 Welcome everyone to this next lecture on tangent spaces and manifolds. So we had just begun with the definition of tangent space and a manifold. We also said what is a tangent bundle and it helps to know a little about some more examples of manifolds. So we saw some standard terms. S1 we said is a circle. Circle meaning we just mean x square plus y square minus 1 equal to 0. What is defined by this equation? S2 was our standard 2 sphere which we prefer calling x square x1 square plus x2 square plus x3 square minus 1 equal to 0. Similarly, we can have Sn, the sphere n dimensional sphere embedded in Rn plus 1. So R stands for the set of real numbers. Super script n plus 1 means n plus 1 components which we will say x1 up to xn plus 1. So this we will say is Sn. Another important object is so called torus. T1, T2, etc. So these are called torus. So what is a torus? This torus is also what can be thought of as donut. T1, T2, Tn are more generalizations but this turns out here in this case this turns out to be S1 cross S1. Any point is denoted as what angle along this torus and then if you cut the torus at any particular angle then we get another circle. In that sense the torus with one hole is embeddable in R3. So this is subset of R3. Let me write this in more detail. Torus with one hole which we should think of as like our donut. Donut embeddable in R3 and this turns out to be S1 cross S1. And why is that? Look at it like this. There is like a ring which we can hold by putting our hand inside this. So this can also be thought of like a ring. A ring whose cross section itself is a circle. So one can tell at any point, any point on this ring can be told at what is the angle with respect to such a frame, let us say. And once it is in the cross section, once we have identified the cross section using this angle, one can say which point on this cross section by another angle. That cross section suppose is like this then with respect to radially outward of the bigger of this frame, one can say what is the angle with respect to this circle. In that sense one can see that it is indeed S1 cross S1. So there are different ways. This particular topic requires a good amount of imagination and also a good amount of rigor to be able to do this systematically and not just keep imagining various objects. So it is easy to see for this example that this is indeed one to one correspondence with the set S1 cross S1. In other words, one angle theta 1, another angle theta 2, these two together describe the precise point on the torus. We are not including the interior of this particular donut. We are just speaking of the torus. This is a torus with one hole. So this is also some standard manifolds that people deal with. This particular thing is very relevant for example in robotics where we have two angles. Two angles, one at one joint, another angle to another joint and one can ask that we are speaking of the evolution on a torus and what kind of dynamical system, what kind of equilibria do we have. This brings us to the next topic that once we are given with a manifold, one can ask what can we say about the equilibria, what can we say about the singular points. So given a dynamical system f of x dot is equal to f of x in which x of t evolves on a manifold. So unless we specify otherwise, in this course we are going to be dealing with only smooth manifolds, what we defined as regular manifolds. So on this manifold we have this dynamics and if it turns out that at point a on the manifold m, if f of a equal to 0, then we will say this a is an equilibrium point. It is an equilibrium point why because if f of a is equal to 0, rate of change of x with respect to time is 0, so x dot is 0, hence x remains at that point, in that sense it is equilibrium. But it being in equilibrium means, so that particular point a is also called a singularity. There is some singularity, it does not mean that some particular matrix is singular. Singularity just means that something is different here and in this particular case all the components of this function f are all 0, it is equal to 0 vector, hence the rate of change of x at that particular point is 0 and in that sense it is an equilibrium point. One can ask what happens about a neighborhood. So this is something that we have seen in detail, so one would have to linearize at that particular point. So at an equilibrium point do nearby trajectories approach, approach which point suppose the equilibrium point was called x naught, do nearby trajectories approach x naught. If all nearby trajectories approach x naught, then we have decided to use the word stable. If all nearby trajectories approach x naught asymptotically, if they converge to x naught, we will call it asymptotically stable. If they do not blow up, if they remain in that small neighborhood, then we will call it just stable. On the other hand, even if some nearby trajectories go far, if some nearby initial conditions, these are things that we already saw in more detail when we were studying Lyapunov stability, but if some nearby initial conditions go away, then we already called it unstable. So these all studies are relevant only at an equilibrium point. Why is it relevant only at an equilibrium point? Because let us consider R2. So if this is a point which is not a singularity, then that point itself is not going to remain there. As a function of time, it is going to evolve further because the vector field at that point is not 0, it is non-zero, hence it will move in that direction. So nearby point also is likely to be non-zero. If this function f is continuous and if it is non-zero at a particular point, nearby it cannot suddenly become 0. So nearby also it will be non-zero. So all those points are going to also anyway move. So it does not ask, do nearby trajectories approach this point. This point itself is not even an equilibrium point. So the question about stability or instability automatically applies to only equilibrium points. But consider this particular point where the trajectories are let us say changing in a way because, so the neighborhood is indeed worth considering, worth studying in detail because that point happens to be an equilibrium point