 Today, we shall study the length spaces, their CW structure and then use the CW chain complex of that to compute its homology. To begin with, we shall define what is a length space in dimension 3, then we will keep on defining we can extend this definition to higher dimension also. So, we begin with two numbers P and Q which are co-prime to each other, they need not be prime number, they are co-prime that is all, put zeta equal to e raised to 2 pi i by P, so that zeta raised to P is 1 and actually zeta is a primitive pth root of infinity, you could have taken any other primitive root also, it would not matter, but let us fix one, no problem. Now, consider the transformation on C2 given by z0 z1 going to zeta z0, zeta power Qz1, this is a shear map from C2 to C2, it is norm preserving also, because zeta is of norm modulus of, the norm of zeta is just 1, zeta Q is also 1, they are elements of order, elements of modulus 1 e power 2 pi i P. So, these transformations are actually orthogonal, since Q is co-prime to P, you can verify that this phi Q is actually isomorphism and its order is P, that means if I apply it to P times and if phi P comes for the phi comes for the phi, you will get identity, because zeta raised to P will come, zeta raised to PQ will come, both of them are identity. Since it is norm preserving, it defines an action of g by Pg, that is the cyclic group of order P here, on the unit sphere S3, because it is norm preserving, so it is not that one if it is inside S3, its image phi of that one will be also inside S3, contained in C2. Let us denote the orbit space by LPQ, orbit space of S3 under this action by LPQ and call it the three-dimensional length space of type PQ. For each pair you have got a length space here, for each pair of integers like this, which are co-prime to each other. Verify that action is fixed at point 3 and this is a finite group action which is fixed at point 3, therefore it is an even action. Remember even action by which we mean what is called as properly discontinuous in some literature, old literature. In any case what you get is the quotient mass phi from S3 to LPQ, so let us denote this by phi here, that is quotient mass, is a P fold covering projection. In particular, since S3 simply connected follows that phi 1 of LPQ is isomorphic to Z by PZ, the cyclic group of order P. So these things we have seen in part one already. More generally now take integers P and Q1, Q2, QR are all co-prime to P, so there are just P and Q1, but now I am taking R of them, Q1, Q2, QR, they are all co-prime to P. So fix such an ordered pair of order triple ordered R to pull off numbers, then take the action you know Z, Z0, Z1, ZR going to Z, Z0, Z1, sorry Z, Z0, Z raised to Q1 Z1, Z raised to QR ZR, when there is only one Q it was just Z raised to Q up, it stopped here. So here now I am taking total of them, each time I am multiplying by corresponding Z raised to QI times Zi, that is the action. Once again this is also C linear, it is a norm preserving map and so on. So it will give you a fixed point free action. By the way once again you have to use the fact that Q1, Q2, QR are co-prime to P to say that this will be a homeomorphism of order P. So it is a fixed point free action. So if you take the quotient let us have just simple notation L, L is L, P, Q1, Q2, QR. So this is a length space of type P, Q1, Q2, QR. So this P has a different role, Q1, Q2, QR and a different role. This P tells you what is Zeta doing? Zeta is a Pth root of unity, is a primitive Pth root of unity. Q1, Q2, QR are coming dummy variables for twisted action. You could have taken all of them are 1111, then this would be diagonal action. That is an interesting action by the way. Q1, Q2, QR could have been just 11111. So there are various interesting special cases here. So look at the Q-fold covering. It will tell you that the fundamental group of L is nothing but Z by Pz. So the fundamental group has nothing to do with all these Q1, Q2, QR. Except that they have to co-prime to P so that the action is a good action. There are no fixed points and so on. So now we want to give a CW structure to these length spaces. So length space is defined as a quotient of something. Therefore what we would like to do is give a nice CW structure on the sphere S3, S5, S7 and so on. Nice means what? So that the action by the group action namely the Zeta there multiplication by Zeta, Zeta free whatever that action is cellular. The cells must be permuted. Therefore so that on the quotient itself there will be the cell structure. So this is what we want to do. Just the simplest case which we have studied for S1 for example. If you take the 3-fold action on S1 by a cube root affinity then you could have taken one omega omega square as your points and arcs from one to omega, omega to omega square, omega square to one as three cells. So three vertices and three cells. When you co-centered out by this action what you will get? You will get just one cell, zero one cell, zero cell which is one and one one cell. It is again S1. So S1 to S1 you know action by group of order 3 or 4 or 5 or whatever you can do all of that. So similar to that we want to extend this one in this case in the length space case. So we shall do it again in a inductive action. The first consider the three dimensional case. For each x belong to a cell let us have some notation here. On the circle let Rx denote