 We are discussing the applications of homology and in particular last time we saw Laxay-Fitzwein theorem and then today we will give you some applications of Laxay-Fitzwein theorem itself. Recall that we have proved the Harry-Wald theorem by just using our computation of the degree of the antipodal map, right? Now here we will do it in slightly different way and using Laxay-Fitzwein theorem. In any case recall that the famous Harry-Wald theorem which has its origin in different topology, it says that you cannot comb your hair without parting at least at one point. The precise statement however is that there is no non vanishing smooth tangent vector field on S2n where S2n denotes the even dimensional sphere, standard sphere. However, you should also see that all odd dimension spheres have plenty of such tangent vector fields. So you can write down here why I have written down one namely x1, x2, x2n. So that is an odd dimensional sphere, okay? Sitting inside r2n, x1, x2, x2n going to x1 minus x2, x3 minus x4 and so on, x2n minus x2n, okay? So this will be of unit vector at each point x1, x2, x2n it is orthogonal to it and the vector is orthogonal to tangent to the sphere at that point, okay? So this is for possible for odd, there are many other possibilities but for even thing you cannot do that is the statement of this Harry Ball theorem, okay? The topological version removes the smoothness part. It says that there is no continuous map F from S2n to S2n such that for each point in S2n, Fx is orthogonal to x, okay? So I do not want to talk about tangent field and so on because this is not a smooth version. This is just a continuous version but Fx orthogonal to x makes sense. In fact, what we will do is we will prove even a stronger form of this theorem from which the theorem that I have stated here follows from which the smooth version of Harry Ball theorem also follows. So that is the theorem that I am going to prove namely for any continuous function from an even damage sphere to itself there is a point x in S2n such that either Fx is x or Fx is minus x. So you see Brauer's 6 point theorem says every continuous function from dn to dn has a fixed point, right? So this is slightly far from that namely slightly away from that namely well Fx is x equal to fixed point if not Fx will be equal to minus x. Such a point will always be there and that is unfortunately only for even dimension spheres for odd dimension sphere it is not true, okay? It is not true because we can have that kind of examples there, okay? For S1 you can just rotate by theta. So it need neither be x nor be minus x you see. So these things are quite exclusive result in that sense. For odd dimension sphere such a thing is not true for even dimension sphere we have. Let us go through the proof of this. Obviously for this one I would have to write a different proof. Of course you can try to do imitate the proof that I have given earlier, okay? Move x to whatever but here it is easier that is the whole idea. So you have different perspective here. The statement suppose the statement is not true that means what? No x has a property that Fx is equal to x or Fx is minus x. So x and Fx are distinct and not anti-portal. The line joining them will not pass through the origin. Is that clear? Whenever you have two points there is line segment joining them but because Fx is not equal to minus x also, okay? It will not pass through the origin. That means what? The entire line segment consists of non-zero elements. Therefore if you write 1 minus t times Fx plus t times x where t between 0 and 1 this will give you the line segment. So that is some vector which is non-zero so I can divide by the norm. So this is now a vector inside 2n itself, s2n itself. So for each t if you have defined like this you get a continuous function from s2n cross i to s2n. When t equal to 0 what is this? t equal to 0 it is Fx. Fx by norm Fx but norm Fx is already 1 so it is Fx. Similarly when t equal to 1 this will be 0 and this will be x and x is also s2n so x by norm x is again x. So this will be a homotopy of F with identity map. So every such map if it has no such points then it must be homotopy to the identity map, okay? That means that the left change number of F is equal to left change number of the identity map which is the Euler characteristic of s2n. Euler characteristic of s2n is very easy to compute. There is only one h0. H0 of s2n is z so that will contribute 1 and s2n will contribute 1. So they are in the same sign so they will add up so that is 2. Therefore the left shift point theorem says that F must have a fixed point but our assumption was it is neither fixed point nor left shift minus there is no fixed point. So that is a contradiction. So you see we computed the degree of antipodal map but we can do the degree in a more general situation also here. Let us give another application of LFT. Recall that for n greater than 1 hn of sn is z I am just recalling this one. We have done it earlier. Therefore for any continuous function from s1 to sn itself the homomorphism at the nth level is given by multiplication by an integer. The homomorphism is from z to z then. So this integer is called the degree of F. So this was the latest definition of the degree. There are several definitions so some of them I have asked you to check whether they are