 Today, we shall now study a number of examples. Before that, I will give you one more definition namely of a sub complex. Like a simple shell complex, a sub complex, so there is a definition of sub complex in the case of CW complex also. A sub complex, let us define sub complex of a relative CW complex. So it will be a relative CW complex, y comma b. It is a reducible sub complex of xa, xa is original thing and yb is a sub complex. What is the meaning? Meaning is that yb itself should be a CW complex on its own wherein y is a subset of x, b is a subset of a and each cell that you are attaching from getting y out of b, whatever cells you are attaching, the attaching maps should come from the corresponding cells from whatever you are attaching from getting x. Each cell in y should be also a cell in x with precisely the same attaching map. So in other words, some of the attaching maps in between along with the cells you may delete. But if you delete arbitrarily, it may not be a sub complex because whatever is remaining, their domains, the attaching maps, their code domain must be there. If you have deleted that part, then these functions should be also deleted. These cells must be also deleted. That is what it means. So there is a very strong restriction on being a sub complex. So sub complex like subgroup, there is a group operation. The set must be a subset of the original thing. But the group operation should be also the same as the original one. So here it is attaching maps and attaching cells. They must be the same as the original one. But the collection of this lambda each time may be smaller. That is the meaning of this sub complex. So if y, b is a sub pair of a relative CW complex xa. It means just b is a subset of a and y is a subset of x. y is the union of some b and some closed cells in x. Then yb is a complex. Take some close of simplexies inside b and then look at that. That will be already provided that boundaries are already there. Then only the cells will be there. The boundary attaching maps will be already there. So if you look at the pictures, then it will be much easier to determine whether something is a sub complex or not. Then just looking at the data given in the, you know, if it is F alpha or this given, then you do not know whether it is a sub complex or not. So this is an easy way of determining if something is a sub complex or not. If y is a sub complex of x, then x, y itself is a relative CW complex. Suppose you start with a pure CW complex x. x is a CW complex on its own. That means that a part is empty, a is empty. Now you take a sub complex. Definition of sub complex for y, it is the same thing. The second part here, namely relative part b has to be empty. So this will be also a CW complex. So y is a sub complex of x, then what happens is x, y itself is a relative CW complex. So then x can be got out of y by attaching all those cells which you have missed in y. More generally, you can put the relative part also. If yb is sub complex of xa, then x, y union a will be a relative sub complex. So you have to throw in a also, it is not just xy, xy, a because y may not contain the whole of a. y will contain b. So y union a then you can attach all those cells which are missing if you get x. This will also relative CW complex. For all k, this xk is a sub complex of any CW complex. If you stop at the kth stage, that is a sub complex. Because what we have done, you have not attached the k cells beyond k plus 1 cells, k plus 2 cells and so on. So that is another example of a sub complex. So these sub complex are named, they are called k-skeleton. For a relative CW complex xa, you just take xk only up to k cell comma a, because a you have to take now, because the starting point was you are here also a, that will be a sub complex. Sometimes you can write it as xa and then bracket k. That is the same thing as xk bracket a. So now we will let us have a number of CW complexes as well as relative CW complexes. You start with any discrete space, it is a CW complex. Any discrete space is obtained by attaching 0 cells to empty set. So any discrete space itself is a CW complex, no relative anything. If you take a subset of that, you can think as a relative CW complex also, no problem. And what are the cells here? 0 cells. What is the dimension? Dimension is also 0. So this is the most easiest example, but it is also important sometimes. Let us take the standard sphere. The spheres and disks, these are our basic objects. If you have left them out, then you will be in trouble. So n dimensional unit sphere is a CW complex. How? Where do you start? Remember if you have a pure CW complex, you have to have a 0 cell there. The x0 cannot be empty. So what is the 0 cell you have to tell? So you can start with a 0 cell to be any point. For example, you can take 0 0 0 1 or 1 0 0 0 1 of them. The single 0 cell you can take, if you remove a sphere, if you remove one point from a sphere, what do you get? You will get an open cell namely homomorphic to RN. RN homomorphic to RN, same thing as homomorphic to open cell DN. So the entire boundary has gone to a single point. Take DN on the boundary, you take a constant map, constant map to a single point, no matter what it is. That is the 0 cell. What is the quotient space? Quotient space is precisely homomorphic to SN. The simplest case is when you take D1 which is minus 1 to plus 1, the open interval, closed interval, both minus 1 plus 1 going to a single point, what do you get? You have a circle. So this is the generalization of that. 1 0 cell and 1 n cell will