 to everyone at ICTP, especially including you, the diploma students, so, okay, and the speaker today, I'm very happy to introduce our very own Pavel Putrov, who's going to speak on some classical invariance of links and their higher dimensional analogs. So, floor is yours. Thank you for the introduction. So, I will talk about links and their high dimensional also, but before I start some systematic presentation, let me mention some motivation and slash applications and why people are interested in links and their high dimensional analogs. So, first, well, in principle, so, there is a big interest in the subject in mathematics, which is maybe called low dimensional apology. There are many places where these links arise there. So, they're interested on their own, but they're also related, very closely related to the topology of three and four manifolds via so-called surgery procedure. Then these kind of links and the invariance, they appear in a certain notion which arose in physics, but it also found many applications in mathematics, and in particular, this is a notion of topological quantum field theory. In particular, there are some and recent interest in this notion related to some particular application to some condensed matter system and also kind of some more fashionable topics such as what is called topological quantum computing, many other things. So, this is what I'm not going to talk about. So, I'm just going to consider these links on their own. So, probably many of you have encountered kind of a notion of the usual notion of link, but let me define it anyway. So, they're kind of different variations for me, so what I will use is the following. So, an L component, so L is some positive integer number, L component link is a, so I will assume that it's smooth to avoid some technical difficulties. Smooth embedding of a, so they're not by link usually, by letter L, smooth embedding of a disjoint union of L copies of a circle into R3, the three-dimensional Euclidean space. And so usually, so we have this following embedding and the image of this map I will usually denote, so the components of the image I will denote by L with subindex I. So, Li is the image of a Ease copy of S1 here. So, let me give you some simple example. So, for example, this is a two-component link. It will be a two-component link, which will look something like this. It will be a component, the image of component one. This is some circle embedded in R3. And there will be a second component, which looks something like this. So, there are two circles embedded in R3, which of course I visualize them on two-dimensional place, but this should be naturally understood as a three-dimensional configuration. Okay. But of course, since we want to consider some, I mean, of course, this embedding, there are too many of them. We want to consider some robust information. So, what we actually want to consider, we want to consider links up to what is called ambient isotope, which I will define in a moment. But morally, you should understand that you want to consider these configurations up to all continuous kind of distortions or deformations where I allow to move things around, but I don't allow the strands of the links to cross each other. More formally, although we're not going to use this definition directly, let me write. So, links, L prime, what is called ambient isotopic and only if by definition. And so, this will define an equivalence relation on all possible embeddings like this. And usually, and what we want to consider links up to kind of links, which I mean as a topic to be equivalent links. If and only if there exists continuous map, let me denote it, g from r3 times an interval to r3 such that for any t in this interval, if I take the restriction of this map to a particular element here, so this would be a map from r3 to r3 is a homeomorphism. So, to remind you, the homeomorphism is a continuous map for which the inverse exists and it's also continuous. And more like such as the following properties. So, g at restricted to zero, so if I take a zero here is just identity map on r3 and g restricted to the end of the interval satisfies the following conditional z. So, L, if I understand links here as the maps from the bunch of circles to r3, the composition of L prime of L, this g restricted to one is L prime. So, more really, again, this just means that we kind of we allow a certain distortion of our ambient space so that we can bring the original link L to a link L prime. And of course, so given some sort of presentation of a link, it's natural to ask are they equivalent or not. And this is the kind of one of the basic questions in the not all links here. For example, one can see the following simple link, two component link. Well, let me use a second component, different color. And so this will be a link L and consider another link L prime, which is So, also, in general, we want to keep orientation. So, on the link components, so there'll be circles with a particular orientation with a particular direction chosen. And we also can see the other link, which is just a disjoint union. It's just kind of two are not separated. One can ask a question, is this so it's kind of, you would say, okay, it's kind of obvious that those link are not equivalent. They're not ambient as logic. So, this is kind of connected. So, you cannot separate. There is no, you cannot shuffle the surround and separate them. But how, I mean, how would they actually prove this? And it's actually not obvious because, I mean, if you just want to use a definition, how would you prove this? Because you have to show that it doesn't exist any map. And there are kind of, a lot of maps like this, so that you cannot bring this configuration to this. So, how you, so what you, I mean, usually it's very kind of, if you want to show that something exists, you can just present it. But if you want to show that something doesn't exist, it's