equivalent or not. H0 of sn is z and for any continuous function from a path connected space to itself h0 is always identity map. Therefore there the trace will be exactly 1. Since all other homologues vanished it follows that LF is trace h at 0 1 and then minus 1 to raise to n times the trace at the point at the nth level. Here the trace will be the degree. The entry in the only 1 cross 1 matrix so it is degree but I have to multiply by minus 1 raise to n. So LF is always 1 plus minus 1 power n times degree of F. So this formula can be used to compute LF if you know the degree or it can compute the degree if you know LF. So either way you can use this one. As in the special case for the antipodal map from s n to s n x equal to minus x since this has no fixed points it follows that L of alpha should be 0. If you take antipodal map of an odd degree thing antipodal map has no fixed points in any case. So LF must be 0 so that means for L of alpha must be 0 therefore the degree will be equal to or minus 1 raise to n plus 1. So this is what take any map be any map without fixed point. Why only antipodal? So this is the extra thing that we have got. There was no way to do this kind of things without LF's fixed point. Suppose there is no fixed point then the degree must be minus 1 raise to n plus 1. Why? Because this is 0 therefore the degree must be minus 1 raise to n plus 1. In particular the degree of the antipodal map is equal to minus 1 raise to n plus 1 which we have done separately earlier also. So more generally if you take a map from s 2 n to s 2 n suppose you say homeomorphism then f star is an isomorphism on the homology. Therefore the degree must be plus or minus 1 because in isomorphism so the generator must go to a generator and there are only two generators there okay one is the other negative of the other. So degree will be plus minus 1. If in addition f s no fixed point then LF is 0 and hence degree must be minus 1. Therefore we have another corollary here. You see we have not done much deeper thing but we are just seeing the different side of the coin the same coin and concluding many other things you know slightly different things. Let G be a group of odd odd acting on s 2 n okay through homeomorphisms any group action automatically the left translations whatever L G G going to GX or X going to GX that will define a homeomorphism for each X right. So for each G and G if there is a group action like this there exists a V belong to s 2 n such that G of V must be V. So you cannot have a fixed point free action of an odd order group on an even dimension sphere. So this is a negative result which is starting point of a big theory anyway okay. So I am just touching it like that G be a group of odd order acting on s 2 n then it must have fixed point okay how do you prove that take G belong to G we are going to G V define a homeomorphism L G from s 2 n to s 2 n okay what is this inverse multiplication by G inverse okay. So if G is of odd order say 2 K plus 1 okay it follows that L G raised to 2 K plus 1 must be identity no matter what it is okay because that is the order of the group itself so G raised to that G raised to 2 K plus 1 is identity therefore multiplication by L G is also identity okay L G raised to 2 K is identity. Now the degree we have seen is just by functoriality degree of G1 composite G2 is degree of G1 into degree of G2 and so on. So degree of L G will be whatever it is suppose it is minus 1 after taking the power okay it must be equal to odd power it must be equal to 1. Therefore degree of L G must be 1 if it if it if it were minus 1 raised to 2 K plus 1 it would have been minus okay so degree of L G must be 1. The degree of L G is 1 the conclusion of all I was saying whatever it is it is it is the cannot be 0 right. So once the degree is not 0 there must be a fixed point. So right now we will take up one of the postponed proofs which is very easy and won't take much time. This is the equivalence of CW homology and cellular homology okay cellular singular homology okay for a CW complex. The CW complex the CW homology itself is is equivalent to the singular homology we have seen. So we have also introduced the cellular singular homology in between the two. So I want to say that that is also equivalent to the singular homology or CW whichever one you want to prove okay which is what you want to prove. So so this was the statement the statement 4.10 don't matter I have just stated it what it is. The basic idea here is that that will be used elsewhere also we will see that again in the homotopy invariance but this is much much simpler there we have more elaborate structure there we will use that one later on. So this is called retraction operator. I have stated it this way so that you can quote it elsewhere. Suppose you want to study something later on this will be very useful elsewhere in algebraic homology. So I have separately stated this one instead of just proving this one directly. So this lemma plays the key role in proving this one and this can be used to do something else also. So let C dot it is a new notation now okay it is not the the chain complex of of the simplicial complex that so that is notation we had used there. Let C dot be any sub complex of the singular chain complex freely generated by some singular complexes. Every sub complex every sub module