give you a CW structure on SN. So that is the simplest way you could have got a CW complex other than the trivial example that we have taken earlier. The attaching map is a constant map again I think so. So this fact follows because if you take the quotient space of DN wherein the entire boundary is identified to a single point, that is homomorphic to SN. Observe that these CW structures even though SN minus 1 is a subspace of SN via the equatorial inclusion, it is not a subcomplex. What I am telling is you can take SN contained inside SN plus 1. For example, SN the circle contained as an equator on S2. Now this is subspace but this is not a subcomplex because for S1 you have to have attached 1 cell. For S2 you are attaching only a 2 cell directly to single point. The single point can be made common to both of them that is fine but the 1 cell which is inside the subcomplex is not present in the bigger one at all. So this is not a subcomplex. So I am giving you an example of a nice picture which may fail to be a subcomplex picture. Equatorial inclusion from S1 to S2 or S2 to S3, S3 to S4 and so on or coordinate inclusions from SK to any higher SK plus 1, SK plus n are at a state. They are not subcomplexes. So here are the pictures I have given. So this is S2 that dot dot dot is representing the equatorial thing S1 that is not a subcomplex. But each of them is a CW complex, single point and a cell. Now we will consider more examples here also. So I will come to that and we will refer to those pictures again. Fix a CW structure on SN, whatever you have done namely point and n cell. Then DN plus 1 you can give a CW structure wherein SN will be a subcomplex. All that you have to do is fill in the DN plus 1 n cell with identity map as the attaching cell from SN minus 1 to SN to SN. DN plus 1 it can be viewed as CW complex by attaching a N plus 1 cell via SN via the identity map. So this follows that SN is then a subcomplex of DN plus 1. So this is a nice example of a subcomplex. The boundary is a subcomplex of the disk DN plus 1. So how many cells are there in DN plus 1? DN plus 1 itself before that there is a n cell SN and before that there is a 0 cell. So there are three of them. So this is the example here. I start with a 0 point here E0. Then I attach E1 I got a circle. Then I filled up E2 with a 2 cell. So this disk D2 is a 3W complex with 3 cells 0 and 1 cell and 1 2 cell. The same picture you can get 0 cell and then n cell and then n plus 1 cell. You do not have to go for 1, 2, 3 and so on. 1, 2, 3, 4 it is not possible by the way. So there will be more complications. So that is what is shown in the next picture. So let us come to that one later on. We can have different CW structures on the same topological space. Like we can have different triangulations of the same space. The topological space is a sub structure over that there can be or there may not be any CW structure. And when they are there may be more than one such structure. For instance SN we have given a CW structure with 2 cells 0 cell and an n cell. But now we will do so that each equatorial inclusions become a sub complex. So consider the usual equatoral inclusions S0 is minus 1 plus 1. Not 1 cell now, not just 1 but 2 points. I start with 2 points as my vertices. Then how do I get S1? Now I attach 2 1 cells. I can attach 2 1 cells for S0 and to get S1. Having got S1 how do I get S2? I will get 2 2 cells upper hemisphere and a lower hemisphere. Like this you can keep on upper hemisphere or hemisphere of higher dimension. You can go on getting S1, S2, S3, S4, SN. Each SK can be obtained by SK minus 1 by attaching 2 K cells 1 above 1 below. Namely the upper and lower hemispheres. And this is the picture here E0, E0 1, E0 2 correspond to minus 1 plus 1. So those are 2 points which are the 0 cells. Then E1 1 and E1 2 make it as a curve. In the top there is one E2 cell and bottom the other E2 cell. So this is the picture of S2. How many cells it has? 2 2 2 and each dimension it is a it has 2 of them. So keep going like 2 2 2 each time. You can go all the way to S infinity also. So this is a nice picture wherein each end damage subspace end damage thing will be a sub complex. This cell decomposition is more useful than the earlier one. It immediately gives a cell structure for the real projective space because this cell structure is invariant under the antipodal action x going to minus x. The 0 cells will be interchanged. The 1 cells will be interchanged. The 2 cells will be interchanged. That is the invariance. So therefore, you know if you identify a plus 1 and minus 1, all that you have to do is identify the corresponding cells that will give you cell structure on the projective space. How many 1 cells, how many 0 cells will be there? Minus 1 and plus 1 will be identified. There are 2 of them. So in P, in projective space Pn or P1, P2 there will be only 1 0 cell. There will be only 1 1 cell. There will be only 1 2 cell. There will be only 1 3 cell and so on. So Pn has 1 cell for each dimension 0 1 2 3 up to n. So that is the structure coming out of the equatorial structure for Sn because it is now it is invariant under the antipodal action. So that is the picture of Pn here. I am repeating this one. Recall that Pn is a quotient space of Sn under the antipodal action X is equivalent to X. With the definition of quotient topology Q from Sn to Pn, suppose this is a quotient map, a subset U of Pn is open if and only Q inverse of U is open in Sn. If we first observe that quotient map Q is both open as well as a closed mapping. This follows easily