a bit hard. It's usually what you want to do is you want to sign some obstruction of this existence. So, this is where the invariance of links enter, one of the questions. Okay, you can also ask a question, you can also say, okay, this is a bit, I mean, here it's an obvious example. But what happens if you have something more complicated? And you cannot, it's actually, it wouldn't be obvious if you cannot disconnect. So, let me try to draw you some other example. Let's see, something like this. Okay, this is a, so you can ask the question, is this link equivalent to this link or not? So, this is, I mean, you need to spend some time to kind of convince yourself, can you separate it or not? But is there some, and it also cannot be, I mean, it's hard to be sure, like what if you are not kind of smart enough, you cannot, so you just cannot imagine a way to disconnect this component. But is there some, so one can ask, is there some systematic way to answer this kind of questions? So, method to show that, for example, two links are not equivalent. So, find an invariant, so an invariant, so invariant means it's, it will be invariant under unbiased entropy. So, for each, for each link, so let me write i, for each link, we are some, some numeric, some value, suppose a numerical value, such that the value of this invariant on a certain link is the same as the value for invariant of L tilde if whenever L is unbiased at topic to L tilde. And, so the first step, and the second step, so if you, if you manage to show, if you manage to find some invariant show, and show that these invariants are different for this particular power, then it follows, of course it follows, and it's not, they are not unbiased. In my talk, I present some simple examples of such invariants, which have, to define them, you don't need a lot of prerequisites, and so they, they can be very just in geometric terms, and they, like if time allows, I will also consider generalization of such invariants for kind of high dimensional generalizations of such links. Any questions? When this, in this case, there is, there is actually not, not really a problem because, yeah, so you can argue, well in this particular case of R3, so we can see links on R3, because any, yeah, any homomorphism of, well, if it's oriented homomorphism, well you have to, yeah, you have to specify that G is preserved orientation. If you specify Z, then you, any homomorphism of R3 into R3 can be homotoped to, to identity. So this is like, yeah, you can, you can require something weaker, weaker condition actually. So before I proceed to, to definition of the invariants, I want to define the certain, the certain important notion of the Eiffel surface. The Eiffel surface of a link component Li, so it's a particular link component, so this is a circle embedded in R3, is an oriented, also connected surface, sigma i, surface embedded in R3, such that the boundary of the surface is, is my link. So for example, if, if my link component just looks like a a not circle, so I can choose the, the Eiffel surface to be just disk, just disk such that its boundary is this circle. So let me, so of course there are, there are, there are questions about existence and uniqueness. So let me give you kind of a brief answer to these questions without going into kind of proofs, detailed proofs of this, of these statements. So the following two remarks, so it's always exist. And so there are different ways to show this. So one way to show that it is constructed by explicit, it can be constructed by an algorithm. I'm not going to go into, into very detail, so just give me, just, just, just give you, illustrate this algorithm on a some example. For example, I can see there's a following oriented link component. I'm sorry, it can be noted in some way. And then what I want to do, I want to, I want to kind of consider this not, not diagrams of the projection on a plane where the intersection, the crossing points I result. So one, one strand goes a bit above the other or under the other. And then using this diagram, I can go, this is, there is something wrong with the orientation. So if I go around, so I start at some point, I go around and I also follow arrow. And on the, on the right of my arrow, I will spun, I will, I will spun some disc. And then I do, then I do the same at other places. So if I pick other point, which is not already a boundary of some disc and do the same here. But of course I don't, this is not really discs. I have to do some sort of resolution here at this point. At this point, there is not, there is not sure way to connect these discs by, or small strips. So if you, if you, you can imagine like this. So if you have a, so here it kind of looks like a just a flat, just a part of the plane. But if I want to transition from here, this part to this part, I kind of, if I take some line here, I rotate this line by 180 degrees. And then I, as I do a transition to this plane. So in any case, there is a certain algorithm which allows you to explicitly construct a zephyr surface by given this diagram representation of an object. And here, as I mentioned, I don't want to go into details here. And so the second question is about uniqueness. And there is about uniqueness, there is a following statement. Any two zephyr surfaces, sigma i and sigma i prime of lambda i can be related by a sequence, by a finite sequence of the following operations. So the first operation is ambient isotope, which preserves Li itself. Okay, let me write it like this. So let me, so again, it's just, so if I have something, let's take me this on a simple case of a node. So I can start with just a kind of flat disk here. And then I can do again, certain distortion where disk kind of continuously deform this disk, but the boundary is still the same. And I do this in a continuous way. Second operation is what is called handle addition removal. And this looks like full. So if I have, I can, some