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Chapter Eight 164 CHAPTER 8 Cell Complexes 1. Introduction We have seen examples of `triangulations' of surfaces for which the rule that two simplexes intersect in at most an edge fails. For exam- ple, the decomposition indicated in the diagram below of a torus into two triangles fails on that ground. However, if we ignore that fact and calculate the homology of the associated chain complex (using the obvious boundary), we will still get the right answer for the homol- ogy of a 2-torus. Even better, there is no particular reason to restrict our attention to triangles. We could consider the torus as a single `2- cell' φ as indicated below with its boundary consisting of two `1-cells' σ; τ, which meet in one `0-cell' ν. Moreover the diagram suggests the following formulas for the `boundary': @2φ = σ + τ − σ − τ = 0 @1σ = @1τ = ν − ν = 0: From this it is easy to compute the homology: H0 = Zν; H1 = Zσ ⊕ Zτ, and H2 = Zφ. A decomposition of this kind (not yet defined precisely) is called a cellular decomposition, and the resulting structure is called a cell complex. The diagram below indicates how this looks with the torus imbedded in R3 in the usual way. Indicated below is a calculation of the homology of RP 2 by similar reasoning. Note that it is clear why there is an element of order two in H1. 165 166 8. CELL COMPLEXES Here is a similar calculation for S2 where we have one 2-cell φ, no 1-cells, and one 0-cell ν. This same reasoning could be applied to Sn so as to visualize it as a cell complex with one n-cell and one 0-cell. 1.1. CP n. The definition given previously for RP n, real projective n-space, may be mimicked for any field F . Namely, consider the vector space F n+1, and define an equivalence relation in F n+1 − f0g x ∼ y , 9c 2 F ∗ such that y = cx; and let FP n be the quotient space of this relation. Thus FP n consists of the set of lines or 1 dimensional linear subspaces in F n+1 suitably topologized. If x = (x0; : : : ; xn), the components x0; : : : ; xn are called homogeneous coordinates of the corresponding point of FP n. As above, different sets of homogeneous coordinates for the same point differ by a constant, non-zero multiplier. Note also that FP 0 consists of a single point. Let F = C. Then the complex projective space CP n is the quotient space of Cn+1 − 0 ' R2n+2 − 0. We may also view it as a quotient of S2n+1 as follows. First note that by multiplying by an appropriate pos- itive real number, we may assume that the homogeneous coordinates n (x0; x1; : : : ; xn) of a point in CP satisfy n X 2 jxij = 1; i=0 so the corresponding point in Cn+1 − 0 lies in S2n+1. Furthermore, 0 0 two such points, (x0; : : : ; xn) and (x0; : : : ; xn), will represent the same n 0 point in CP if and only if xi = cxi; i = 0; : : : ; n with jcj = 1, i.e., c 2 S1. Hence, we can identify CP n as the orbit space of the action of the group S1 on S2n+1 defined through complex coordinates by c(x0; : : : ; xn) = (cx0; : : : ; cxn): The student should verify that the map p : S2n+1 ! CP n is indeed a quotient map. It follows that CP n is compact. We also need to know that it's Hausdorff. Proposition 8.1. CP n is compact Hausdorff. 1. INTRODUCTION 167 Proof. It suffices to prove that p : S2n+1 ! CP n is a closed map. Then by Proposition 3.9 we know that CP n is compact Hausdorff. To show that p is closed consider a closed set A ⊂ S2n+1 and look at the diagram S1 × S2n+1 - S2n+1 3 6 S1 × A The horizontal map is the action of S1 on S2n+1, which is continuous. Since S1 × A is compact, the image of S1 × A under this action is compact, hence closed. But this image is just p−1(p(A)), which is what we needed to show was closed. We shall describe a cellular decomposition of CP n. First note that CP k−1 may be imbedded in CP k through the map defined using ho- mogeneous coordinates by (x0; x1; : : : ; xk−1) 7! (x0; x1; : : : ; xk−1; 0): With these imbeddings, we obtain a tower CP n ⊃ CP n−1 ⊃ · · · ⊃ CP 1 ⊃ CP 0: We shall show that each of the subspaces CP k − CP