from the fact that for any subset f of Sn, f union minus f is open. If f is open, if f is closed, it will close. If f is open, it will close respectively because minus f is what its homomorphic copy. The closed union closed set is a closed set. This will be the inverse image of Qf. That is why it is an open mapping as well as closed mapping. From this, many of the logical properties of Sn pass on to the quotient space. You can use this to prove that Pn is a hostile space. This is a wonderful thing because quotient spaces are quite often not hostile, even under group actions. So you have to be careful. This is easy to do. So, for example, you can use Pn is second countable, Pn is compact of course. A Cw structure Sn, as described in four, is compatible with action in the sense that action preserves each skeleton and merely permutes the various cells by homomorphism. In such situations, the quotient space acquires a natural Cw structure, which you can call it as quotient structure. So we begin with X0 as a single 0 cell in Pn. This is a general remark. Now in particular for Pn what is happening, a 0 cell in Pn is the image of S0 under the quotient map. Inductively having defined Xk minus 1 whose underlying space happens to Pk minus 1, we attach a k cell, you have to choose only one of them. But what is the attaching map? Attaching map is now the quotient map from Sk minus 1 to where Pk minus 1 double covering. So, one single sphere Sk minus 1 wraps around twice in some sense. Pk minus 1, that is not a sphere of course. So, you have to understand what is the attaching map carefully here. It is no longer an identity map. In the case of sphere Sk minus 1 upper hemisphere, the boundary is precisely Sk minus 1. So, attaching map was identity, but the quotient space structure in this P again, attaching map is the quotient map. So, for definiteness let us choose upper hemisphere and take the attaching map to be the Q restricted to Sk minus 1. The space obtained is equal to Pk. So, Pk is obtained from Pk minus 1 by attaching a k cell. The boundary of the k cell maps Sk minus 1 to that one is the quotient map. So, what I want to tell you here is that this process can be carried on all the way to S infinity. S infinity is what? Union of all these increasing sequences of spheres. What is the topology? The topology has to be defined by taking a set to be closed if and only if intersection with each SN is closed in SN. The coherent topology is obtained. That S infinity is a subspace of R infinity with R infinity is a direct sum of copies of R, not direct product. Do not make that mistake. The standard Euclidean inner product can be taken, but the topology is not the metric topology. So, you have to be careful about that. Topology is weak topology. The cell structure on S infinity is compatible with antipodal action. The vector space structure is there on R infinity. So, in particular x equal to minus x makes sense there also. This cell structure is compared to the antipodal action. Hence, we get induced cell structure on the infinite projective space P infinity. What is its k skeleton? It is just pk, the projective space of dimension k. In homotopy theory, this is a very important space. Its fundamental group is Z2 and all the higher homotopy groups whatever you do not know, whether you know, they are all 0. So, such a thing is called Eilenberg-McClain space of type Z2, 1. Pi 1 is Z2 and all other things are 0. This is a very, very important space. Exactly same way, you can define the infinite projective space and before that all the projective spaces are cpn. Remember, cpn is the quotient of cn plus 1 minus 0 under the scalar multiplication. Z0, Z1 Zn is equal to lambda times Z1 Zn. Lambda must be nonzero scalar. So, scalar are now complex number that is all. However, the cell structure is a different story number, it is a very interesting story here. So, we will we will study this one carefully and perhaps that is the last example for today. So, we start with C contained inside C2 etcetera just like R contained R2, R3 and so on. And then, you see multiplication by scalar, these are all subspaces. So, it is compatible, whether multiplying by scalar here, treated as element here, they are same thing. Hence, we can use the same notation Q for all the quotient maps Q from cn plus 1 minus 0 to cpn. And then, we can restrict it to the unit sphere here. The unit sphere here is s2n plus 1 because this is real 2n plus 2 dimension. So, the unit sphere there will be s2n plus 1. You can restrict Q to the sphere here that will surject because after all every vector nonzero vector is equivalent to a unit vector. There may be many of them namely, if you multiply by unit complex number, we will get a whole set of a circle of those elements. They are all representing the same element in the projected space. In the case of real place, you had only two of them or two unit vectors namely, x and minus x. Here, you will get take any vector, take multiply by unit vector, multiply by a complex number of unit length. It will be still a unit vector. So, cpn as a quotient of s2n plus 1 modulo the same relation that namely lambda restricted to being a unit vector, unit length. In any case, what is cp naught? cp naught is complex line, one single line. Namely, one unit vector in c up to equivalence. Any two unit vectors are related by a complex number. So, cp naught is a single point. Just like rp naught is a single point. What is cp1? It will be a quotient of s2 minus 0 or s3 by the s1 action there. That is an interesting object. First, you have to study what is cp1. The quotient map Q from s3 to cp1 sends the center circle z1 equal to 0. See, z1 equal to 0 defines a circle in s3. It is intersection with a plane to a single point which is our cp naught. 