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Differential Topology: Exercise Sheet 2

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Differential Topology: Exercise Sheet 2 Prof. Dr. M. Wolf WS 2018/19 M. Heinze Sheet 2 Differential Topology: Exercise Sheet 2 Exercises (for Nov. 7th and 8th) 2.1 Real projective space Consider the equivalence relation x ∼ y :, 9λ 2 Rnf0g : x = λy (1) on Rn+1nf0g. Then RP n := (Rn+1nf0g)= ∼ is called the (real) projective space. (a) Show that RP n is homeomorphic to Sn= ∼ where the relative topology and the equivalence relation x ∼ y :, x = ±y (2) on the unit sphere Sn ⊂ Rn+1 are used. This provides an alternative definition of RP n. In the following, we show that RP n is a smooth n-manifold: (b) Show that RP n is Hausdorff. (c) Show that RP n is second countable. n+1 n (d) Define Uj := fx jx = (x1; : : : ; xn+1) 2 R n f0g; xj 6= 0g and let q : Rnf0g ! RP n be the quotient map associated to the equivalence relation (1). Show that fq(Uj)gj=1 n n is an open cover of RP , and that each q(Uj) is homeomorphic to a subset of R . (e) Show that RP n is a smooth manifold. Show that the quotient map q : Rnnf0g ! RP n used to define RP n is smooth. Solution: (a) In this exercise, we will construct a homemorphism between RP n and Sn= ∼ . Let us first show the following: Lemma 1 Let f : X ! Y for some topological spaces X; Y , ∼ denote an equivalence relation on X and q be the quotient map of the quotient space X= ∼. Let f be constant on each equivalence class, such that it induces a map f : X= ∼ ! Y by f(q(p)) = f(p) for all p 2 X. The induced map f : X= ∼ 7! Y is continuous if and only if the map f : X 7! Y is continuous. 2 Proof : Let f be continuous. Since q is continuous by definition, also f = f ◦ q is continuous as a composition of continuous functions. For the converse direction, assume that f is continuous. Let U ⊂ Y be an arbitrary open set. By assumption, −1 f −1(U) = q−1(f (U)) is open and by the definition of the quotient topology, also −1 f (U) is open. Since U was arbitrary, it follows that f is continuous. We consider the following situation: f Rn+1 n f0g −−−! Sn ? ? ?q ? 0 y yq fq0 RP n −−−! Sn= ∼ Let f : Rn+1 n f0g 7! Sn be defined by x f(x) = jjxjj This is clearly a continuous function. Let q0 : Sn 7! Sn= ∼ be the projection onto equivalence classes, which is also continuous. Then q0 ◦ f is continuous as well. We show that q0 ◦ f is constant on equivalence classes. Let q the projection map q : Rn+1 nf0g ! RP n. Let y, z 2 Rn+1 nf0g such that q(y) = q(z). Then 9λ 2 Rnf0g such that λy = z. Since λy f(z) = = sgn(λ)f(y); jλjjjyjj it follows that f(y) ∼ f(z) and hence q0 ◦f(y) = q0 ◦f(z). By the above Lemma, q0 ◦f n n 0 induces an fq0 : RP 7! S = ∼ , which is continuous since q ◦f is. To find an inverse, consider g : Sn 7! Rn+1 nf0g given by g(x) = x. This is clearly continuous, as is q ◦g. We show again that q ◦ g is constant on equivalence classes. Let u, v 2 Sn such that q0(u) = q0(v). Then u = ±v and hence also g(u) ∼ g(v), such that q ◦ g(u) = q ◦ g(v). n n Hence q ◦ g induces a continuous gq : S = ∼ ! RP . We are left with showing that indeed gq is an inverse of fq0 . However, since 0 0 0 0 fq0 ◦ gq(q (x)) = q ◦ f ◦ g(x) = q (x=jjxjj) = q (x); because jjxjj = 1 for x 2 Sn and gq ◦ fq0 (q(y)) = q ◦ g ◦ f(y) = q(y=jjyjj) = q(y); we have verified fq0 ◦ g = idS=∼ and g ◦ fq0 = idRP n . Thus fq0 is the desired homeo- morphism. (b) Take two points x, y 2 RP n such that x 6= y. Since we have shown in part 3 of n n this exercise that RP and S = ∼ are homeomorphic, consider corresponding sx, n 0 sy 2 S such that gq ◦ q (sx) = x and the same for sy and y (gq given as in the last exercise). Since Sn is Hausdorff (since it can be made into a metric space), we can find neighborhoods Ux, Uy of sx and sy, respectively, such that Ux \ Uy = ;;Ux \ −Uy = ;; −Ux \ Uy = ; and − Ux \ −Uy = ;; n where −Ux = fs 2 S j − s 2 Uxg and likewise for −Uy. This is true since we can construct neighborhoods for each relation (using that sx 62 {±syg as x 6= y) and consider intersections to find neighborhoods which satisfy all the equations at once. 0 0 n 0 Then also gq(q (Ux)) and gq(q (Uy)) are disjoint in RP , since for p 2 gq(q (Ux)) \ 0 0 0 gq(q (Uy)), we would have fq0 (p) 2 q (Ux) \ q (Uy), which means that fq0 (p) would be contained in one of the intersections above, which were constructed to be empty. 