and one can ask that can we stabilize this? If there were an input, one can say that we will like to stabilize this. So stability is a question that we also apply only to equilibrium points and it depends on the manifold. It is possible that certain manifolds allow, certain manifolds allow no equilibrium points. Certain others, certain others require at least one. I am now speaking about a global property of this manifold. What is global about it? Not just locally, it is, it turns out that certain manifolds force you to have some equilibrium point at least. If you want the dynamical system f, x dot is equal to f of x, if the function f should be continuous then certain others require at least one. So this is what we will see in little more detail in this lecture. So let us take an example. So suppose we speak about R2. The question is, if somebody tells us, can you draw a vector field on R2 in which there is no equilibrium point? Does there exist f such that x dot is equal to f of x, f is smooth, smooth meaning in this case it is just continuous and differentiable. So more generally smooth word could also mean infinitely often differential. Any number of times the derivative exists, f is smooth and no equilibrium points at all. The answer is yes. One can just make all, one can say x dot is equal to 1. So that no point is an equilibrium point. On the other hand, the same question, the answer is no if you want this f to be smooth and x evolves on a sphere. So on a sphere it turns out on S2. On S2 it turns out the answer does not exist. So we cannot have a vector field that is continuous and there is no point where the vector field is 0. In other words, there is no equilibrium point at all. Such a situation cannot happen as far as the sphere S2 is concerned unless you let this f to be discontinuous. So this turns out to be a very important result which we will see in more detail today. That result is called Harry Ball theorem. So this requires us to develop a little more concept. But why I am trying to draw your attention to this is that it is a property of the manifold even though stability of the equilibrium point, the equilibrium point itself appears to be of a very local nature. The fact that f has to be continuous and it has to eventually cover the entire manifold forces some properties on the manifold itself. It requires something on the manifold for existence of an f whether or not equilibrium points should exist for that f. Let us ask about a circle. On the other hand, S1 circle, is it possible to think of an equilibrium point? Is it possible to think of a dynamical system in which there is an equilibrium point? Yes, we can just have theta dot equal to 1. So that continuously it is rotating like this and it is going on rotating. So at no point there is an equilibrium point. So Harry Ball theorem speaks about S2 and it says that one is forced to have an equilibrium point. In fact, if one requires only simple singularities, then one will in fact have at least two singular points and this turns out to be related to a very celebrated and familiar result to all of you, which is that the number of phases, edges and vertices of any polyhedra satisfy a relation. So let us come back to classification of equilibrium points on a plane. On a plane, we already saw some examples. Stable, unstable node. We saw a node. We saw a saddle point. We also saw a center. What were these? If all trajectories were coming inwards, which turned out to be the case, which turned out to be the case if the matrix A had eigenvalues, which were both real and negative. This was what we called a stable node. We saw another case where when both eigenvalues were real and positive, that time this one was an unstable node. On the other hand, a saddle point was where eigenvalues are real but one is positive, one is negative. For example, this is an eigenvector corresponding to positive eigenvalue, let us say. So everything goes away and if this is an eigenvalue corresponding to negative eigenvalue, we are speaking of a plane and we have linearized about the equilibrium point and we are considering the eigenvalues of the matrix, the linearization at that point and at all other points it is like this. So we also saw center. We saw the situation where the eigenvalues of A are complex but depending on whether the real part is positive or negative, we can have oscillations that are coming inwards or going out. This was the situation where the eigenvalues are complex but on the imaginary axis, so the oscillations are neither coming in nor going out. That is as far as the linear system is concerned and in this case it turned out that the linear linearized system, the real part is 0 but the second order nonlinearity might cause that the oscillations come inward or go outward and that is why one cannot use the linearized system's conclusion about it being a center for the origin nonlinear system also for the equilibrium point of the nonlinear system. So as I said for the case that eigenvalues of A are complex, if the real part is nonzero, if the real part is positive then these oscillations are going out. On the other hand, if the real part is negative it comes inwards. So these all we will like to classify as something called the index of the vector field. So it will turn out that the stable and unstable node, the index is 1. So we are going to very soon define the notion of index of an equilibrium point which will turn out to be plus 1, plus 1 for stable node, unstable node. For the saddle point it will turn out to be minus 1. We will verify it for a few examples. For center also it will be plus 1 and for the stable and unstable focus also it will turn out to be plus 1. So only the saddle turns out to be little special for which it will be minus 1. So we are going to see in more detail what is the meaning of the index of the vector field. So let us take equilibrium point. Suppose this is an equilibrium point and we have vectors all around the vectors themselves. First we will consider an isolated equilibrium point. What is the isolated about it? There is some neighborhood within which this is the only equilibrium point. We are able to find some circle, some curve