vicinity of some point, my zephyr surface looks like this. I can change it. I can add a handle here and it will look like this. Essentially what I do, I take a tube, a circle times an interval, and I cut out two holes here and attach this tube like this. And the third is what is called infinity passing move. And again, kind of schematically on the example of a, of a knot. So suppose I start with just the ordinary disk and then I take, I take a big sphere. So in general, I take a big sphere which surrounds everything. And I, so I have this and I, and then I attach this handle, sorry, I cut out some circle here and cut out some circle on my sphere and attach a handle, attach a cylinder, connecting these two cut out circles. So it would look something like this. And then there's a hole here. This is essentially artifact of what we are, because we are doing things in R3. I'm not on S3 questions. Yeah, I'm not here. I'm not kind of very precise, but on the geometric level, I hope it's clear what's kind of possible. And one kind of, well, the proof of this statement is quite hard, but kind of, on an intuitive level, it's kind of, every kind of change, you can kind of argue that this is any change you can imagine. It's just these three basic operations. At the same time, the simplest possible link invariant can be defined in terms of this, the effort surface as follows. So many of you are probably familiar with this. So linking number of two component link in all our links are oriented here is defined as follows. So it's denoted by Lk, L1, LT. So my link has two components, L1 and L2, and it's defined as what is called intersection number of the effort surface associated to the first link component, and the second link component. I have to assume that for generic choice of C1. So they assume that they intersect transversely. So what exactly this means? The intersection number. So exactly this means the fall. So I have to do a sum over all points. So assuming this intersection is just a finite number of points, I do a sum of all points here, and for each point I add plus or minus one, depending on a certain orientation. So let me elaborate on this choice. So if locally the situation looks as follows, so I have a sigma one. So of course the choice, as I mentioned, so this is a oriented surface. So the orientation should be in agreement with the orientation at the boundary. So in particular, if I rent, so if there's only this example, if I rent the boundary like this, so the orientation of the surface which can be determined by choice of normal vector is will be upwards. So this is the kind of exterior, the upper part is the exterior part of my surface. So suppose the surface here is oriented like this, and the vicinity of my point A looks as follows. So it kind of pierces the L2, pierces the surface like this, and then this should be counted with sine plus one. And if it's the other way around, it goes like this, this point should be counted with sine minus one. So again this definition, it, so first of all it's not, so one has to do some checks. So first of all we kind of used a choice of the surface here, and so one can, so for this to be well defined one has to check that this is independent of the choice of the surface. And the second question is that actually invariant one has to check that this is, it did invariant on the ambient isotopia that we can use to distinguish those examples. Again I'm just going to give some kind of generic geometric idea of why this is the case. So the first remark about this definition is it does not depend the choice of sine one. So we know, well you have to believe me, but those are at least, I mean I will assume those facts to be true that how, what is the ambiguity, we know what is the possible ambiguities of the choice of the surface. And so we have to check that this is invariant on those basic relations. So in particular, essentially the most non-trivial part is, well it's also trivial, but so for the first part we want to check that this invariant on the ambient isotopia which kind of deforms my zephyr surface. So suppose, so you need, I want to, so essentially what can happen here is that we have to be afraid that there are kind of, once we do some deformations there are some points here which can disappear or appear again. But essentially what the only thing which happens is that they can disappear always in pairs or appear. So we can have an example situation like this. So we start kind of moving the zephyr surface, so this is L2, it's kept the same, but we kind of deform the zephyr surface. So this is no longer intersect L2. And so we can lose one point, but we always will lose it together with another one. And they come and as you can see, so if this is orientation like this, this comes, they came, this point came into the sum with minus one and this came to the sum with plus one. So the total sum is still, is still zero. So this is unchanged. And similarly, you can also argue that the adding handles in this infinity passing move are also can be performed such that it doesn't change because you can essentially always this tube, you can make it essentially very thin, or similar this very thin and you can always move it around so it just doesn't intersect. This three tubes that should be at doesn't intersect with the second component at all. So this is some basic idea of why this is an invariant under the choice of sigma one. And another property and this is also invariant under ambient isotope of link itself. So essentially the argument will be similar. So instead of moving kind of for the surface, we can move one of the link components. And the argument essentially will be very similar. So I'm not going to repeat it. And another nice property that although this is defined in some sort of, so it's defined for two component link, but it's defined in a certain asymmetric way. So it has some preference between first and second