Introduction to Vassiliev Knot Invariants

インデックス作成: 2026-05-17

[00:00:00]Okay, great. So, I'm going to talk about vasil of knot invariance. I'm going to start with some more general knot theory. Um, but it's mostly based on this book over here, which you can read for free on archive.org with an X. So I'm going to write recall uh even though this bear foil may be a new definition to you but so knots are things like we have the unnot we've got ourselves the tfoil this thingy over here is the figure 8 knot etc loops uh in threedimensional space essentially links s are like knots but potentially with more than one connected component. So uh things like this doodad which is the simplest link and uh other than the unlink. So this is called the hop link. You can have two completely disconnected things. That's that's an unlink.

[00:01:11]Just some more examples.

[00:01:18]Let me see if I can freehand a Bomian rings. Um, I'm really challenging myself here. All right, I think that worked.

[00:01:34]And then oriented knots.

[00:01:39]What What do I mean by an oriented knot?

[00:01:45]Uh so here's a regular knot and then I'm going to add like an arrow essentially uh directing flow of traffic around the knot. So for the purposes of this talk when I say not I'm going to mean an oriented knot by default.

[00:02:12]So not diagrams are pictures of knots.

[00:02:23]The same uh uh the same knot can have more than one diagram. For instance, this is the unnot. Um this is also the unnot.

[00:02:39]This is also the unnot which might be a little bit harder to see if you sort of imagine pulling in the sides. So we count uh these as being the same because you can deform one to the other by uh without passing it through itself.

[00:02:54]Um and we like not diagrams because we can compute with them. you know, uh, I can describe a knot diagram to you and they have more interesting anatomy, if that makes sense. Every part of a knot looks the same as every other part of a knot. But not diagrams have strands, they have crossings, they have regions.

[00:03:20]So crossings of an oriented knot diagram have two flavors.

[00:03:40]They look like this and like this. So essentially orient the crossing so that both strands are going up. uh and then look at the slope of the overcrossing.

[00:03:53]If the slope of the overcrossing is positive, we say that it's a positive crossing. Otherwise, we say it's a negative crossing. And we call that the local wythe or the sign of the crossing.

[00:04:09]It's going to be plus one or minus one.

[00:04:12]So an example, let's say I draw this figure eight over here.

[00:04:26]Uh an exercise for the viewer is to check that these two crossings up here are both positive and these crossings in the middle are both negative.

[00:04:38]All right, some more operations to do with knots.

[00:04:44]Um, given two oriented knots called K and L, we can conform sorry, we can form the um connect sum or the not sum k# L.

[00:05:02]And the way we do that is let's say uh I've got my K over here, my L over here. I've I'm sort of abstracting away most of the detail of these two knots.

[00:05:21]We cut out uh we snip the knot at these two places and we join them.

[00:05:32]Like so.

[00:05:34]A natural question is does this depend on where we snip the two knots? And the answer is no. Essentially imagine imagine K is some complicated thing. We can imagine sipping it at lots of places.

[00:05:51]If we shrink L down so that it's very small, then we can sort of slide it around K.

[00:05:58]So, it doesn't matter where we do the connecting. I don't know if that makes sense. I'm sort of skimming past this.

[00:06:06]Um, but what this Yeah, I thought someone said something.

[00:06:15]So, what this does is it makes the set of knots a monoid.

[00:06:27]What's a monoid? It's a set with an operation satisfying. I think the only condition is associivity.

[00:06:36]Um, in fact, it's a commutive monoid.

[00:06:38]You can show that K-L is the same as LH K.

[00:06:44]And what we can do with that is we can form this bigger set called ZK or this other bigger set called CK.

[00:06:55]Um, and I'll define these in a second.

[00:06:59]So these form rings called the algebra of knots.

[00:07:12]uh how did that work? So we already have a version of multiplication for knots.

[00:07:18]What I'm doing here is I'm formally making a notion of addition and the addition doesn't have any uh polomials with not exponents eh sort of I suppose you can think of it like that. So, what I'm I'm reading the chat there. Uh, so where's my good eraser? All right.

[00:07:51]If I imagine like this plus two this that would be an element of my thing sort of the addition doesn't mean anything it's just formal and then multiplication is going to be the um the notum 120 cell asks would the monoid of knots be a free monoid? Uh yes it is.

[00:08:20]Uh it turns out that no knots have any inverses. So we do have an identity that's going to be the unnot. Um it can be shown that if a and b are each not the unnot then a hashb is also not the unnot. So we don't have any identities and also factorization into prime knots is unique. So we don't have any non-trivial relations. uh and a prime nod is one that can't be written as the connect of uh non-trivial knots um all right I have in my notes that I want to talk about G diagrams this is going to be I don't know that this is going to be so super relevant but it is a cool thought of how can we describe knots to a computer So given a not diagram we can form its G diagram.

[00:09:28]So I'm going to draw yet another figure eight.

[00:09:40]Uh if you remember these are positive, these are negative. I'm going to label the crossings. So I'm going to call the two crossings on top one and two. The crossing in the middle I'm going to call three. The crossing in the bottom I'm going to call four. To form the G diagram, I'm going to open the knot up into a circle.

[00:10:01]And I'm going to ask in what order do I visit the crossings? Well, starting here going around the orientation or direction I visit crossing one then two 3 4 2 1 43.

[00:10:17]So, okay, I'm going to mark those all on here. So, 1 2 3 4 2 1 43.

[00:10:31]I'm gonna connect points with the same label.

[00:10:39]I'm gonna make them point from the undercrossing to the overcrossing.

[00:10:47]And I'm also going to label them. Uh so each of these uh chords essentially corresponds to a crossing. I'm going to label the chords with the sign of the corresponding crossing.

[00:11:03]And you can imagine, oh, if I want to describe that to a computer, maybe I can uh just write the string 1 2 3 4 2 1 3 4 and also somehow tell it about uh the directions of each of these chords.

[00:11:22]And this isn't limited just to knocks. I've also defined links. So links also have G diagrams.

[00:11:48]Going to call this crossing one. Going to call this crossing two. I believe these are both positive crossings.

[00:11:54]Actually, any guess on how what this G diagram is going to look like?

[00:12:00]See if I can have these.

[00:12:06]>> Well, I guess I'll uh take a guess. I mean, I don't actually know what Gaus drawings or G diagrams are uh besides what you've just explained, but I'd imagine you're going to draw two separate circles. And this time you'll also have the overlaps like like you'll you know the first one will have uh two two spokes on it and the second one will also have two spokes on it and they'll also have the numbers one and two. Um in I guess both I mean it could be any order for either of them I guess. Um and then you would Oh yeah yeah yeah. Okay. So that's how they >> Yeah. something like this.

[00:12:47]>> Yeah, something like that.

[00:12:48]>> Yep, perfect. That's exactly what it is.

[00:12:54]Um, so when I get to villa of invariance, it's unfortunately it's not in the notes that I have prepared the relevance of Gaus diagrams to avarian, but if I have time, I might talk more about it. Um, so what is an invariant?

[00:13:11]Uh, an invariant is a function. It takes in knots and it gives you in general some mathematical object. Um for our purposes it's generally going to be a number. Uh it could or potentially a polomial and uh our first uh link invariant.