such that that curve contains this equilibrium point and that curve contains this equilibrium point in the interior and this is the only equilibrium point inside it. If such a curve can be found small enough then we will call that that equilibrium point is isolated in the sense that it is not sticking to any other equilibrium point. In that sense it is isolated at least some sufficiently small neighborhood contains only this one. Now we will like that there is such a curve which doesn't contain any equilibrium points on it. So choose suitable curve with no equilibrium points on curve. Let us call this curve. So this curve is a closed curve. It is also a simple curve in the sense that we are not going to allow this curve to have self-intersections. Starting point and end point are the two only points that are common. No other intermediate points are repeated. So that is indeed the case and we can also give it an orientation. We can either clockwise or anti-clockwise that is not the issue. So now we will ask that look as we go along this curve we can the vector field at every point on this curve has a unique direction because there are no equilibrium points on the curve. If there are equilibrium points on the curve then the vector field has length 0 and hence it would have no direction and that would cause a problem to us. So we chose a suitable curve which has no equilibrium points on the curve and we chose that this curve has this equilibrium point only and that is possible because this equilibrium point is isolated. So now this vector also goes through our rotation. Notice that this vector is pointed like this and as we go along that curve this vector is also turning and if this curve is traversing clockwise it turns out that this vector also has turned clockwise. So vector field on the curve has unique direction because the curve has no equilibrium points on it has unique direction as curve traverses one rotation say clockwise. As I said the direction clockwise and clockwise will not matter. Let us choose this clockwise then we can ask whether that particular vector also has rotated how many number of times and whether it is a whether the orientation of that vector also has remained the same or not. So if we have chosen the curve to be clockwise has the vector rotated clockwise is the first question and has it rotated say how many number of times once or more. So index is defined as plus 1 if rotation once in the same direction plus 2 if it is rotation twice in the same direction minus 1 if it is rotation 1 but in the opposite direction than the curve. The fact that it is same or opposite is what decides whether sign and the number of rotations of course is decided by how many times it has rotated. It remains a question do there exist equilibrium points where you have two rotations. It seems unreasonable that the vector rotates in opposite direction as you rotate along the curve in a particular direction as a clockwise. Most easy to think of is plus 1 where the vector field rotates in the same direction as the curve itself. So notice that if there is no equilibrium point inside if there is no equilibrium point inside then the vector field does not rotate any net rotation it might just change signs like this it may not complete a rotation so 0 is also possible. In fact if you have multiple equilibrium points isolated equilibrium points inside then the curve will add all these indices and it will be an algebraic sum that turns out to be a extremely neat concept that all these indices of isolated equilibrium points add up for a curve that contains all of these. So if the curve contains none if it contains no equilibrium point then the index of that curve will also automatically be 0 that means the vector field and there goes no net rotation. So let us see an example now. Let's take for unstable equilibrium point for unstable node we have only verified let's take a center and let's take that particular periodic orbit itself as a curve C. So we see that at every point the curve C the vector field is tangential to the curve itself and hence when the curve undergoes one rotation the vector also undergoes exactly one rotation in same direction. So now notice that even if the even if I considering sorry this particular arrow should have been like this even if the curve was chosen anticlockwise positive that does not change the directions of the vector field. So vector field decided by f of x dot is equal to f of x is decided only by f. Curve chosen pretty arbitrarily only important constraint is that in order to decide the index of an equilibrium point once you choose a curve that contains only this particular equilibrium point and that curve should not have any equilibrium point on it. This curve should be should contain precisely one equilibrium point inside it precisely the equilibrium point for which we are trying to find out the index. Moreover this curve should be a simple and closed curve except for these constraints it's to be chosen arbitrarily. One can ask it need not be the periodic orbit that's decided by because this being a center could be in fact the periodic orbit between opposite orientation also and still it turns out to the index of that equilibrium point is independent of which curve has been chosen as long as it satisfies these conditions. So center one can verify that it is indeed index index of a center equal to also plus 1. So now let us verify a saddle point. So now let's take a curve like this let's orient it positive. So this particular point is equilibrium point. So we see that when we start at this point on the curve it is pointing upwards as so this is how we should fill. So now you can check that as you go along this curve in fact at this point it is tangential and opposite to the curve and this point it is inverse like this. So we see that as you go along that curve this particular pen that I was showing has rotated by one number but in opposite direction. So index of saddle equal to minus 1. Why because if you chose a curve clockwise when you took the vector field along this curve as you are traversing along the curve in the clockwise direction that particular vector turned out to rotate by one number