 Today, we start a new topic, G Coverings. As we have told earlier, the Covering Space Theory has three main points of view. Among them, the viewpoint of group action is the most ancient. Due to people like Gretchen Daig, this ancient point of view has become in the forefront now. In this last chapter, for this course, we should exploit this viewpoint and reap some wonderful harvest. Among these, proof of various forms of Seifert-1-Kampan theorem is the foremost. Let us introduce some convenient terminology this time bringing out the group action before. By a G covering, we mean at the total space in the projection map P, the bottom space D, wherein P is a covering projection. And this map is the quotient map of an even action of the group G on E. We have already seen that whenever a group G acts evenly on a topological space through defiant morphisms, the quotient map is a covering projection. So far, we have been studying covering projections without much regard to group action. Now, we want to bring the group action in the forefront. Though in principle, all the examples that we have discussed are G coverings. They come out of group action of a group. So, let us make a few definitions here. Suppose you have two G coverings. One is E, B, E, P, B, another one E prime, P prime, B prime. So, I am denoting them. A map between them earlier was just a map from E to E prime, which commuted with the projection maps. So, what we are taking here is the base space is the same. The map should be such that it is respect in the G action on both sides. It is G equivalent map, alpha of G z equal to G of alpha z. This should happen for every G in G z in A. Such a thing will be called a G map. Once it is a G map, automatically it happens that P prime composite alpha could be equal to P. Why? Because both P and P prime are quotients. They are taking the equivalence classes by the G action to the same element. So, they are quotients given by the G action. So, automatically if alpha is respect in G, so it will have P prime. Once we have this, you can talk about another map, say beta from E prime to another G covering, say E develop prime, then the composite will be also G map. The composite of G maps is G maps makes it into a category, whatever it is. So, this you have to just remember that you can take composites and identity map is there and composition is associative. These are the basic things that we have to bother about. Now, two G coverings will be called G equivalent. If there is a G map between them, which is a homeomorphism. If it were not a G map, just a homeomorphism, then remember that was the meaning of covering equivalence. Equivalence classes of coverings have been studied thoroughly and we have even classified them earlier. Classification of covering projections was that topic. Now, we are putting extra condition, namely the covering transformation that we are taking must be respect in the action of G. So, it must be G map, then we call it a G equivalent. Clearly, G equivalence is also an equivalence relation. It is a stronger equivalence relation. If two things are G equivalent, then they are covering trust, they are as covering projections, they are equivalent. But here is somewhat unexpected gift, you can say unexpected and very important one. Namely, every G map of G coverings, the base space is the same, remember all the time, over a single base space is automatically a G isomorphism. You do not have to assume that in the definition where is a homeomorphism, this is not necessary at all, automatically this homeomorphism will come. That is the meaning of this one. Not only that, once it is homeomorphism, inverse is there. Inverse is also G map, that is very easy because that is algebra. Say, if a group homomorphism is invertible, automatically the inverse is a group homomorphism, it is just like that. But why it is a homeomorphism? That is the beauty here. It is not very surprising, but it is a mild surprise, but present one. So let us go through this proof. Take alpha from you to be a such a map means it is a G map. As we have seen it, once you have G map, it goes from fibers to fibers, right? Because P, P prime composite alpha will be P. The same thing as alpha of P inverse of B is same to P prime inverse of, contained in P prime. So every fiber is an orbit. For both E and S E prime, we have alpha coming from G E. G E is an orbit to G of alpha E. But now alpha of G E is G of alpha E, it follows that alpha is surjective. Every point here is G of alpha something. So if that is a case, you take G E, G E would go to that element under alpha. So alpha is surjective by this definition. But now the action is E 1, that will tell you alpha is injective also, right? Because if alpha G 1 E equal to alpha G 2 E, then what happens? G 1 now alpha E will be equal to G 2 of alpha E. So G 1 and G 2 are given the same element because of fixed point freeness. G 1 must be equal to G 2. So this means that what we have got here is a bijection, okay? Finally, the even of action, it also tells you that it is an open mapping. If you choose an open subset which is evenly covered in the sense which translates or disjoint, okay? Alpha of that will also have the same property and so that will give you an open subset which is disjoint. All of them must be, you know, disjoint open sets should have been mapped to disjoint open sets, okay? So this shows that it is a homeomorphism. Clearly G covering is a special type of covering projection. What I mean should say it is a covering projection but with extra structure. Right? That is all. So also a G equivalence of even 2 E 2 of 2 covering obviously defines usual equivalence relation. Two