[00:13:41]Oh, and sort of the main goal of these is hopefully they're going to be computable from the diagram so that if two diagrams have different values according to the to an invariant then they represent different knots or different links. So our first link invariant is the linking number LK AB where A and B are the components of a link.

[00:14:07]So we've got a few uh equivalent definitions.

[00:14:11]One is it's the signed count of oriented intersections of A, which I'm thinking of as the knot itself, you know, in threedimensional space with a surface spanned by B.

[00:14:38]So this isn't diagrammatic at all. This is threedimensional. We imagine, you know, here's the hot flink. I imagine filling in one of these, let's just call this one B, with a surface. And I ask how often does A hit that surface. And when I say oriented count, I essentially mean depending on which direction it intersected from, we're either going to count that as positive 1 or negative 1.

[00:15:04]I'm not going to go into a ton of detail about that. There are questions of like does it matter which surface we choose?

[00:15:11]Turns out no.

[00:15:13]Um, oh, Vleta says an invariant is a function with domain not diagrams which preserves the equivalence relation.

[00:15:20]Yeah, you can think of that. You can either think of the domain as knots themselves or you can think of them as not diagrams up to uh what are called the righteister moves which are moves that don't change the knot but do change the diagram which I don't think I'm going to talk a ton about. Um so two second definition is the sum over all crossings P.

[00:15:45]So sum over P a crossing of A over B of the wythe if you remember uh wythe of a crossing is either plus one or minus one. I also called it the sign.

[00:16:03]This is symmetric. It's also the sum of the rise of crossings of B / A.

[00:16:12]And therefore, it's half the sum uh of the rides.

[00:16:22]So P of a crossing involving both A and B.

[00:16:32]So crossings between A and itself do not count for this. And to look at this thing over here, this is going to have uh linking number one.

[00:16:42]We see uh looking at uh crossings of A over B, it's just this one. Crossings of B over A, it's just this one. If I include both of them, divide by two, it's going to be 1 plus 1 divid by two.

[00:17:00]And you can imagine what would it mean to have linking number two? It would sort of have to I can't think of any way of describing that in words other than it would link with it twice. So hopefully that makes some sort of sense.

[00:17:16]Um three, this is going to go into calculus a bit. If you don't understand what I'm talking about, uh just ignore it and I'll go back to something more elementary in a second. But it's the degree of the map from the Taurus S1 cross S1 to the sphere defined in the following way. So a Taurus S1 cross S1 S1 here means um a circle. So this is a map that takes in two points on the circle. And what does it do?

[00:17:56]It looks at where B thinking of B as a function from a circle to threedimensional space. We see where it sends that second point. Look at where A sends that first point. Uh we have to land in the sphere. So we're going to normalize it.

[00:18:12]This gives us what's called the Gaus map.

[00:18:15]So every length gives us some map that each functions you know pairs of points on the circle or equivalently points on a Taurus gives us a point on a sphere.

[00:18:26]The degree of a map is essentially the number of pre-im images of a generic point counted up to orientation.

[00:18:35]So it turns out that that's equivalent.

[00:18:36]And the reason we care about that is because it gives us a final formula.

[00:18:51]Going to write this down. Then I'm going to explain the notation.

[00:19:02]I think there's a cubed in there. Um I could be wrong. Um so what's this up top? I have three vector valued uh things. These two are like infinite decimal vectors. Um, and I'm sort of defining the triple product of three vectors to be like the determinant of the matrix whose columns are these things or equivalently U dot V cross W. Sort of a relatively common vector operation.

[00:19:41]This is just to say there are um there is a an integral that gives you the linking number. If I have enough time, I'm going to talk about a much more powerful uh integral called the convage integral. Um so this is sort of setting that up, but I frankly I don't actually know if I'm going to have time. So sorry about that.

[00:20:04]But on to more uh slightly more exciting invariants. So here's a theorem.

[00:20:14]There exists a unique link invariant.

[00:20:24]Links include knots, by the way. Um I like to say that a link is a knot with many components and many means zero or more. So there's this unique link invariant C that takes links as input gives us polomials with integer coefficients as output satisfying the following conditions.

[00:20:49]One of this is going to be the Conway polinomial by the way.

[00:21:00]So C of the unnot is going to be one.

[00:21:03]C of and I'm going to have to explain my notation again in a second.

[00:21:09]C of a positive crossing minus a negative crossing is T of is T * C of this smoothed crossing. When I write a portion of a not diagram in a formula, I'm doing the convention where the rest of the not diagram is arbitrary and where the same rest of the knot diagram appears in each of these three cases. So I'll give you um an example an example computation.

[00:21:49]How would I find the Conway polomial of the unlinkling?

[00:21:58]Well, this middle section sort of looks like the right hand of this second rule here. So dividing by t substituting that in that's 1 / t * c of this u minus 1 / t * c of the other way but these are both uh the unnot you know I said that the domain is links I didn't say the domain is link diagrams so I'm allowed to untwist these So that's the same thing minus itself.

[00:22:41]It's zero. All right.

[00:22:45]Um here's another example. Let's find the cone polomial of the hop link. Technically, there's two versions of it depending on how I orient these um depending on whether the linking number is plus one or minus one.

[00:23:04]But okay, what I'm going to do is I'm going to focus on this bottom crossing here. Okay, this bottom crossing is I believe a negative crossing. So, it corresponds to this thing over here. I'm going to isolate this, solve for it.

[00:23:25]So that'll be C of this link link diagram with the opposite uh version of this crossing on the bottom minus t * Oh wait, sorry.

[00:23:50]I'm looking at my notes. I'm doing I'm doing the top one actually. Sorry. Um I I I'm just going to try to be consistent with my notes. So So I've switched the top crossing and now I'm going to smooth the top crossing which is going to look a little bit strange.

[00:24:20]This over here is the unlink. We can separate the circles. So this vanishes.

[00:24:26]This is the unnot rule. This is one.

[00:24:32]So that gets us minus t c of the unnot which is minus t. And then the last example I want to show you will be the conway polomial of the tfoil which is what you call this knot with three cross crossings over here.

[00:24:49]So now I'm going to focus on the bottom crossing.

[00:25:16]I think that looks right. No. No, it doesn't.

[00:25:20]Um, sorry.

[00:25:25]It's smooth the other way. Um, the way you can tell which direction to smooth it is by the orientation.

[00:25:31]So over here, it's got to go this way.

[00:25:37]Otherwise, the orientation would clash.

[00:25:39]Oh, I'm sorry. It was off the screen.

[00:25:40]Okay, there we go. So given what what we know so far can anyone finish off this computation for me? Uh hopefully we should be able to figure out what the common polinomial of this and the common polinomial of this are.

[00:25:58]Well the second one's - t and the first one looks like the unnot. So that should be one. So it's 1 - t * t. So 1 + t ^2 perfect.

[00:26:15]All right.

[00:26:18]So uh this now gives us a proof that the unnot and the terfil actually are different uh because their value according to the conway polomial is different that is I've proven it up to the fact that it exists in the first place. So by sort of continuing this logic you can probably convince yourself that if such a link invariant exists then it has to be unique. You can compute it for anything. Um the existence part is harder. I'm going to skip it. There is a way to think of it in terms of topology things like if I say the words homology of the infinite cyclic cover of the not complement then maybe if that means something to you there's a way to derive the common polomial from that uh for for what I'm going to do now I'm just going to sort of treat it as a thing that exists all right so now let's get closer to defining what a vicilia invarant is. And to do that, I need to broaden our conception of what even counts as a meat.