of times but in the anticlockwise direction. That is why the index of the saddle point is equal to minus 1. So now that brings us to the question that do there exist equilibrium points with index 2. So this particular question sure would have haunted the dynamical system community for many years and indeed there is this was told to me by my teacher. So it requires some effort to construct one. So one can check that this particular equilibrium point of course trajectory is for smooth vector fields. Unless you have non-lipchitz properties trajectories do not intersect. It is just that they are very close by and they eventually separate like this. So for such a vector field you can draw a curve and check that the vector rotates by 2 times when you go around this particular equilibrium point once. So such a vector field has index equal to 2. So it turns out that this is a little on a special side. This is a special vector field for it to have such an index and we will let to see what is special about it that it is not a simple singularity. It is not simple. It is a singularity because it is an equilibrium point. What is not simple about it that we will see that the linearization if x dot is equal to f of x of this then you take the derivative of f with respect to x then it will be square matrix when you evaluate it at the equilibrium point x equal to this particular equilibrium point and this matrix Eigen values is what decided everything. So you look at the, suppose you call this matrix square matrix. It is a 2 by 2 matrix. Determinant of A non-zero will decide that that equilibrium point is simple. What is the meaning that the determinant of A0 for the linearized system if the determinant of A0 it means that there is the equilibrium point is not isolated as well as the linear system is concerned. The non-linear system equilibrium point might be isolated but the linearization is suggesting that there is a continuum of equilibrium points and that is indeed what happens for linear systems if the matrix A is singular. So if the determinant of A is non-zero then the matrix A is what we will like to say as a non-singular matrix and for such a situation that isolated equilibrium point we will say is a simple singularity if the determinant of A is non-zero. If the determinant of A is zero then that equilibrium point even if it is isolated we will say is not a simple singularity. So only with non-simple singularities one can have index more than 2 more than 1. So index equal to 2 or more is possible only for non-simple singularities. One can check that the saddle point stable and simple focus they all have simple there are simple singularities and hence the indexes are in as a plus or minus 1 only. So this brings us to the one of the last topics of this one of the last subtopic of this topic that is about the hairy ball theorem. So one can ask now now suppose we are given with a sphere given a sphere find f to have no singularity or if inevitable only simple singularities. So this is the question given a sphere find a function f what is this function f because we are trying to construct dynamics like this. So we see if there are no singularities it means that you can never stabilize the system at any point. There can't be any equilibrium point itself that is a consequence if there is no singularity. If there are singularities then we will like that they are simple singularities because they are linearized because we will ideally like linear system linearization and linear control to operate there. So if it is not a simple singularity then we need more complex systems because the Eigen value at the origin suggests that at steady state there is a non-zero value it is not converging but it is staying close by only asymptotic stability requires that all equilibrium points are in the left half linearization at every equilibrium point is in the has Eigen values and left half complex plane. So simple singularities are good in that sense. So it turns out that there is this person not there is a hairy ball theorem that tells that no singularity is not possible one requires at least two simple singularities if you allow the singularities to be non-simple then one would suffice. Hairy ball theorem says that continuous combing of a hairy ball leaves at least one singularity. What is combing of a hairy ball? Suppose we are given with a ball and suppose we are this ball has a lot of hair along it this hair each hair hair is like a regular hair HAIR yeah if this hair denotes a vector field at that point at each point there is some hair that starts starts at the origin of the of the tangent space at that point and it is it defines a unique direction and we are now asked that we are required to comb it in a continuous way. So this tells that the F has to also be continuous in X. So what is combing about it because we want that hairs are all in the tangent space in that sense they are tangential to the ball they can't be standing out the hair can't be sticking out like this that better get combed on that ball so combing means that that particular vector is in the tangent space continuous the function F is continuous hairy ball meaning the ball has hairs and each of this hairs are nothing but vectors in that tangent space then tangent space is one that is getting forced because of this combing process now we will like that there is at every point there is one hair that is non-zero length that is a meaning that there is no singularity so is it possible that we can continuously comb without a singularity yeah so it turns out that the answer is no the hairy ball theorem says that there will at least be one singularity this is the meaning that when we comb there is a at least a point on the head where all the hair are going away or going round and round these are what is well known when we comb the comb our own heads hair for example yeah so what is this one singularity this one singularity is inevitable because the sphere the fact that we are given with a sphere the sphere as a manifold is forcing that the manifold all the isolated singularities when we add the indices we end up getting number 2 yeah that is the sum of all the indices of every equilibrium