this chain complex when you've set V equal to zero. Okay, right, so this thing has a grating, this is a Z graded, the U graded. So in particular, well this is a finitely generated module over a PID, so we have the structure theorem that it's always gonna look like a three-part plus some torsion part. This is a finitely generated graded module over a PID. So it's always going to exactly one free part that's because it's a not in S3. I'm always gonna have this form. Yeah, so the question in chat is HFK minus of K always the HF minus of Y of some three-manifold Y like the example I showed today. I'm not positive of the answer, but maybe not, I actually don't know. It's sort of a tough question, sort of which of these modules can be realized or not. Maybe if I, yeah, I'm not sure. Okay, that's it, yeah. So we've set V equal to, yeah we set V, right, so remember when we set V equal to one, that's saying, okay, well now we can cross the Z point base point and it's like the ZBF support isn't there. When you set V equal to zero, that's like saying the Z base point is like a big like do not cross zone, like don't like, so discs that used to cross Z, now you say when you set V equal to zero thing, don't count those anymore. And then very tau of K, so this is gonna be minus one-half that's just some normalization convention. The maximum U grading of an element X is in H of K minus of K and U to the N acting on X is non-zero for all say large N. And maybe if you did the exercise from last lecture where I asked you to check that D was well-defined, maybe this should look kind of familiar. Exercise for you to check is that shows that rating of one in H of K minus of K in here is minus two tau of K. So certainly if you already have this split in here, right, and you look, remember U has grading minus two, so if you look for the highest graded thing such that like any amount of U acting on it is non-zero, well this part's all U torsion, so it can't come from here and you want the highest graded piece and U has degree minus two, so it's gonna be the element one in here. So I guess I'm doing the exercise for you. Wait, so this is, so it means that this has to, and because of this weird normalization, this has to be in grading minus two tau. And then, so if you have some splitting, you can just look at the grading of one in here, but if you don't have a splitting, you can just use this definition. Great, so I've defined right now what is just a non-variant, but due to Ashraf and Sabo, although I guess the way I'm telling this story, it's a corollary of the Zemki result that I put on the board, but the Zemki result was sort of proved later. In the Ashraf and Sabo, that tau, it's a concordance invariant, so if two knots are concordant, they have the same value of tau, but even stronger, it's a homomorphism from the non-concordance group to the integers. So, but being a homomorphism, concordance homomorphism is stronger than being a concordance invariant, in some particular respects connected sums. Oh, and moreover, this is in fact a subjective homomorphism. Questions. Great, so the question was, is it easy to see that slight knots should get mapped to zero? Guess it depends on your definition of easy, right? So, you should just check that, well, if you, from the Zemki result, it's an exercise to check that indeed if a knot's concordant to the unknot. Pull the trefoil on the board, so let's compute tau of the trefoil. But what was, what did I say to do to compute tau of V equal to zero? Wait, so we're gonna set, let's set V equal to zero. Wait, so this part goes away. All right, so then hfk minus, hfk minus of the right-handed trefoil. Okay, so, well, what's in the kernel? A and C are in the kernel. And what's in the image of the boundary? Well, Ua is in the image. All right, and this is all over f for join U. Great, well, the C generates like a free summoned, right? Cause there's no C appearing down here. This is generated by C. And then, what we're gonna get is an f for join U and this is generated by A. But I guess the part that we care about is you care about the grading of this, right? And we care about the U grading of this part. And so C was in U grading minus two. So this, the element one in here is in grading minus two. I'll make that larger. And so by the normalization over there, this tells us of the right-handed trefoil. Is one. Questions about that? No question, great. Okay, so, Zemki result implies that tau is a concordance invariant. So, how is it a subjective homomorphism? Great, so I guess the first step is that, so first we wanna prove that it's a concordance invariant, it's a K one. But we had this Zemki result where, well, tau is defined where we set V equal to zero. So we can just set V equal to zero in the Zemki result. So, set V equal to zero in Zemki. Great, so this is gonna tell us that there exists grading, preserving, f for join U, equivariant, since you can turn up concordance around, we get them in both directions. So we get a map, say, f this way and g this way. And these induce isomorphisms once we've inverted U. That's our only variable that we still have around. Great, so we know that hfk minus, so I'll leave this part of the proof as an exercise. So check that the existence of f implies that tau of K naught is less than or equal to tau of K one. So the idea is, well, these are U-equivariant maps. They induce isomorphisms when we've inverted U. So the key is to think about what has to happen on the free part of hfk minus. Basically, the sketch of this is that the free part has to map, the free part of hfk minus of this side has to map non-trivial of the free part here in a U-equivariant way, and that's gonna imply this inequality. But