[00:27:32]There is a word for a kind of adjective that broadens a category where like a toy boat isn't a boat, right? Or a dwarf planet isn't a planet potentially controversially. So a singular knot isn't really a knot. What it is is uh the same way that a knot is a map from a circle to three space. A smooth knot sorry a singular knot is also a smooth map from the circle to threedimensional space. But it fails to be an embedding at finitely many double points.

[00:28:23]What do I mean by that? I mean something like this. This this knot passes through itself and therefore it is not an embedding of the circle into threedimensional space.

[00:28:39]We have one double point over here. A terminology thing. I'm going to call this a one singular knot because it has one singularity. If it if I talk about like a 17 singular knot, it's going to have 17 singularities.

[00:28:54]And Vicilio's idea in I want to say like the '9s or something I I don't actually know when this happened.

[00:29:05]uh was that any not invariant and for us um the uh range of a non- invariant is going to be something that can be added or subtracted like numbers or polomials. So any not invariant uh can be extended to knots with double points aka singular knots by the vicil the vicil of relation you could in theory have triple points right like there's nothing >> you could I'm not going to consider them uh but they certainly have been considered.

[00:29:49]>> Okay.

[00:29:56]>> I mean, I guess in theory you can have you can go crazy and have points with like infinitely many pre images, but that might be a little harder to handle.

[00:30:07]Anyway, um what is the facility relation?

[00:30:12]If I have an invariant V and I'm trying to evaluate on a singular knot, I'm going to orient it so that both oriented strands are pointing upwards.

[00:30:24]First, I'm going to pull it pull the uh strand with positive crossing towards me and then I'm going to sorry with positive slope towards me. Then I'm going to pull the strand with negative slope towards me and I'm going to subtract them.

[00:30:42]So we can think of this as like the positive resolution minus the negative resolution where when I say resolution I mean I've uns singular if I it um what about a knot with two double points? You can convince yourself that it doesn't matter which order you resolve them. So for instance, if you imagine a knot with two with two double points, it's going to be the value of V on it is going to be V of the resolution where they're both positive plus V of the resolution where they're both negative minus V of each of the two resolutions where one is positive and one is negative. Hopefully that makes sense. Uh if you want I can draw out an example but now I'm going to talk about what it means for a an invariant to be vicilive.

[00:31:41]So a vicilive or finite type invariant of order at most n is one whose extension to singular knots vanishes.

[00:32:14]uh a knots with strictly more than n double points.

[00:32:26]All right. And then we say that it's order uh exactly n if it's order at most n but it's not order at most n minus one.

[00:32:38]Um, I think it's time to go on to new sheet of paper.

[00:32:49]I kind of want to have that still on the screen. All right.

[00:32:56]Question.

[00:32:57]>> Yeah.

[00:32:59]If a if an invariant vanishes for a certain number of double points, will it vanish regardless of which knot has that many double points?

[00:33:12]>> Um, by definition, yes. When I say advantages on knots with at most sorry with more than n double points, I mean advantages on all of them on all n plus one singular knots and >> but is it the case where it will be zero for some but not for others.

[00:33:31]Um you could certainly invent invariance where that is the case and then uh the order of it if if it is a finite number is that is going to be the smallest n for which that is the case for all of them. Okay, there's an analogy by the way I won't develop this but there's an analogy with polomials and the degree of a polomial is how many times you can do the finite difference operation before it vanishes um so I'm going to say v subn is going to be the set of vicilia of or invariance of order at most n.

[00:34:27]Um I suppose I can put a little parameter here for like what ring these take values in. So it could be the integers, it could be the reals, it could be the complex numbers. uh I'm generally going to take uh RT to be the complex numbers.

[00:34:53]So this gets us what's called a filtration.

[00:34:57]So a filtration of I don't I don't know if people here know what the word module means. Um, have people seen like vector spaces over fields other than the reals or the complex numbers?

[00:35:19]Yeah, I'm just going to say um vector spaces here or our modules in general. A module is like a vector space, but you have it over a ring instead of a field. Um, anyway, this isn't all that important. I'm just realizing that that might not be terminology that's universally understood. Um, but what is a filtration? It's an increasing sequence of sets.

[00:35:49]And the union of all of these uh is what I'm going to call just V. So V is defined to be the set of facility variants of any order.

[00:36:04]Okay. Do any of these actually exist?

[00:36:07]Yes. Here's an example.

[00:36:10]The nth coefficient of the Conway polomial which I'm going to write as C subn is vill of order at most N.

[00:36:31]All right. How do we prove this? I'm gonna Where do I have the definition of the second? Where do I have the definition of the common polomial? Okay, I have it over here.

[00:36:43]This should look uh similar. This is the um right hand side of the facility relation. So um proof by definition.

[00:37:00]So con polomial of the unnot is one and the conary polomial of a singular not is going to be a multiple a multiple of t it's t * the common polomial of this smoothing. So therefore if we have something with lots of singularities with lots of double points it's common polomial is going to be a multiple of a large power of t.

[00:37:40]this notation means divide. If k has n singularities then t to the n is a factor of or divides c of k.

[00:37:50]Um, looking at my notes, I'm realizing here's something that might I might want to skip a little bit. Um, okay. So, okay, >> maybe I missed it, but did you explicitly say what the codway uh polomial is for like singularities?

[00:38:20]Um I said what the what any invariant under the sun is for singularities.

[00:38:29]If v is any invariant we're extending it to singularities by this relation.

[00:38:37]>> Gotcha. Thanks.

[00:38:45]>> I frankly I would have to check. I believe it is exactly n. Um but I don't know how that's proved.

[00:38:59]Um it is the case that C sub one uh vanishes for all knots.

[00:39:06]Um which will fit in with what I'm about to prove which is that there aren't any interesting facil variants of order zero or one. Um C sub2 is interesting. In fact, C sub2 by these computations here take the value of zero on the unnot and the value of one on the trifoil.

[00:39:28]Okay, here's uh just a little peak into the structure of these.

[00:39:40]Um a vill invariant is primitive if V of the connect sum of K and L is V of K plus V of L. Um a theorem is that if f and g are vicil invariance of orders at most m and atmost n then f * g is a vicil of invariant of order at most m + n.

[00:40:29]This means that the set of all of a silicon variants is what's called a graded algebra and there's a 10% chance that the theorem I'm about to write is wrong but hopefully it's right. um theorem.

[00:40:48]The set of vasilic variants is freely generated by primitive basilic variance and I do want to talk about this second theorem over here because uh you might think that it's trivial. So why would would you think that it's trivial? Like let's imagine that f and g are um vill of order two.

[00:41:21]Well let me take a random uh two singular not sorry vility of order two means that it vanishes on everything with three or more singularities.

[00:41:37]So by the definition of a product of two functions you would think that you can write this And then these two vanished because F and G are of order two. And therefore FG also has order two. Right? So you might think the product of two things with order two is also order two. So why am I talking about adding them? Um the reason it's not uh the case. Um in fact this equality is the one that's wrong is I'm thinking of f and g as being defined on knots and then I'm extending them to singular knots by the facil the vasilic variant sorry the villic relation.