things are equivalent. If they are G equivalent, but other way round may not be true. The question is now how far the converse is true? To understand this properly, let us do some artificially looking construction here, but that seems to be the final answer. So let us see. Start with an action of G on a spacey, associate quotient map P from E to B. Now you take an automorphism of G, self-automorphism. Define a new action of G on the same E by this formula, namely G of Cirque E. I am writing it as different notation here, G of Cirque E equal to phi G of E. This action on the right, it is the given action, but after taking phi, so it is twisted reaction, phi G of A. And that is the action on the left side here. Now the quotient map is the same quotient map P because whether they are related by phi G, phi is an automorphism, it is a bijection. Whether they are related by G1, G2 or phi G1, G2 or something, they will be related. So the same relation will hold for the same set of things. Therefore, the orbits are the same. Therefore the quotients are the same. The topological space E was the same, quotient is the same, orbit is the same. So B is the same, the map is the same. But I am thinking of G acting on once this way and once this way. So they are different. Are they really different? To see that, I should see that there is a map from this G covering to that G covering which is a G map. If I find that one, then they are well, they are the same in the sense of they are equivalent. The strange thing is there may not be any such equivalence. For example, you think everything is same. So maybe identity map itself will do. Check identity map from E to E. Is it a G map? On this side, the action is alpha of G, whatever it is, by doing phi G of alpha. On the other side, it is just G of alpha it should be. G of E it should be. So that will happen only if phi G equal to G for every G. That means your phi which is an automorphism must be identity. So identity map is not a G map whenever phi is different from identity. But there may be some other. So answer is not clear. There may be some other map. So finally what I want to say is there is this very simple example wherein no covering transformation will be there which is a G so let us see that example. So let us see that example. Take the three-fold covering S1 to S1 given by phi Z equal to Z cube. What is the action here? Action is by the cube roots of unity 1 omega omega square which forms a group of order 3. The action you go by a modular action which is an even action what you get is this map P Z equal to Z cube. You get a quotient view against S1. What if the Galois group of this, what are all the coverings? Obviously let us say omega is the cube root of unity. Z going to Z, Z going to omega Z, Z going to omega square Z. So there are three different maps. They are covering transformations. Right? There cannot be anything more because the order of the covering itself is 3. The fiber has only three elements. We have proved once that the group of covering transformations injects into the fiber. Therefore what we have is there can be at most three such covering transformations and we have already produced three. So it must be the full group. The Galois group is exactly equal to the group of what? Two groups of unity. Right? So we know all the covering transformations. Now you just check that none of them is a G map under what I am going to produce that namely I have to take an automorphism of the few groups of unity, group of three elements. Well that is very easy. Namely omega going to omega square is an automorphism. There is only one other identity automorphism. This is a bad enough already omega going to omega square is an automorphism. Under that none of this Z going to Z, Z going to Z square, Z going to sorry Z going to omega square, Z going to omega Z none of this is a G map. Okay? Check that. Therefore the same covering transformation can be thought of as a G covering in more than one way. Okay? This was totally ignored in the usual class, usual study of covering transformations that we have done so far. All right? So why this is so important is precisely the question here that we are going to study namely homomorphisms of one group to another group will be taken care here. This is an automorphism is taken care so from same group to same group. So first simple example, counter example we have given. This is the key namely the next theorem says so that is what I have completely given here. This theorem says that nothing else will be wrong. This is all that is going to happen. In the case of same covering transformation, same covering transformation here, same covering projection you have taken and only action could be different. How they are related in what way they are related is precisely this namely start with a connected space B and connected coverings E1 and E2 which are G coverings. They are equivalent as G coverings as covering transformations. Okay? So they are equivalent as covering transformations. Since only if you have an automorphism of G and a covering transformation F from E1 to E2 such that this F becomes a G map after you twist it by F EG. On the left side E2 you have to treat, you have to take a different action. What is that action? It corresponds to an automorphism. Both E1 and E2 are given an action. Okay? So if we take F of GZ here equal to G of FZ that would have been a G map of that. But what we get is FG of FZ and this FG is an automorphism. So this is the theorem. Okay? This is the, this is the difference between covering transformations which are both G coverings over a connected space and that is all. E1 E2 connected, B is