we also have a map the other way, right, right, and existence of g implies that tau of K one is less than or equal to tau of K naught. So this shows that, well, this shows that tau is a concordant invariant. If K naught and K one are concordant, well then tau has to be the same. And to show that this is a, wait, so to show it's a concordant homomorphism also has to behave nicely under connected sum. So to see that K one connects some K two is tau of K one plus tau of K two, I said I was gonna prove this, but I'll, I'll state this as an exercise. Use the Kerneth formula, right? So the Kerneth formula says that, well the not flow complex of the connected sum is a tensor product of the respective not flow complexes that still holds once you set V equal to zero, and then I guess maybe there's, to understand sort of, but the statements of the Kerneth formula is on the chain level, and so you can either do things on the chain level or you can use some universal coefficients or something to sort of figure out what has to happen on the level of homology. But sort of the point is, well, per tau we just care about the sort of the free parts and sort of when you tensor those free parts together, that's how it's additive. So I've just sketched the exercise. Oh yeah, yeah, you get chain maps, they descend to the homology, that's right. Yeah, I guess the statement about the Kerneth formula, the Kerneth formula is stated on the chain level and then that, since we're working over polynomial ring, the tensor product, the homology of the tensor product isn't going to necessarily agree with the tensor product of the homologies. You get some tau terms or something, yeah. But somehow, the tau terms don't really affect this part that much because tau is just sort of a statement about like the free part, really. To show, right, so we want to show this the subjective homomorphism. Well, okay, so show is subjective. Well, we already basically did that, right? Because we computed that tau of the cheff oil is one and we know that it's additive under connected sum. So, well, tau of n times the right hand of cheff oil is equal to n for any. And so, you can already see that, well, by setting v equal to zero, we throw out some information, but we still get this really great concordance invariant. And so, one thing that I spend a lot of time researching is, well, if you keep v, if you don't set v equal to zero, you can actually get lots more concordance homomorphisms. In fact, in work with Irvin Dye, Matt Stauffer and Lynn Truong, we showed that, well, if you keep v and you do some algebra stuff, you can in fact get infinitely many more homomorphisms, subjective homomorphisms to the integers. In fact, they're linearly independent. If you're curious about that, I'd be happy to talk more about that. This is what I wanted to tell you about. The question was, what is known about the structure of the concordance group? Great, so, it's infinitely generated. There's two torsion, right? So, any knot that's negative-infra-chiral, isotopic to the reverse of its mirror image, is going to be torsion, and then you need to check that it's not already sliced, but there are lots of non-sliced, negative-infra-chiral knots, like, starting with the figure eight. Great, so it's known that the concordance group, so I guess, yeah, and there's, right, so there's sort of infinitely much two torsion. So, the simplest thing that the concordance group could be is this, that's like the simplest thing it could be. However, no one has obstructed, for example, like a Q subgroup. So, no one can prove that, like, it seems hard to have a knot that's like infinitely divisible in this group, but no one's been able to obstruct that. So, the question is, do we expect that there are other types of torsion? It probably depends on who you ask. So, so, Levene, Jerry Levene in the 50s defined a surjective homomorphism to this group, which in higher odd dimensions is actually an isomorphism. So, by higher dimensions, that just means knotted SNs and SN plus twos. And so, in higher odd dimensions, you can define the analogous homomorphism and it's actually an isomorphism. In the classical dimensions, so S1s and S3s, this has a non-trivial kernel and all known examples of things that hit this part. Oh, the question is, how do we know that Hfk minus has rank one? And that's an exercise from the fact that this is a knot in S3. So, if you have a knot, you can do this for null homologous knots in any three manifold in which case, it might not be the case that Hfk minus has rank one. So, the question was basically about the kernel of this homomorphism. Can you say things about the kernel of this homomorphism? It's very non-trivial, big. The question is, do we have a more geometrical description of this tau invariant? Not really, tau, it behaves nicely though in lots of ways. So, for example, tau behaves nicely under crossing changes. So, for example, if we do a crossing change, it can change tau by at most one and the direction in which it can change depends on the sign of the crossing change. Oh, are we able to find a basis for the free part of the knot concordance group? You can find a basis, yeah, yeah, so I mean, you could just like, right, ignore this part and then you could, yeah, this part relies on looking at levine system signatures. So, I believe to show that this part is subjective. I think even like some family of tourists knot probably works, yeah. So, I guess, maybe I personally could not give you a basis, but I believe it's in the literature. The question was, could a periodic knot, let's say periodicity, they may be sliced? I haven't thought about periodic knot, but you can enter one in the audience.