[00:42:21]So what I mean by that is FG of this is going to be FG of like the sum over all eight resolutions of it and then I I evaluate them.

[00:42:52]Um, does that make sense? I'm I'm sort of defining FG on knots first and then trying to extend it to the singular knots instead of uh the other way around.

[00:43:08]>> I'm holding on. I think >> I'm just trying to convince you that this is not actually a trivial theorem.

[00:43:16]So yeah, I mean it's frankly it's not impossible to prove. It's if you think about it hard enough, you'll be able to prove it, but it's uh you do need to have a little bit of care when you verify it.

[00:43:30]Um in seconds, think about what I want to do now. Okay.

[00:43:43]So now I'm going to find all villarants of order zero.

[00:43:57]The set of villar variants of order zero is a one-dimensional vector space. Uh it corresponds just to the constant maps.

[00:44:06]Right?

[00:44:07]In other words, there aren't any interesting ones. uh it's just the just the constant.

[00:44:13]So proof what I want to do uh I'm claiming that if f is visible of order zero then f is constant on nuts and I'll sort of prove this by example.

[00:44:38]Let's say f is in v 0 and I'm going to try to evaluate it on the tfoil.

[00:44:48]Let me pull up the the facility relation again. Um which is on one of these sheets of paper in theory. Um uh what this means is I can bring this over to the other side and express this in terms of this and this.

[00:45:10]So that's f of the tfoil with one of the crossings changed plus f of the tfoil with one of the crossings made into a double point.

[00:45:30]This vanishes because it's vill of order zero. It vanishes on anything with strictly more than zero double points.

[00:45:39]And therefore, it's the same as f of the unnot.

[00:45:44]What can we learn from this example? We learned that f doesn't care when you change a crossing. And it's a theorem that all knots can be converted into the unnot by a series a finite sequence of crossing changes.

[00:46:01]Therefore f of any not k will equal f of the unnot. Therefore f is constant.

[00:46:12]Speak up if you have questions about that proof because I'm going to go on to order one and I'm going to say order one doesn't give us anything new either.

[00:46:27]Um, claim V1 equals V 0.

[00:47:02]by a very similar argument to what we just did. If I take any one single knot F and F has order one, f does not care about crossing changes in uh in the one singular knots.

[00:47:20]Why? The difference between the values of f before and after the crossing change will equal f on some two singular and then it's going to vanish by the fact that f is order one.

[00:47:35]So, I'm going to take uh K, this one singular knot. I'm going to allow it to pass through itself because F doesn't care about crossing changes. I can sort of separate out the part of K uh that's on one side of the double point. I can separate out the part of K that's on the other side until it looks like this.

[00:47:59]And then finally f of this simplest looking one singular knot the one that looks like a figure 8 equals z by the vill uh relation applied in the following way.

[00:48:20]We resolve it in the positive way and we resolve it in the negative way. We subtract we get the same thing twice because these are both the unnot equal z.

[00:48:32]And the conclusion is that we thought f had order at most two. Sorry, we thought f had order at most one. It turns out it actually has order at most zero because it vanishes on everything with one double point.

[00:48:45]Therefore, v1 equals v 0.

[00:48:50]All right.

[00:48:52]Um, I should call this a theorem.

[00:48:57]So why doesn't this argument work? Uh, one step up.

[00:49:08]How is it that v sub 2 has non-trivial things in it? So first I'm going to claim that the dimension of v sub 2 is at most two.

[00:49:39]All right, here's my claim. If I'm allowed to change crossings on two singular knots, I can bring it to exactly one of these two forms, but I cannot bring this thing into this thing. And I'm going to open that up to the audience if you can convince me that no sequence of crossing changes will turn this into this.

[00:50:22]We can look at how many strands come out of each point. Well, any double point is only going to have four four strands coming out of it.

[00:50:32]Right.

[00:50:38]How many Sorry, I'm not sure I understand that.

[00:50:46]How many how many what there are between the two?

[00:50:51]>> How many arcs are there between the two crossings?

[00:50:55]Um yeah, I Well, >> it looks like there's two in both of them, right? Yeah, actually there are two in both of them.

[00:51:01]>> Oh, wait. No, there's four in the first one.

[00:51:04]>> Oh, you're right. You are right. There are four arcs connecting the two of them in the first one and then only two in the second one. Okay, I think that works. That wasn't the proof that I had in mind, but it is it does work. Um, what I had in mind is imagine we label these double points. Uh, we call them one and two.

[00:51:25]If we travel along the knot in this one, we encounter them in the order one two one two you know repeating because it's a loop. So 1 2 1 2 1 2 1 2 Whereas here if we travel along the knot we hit them in the order 1 22 2 1 22.

[00:51:45]So they are genuinely different.

[00:51:49]Um, and if k is a two singular, not f of and let me actually sort of I I I'm I'm using an unspoken rule uh an unspoken conclusion in all of this that probably should be written down. So if K and K prime are two singular knots um differing only in one crossing change and F is vicil of um >> what's the definition of crossing change?

[00:52:35]Hm.

[00:52:35]>> What is the definition of crossing change?

[00:52:40]>> I start with a knot that has something like this and then I do that.

[00:52:45]>> Okay.

[00:52:52]So yeah, so this is the bit of logic that I was using without writing it down. Um f doesn't care about crossing changes in two singular knots. Um and the reason for that is because f of k minus f of k double k prime equals f of k uh double prime for some three singular kn uh and then that equals zero because f only has order two. Um and again that that comes from the way we're extending things to uh singular or not is we subtract before and after crossing change and it's equal to the value at uh the double point.

[00:53:46]So why does this prove that the dimension of V sub 2 the set of villain variance of order at most 2 is at most two? Well actually this is um more than one step of reasoning is left actually. So hold on.

[00:54:15]So thus we have a map I'm going to call it alpha from pil variance of order nos 2 to um c^ squ where we take an f as input we get f of these two things as output.

[00:54:42]C^ square being the set of pairs of numbers.

[00:54:46]Um the kernel is V1. Kernel means the set of uh functions that get sent to um 0 0. Why is kernel V1? Because by definition that means that f also vanishes on all two singular knots which is what it means to have order one. So but uh the dimension of the image of alpha is at most one. Why?

[00:55:20]Because f of this thing the one where going around it encounters the crossings in the order 1 221122 that's always going to be uh zero.

[00:55:40]And that's just a computation.

[00:55:43]Both of the resolutions of this are the same one singular knot.

[00:56:00]So therefore the dimension of v2 is at most the dimension of vub1 + 1 which is 1 + 1 which is 2.

[00:56:15]Uh and then finally the claim is that the dimension is not just at at most two but it equals to and then the proof is C2 is in V2 but C2 is non-constant where in this case C2 means the second coefficient of the cone polinomial.

[00:56:44]I feel like that was a little complicated. Um, but the upshot is now we know all of the vicilian variance of order at most two. We have the constant function. We have the this weird C2 thing where you compute the Conway polomial and then you grab the coefficient of t^2. Uh we have all linear combinations of these and that's it.

[00:57:24]And in fact this is a big part of why um villic variants are called finite type. It turns out that the set of villar variants of any given order is finite dimensional. you can only have finitely many independent villarants of a given order.