link component. But what is actually true is that this number is symmetric. And again, let me give you idea why this is the case. So you have to check that the difference is zero. So the difference is intersection number of sigma one with l two minus intersection number of sigma two minus this l one. But since l two is the boundary of sigma two and l one is the boundary of sigma one, this is a condensed suit. So all points here together with these points here, including taking account of proper signs, are points which are contained in the boundary of sigma one intersected with sigma two. But this is just an intersection of pair of surfaces. So this contains either circles, so this joint union either circles which don't have any boundary or it can contain intervals whose boundary a pair of points, but this pair of points is always contribute with opposite signs because they will come with opposite orientation. And so they will always cancel in this sum of the points taking account of the notation. So this is zero. Questions? One of the components is what? I mean, one of the components is not going zero. I mean, this intersection can be empty. Is there many definitions of this? Yeah, this can be given by integral, by Gauss integral, explicit. So they can be given by some explicit formula. So you kind of integrate over pairs of points like this. So you put a point here and point y here on the second component and you can write some sort of integral. So this kind of, I understand this point as a vectors in Euclidean space. So you write some, I'm probably going to mess some of this. This is probably wrong, but something like this. Or is it, maybe this is even, I think this is, sorry, what, yeah, I'm not, I'm going to write something wrong here. Yes. And the Gauss definition actually appears to something in physics. Okay. So suppose we convinced ourselves that this is a well-defined environment. Okay. So can we use to figure out the simple examples? So here, let me choose this to be a, let me choose the surface for the second guy here. So here there is no intersection. So the linking number is just zero, obviously. So here, like this. So the orientation is such that the normal vector points at us. And so this comes so that way. So the linking is plus one. So, okay, this is a difference. So they're necessarily not equivalent. And what about this guy? I choose again the second component to pick the surface. And after a bit of, well, this is, if you move this a bit around, so you can only, yeah, so, yeah, you can imagine, so you have intersection here and here. And everything else, yeah. So this goes up and this goes below. This goes below and this up. Everything else is up. So this is a, this is plus two. So they're all, so we, I mean, suppose we have this definition, then they proved that they are, they're easy check that they're indeed non-accurate. Yes, that's right. So you can, the other, yeah, but yeah, kind of, I wanted to do this more geometrically, but you can use just a plane projection. And essentially if you, so suppose you go, so you go around one of the link components and you look at the crossings. And you look at the crossings and suppose I pick my, I go around L1 and I only look, actually it's enough to only look at under crossings. So if my under crossings is like this, if the second, if the second component passes below in this direction, this is, gives me minus one. And if I have a, if the second component under crosses and goes to the left, this gives me plus one. And I just do the sum of, sum of all these crossings when I go around L1 here. Yes, you can check. Yeah, I mean, yeah, here it's obvious that the, the Lincoln is zero, but here, well, you have to show that, you have to prove that you cannot deform it in any way because the diagram is not unique. You can always change your diagram. There can be different diagrams which represent, well, for once, so this is specifically defined for a pair, for a link component with two components. You can, you can define what is called like linking, self-linking number, but for this you need to use some additional, additional structure on your, on your, on your link component, that's called framing. Then you can define linking number with itself, but you need to use kind of sort of the framing. Or in other words, you need, consider this to be a sort of circle, but a ribbon, a small ribbon. Then you can define, then you can set the definition of the link into a self-linking number. But, but I don't, I'm not using this here. So now, so this was something which, and the basic invariant which is defined, yeah, for kind of, for a two component link and is a basic abstraction, simple, the simplest abstraction which tells you that these cannot be, the components cannot be separated. Let me consider, well, suppose you have something like this. This is what is called Boromian rings. So, so Boromian is actually was a, some wealthy family living in the north of Italy. They had, they had this, was a part of the coat of arms. And so this is a three component link. I don't remember if I mentioned, but this is also has, this link, this particular link has a nice name. It's called Hopflink. So here, so if I, so let's me, component one, component two, and component three. It's easy to see that if I just take a pair, each pair, they can be, they can be easily separated. And indeed, there's also, which is also an agreement with the fact, if you click the link in number for each pair, this would be zero. So either, either some simple, either a kind of similar simple environment which gives, which is something, gives something on trivial here and in a zero, when they are all, when, in a zero for the situation, when they are all separated so that we can, we can, we can, we can find some sort of quantity which tells us how the triples of circles can be, can be linked together. And so the answer to this question is, can be given by the