connected. Okay? Let us prove this one and that is some kind of satisfaction we have about introducing G coverings. Obviously we have to prove only the if part here. Okay? Sorry, only if part we have to prove. Once it is, if this is satisfied automatically, all day it is a covering transformation. Okay? So just start with a covering transformation. You should produce F that is the part which you have to prove. Okay? The only if part. Given a homeomorphism F from E1 to E2 such that P2 composite F equal to P1, we must produce an automorphism P from G to G with the property that F of GZ equal to FG of FZ. Okay? So fix a base point B E1 and E1 and E2 in E2 but E2 I will take it as P1 of E1. Okay? Sorry. P1 of E1 is B I will take that. Then E2 I have taken F of, there is already a F which is a covering transformation. E2 I am taking it as F of E1. Obviously P2 of E2 will be again B. Okay? This much is fine. Right? Now it follows that for each G in G there is a unique FG. I am going to define it such that F of GE1. Look at GE1. GE1 is the same fiber as E1. So F of that will be the same fiber as FE1. Therefore FE1 times whatever reason element must be differing by some element of G. I call that element as FG. So this FG is defined by this and it has to be unique because of the even action. So F of GE1 equal to FG of FE1. This is happening at one single point namely the point that we have chosen as base point. Okay? Just like in all other covering space theory this will tell you the whole function is very defined. Okay? Now for a fixed G, if fixed G consider the two maps even to E2 given by one thing is E equal to F of GE. Another one is going to equal to PG of FE. What I am doing here? First I fixed E1 and got a G. Now I keep this G fixed but vary the point E. Point E is varied on the entire of E1. Okay? Then I get two maps. One is E equal to F of GE. G is fixed here. E is varying. Here again G is fixed. So FEG is fixed F of E. E is varying. So I have two maps. At E1 these two entries are the same. So there are two functions that causes some alpha and beta. Beta at E1 is equal to beta at E2. Right? But what are these, sorry, what are these alpha, alpha and beta? Alpha of E1 equal to beta of E1. That is what, that is what we have. So not E2. Sorry. Alpha of E1 is equal to beta of E1. That is the meaning of this one. Okay? But both are the lifts of P1. Okay? And agreeing at a point, even to E2 there is a P1 here. There is a P2 here to B. Right? You take this, these are two maps of even to E2. You apply P2, what is it? It is P1. Apply P2 here. P2 composite F is what? P1. P1 of GE is PE. G, oh, sorry, it is PE. P1 of E. Here look at it. P2 of Fiji of whatever is, whatever P2 goes, Fiji goes away. So it is P2 of F. But P2 of F is again P1. So these are the P1, these are the P1. So they are lifts of the same map P1. Okay? They agree at one point. Therefore they must agree everywhere. Why? Because we have assumed that even is connected. So they agree everywhere means just look at this one. F of GE equal to Fiji of FE. Okay? So F becomes a G map, Fiji map. But now we have got a Fiji. We have got a function. We have to verify that it is homomorphism. We have to verify that it is isomorphism. Right? This part always says that F of GE, GZ equal to P of GFZ for every Z inside. All right? So we have to verify that this G is, Fiji is, G going to Fiji is an isomorphism, an automorphism of group G. Let us prove that FE is a homomorphism first. So given G1 and G2, we have to show that FE of G1, G2 equal to FE G1 into FE G2. So start with FE G1, G2 operating upon FZ. This is equal to F of G1, G2 operating upon Z. But G1, G2, Z can be written as G1 into G2Z by associativity. By the property of F again, it is equal to F of FE of G1 into F of G2Z. Okay? Now again apply the property of F. G2 will also come out now. So that will be equal to FE G1, FE G2 into FZ. Now use the evenness of the action. In particular, switcher point to be, therefore, what we get FE of G1, G2 must be equal to FE G1 into FE G2. So that proves that FE is a homomorphism. It remains to prove that it is injective and surjective. So let us now prove that FE is injective. If FE G1, FE of G is equal to 1 for some G, then we get F of GZ equal to FE G into FZ by definition. But now we have taken that FE G is 1. So it is equal to FZ, which implies by the injectivity of F, F is a homomorphism. Therefore, GZ must be equal to FZ. But then the evenness of action which is particular point free, G must be identical. That proves that FE is injective. Now finally, to show that FE is surjective, given any H in G, look at the element H times H operative upon E prime, where E prime is any element. So what you want to do is the following. Given any H in G by surjectivity of, there is an E prime in E, such that F of E prime is H of E2, where E2 was the base point. But then E prime will be also in P1 inverse of B, because all of them are going into E2, going into B. Therefore, E prime itself will be equal to G E1. Okay. I started with some E prime which goes to H of E2 under F. But this E prime must be the same fiber with E2. Therefore, E prime could be G E1 for some G inside G. Therefore, H of E2, which was F of E prime, but E prime is G E1. So it is F of G E1, which is FG of F of E1. Now, FG of, F of E1 is just E2. Okay. Therefore, this is equal to H of E2 is equal to FG of E2, which means by evenness of the action, H is equal to FG. So we have produced a G such that FG is H. So this proves a surjectivity and therefore, P is an isomorphism. So this completes the proof of that P is a isomorphism. He is a racomorphism of P and hence completes the proof of the theorem. Thank you.