[00:57:45]But to prove that a little bit more formally, I'm going to have to define something called a chord diagram which I can define informally as a circle with chords.

[00:58:08]Okay. Um, here are some examples.

[00:58:13]Um, chord diagrams with only one chord. You have this the set of chord diagrams with two chords.

[00:58:24]You have a circle with two non-intersecting chords. A circle with two intersecting chords.

[00:58:32]You can find five different chord diagrams with three chords.

[00:58:43]Let me see if I can draw all of them.

[00:58:48]Um I'm missing one.

[00:58:52]>> I think they all intersect.

[00:58:54]>> Ah yeah. Yeah. Okay. This one. Yep.

[00:58:57]Great. Um, by the way, a small note. Um, I am considering mirror images to be different. So, here's Okay.

[00:59:17]>> Yeah, that one does look chyro.

[00:59:21]>> Is it actually? Okay. I have it in my notes as being Chyro, but I >> I'm realizing at the last second that it actually might not be.

[00:59:29]>> I'm pretty sure it is because like the one in the bottom >> right >> like like Okay, you could you could use a similar justification to before where you like did the like the counting the the incidences along the path of the outer circle >> and like you would have like >> I guess I don't know. I can't point at anything for you to see it, but I I it goes different in opposite ways.

[00:59:54]>> I sure sure I believe you. Anyway, my my point here is just that um I'm not considering chord diagram up to reflections. I'm so the the sets are slightly larger than they might be otherwise. Um and why am I making this definition?

[01:00:13]If k which I'm thinking of as a function from the circle to threedimensional space is an n singular knot meaning it has n double points or n singularities then what I'm going to call sigma of k in the set a subn is defined by joining points in S1 uh with the same value. So here's an example.

[01:01:00]Sigma of this two singular knot has two chords that don't cross.

[01:01:09]um sigma of this other two singular knots not has two chords that do cross.

[01:01:18]Hopefully that sort of makes sense. I'm imagining I'm I'm following along the knot. I'm encountering the points in the sequence 1 2 2 1 and then sort of joining them up like that.

[01:01:33]>> Makes sense to me.

[01:01:35]>> All right. So proposition my pencil is being annoying. Okay.

[01:01:43]Proposition if K1 and K2 are N and singular knots.

[01:01:55]Um then for V of a silicon variance of order most n if k1 and k2 have the same chord diagram then they have the same value according to v and it's the same logic as before like let's say v has order two and I'm evaluating it on why did I choose a topic that would involve me trying to draw so many knots.

[01:02:47]>> Sorry.

[01:02:50]>> Thank you. Yeah, I before I knew that you had notes that you were looking at, I was like, "Holy crap, there's no way this is real.

[01:03:03]I wish I were doing the the more impressive version where I kept my secrets." Um, anyway, uh, this is by the facility relation. This is by the definition of having order 2. Therefore, V2 doesn't care about crossing changes. And the claim is that if two knots have the same chord diagram, then we can get from one to the other by a bunch of crossing changes. In theory, that needs proof, but I'm going to skip it. I'm going to sort of have you trust me that that's true.

[01:03:36]Um wait so the claim is that through deconstruction like through um untangling the knot via crossing changes we can reduce that to one of the two uh one of the two knots that we saw above.

[01:03:53]>> Yeah.

[01:03:54]>> Okay. Um, and therefore like even with larger values of n, if I have if k1 and k2 are both 17 singular, they both have 17 double points. As long as the corresponding chord diagrams are the same, I can trans uh transform one into the other by a bunch of crossing pages.

[01:04:16]>> Makes sense.

[01:04:20]So, thus we have a map um going to call it the I'm going to call the map sim for symbol takes in a function this yeah takes invariant and spits out a function on chord diagrams.

[01:04:48]So, here's a little currying thing that I'm doing.

[01:04:52]Um, sort of we can evaluate uh V the symbol of V on a given chord diagram by evaluating it on a knot whose chord diagram is C.

[01:05:28]Um and the kernel of this map is the set of things with smaller order.

[01:05:41]Uh why is that the case? Because if uh the symbol of V is sort of the all zeros map then that means that V vanishes on everything with n singularities. Therefore it has order n minus one.

[01:05:59]Um this is good news for us. It means correlary.

[01:06:06]How the hell do you spell corary?

[01:06:09]There's two L's.

[01:06:11]Okay.

[01:06:13]And has two Rs, but they're not next to each other. Um, V I'm going to stick with vector spaces because I'm going to stick with um working over a field. Okay. The vector space of vasil invariance of order at most n is finite dimensional.

[01:06:47]Uh this follows sort of by rank nullity.

[01:06:50]Essentially this thing over here is uh a finite dimensional vector space.

[01:06:56]Its dimension is the number of things in a subn. Um this kernel is finite dimensional by induction.

[01:07:05]Uh therefore the dimension of this is the sum of those.

[01:07:16]Okay. And here's uh an example.

[01:07:22]So the symbol of C sub2 is the function that sends this crossing diagram to one and this non-crossing diagram to zero.

[01:07:38]Uh there exists an interesting villain invariance of order three. I'm going to call it V sub3.

[01:07:45]uh and it can be shown that what it does is it sends this crossing diagram to -1, this crossing diagram to -2 and then the other three all get sent to zero.

[01:08:05]And I believe this is actually sort of the only interesting visil of invariant of order three up to linear combination with invariance of smaller order. Um so this is in some sense unique and that leaves us with a question.

[01:08:29]What is the image of this symbol thing?

[01:08:39]What maps on what functions from chord diagrams to numbers actually arise as symbols of vasilla inversions and uh my notes actually postpone answering this for a little bit. So I'm going to leave you with that uh question for a little bit longer to build up this event because I want to talk about things called actuality tables and these are used for computations. All right.

[01:09:16]To define um a specific villic variant of order at most n.

[01:09:30]What we need to do is choose a representative of each cord diagram.

[01:09:38]All right. So, how what am I talking about?

[01:09:53]Um, so we work. Okay. So, a subzero set of chord diagrams and zero chords is just this. Uh, a sub one is just this.

[01:10:08]A sub 2 are these.

[01:10:11]We had the three thingies from AS 3.

[01:10:24]I'm going to choose uh this not to have this symbol or this cord diagram.

[01:10:34]uh this not has this chord diagram and here I am making like genuine choices uh because there are lots of knots for the given chord diagram. I'll explain a little bit why I'm making these choices, why I'm finding these representatives.

[01:11:06]Give me a little bit though.

[01:11:30]Okay.

[01:11:34]And now if I let's say I run up to you and say, "Oh, I have a nice vasilic variant of order two." What I should tell you is what it does to this thing uh to this knot. What it does to this knot, what it does to this knot, what it does to this knot, and then uh I don't actually need to tell you what V2 of any of these things are because it's uh it's got more double points than the order.

[01:12:01]So, you know that they're all going to be zero. If I want to tell you about a cool uh villain variant of order three, well, I need to tell you about uh the value on everything up to here and then uh sort of an exercise. It follows from everything we've done up to this point.

[01:12:32]Given this information, you can find V2 and V3 of any knot that you so wish.