1 z1 equal to 0 is a plane. There are in s3 contained inside c2. So, there are two independent planes z1 and z2. z1 equal to 0 is a plane and that plane goes to one single point after every plane here is a line in terms of complex vectors. The subspace, now I am having e2. I am taking a subspace here. The first coordinate is z naught. I am writing z naught comma z1 for elements of s3. z naught is a general vector, complex vector, but second coordinate is a real part z naught comma r. I am just defining z naught square plus r square equal to 1. So, that will be an element of s3. The first coordinate is any complex number. Second coordinate is real. Take such a space that naught square plus r square equal to 1, r positive. Okay. It is clearly homeomorphic to d2. Why? Take any mod z. Since square plus something is h to 1, that must be less than equal to 1. Take any mod z less than equal to 1. Whatever it is, okay, the second coordinate you take it as the modulus of that. Okay, sorry, 1 minus square root of 1 minus modulus square. So, r square is defined to be 1 minus mod z square. So, take that number in the second coordinate. So, okay, so it is like a graph. So, this is clearly homeomorphic to d2 because r I have taken positive. So, r should be positive, non-negative, non-negative solution of this equation. Okay. So, then the variable is only mod z. It will be homeomorphic to d2. Okay. And its boundary is given by z1 equal to 0. What is the boundary? Mod z naught itself is of square of mod z naught equal to 1. Mod z naught itself is 1. Okay. So, that is the meaning of the boundary here. So, boundary is given by z1 is 0. Note that if you take q of that, that will cover the whole of cp1. Every point here is now equivalence class of z naught comma z1 up to scalars. Okay. One point we have taken namely the second coordinate being 0. There is a single point. All other points, second coordinate will be non-zero. Right? Once it is non-zero, you can multiply out by its real, its complex part. Any complex number is r times e power i theta. So, divide by e power i theta, you have left it just r. That is this equation. That e power i theta you have divided, you divide by this one also. So, that is the equivalence class. So, I have proved that this is an all to map. Okay. q of e2 is cp1 and q restricted to the interior of this two cell namely when r is strictly less than 1, there is a unique solution. Okay. So, this will be and r positive is a unique solution. Okay. When r is 0, there is a problem. But then you get only one point. So, in the interior, it is an injective thing. So, this is precisely what we wanted for a characteristic function to have. Injectivity on the interior, on the boundary is some continuous function. That continuous function, you can take it as an attaching thing and the interior gives you the characteristic. Therefore, cp1 is nothing but a two disk this e2. Okay. It is boundary collapsed to a single point. Hence, is homeomorphic to what? If you take a disk and collapse is bounded to single point, it is nothing but s2. Therefore, we did not know what is cp1 by getting in the pretext of getting the CRW structure on that. We actually showed that it is actually homeomorphic to s2. Okay. It has a better structure. You can think of this as the so called what? Extended complex plane because these complex numbers are involved. So, that is left to you to decide. This map s3 to s2 now, I have identified cp1 with s2. This is a very, very familiar and very, very important map. Okay. It is called half vibration. This was used to get a big landmark result that pi3 of s2 is not trivial. At the time of half, that was a very, very big invention. Okay. It is a landmark invention with milestone invention in algebraic topology. So, that is the name half vibration. Okay. Inductively having established that cpk is obtained by attaching a 2k cell to cpk minus 1 via the quotient map s2k minus 1 to cpk minus 1. Okay. You see that the subspace e2k plus 2 contained in the s2k plus 3 given by e2k plus 2 equal to z0 z1 zk comma r, i range from 0 to k zi square plus r square equal to 1. Okay. That e2k plus 2 will be homomorphized to t2k plus 2. The proof is exactly same. There are k plus 1, there are k plus 1 coordinates which are free here. And r is completely determined by the values of these by this equation. Okay. So, checking again that q of e2k plus 2 is cpk plus 1 is also exactly the same proof. Okay. Look at the last coordinate. If it is 0, already you are in cpk minus 1. If it is non-zero, you can solve for this equation. That is it. So, therefore, cpk plus 1 is obtained by attaching a 2k plus 2 cell e2k plus 2 cpk via the map s2k plus 1 to cpk. Okay. Keep doing this all the way cp1, cp2, cp3, cp infinity. Okay. This will give, will give the even dimension cells only each time. e2, 0, there is a 0 cell. Then there is a e2. Then there is a e4 and so on, e2k plus 2. All the way to c infinity. And this is another important space cp infinity. This is again a Eilenberg-McLean space. This time it is simply connected. So, pi1 is trivial. So, pi2 will be infinite cyclic group and all other homotopic groups will be trivial. So, this is what is called as Eilenberg-McLean space of type z comma 2. Okay. So, one cell in each even dimension, 0, 2, 4, 6 and so on. Okay. So, we will study more examples next time. Thank you.