quote Milner triple linking number. If some historical reasons it's usually denoted by mu bar, they define for three component links. And so, but actually to define it, we want to assume that all pairwise linking numbers are zero. And so this is the case in this example. But this can be defined for any, for any, for any three component link which satisfies this condition. So the pairwise, they can be, they can be separated. Well, the pairwise linking, not necessarily, but the pairwise linking number is zero. So this will be, let me write in this kind of first, kind of mysterious way, but I will elaborate. So the first part is, will be a number, again counted with signs, particular signs, a number of intersection points of the triple of the effort services. So this is okay because these, the two surfaces in general, they transversely intersect over, over a line, as we've seen. But if you take the, a line, generally intersects with the third surface over a point. And so the choice of signs is determined by, the choice of signs is determined by, again, by orientations. So let me try to write some picture, draw some picture. So this is how vicinity of a intersection point can look like. So this is, will be a surface surface sigma one. And the blue will be a surface surface sigma two. Violet will be a surface surface sigma three. This is my section point. And suppose the orientation is like this. So this is a orientation for sigma three. This is orientation, this normal vector to sigma two. And this will be a normal vector sigma one. So this is a first, this is a second, and this is a third. So this, in this particular case, this will come, could reach plus one. But if I have a kind of, if this is a standard ordering of the basis vectors in R three, but if this, if I have something like this instead, then this will come with minus one. So the other terms point as follows. So we'll sum over pairs of points. So the first point will be in the intersection of l one with sigma two. The second point will be in the intersection of l one with sigma three. And I only over sum over pairs of points which satisfy a certain ordering condition, which I will explain in a second. And each pair contributes with plus and minus one. And this plus and minus one is determined by, by the signs associated to this pair of points in the same way, the pair of points in the intersection of the link components with the surface, in the same way as we had it before. And the ordering is defined as follows. So I take, I consider my points in, so this is, so I take my link component, so link component l one, and I pick a particular base point. And by picking this point, I can do, I can order points here. So starting from here to here. And so the correct ordering is, for example, the correct, I do, I get a contribution from the, from the pair of points if B is, goes later than A. So I have a configuration, for example, a pair of points can contribute if I have configuration of this type. So this is, I have something l one, this l one goes, pierces through sigma two here and sigma three here. So for example, if, if orientation on l one is like this and rotations on sigma two and sigma three are in agreement, then this pair of points will contribute plus one. And similarly, I will define contributions t two, three one and t three, t three, one, two just by cyclically permuting indices one, two, and three here. And so what I can argue by doing, so there is an, by doing this kind of similar argument, so I don't have time for this, but I can argue that this is indeed invariant. So it's invariant both on the, so it's independent of the choice of that surfaces and also invariant on the, under, under ambient as a topic of the link itself. And the way to, to check this is essentially, so there can be a certain different, you can consider certain basic deformations. And the, you can show that the, the, the contributions to different pieces here can, can be traded off when you, once you do the deformation. But the whole, the total sum is invariant. And yeah, unfortunately, I don't have time to, to, yeah, I was planning to do this to, I mean, to argue why this is invariant, but I don't have time. But let me just give an example. So I have to finish, just give me, let me give you example of the calculation of this invariant for this Boromian rings. So there is a way to represent the Boromian rings as follows. And so I'll just do this and I stop. So instead of writing durian like this, I can grow, and grow it like this. I can, so that essentially in each, each component is a serve, is a oval line in a, in a particular x, y, or z standard planes. So then the, this will be the surface just interior of the, of the, of the, of the oval line in this kind of horizontal z plane. And this will be one and they intersect of course. So they, they intersect along this line. And this, the last guy interior of this oval. And they will act like this. And so of course, the, the violet oval and the green oval, they will intersect along this. So there will be a single point in the intersection, in the triple intersection of remands, of the surface. And it's easy to see there is no, so if you go, for example, along this link component, we only have intersection of this link component with points with sigma two, but not with sigma one. So there will be no actually pairs of intersection of points, both in sigma two and in sigma three. So there will be all other contributions will be mu is one. And of course, for this, in this case, the mu is obviously, mu bar is obviously zero. Okay. Of course, I didn't have time to go into high dimensional analogs. So there can be analogs of surface links kind of for the, the, you can see some limon surface embedded into R four, four dimensional space. And so they're also quite natural object. And you can construct new variants in a similar way. And but yeah, I'm sorry, yeah, I'm out of time. So I stop here.