[01:12:41]And the idea is you do a bunch of crossing changes to convert it into something you know about. Um all of the changes in the value are recorded by um knots with one singularity. You do a bunch of crossing changes to convert them into things on this table.

[01:13:00]um all of the changes are recorded by not with two singularities etc until you exceed the degree.

[01:13:11]So I don't know that I want to actually go through an example just because I don't have one prepared and it would be annoying to come up with one on the spot. Um, >> in general, you're not making the claim that like there's always uniquely one correct row here, are you? Like you could have two V4s that are different from each other. I'm imagining for four.

[01:13:34]>> Um, in fact, um, so remember that like, okay, so I was talking about this is actually the same as the C2 that I was talking about earlier. um C2 time C2 turns out to have order four uh because of that um multiplicative multiplicativity thing I was talking about before you can multiply together two different facil variance you get another facility of variance uh of higher order um so I'm going to tell you about the dimension of the primitive invariance Okay. So, I'm going to say n equals 01 23.

[01:14:21]Okay. So, I'm going to I got these off of OEIS, but I might be like off by shifting by one uh by accident. I guess it's 1 2 3 5 8. Um, everyone say it all at the same time. 12 18 >> 27 uh 39 55. And then I believe that's as far as we know. I think that is >> the extent of >> Fibonacci or something. It looks very Fibonacci.

[01:14:51]>> Yeah. Well, that's that's why I did that.

[01:14:53]>> Yeah, I know. That's what I thought too.

[01:14:56]>> I'm juking you out. Um and then the dimension of all invariance. It can be computed from these um because we're going to have like another visil invariant of order two.

[01:15:10]That's C2 squ. So it's like zero. Hold on.

[01:15:15]Why do I have one? Oh, I guess the constant map isn't considered primitive.

[01:15:19]It's like too si uh too simple to be considered primitive. Anyway, 3 4 9 14 27 44 80 uh 132 232.

[01:15:36]I believe it's possible that I was copying down a a related but not exactly the same sequence from OIS. So, apologies if my numbers are slightly wrong. Um, but the point is beyond order 12, we actually don't know how many of these there are.

[01:15:56]That said, we still have a very powerful theorem that tells us about um what it takes to be a basilic variant.

[01:16:07]So, here's a definition.

[01:16:15]A function um from chord diagrams to numbers satisfies the four term relation.

[01:16:32]Um, this is not just me abbreviating for term by the way. People write it like that in books. Um, but T stands for term. If Okay, I'm going to have to explain my notation again in a second. Um, so just like not when I write part of a chord diagram, the rest of the chord diagram is arbitrary. I just ask that it's the same uh in all four cases in this equation.

[01:17:26]So it satisfies the fort relation. If whenever we have four core diagrams that are related by uh the pictures, then the alternating sum of the values under f is zero.

[01:17:37]Um and we also have the one-term relation.

[01:17:45]If f of c equals z when c has an isolated chord, what does an isolated chord mean? It means a chord that's not intersected by any other chord. So, eg f of this equals z.

[01:18:15]All right.

[01:18:21]If f satisfies the one term and the four-term relations, then we call it a weight system.

[01:18:39]And the set of weight systems is called WN. So it's a subset of the set of functions here.

[01:18:53]And now I'm ready to state the fundamental theorem of vicilian invariance.

[01:19:10]And based on what I've done so far, can anyone guess what the fundamental theorem is going to be?

[01:19:18]Hopefully I've set it up reasonably well.

[01:19:26]Um, is it that claim that you didn't uh that you that you were going to leave us on a cliffhanger on?

[01:19:34]>> Um, which hold on. Where did I >> I forgot the statement already.

[01:19:39]>> Where where the hell did I one of these papers has it? Oh, it's on top of this one.

[01:19:51]It was about the image of this symbol thing, right? So remember the the whole point of the symbol thing is that we can think of the invariance as acting not just on knots but on chord diagrams.

[01:20:06]Um the function on chord diagrams that you get is called it symbol.

[01:20:11]Um if you if we if I have two different facil variants where like the difference between them is a facility variant of lower order then they're going to have the same symbol. So it's not an injective map. Um but its kernel is well understood. Um in fact let me see if I can find exactly where I wrote this down.

[01:20:36]Uh sorry I have a lot of papers now.

[01:20:44]Uh yeah, hold on. Sorry. We have a map from VN to the set of functions and the kernel is VN minus one, right?

[01:20:54]Okay. So that's that's the deal. Um and understanding the size of the image of symbol will tell us a lot about the dimension of V subn. It's going to be how much bigger is the dimension of V subn than the dimension of V subn minus one.

[01:21:16]So now the fundamental theorem is the map um symbol Sorry, a lot of writing and then I'll read out what All right. So this map, this symbol map uh identifies v subn mod v sub n minus one with the set of weight systems which as a reminder are functions satisfying this.

[01:22:41]In other words, that set of weight systems is exactly the set of possible symbols.

[01:22:48]All right, so this theorem has an easy direction and a hard direction. So the easy direction was proved by Vicilio.

[01:22:58]The hard direction was proved by Kvich.

[01:23:02]Um I suspect I won't have time to do the hard direction. But let's uh do the easy direction starting with Okay.

[01:23:35]The four-term and one-term relations are actually necessary. The symbol of any vasil variant satisfies them. So what's with the one-term relation?

[01:23:46]If a chord diagram C has an isolated chord, which again as a reminder just means a chord that's not intersected by any other chords, then there is a singular knot uh corresponding to it.

[01:24:12]Oh my god, running out of graphite.

[01:24:21]um where K has the form uh this we can represented we can represent C by a knot that looks like this and then um one the symbol of any facility convant F at C will equal F of K by definition, which will be F um resolved one way minus F resolved the other way.

[01:25:08]But you can flip each of these two things over to reveal that you're actually subtracting something by itself.

[01:25:15]So that's going to equal zero.

[01:25:19]So that's the origin of the one-term relation.

[01:25:23]What's up with the four-term relation?

[01:25:25]And maybe I should actually point out that this explains where did I put my table? Yeah. Um, this explains all of these zeros over here because all of these have isolated chords.

[01:25:47]Um, and it also explains why this zero has to be here.

[01:25:53]Um, great.

[01:25:57]What's up with fourterm relation?

[01:26:01]This follows from the following set of pictures.

[01:26:11]I was sighing because it's a lot of drawing again.

[01:26:51]Wait.

[01:26:53]Crap. I might have to redo this. Um, sorry. One moment.

[01:27:15]I think these need to be overcrossings.

[01:27:26]Okay. So, uh the point is there is some relation between uh these various singular knots um which I am like 80% sure that I've now drawn correctly and that itself follows from the following picture.

[01:28:11]Yeah. So I don't I don't expect you to have uh 100% followed this argument but just to explain a little bit more in this last picture I'm sort of pulling a bite of string sort of out from behind a crossing and then in front of it and then this if you expand all of the uh resolutions on on like this traveling singularity around it it ends up sort of telescoping and equaling this minus this and then finally once you convert it into the language of chord diagrams by taking the the symbol map uh you'll get sort of trust me that this happens you'll get exactly the four-term relation which which I want to find so that I can put it in front of the camera again.