0 0 n Furthermore, q (Ux) and q (Uy) contain open sets, since −V ⊂ S is open if and only 0 n 0 if V is and q (V ) is therefore open for every open set V ⊂ S . Thus, gq(q (Ux)) and 0 gq(q (Uy)) are disjoint neighborhoods of x and y, respectively. (c) We will exploit that Rn+1 is second countable. Let U ⊂ Rn+1 nf0g be open and define λU = fλu 2 Rn+1 nf0gju 2 Ug. Obviously, λU is open for all λ 2 Rnf0g. Therefore, q−1(q(U)) = S λU is open, which implies that q(U) is open as well. Let fU g λ2Rnf0g i i2N be a countable basis of Rn+1. Let V ⊂ RP n be open. Then q−1(V ) is open and there −1 S S is an index set I ⊂ N such that q (V ) = i2I Ui. Then also V = i2I q(Ui) and we have already argued that q(Ui) is open for all i 2 N. Therefore, fq(Ui)gi2N is a countable basis for the topology on RP n. (d) We have already seen in part (b) of this exercise that q(Uj) is open because Uj is. Sn n Sn n+1 n Furthermore, i=1 q(Ui) = RP , since i=1 Ui = R n f0g. Define Φi : Ui 7! R by Φi(x) = (x1=xi;:::; x^i; : : : ; xn+1=xi): n This is constant on equivalence classes and therefore induces a map Φi : q(Ui) 7! R −1 by part 2. Φi is continuous because Φi is. We can define a map Ψi :Φi(Ui) 7! Ui given by Ψi(x1; : : : ; xn) = (x1; : : : ; xi−1; 1; xi; : : : ; xn): Then q ◦ Ψi is an inverse for Φi and as a composition of continuous maps it is again n continuous. Thus q(Ui) and Φi(q(Ui)) ⊂ R are homeomorphic. (e) We already know that RP n is a topological manifold, thus we need to find a smooth structure. Let q be the projection on equivalence classes and define again q(Ui) = n fq(x1; : : : ; xn+1)jxi 6= 0g and Φi : q(Ui) ! R as x1 xn+1 Φi(q(x1; : : : ; xn+1)) = ;:::; x^i;::: xi xi (^xi means that xi is omitted) with inverse function −1 Φi :(y1; : : : ; yn) 7! q(y1; : : : ; yi−1; 1; yi; : : : ; yn) We have already seen that these definitions give us charts, so we have to see that the transition functions are smooth. For simplicity, let us consider q(U1) \ q(U2). Then for x 2 Φ1(q(U1) \ q(U2)) −1 1 x2 xn Φ2 ◦ Φ1 (x) = ; ;:::; : x1 x1 x1 This is smooth since x1 6= 0. On any other q(Ui) \ q(Uj), analogous statements can be made. Hence this defines a smooth structure. 2.2 Boundary and interior of a topological manifold Let M be an n-dimensional topological manifold with non-trivial boundary @M. Show that (a) int(M) is an n-dimensional manifold, (b) @M is an (n − 1)-dimensional manifold. Solution: (a) For x 2 int(M) exists an open neighborhood U ⊂ M of x that is homeomorphic n to an open subset V ⊂ R+ via a homeomorphism φ : U ! V . By definition of an n interior point, we have that φ(x) = y 2 R+ where yn > 0, i.e. there exists an open 0 n −1 0 neighborhood V ⊂ R+ n fyjyn = 0g of y such that (φ (V ); φjV 0 ) gives a chart at x. As this works for arbitrary x 2 int(M) the proof is done. (b) Let x 2 @M. By definition of a manifold with boundary, there is an open neigh- n borhood U ⊂ M of x that is homeomorphic to an open subset V ⊂ R+ via a n homeomorphism φ : U ! V . Since x 2 U \ @M we have φ(x) 2 V \ @R+. The −1 n restriction of the homeomorphism φ to φ (V \ @R+) = U \ @M is again a homeo- −1 n n n morphism φj −1 n : φ (V \ @ ) ! V \ @ . Since V \ @ is open in the φ (V \@R+) R+ R+ R+ n subspace topology on @R+, it is an open neighborhood of φ(x) 2. Hence we showed −1 n that the open neighborhood U \ @M (=φ (V \ @R+)) of x 2 @M is homeomorphic n n to an open subset of @R+.
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