the closed infinite ray namely R times x R greater than or equal to zero. So x is fixed at one single vector in S1. Look at all the zero to infinity the ray Rx, R greater than or equal to zero. That is a subset of C or I can say R2 none matter. The collection of all these all these rays so let me R xi where xi is some power of zeta. k equal to zero it is just on the real axis the positive real axis and k equal to one it is passing through xi and xi square xi q and so on xi p minus 1. So there will be p of this rays which will divide the entire plane into p sectors. Each of them of angle of each of this sector will be e raised to 2 pi i by p sorry just 2 pi 2 pi by p that is the angle. So so you have the sectors inside R2. Now you take product pixie C cross R xi's okay these will be what these will be now half three dimensional planes half half of them okay because these were rays it is not as a plane but it is only it is not a line full line it is a ray. So similarly when you take C cross this one these will be the half half planes they will cut the whole of C cross C or R4 into p sectors I can call them as sectors they are sectors cross they are sectors cross C or C cross sectors okay. So I do not want to draw any pictures here because I am doing everything very rigorously so you can draw your own pictures if you draw it correctly which is very difficult in any case okay you will understand it. The final thing is everything should be done without the help of pictures by rigorous arguments. So arguments should be converted into pictures for getting familiar for your own sake that is what you have to do each time okay yeah. So we have cut down the entire of R4 or C cross C whatever you want to say into p sectors okay close sectors. So we now we shall start constructing the the cell structure on the sphere first on the S3 okay. So we declare all the points namely 1 comma xi comma xi square etcetera x p minus 1 as to 0 cells and arcs of S1 from xi k to xi k plus 1 as a one cells so this is similar to when p is 3 take 1 omega omega square where omega is the cube root of infinity okay cut the circle into three portions that is a seated structure. So similarly I am doing it into p portions where p is any integer any positive integer okay. So you have p 0 cells and p 1 cells okay so this will already describe the one skeleton over the one skeleton we have to attach two cells. So what are the two cells I am telling you now. So they are all subspecies of S3 so finally they will constitute the entire of S3 they will build up the entire S3 they are trying to build up S3 here. So for two cells we take e to k where carrying from 0 to you know 0 up to 1 p or 1 up to p whatever you want to say 0 to p minus 1 is the indexing set here each time okay e to the k as the unit half sphere in the half space C cross C cross R okay you have you know the these half spaces are there right this is a three-dimensional space in that it is half space it is just like upper half plane and so on okay in that you take the half spheres that means you are intersecting with this the standard S3 itself okay so look at the intersection of that S3 with this one so you call that as e to k okay so note that each e to k has S1 cross 0 see S1 cross 0 is contained inside that one okay this S1 here cross 0 0 is there R and R xi k right 0 to infinity it is the ray 0 to infinity so that S1 cross 0 will be the boundary when you keep moving R positive and so on the sphere the circle becomes smaller and smaller and ultimately when xi k is itself is of unit 1 is come that will be this part will be just 0 okay circle will become smaller so that is the picture for for this e to k okay so this has boundary S1 cross 0 and okay as its boundary which is constitutes their intersection of any two of them to take xi this e to k and e to k plus 1 their intersection will be precisely S1 cross 0 because that is what it is because xi k's are different okay so second coordinate will be always different unless second coordinate is 0 they will not have intersection at all and then second coordinate is 0 the first coordinate here is S1 okay therefore so what you have is two disks with common boundary S1 the common boundary the whole thing is constitutes a two sphere embedded inside S3 okay therefore the inside of that is actually a D3 but if you do not have to use any general theorem here you can directly show that the bounded part between any two consecutive e to k and e to k plus 1 is actually D3 as follows okay indeed the portion ak let me write as ak of S3 bounded by e to k and e to k plus 1 is the union of arcs on the sphere S3 that means they are great arcs okay from a point S1 cross 0 to 0 comma z1 in 0 comma S1 where z1 itself lies between xi k and xi k plus 1 on the 0 cross S1 on the circle okay when z is equal to z1 is equal to xi k you get once one disk and when z equal to xi k plus 1 you get another disk okay therefore it follows that each e k is homeomorphic to because they are lines from S1 one point in S1 another point in the arc arc between xi k to xi k plus 1 therefore it is homeomorphic to the closed interval line so the entire thing is it is a join of this S1 and a an interval so this you can show is easily homeomorphic to D3 okay for the three cells so we have already already done this we have already described it but we take these as three cells now this ak so three cells we just take these