[01:29:22]I have no one to blame for myself. I think uh I have no one to blame but myself, I think, for how much paper I've generated now. Um but yeah, just as a reminder, this is what it looks like. Um the way to think about this actually is think about this second chord as being fixed and this first chord as moving around. So or actually even like think of this end point as being fixed and this end point is moving around. So, first it travels to the opposite side of of this chord. Then it travels up here.

[01:29:53]Then it travels uh around. And we alternate sign as we do that.

[01:30:00]And that ends up corresponding exactly to this.

[01:30:06]It's okay if your takeaway is just it can be proved. Um but I I wanted to include that for the sake of uh completion.

[01:30:20]So I can go into a little bit of algebra now. Um okay I have two different directions that I can go in right now. I can talk a little bit more about the various algebraic structures that the set of uh chord diagrams get because of these things. Or I could define the concavage integral which is going to take up like a lot of time. So I don't know what what you're more interested in.

[01:30:56]>> I'm interested in algebraic structure.

[01:30:58]>> Okay, great. Um, so I'm going to define I'm going to do like the sort of cardinal sin of typography, which is when the same letter means two different things depending on the font.

[01:31:15]Um but I'm just following what the book does. Uh script a subn is the vector space generated by uh these chord diagrams mod the 4t and 20t relations.

[01:31:40]Um the therefore the set of weight systems which if you remember is functions uh on core diagrams satisfying these relations can be thought of as functions from this quotient space.

[01:31:59]Hum for those who aren't familiar with it means uh the set of linear maps essentially the set of maps um preserving the structure you care about which in this case is the vector space structure.

[01:32:15]What's it stand for? Does it stand for homorphism?

[01:32:20]I don't know if any of the people here know that.

[01:32:22]>> I would have guessed but I don't know.

[01:32:24]>> Yeah, that's probably true. I can't think of anything else it would work. It would be um and then my claim here is that a which is uh the sum of all of these the direct sum of all of these vector spaces is a by algebra. It's a graded by algebra. All right.

[01:32:55]So what is a by algebra? It means something that is both an algebra and a co-algebra. What is a co-algebra? It's the opposite of an algebra and you'll see it shortly. Um, okay. So, first off, a regular old algebra in this case means uh a ring containing the complex numbers. Um, so basically I need to tell you how to multiply two different diagrams.

[01:33:25]So definition the product of diagrams.

[01:33:42]B1 um in A sub M D2 in A sub N is D1 D2 in A sub M + N um defined by okay I'm going to switch to a different this one is uh not sure There we go.

[01:34:14]Let's say I'm going to try to multiply this basketball looking diagram with itself.

[01:34:31]I'm going to cut them open and glue them.

[01:34:41]And at this point there's a question that you need to ask me.

[01:34:45]>> You rotated it. How do you know which way to rotate it?

[01:34:50]>> How do I know which way to rotate it? Or in other words, why is this or even is this well- definfined?

[01:34:58]>> All right.

[01:35:00]>> Yep.

[01:35:02]So lema this is well defined mod the 4T relation. So remember I'm doing this script a which means that I'm um if I've got like these four chord diagrams that are related by the four-term relation then I'm quotienting out by this alternating sum.

[01:35:31]So this thing has to this is a zero.

[01:35:36]Um this thing has to equal zero.

[01:35:41]Uh, I'm not going to prove this because it's actually uh a little bit annoying, but it's in the book and also you can try to do it yourself. The key thing to do is uh you want to show that sort of rotating one of these by one unit essentially uh gives you something that differs by a sum of things of that t form.

[01:36:11]And uh also the onet relation isn't needed for this. You get a slightly different vector space if you only mod out by 4t and not by 1t. And it also is a bio algebra. Um I'm just not talking about it because it's less important for us.

[01:36:27]Okay. Now definition the co-product this this is what turns it into a co-algebra delta.

[01:36:47]Okay.

[01:36:51]um that you kind of a little bit need to know what a vector sorry what a tensor product is. Um a tensor product is a kind of operation that eats in two vector spaces and gives you another vector space whose dimension is the product of the original dimensions. Um for us what it is is if you give me like a basis for a subn then the basis for this thing is going to be the set of all pairs of things in that basis.

[01:37:25]Um and then the stuff about k plus l= n is just that the total number of chords has to equal n in that pair of bases.

[01:37:37]So it's defined by uh delta of d here being a diagram mod 4 t and 1t and I'll explain this notation in a second.

[01:38:02]Okay.

[01:38:04]Um, here D within brackets means the set of chords. Uh, so J is some subset of the chords. And I'm let me just draw you an example. Actually, that'll be easier.

[01:38:16]Here's delta of this basketball.

[01:38:25]First, I'm going to consider J to be the empty set. So, I'm going to have all of the cores over here. Then I'm gonna consider J to be that set.

[01:38:43]Sort of all ways of choosing some subset of the chords and then the remainder go into the second uh diagram.

[01:38:53]And actually because of the onet relation this is going to vanish because that um all of these are like isolated chords. This is also going to vanish because this is an isolated chord. So this case actually simplifies really nicely. I think the only terms that survive are these two terms.

[01:39:15]Um you still do need to check that delta is well defined modulo one fort and one t um because there's more than one linear combination of diagrams that equals a given thing. So lema well delta is well defined mod 4 and 1t um proof see the book uh and then what's a by algebra basically it's something with both a co-product and a product that play well together. I'm not going to write all of the um the things that it needs to satisfy. Oh, question from multi- tube. Can the 4t relation be viewed as some kind of commutator relation?

[01:40:11]Um well, I guess in theory you could try to think about commutators with respect to this product over here. But the thing is it's this is a communive algebra.

[01:40:29]So the commutator would always be zero by commutativity. Um oh I guess because we're quotient out but maybe you would expect that. Yeah I don't know any good way of thinking about it in terms of uh commutators. Uh actually sorry there is a massive um connection to the theory of le algebbras which the book spends several chapters talking about. I'm not going to talk about that at all here but uh you uh it is it is a thing. Um, so yeah, that's not as good of an answer as perhaps you were hoping, but I'll just leave it at there exist links to Lee algebbras and you can check it out on your own time if you so wish. Um, because I don't feel qualified to explain them. All right.

[01:41:29]Now remember weight systems are functions from a subn to uh numbers.

[01:41:37]It turns out that um this set of all weight systems is also a bio algebra um and in fact a graded by algebra.

[01:42:00]How do I define the product of two weight systems?

[01:42:06]Well, what I do is I view it as the dual of the co-product.

[01:42:12]So in other words, to find W1 * W2 of this thing, I do like W1 of this time W2 of this plus W2 of this time sorry W1 of this time W2 of this etc. Um and then what's the other one? I have to define a co-algebra co-product of a weight um it's going to be defined by this formula also This symbol map from basilic variance to weight systems commutes with multiplication and coiplication uh h sorry this also turns out to be by algebra I'll maybe I'll explain uh out in a second. Um but I do want to finish this sentence. So, running out of room. Okay. Uh this is just to say we get nice um algebraic properties for the symbol map.

[01:44:18]Uh with respect to multiplication and co- multiplication. Uh addition is a little bit of a bummer though because it doesn't commute with it. Um, so I don't have this in my notes unfortunately, but I should define how to multiply and co multiply for silicon variance. Um, I think it's probably going to be self-explanatory.