over a PID would be free but I am stating it very clearly freely generated by some simple x's okay. Assume that to each singular simple x sigma from delta q to x this is this is remember I am stating the whole thing in a unrelated way but relative things also it is possible okay anything any sigma is whether it is in A or x it is a it is inside x so take a singular simple x sigma inside x those are the generators for sx remember that there exists a singular prism okay so what is the meaning of prism here it is a map P sigma denoted by P sigma which continues where from I cross delta q okay so that is the prism I cross delta q to x with the following properties okay so what are the properties at the 0th level P sigma is just sigma at the 1th level it is a it is inside this one inside C okay where C is a sub complex of sx okay that is the hypothesis at 0th level it is sigma at 1th level it is inside C P of sigma composite fi when you when you take it restricted to someone of the phases that is composing with fi it is P sigma composite 1 cross fi okay so at the top level for each phase operator ff okay whatever happens is the top level it should be it should have this property if P is in already in sigma already P is already in P in C sorry if sigma is already in C dot then this must be just the identity sigma P sigma T Z must be sigma Z for the all of T and all of Z here okay finally if P is if a sigma is in the relative thing SA then P of sigma I cross delta should be completely inside A that is the relative part now comes so such a thing will be called a retraction operator suppose you have such a retraction operator P okay for each sigma in delta Q that is defined on the generator you have a thing coming into C okay it is a retraction why because if it is already inside C it must be identity that is why the name retraction operator is justified okay conclusion is that then there is a chain deformation retraction from S X SA to C SA intersection C see some sub module remember sub complex so in particular the inclusion map this is a chain retraction remember chain deformation retraction that means it is retraction and there is a chain homotopy which is identity on on the sub module C in particular the inclusion maps C to S X induce isomorphism in homology so this part last part is clear because it is a chain retraction okay it is a deformation retraction that chain homotopy but in addition it must be identity on C okay so the proof is very easy all right proof of this one is very easy because all that you have to do taking tau of sigma equal to P sigma on the boundary on the on the 1th level you see this define I cross delta Q on 0 cross delta Q to sigma take tau to be 1 cross delta restricted to 1 cross delta P of 1 cross delta okay so then you have a map defined on the generating set to a module C dot therefore you can extend it linearly over all of S X A okay we get tau we get a chain map tau from S X A to C this map follows that it is a chain retraction okay all that you have to say is this part you have to use so you have to verify boundary composite tau or r boundary is equal to this minus that okay so all that is absolutely trivial verification they have been these hypotheses have been made into such a thing so this the boundary operator comes here on the face on the top face it is it is P sigma composite 1 cross f i okay restricted to that one so use that to show that sorry so it is a chain retraction now a chain homotopy D from identity tau is defined by D sigma equal to H sigma where H is the prism operator which we have defined earlier okay so I will not go into this one now so when you when you come across with this one again we can explain that one now what we have to prove the this equivalence CW CW equivalent all that I need to do is construct such a P okay where this C will be now CW CW of X A and S X A S X A is as it is the usual chain complex singular chain complex take the special case the C sorry CW for a CW complex you take cell complex C cell right we had that notation yeah so that is what I will do okay what I do to complete this one by appealing to cellular approximation theorem given any continuous function from delta n to x all we do is choose a cellular approximation plate and a homotopy of the original map with the approximation which gives the prism P sigma a cellular approximation comes with a homotopy that homotopy is P sigma start the sigma take a cellular approximation for that what you have to do nothing you don't have to do anything there is the cellular approximation theorem okay so so what but there is a homotopy also just take that homotopy when it's the last thing is a cellular map therefore it is in the C cell okay so sigma is already a cellular map if it's already a cellular map don't do anything you have to choose for each sigma separately see there is no continuity argument here at all because for each sigma which is a generator I have to do it separately for that okay so what I do if it's already cellular I keep it P as identity cross sigma z that's all so we can just take P z t equal to sigma z in that case so this automatically satisfies me that's why this is so easy next time we will continue with the applications of the Somology namely the big promise Jordan Brauer separation theorem Jordan Brauer invariance of domain and so on thank you