k−1; k = 1; : : : ; n is homeomorphic to an open 2k-ball in R2k. To prove this, assume as above that the homogeneous coordinates (x0; x1; : : : ; xk) of a point in k Pk 2 iθk CP are chosen so i=0 jxij = 1. Let xk = rke where 0 ≤ rk = iθk jxkj ≤ 1. By dividing through by e (which has absolute value 1). we may arrange for xk = rk to be real and non=negative without changing P 2 the fact that i jxij = 1. Then, v u k−1 u X 2 0 ≤ xk = t1 − jxij ≤ 1: i=0 Let D2k denote the closed 2k-ball in R2k defined by 2k X 2 yi ≤ 1: i=1 2k k Define fk : D ! CP by fk(y1; : : : ; y2k) = (x0; : : : ; xk) where x0 = y1 + iy2, x1 = y3 + iy4, ::: , xk−1 = y2k−1 + iy2k and p 2 xk = 1 − jyj . The coordinates on the left are ordinary cartesian 168 8. CELL COMPLEXES coordinates, and the coordinates on the right are homogeneous coor- dinates. fk is clearly continuous and, by the above discussion fk is 2k k onto. Since D is compact and CP is Hausdorff, it follows that fk is a closed map. We shall show that it is one-to-one on the open ball D2k − S2k−1, and it follows easily from this that it a homeomorphism on the open ball. 0 Suppose fk(y ) = fk(y), i.e., 0 0 1 (x0; : : : ; xk) = c(x0; : : : ; xk) c 2 S : 0 0 p 0 2 Suppose jy j < 1. Then, xk = 1 − jy j > 0, so it follows that c is real and positive, hence c = 1, which means jyj < 1 and y0 = y. Note that this argument shows a little more. Namely, if two points of D2k map to the same point of CP k, then they must both be on the boundary S2k−1 of D2k. 2k−1 We now investigate the map fk on the boundary S . First note k−1 that jyj = 1 holds if and only if xk = 0, i.e., if and only fk(y) 2 CP . 2k 2k−1 k k−1 Thus, fk does map D −S homeomorphically onto CP −CP as 2k−1 claimed. Moreover, it is easy to see that the restriction of fk to S is just the quotient map described above taking S2k−1 onto CP k−1. (Ignore the last coordinate xk which is zero.) The above discussion shows us how to view CP n as a cell complex (but note that we haven't yet defined that concept precisely.) First 0 2 take a point σ0 to be viewed as CP . Adjoin to this a disk D by identifying its boundary to that point. This yields CP 2 which we see 2 4 is homemorphic to S . Call this `cell' σ2. Now attach D to this by mapping its boundary to σ2 as indicated above; call the result σ4. In this way we get a sequence of cells σ0 ⊂ σ2 ⊂ · · · ⊂ σ2n each of which is the quotient of a closed ball of the appropriate dimen- sion modulo the equivalence relation described above on the boundary of the ball. In what follows we shall develop these ideas somewhat further, and show that the homology of such a cell complex may be computed by taking as chain group the free abelian group generated by the cells, and defining an appropriate boundary homomorphism. The definition of that boundary homomorphism is a bit tricky, but in the present case, since there are no cells in odd dimensions, the bound- ary homomorphism should turn out to be zero. That is, we should end 2. ADJUNCTION SPACES 169 up with the following chain complex for CP n: C2i = Z 0 ≤ i ≤ n; C2i−1 = 0 1 ≤ i ≤ n; and @k = 0 all k: Hence, after we have finished justifying the above claims we shall have proved the following assertion: Theorem 8.2. The singular homology groups of CP n are given by ( Z if k = 2i; 0 ≤ i ≤ n; H (CP n) = k 0 otherwise: 2. Adjunction Spaces We now look into the idea of `adjoining' one space to another through a map. We will use this to build up cell complexes by ad- joining one cell at a time. Let X; Y be spaces and suppose f : A ! Y is a map with domain A a subspace of X. In the disjoint union X t Y consider the equivalence relation generated by the basic relations a ∼ f(a) for a 2 A. The quotient space X t Y= ∼ is called the adjunction space obtained by attaching X to Y through f. It is denoted X tf Y . f is called the attaching map. Example 8.3. Let f : A ! fP g be a map to a point. Then X tf fP g ' X=A. Example 8.4. Let X = D2 be the closed disk in R2 and let Y also be the closed disk.
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