[01:44:58]That's what it is. You have to talk. You have to use the um connect some Okay. So, why do we care about this whole algebra thingy? Um, this pogebra structure, it gives us an opportunity to use a nice theorem called the Milner Milnor Moore theorem.

[01:45:25]Uh any commutative co-commutative what that means is you take the co-product and then you swap the uh order of each of the tensor things you get the same thing uh connected.

[01:45:52]What that means is that the zeroth order things are essentially your set of scalers graded means it satisfies the um things you would want it to do. It's sort of just just trusted that trust me that it's a reasonable condition by algebra is isomeorphic to the symmetric algebra on the primitive elements. What is a primitive element?

[01:46:44]An element is called primitive if it satisfies co-product of P equals 1 tensor P plus P tensor 1.

[01:47:00]Um and by the way in the chord diagrams one is this empty chord diagram which looks like a zero unfortunately.

[01:47:15]Um okay so what does that mean?

[01:47:22]Uh it can be shown that uh these two diagrams are primitive.

[01:47:39]In other words, you use the algorithm I described before for finding their co-product. Um you end up with uh this formula.

[01:47:52]But this thingy is not.

[01:48:01]And what does that mean? According to mil nor it means that uh I can express this in terms of these.

[01:48:11]The symmetric algebra is the set of polomials whose variables are the primitive elements and where multiplication is what you expect and where co- multiplication looks like uh this sort of where x and y and z are primitive.

[01:49:05]Okay.

[01:49:09]Uh long story short, the primitive ones are our generators. There are no non-trivial relations between them. The non-primitive ones can be expressed in terms of them. That gets us a better sense of what the structure of a subn looks like even just as an algebra. Um and here's a uh another remark.

[01:49:36]If I define a uh framed as the um by algebra of chord diagrams mod 4t but not one t.

[01:49:57]uh everything I've said so far still holds, but the the primitive spaces kind of look a little bit nicer or a little bit more interesting actually. So P1 I'm going to say is the set of primitive things of order one that ends up being generated by this thing over here.

[01:50:21]Primitive things of order two is this difference.

[01:50:30]Uh and then primitive things of order three is this strange uh linear combination.

[01:50:48]So yes, the structure of our algebra is nice. It's generated by a bunch of things without any relations, but the things that is generated by are a little bit funky. And in fact, the book uh goes into a lot more detail about this. there is a good way of thinking about why these happen to be primitive uh sort of geometrically but I won't get into that. Um okay before I go into conseage there's a thing I skipped a little while ago that I might want to talk about.

[01:51:29]Uh are there any questions up to this point?

[01:51:35]Looks like Sawdust might have a question.

[01:51:38]Oh, sadly I have to go, so I'll have to finish watching the VOD. Yeah, I'm sorry. I knew that this was going to take a long time. Um, but apologies.

[01:51:49]>> Would um in the uh is there any reason that that looks slightly binomial like like I'm imagining >> Oh. Um, >> right. as in like one one two one >> something like that. Yeah. Or like the first one being like maybe negative one uh like up to up to multiplying by a factor of negative 1. The second one is like x - one and then the next one's like x^2 - 2x + one.

[01:52:17]>> Not to my knowledge.

[01:52:20]Um, the way this ends up being explained is you extend your view from just chord diagrams to things that look like this.

[01:52:37]Um, and you have like a bunch of relations relating these things to court diagrams and these all end up being equivalent to uh sort of connected diagrams over here. Um, Sada says, "I'm a huge fan of the pencil and paper style presentation. Physical media for the win. I appreciate it. Um, I was worried that it was going to uh bore you by how much time it was taking me to write stuff, but it probably slows me down in a way that is helpful for comprehension, maybe." Um, okay. Just a little um miscellaneous thing that I skipped from before. um two knots uh k1 and k2 are called n equivalent if they can't be distinguished um by vicilian invariance of order uh at most n K is n trivial if it's n equivalent to the unnot.

[01:54:06]Let um gamma subn of script k be the set of n minus one trivial nuts.

[01:54:26]So we get uh what's called the gustarov filtration that is We start with K and then we get smaller and smaller sets as we look for things being equivalent to the unnot to higher and higher orders. Um a natural conjecture is are there any non-trivial knots that are inequivalent to the unnot for all n open problem? No idea. It would be wild if that were the case. Uh, nobody knows.

[01:55:20]I'm just writing what I said in symbols.

[01:55:23]And then here's a wild theorem. So remember earlier I said knots form a monoid using this connect sum relation.

[01:55:31]Uh, and no knots have any inverses.

[01:55:34]Right? You can't have like a connect sum b equal the unnot if a and b are individually non-trivial.

[01:55:43]So, Guster says that um the moderate of knots mod this set of entrivial knots is uh it's not just uh an Aelian monoid aelion meaning commutative but it's also though an aelion group meaning everything has inverses mod uh mod this gusar subset so I'm not going to prove that because I don't really know how it's proved but fun fun thing that happens to be true okay um from now on I I think we're going to need uh a little bit of stamina. So, okay. Actually, of the of the three pe no, of the two people here, which is two swap and multi tube, uh do you think you'll be able to sit for another like or while I talk about this thing or do you think I should postpone that to a different talk? I think you should postpone it. So, actually, um, with the recording, I realized halfway through that like the way that I have it set up in OBS, I think it's whenever like nothing changes in Gnome or something like that, it doesn't actually update the frame. So, I'm literally shaking my mouse and if I don't, it doesn't record anything.

[01:57:30]>> Oh, >> so I'm I've been shaking my mouse back and forth for the past hour.

[01:57:36]>> You can't see it on the recording because it's on a different monitor, but it's fine.

[01:57:40]I'm about to lose it.

[01:57:43]>> Okay. If that's the case, then I should I should uh devote a part two to this.

[01:57:48]>> I'd appreciate that.

[01:57:50]>> Yeah. Um but thank you all for commenting. I hope this was >> Thank you. That was awesome.

[01:57:56]>> Uh comprehensible. It was uh it's a nice set of invariants that sort of each one only cares about like finally many points of the knot at a time in a sense.

[01:58:10]Uh I haven't really talked about that perspective but it's like a relatively natural set of invariants that happen to be uh pretty strong. Uh natural conjecture is are all not distinguished by facil variance. Um it's that's an open problem.

[01:58:30]Um but uh yeah that's where I'm going to leave it. The conservative integral by the way uh pro is used to prove that hard direction of the fundamental theorem that I was talking about of like you know we saw that the 4 and the one relations are necessary but how do we know that they're sufficient?

[01:58:49]>> Uh all right.

[01:58:50]>> Oh cool.

[01:58:51]>> I'll see you possibly later. All right.

[01:58:53]>> Yeah I'll see you later. I I kind of want to implement some of these things.

[01:58:56]This is this is quite neat.

[01:58:59]>> Yeah, there's there are some like nice formulas for um villian variance that I didn't get to talk about. Like I talked about these Gaus diagrams earlier. Um there are ways of evaluating villian variance that are very efficient that where like the first step is convert your knot into a G diagram. Um, I forget the name of the people that that victim is named after, but that's that's a thing that I might send you uh some reading on if you're curious. Um, but yeah, great. Later. Bye.

[01:59:34]>> Bye.

[01:59:35]>> Bye.