point isolated equilibrium point and we have defined index only for isolated equilibrium points so let us now say if simple if simple singularities at least two are inevitable at least two required so this number 2 is extremely special what is special about it for example we can now think of a ball like this let's say we have a north pole and a south pole so we can think of vectors leaving from north pole and all converging towards the south pole yeah this is an example of a vector field which is continuous leaves the north pole all comes towards the south pole so two simple singularity two singularities only both are simple why because this one the north one is a unstable node by the south one south pole is a stable node so they both are simple singularities they both have index two index one each so the total sum of all indices for all equilibrium points is exactly two so we can see that we can construct such a case one using that particular non non simple singularity we can also think of this on the sphere and this goes all along like this so one can think of a non simple singularity a singularity itself of index two defined like this on the sphere yeah so now what is special about two so that brings me to one of the last very good relation with for a polyhedral so for polyhedral the Euler's theorem for polyhedral says that the number of faces minus the number of edges plus the number of vertices is equal to 2 yeah for any closed polyhedral with no holes importantly with no holes so for example for a cube so then faces the number of faces is equal to 6 the number of edges is equal to 4 on the top 4 on the bottom and 4 vertical that's 12 and the number of vertices we have 4 on the top for the bottom that's 8 so we get that 6 minus 12 that's minus 4 6 minus 12 minus 6 plus 8 that's plus 2 we get 2 yeah so what is the relation within this and singularity is on a sphere so one can now think of a sphere and we can try to sort of look at this having a faces edges in which we put these points all the edges are marked and the faces are like different regions on the sphere so polyhedral we can think of is actually very similar to a sphere in some topological sense it is nothing but a sphere with all these lines marked on the sphere yeah now inside each face one can have let's say for example a stable node and at each node at each vertex we can have an unstable node everything going away and between two vertices it will turn out that there will have to be a saddle point yeah so let me draw this figure a little larger so let's take this cube so this is at the center of each face we decided to have stable node stable node at each vertex we will have an unstable node at the center of each edge we will have this saddle point so that so we see that at the center of every edge we need to have a saddle point if we have to be able to do this systematically continuously more precisely sorry this arrow should be inverts because all of our vertices have been decided as unstable nodes so it's all going away from from vertices at if it goes away from every vertex then between two vertices there's exactly one edge that center of that edge is where both seem to be coming in so why don't we make that stable as well as this edge is concerned but if we have the center of every face to also be a stable node then it will be going away from this particular point so fine that way let's allow this to be a saddle point yeah so notice that this is a systematic continuous consistent way of placing the saddle points how many saddle points would we have placed e number and the number of edges how many stable nodes would we have put f number of stable nodes and how many unstable nodes would we have put number of vertices v number of unstable nodes so now we know that this we can add the indices for all of them so we know that indices of saddle point is minus one so the sum of over all equilibrium points will in fact give us f minus e plus v because both stable and unstable nodes have plus one as their indices so f and v both come with plus sign and e on the other hand because it corresponds to a saddle point which has index minus one it corresponds to minus sign here so this one is what we saw for polyhedral is equal to 2 and that is exactly what we also saw for that north pole and south pole there also it turned out to be 2 so what it says is that the sum being 2 is a property of the sphere it is not a property of whether you take a cube or a pyramid the fact that every face you associate a stable node and every vertex you can have an unstable node and then at the edge you are forced to have a saddle point and then some of all vertices will some of all the indices of equilibrium points will exactly turn out to be 2 by this particular formula and the 2 is like an invariant of this particular topological object called the sphere and all this polyhedral are in that sense homotopic to a sphere so this brings us to the end of seeing how the theorem says that it is inevitable that for a sphere we cannot have a situation where for a continuous function f we don't have any equilibrium points at all we don't have any singularity such a situation is not possible one last question is somebody can ask can you interchange the role of stable node unstable nodes in other words can you have a stable node at every vertex and an unstable node at every phase yeah that is still possible still at the edge you will require a saddle point as we just some arrows reversed but then the formula will still turn out to be 2 that is because some of all the indices of equilibrium points will still turn out to be 2 because we know that both stable and unstable nodes both have index plus 1 so this is one of extremely important extremely enchanting topic within non-linear dynamical systems about how it is related to Euler's polyhedral formula and through that to something called algebraic topology but then I do not work in this nor do I know enough I want just you to know about this and tangent spaces on the other hand finds find wide applications in control when dealing with non-linear systems especially on manifolds with that we will end this lecture on tangent spaces and manifolds thank you