three disks e3k equal ak okay e3k as carrying from 0 to p minus 1 okay so that will cover the entire of S3 so the the partitions that we have made inside r4 restrict to a partition inside for S3 it says 3 say unit sphere inside r4 okay so the largest these partitions are the disks here their boundaries are this e2k e2k plus 1 and so on their boundaries are just the circle S1 okay the S1 is cut into k parts and with k vertices so there are k vertices k edges sorry p vertices p edges p two cells and p three cells so this completes the CW structure for S3 okay now you observe that if you take the action of action of this z by zp by multiplication by zeta and zeta power q what happens the points on the circle they will just permute the arcs will permute the cells will permute some e2k may not go to e2k plus 1 no it will go to some other thing nothing matter okay they will permute all right the three cells permute so the action just permutes these cells that means it is actually a cellular action and what is the orbit of each action once one cell once zero cell if you take all other zero cells you get one arc you take all other arcs you get one two cell like e2k you take you get all the other k cells two two cells like these three cells you get all other these cells because the whole action of this one is transitive action in this sense okay for on the zero cells one cell therefore in each dimension you have exactly one orbit in each dimension you have exactly one orbit so for each one orbit so one cell you take one cell there zero cell one one of them one p cell one two cell and one three there that will cover the the LPQ okay the under the quotient map the LPQ is covered okay now what are the attaching maps that is what you have to understand okay attaching maps have to be described so before that let me complete now what I am going to do for higher dimension so inductively for s3 we have done so suppose you have done it to s2 s-1 s of s2-1 s of 2s-1 okay then I want to do it for 2s-1 that means what I do I take this this is lying inside the c raised to s so I am taking one more cross c and doing the same trick okay so put tau k equal to 000 psi k okay in c cross in c c raised to s plus 1 and let p 2s plus 1k be cs cross this xi this is r xi k just like that we have done okay so actually this this tau k is not necessary at all I will directly take this cross this one so these will be the partitioning planes now these are of dimension 2s 2s plus 1 in 2s plus 2 dimension real 2s plus 2 dimensional space okay that is why 2s plus 1 I have written down okay so these half spaces are there and they will cut the c raised to s plus 1 into p parts which which I can call the sectors okay just I have to now introduce what are the 2s cells okay what are 2s cells and 2s plus 2 cell okay after after going up till here okay 2s cells and 2s plus 1 cell so e 2s k is by definition this p 2s plus 1k the half space intersection with the sphere exactly same way as we done with the three three dimensional k's okay how we define 0 and 1 were defined separately okay the 2 cell was defined exactly like this if you put s equal to 0 what you get is the definition of the s equal to 1 what you get is the definition of e 2k here right so these be the two dimensional half half sphere all right check that the boundary of this one is precisely the sphere here s 2s raise to 2s minus 1 okay for all k the same argument will give you that we can take e 2k s 2s plus 1 to be the portion of s raise to 2s plus 1 in this one bounded by e 2k and e e k plus 1 e raise to s 2s k plus 1 cell okay this picture is exactly the same thing okay once again verification that this extends the cell structure of on the lower dimension sphere namely s raise to 2s minus 1 to a cell structure of s raise to s plus 1 and that it is invariant under the third p action is straightforward invariant means what the action just permutes the cells up to 2s minus 1 you have already verified so the only thing is the new things you have to verify but definition is identical okay you have to look at this one the the ones which were dividing the whole thing that partition that partition is is the invariant and that is why all these things work okay so it follows that the quotient space has a cell structure with precisely one cell in each dimension 0 less than or less than 2s plus 1 okay the in particular the cw chain complex of this cw structural has the property that all the k dimensional k dimensional modules there are just z so z z z z z z and so on how many of them are there 0 to 2s plus 1 so 2s plus 2 of them are there after that everything is 0 okay so what remains is to describe the decay from ck to ck minus 1 and that as we have pointed out will depend upon the attaching maps the how the attaching maps are what are the attaching maps here if you describe that then get the degree then decay will be defined okay so let us do that one which we will do again inductively one by one okay for d1 d1 is from where to where from c1 to c0 what is c0 c0 is just z one single 0 cell and infinite cycle group generated by that and what are the n one cells there is only one one cell and both the boundaries of this one cell are identified with this point therefore the boundary of one cell is u v minus u right both v and u are going to the same point so what will be the what will be the effect the map d will be 0 it is v minus u but v is equal to u when you go to the boundary so the map d1 is 0 from z to z c1 to c0 the map is 0 okay so this is a picture for the cw structure for s1 okay one 0 cell and one one cell okay since one cell is attached to a single point single one the boundary of any of the oriented two cells by the way orientation here nothing but you are choosing phi alpha okay as a map from delta n to that when you choose the choose that phi alpha as a generator that is the orientation when you change change the the with a minus sign you can do the minus sign that will be opposite orientation the generator there are only two choices there okay so that is nothing more than that in the the meaning of oriented here its orientation has more geometric meaning but that is not needed here so don't worry about that right now so the oriented two cell means that I have already chosen the generator that psi alpha as a as a attaching map right so two cell is is in you see s3 is oriented circle s1 s1 you know what we can think of anticlockwise or clockwise there is no problem so that we can say that the boundary of the two cell comes to the oriented s1 here that's all which is clearly wrapped on to the two cell you see each two cell the boundaries exactly identically identically s1 so I can think of this as oriented in the counter clockwise direction okay the sign is not so important now in the quotient map all the p of these arcs are going to the circle so what happens first one goes the whole circle second one also goes to whole circle therefore this circle maps on to the quotient circle namely in the LP LP cube exactly p times right therefore the degree of this map when you compose it with the quotient will be degree will be of p okay it follows that d2 is just multiplication by p the generator goes to p times that if you have understood this one the story just repeats now okay let us do one more thing wherein something else comes and again the procedure will repeat so let us see how okay now look at the oriented three cell remember how the three cell was got it was got by two loons namely the half discs okay half two discs which are both of which have the same boundary s1 all right yeah so they were e2k e2k and e2k plus one is bounded by these two things okay so if you take boundary of e3 it is e2k plus one minus e2k it is like in the product if you take cylinder the boundary will be one circle minus the other circle if we take a single edge the boundary will be this minus that this vertex minus that vertex similarly the boundary of this e3 will be just e2k plus one minus e2k so if you have doubts it could be e2k minus e2k plus one does not matter it can happen because if you have chosen the other way orientation no problem since the quotient map restricted to the interior of each of the 2k cell look at the 2k okay and go to the quotient map both of them go to the same cell right under phi under this action zeta multiplication by zeta right the phi action and then quotient out so both of them are made in the same way to the 2 cell in the the quotient map one of them which is that is the choice here the equivalence class or whatever way you want to think of that therefore when you say e2k goes to some e e2 in the in L it also comes to e2 this minus this will be 0 therefore d3 on each 3 cell is 0 the whole d3 is 0 okay whichever cell you take as a as a representative there is only one cell after all when you come to the quotient okay you you choose a representative above and then work out what happens to that when you come down so that d3 is 0 right now in the general construction the same d happens d d4 will be p times there is a p of them one one after another d5 will be 0 the same argument okay therefore what we get is this exact sequence so this is c1 to c0 is 0 c2 to c1 there will be multiplication by p and so on one is proud multiplication by p other one is 0 multiplication by p other one is 0 this is the chain complex that we have okay the homology of this one is easy to compute now kernel of this is whole thing the image will be p times that okay image is z to z multiplication by p so it is p times z so z by p z will be the homology here is that clear so that is why we can immediately conclude this following thing hi of this lz is nothing but z for i equal to 0 because it is connected at the top it is again z because below behind this there is nothing okay here there is nothing and this is 0 map so this is that okay what is this one this is multiplication by p so it is injective so it is kernel is 0 so this is 0 what happens here kernel the whole thing image is p times z by p 1 0 z by p 0 z by p that is what you get here your z by p here is z okay the last one you do not take the the you take ordinary homology the h0 is z if you take the reduced homology this would be also 0 so that would be nice pattern okay so 0 otherwise this means what in all even dimension it is 0 in odd dimension it is zp I have written zp here I also use this is a slightly north standard notation because it can use some other it may be some other thing so z by zp is the right notation okay so so this brings us to something very satisfactory thing cw chain complex have been used to compute at least something not trivial okay so next time we will take care of a few assorted things that we have missed because of time constraints and after that we will start applications of this homology thank you