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 :⇔ ∃λ ∈ R\{0} : x = λy (1) on Rn+1\{0}. Then RP n := (Rn+1\{0})/ ∼ 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 := {x |x = (x1, . . . , xn+1) ∈ R \{0}, xj 6= 0} and let q : R\{0} → RP n be the quotient map associated to the equivalence relation (1). Show that {q(Uj)}j=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 : Rn\{0} → 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 ∈ 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 \{0} −−−→ Sn   q  0 y yq

fq0 RP n −−−→ Sn/ ∼ Let f : Rn+1 \{0} 7→ Sn be deﬁned by x f(x) = ||x|| 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 \{0} → RP n. Let y, z ∈ Rn+1 \{0} such that q(y) = q(z). Then ∃λ ∈ R\{0} such that λy = z. Since λy f(z) = = sgn(λ)f(y), |λ|||y|| 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 \{0} 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 ∈ 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/||x||) = q (x), because ||x|| = 1 for x ∈ Sn and

gq ◦ fq0 (q(y)) = q ◦ g ◦ f(y) = q(y/||y||) = q(y),

we have verified fq0 ◦ g = idS/∼ and g ◦ fq0 = idRP n . Thus fq0 is the desired homeomorphism. (b) Take two points x, y ∈ 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 ∈ 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 = {s ∈ S | − s ∈ Ux} and likewise for −Uy. This is true since we can construct neighborhoods for each relation (using that sx 6∈ {±sy} 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 ∈ gq(q (Ux)) ∩ 0 0 0 gq(q (Uy)), we would have fq0 (p) ∈ 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 \{0} be open and deﬁne λU = {λu ∈ Rn+1 \{0}|u ∈ U}. Obviously, λU is open for all λ ∈ R\{0}. Therefore, q−1(q(U)) = S λU is open, which implies that q(U) is open as well. Let {U } λ∈R\{0} i i∈N 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 ) = i∈I Ui. Then also V = i∈I q(Ui) and we have already argued that q(Ui) is open for all i ∈ N. Therefore, {q(Ui)}i∈N 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 \{0}. Deﬁne Φi : Ui 7→ R by

Φi(x) = (x1/xi,..., xˆi, . . . , 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 {q(x1, . . . , xn+1)|xi 6= 0} and Φi : q(Ui) → R as x1 xn+1 Φi(q(x1, . . . , xn+1)) = ,..., xî,... xi xi

(ˆxi 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 ∈ 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 ∈ 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 ∈ R+ where yn > 0, i.e. there exists an open 0 n −1 0 neighborhood V ⊂ R+ \{y|yn = 0} of y such that (φ (V ), φ|V 0 ) gives a chart at x. As this works for arbitrary x ∈ int(M) the proof is done. (b) Let x ∈ ∂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 ∈ U ∩ ∂M we have φ(x) ∈ 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 φ| −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) ∈. Hence we showed −1 n that the open neighborhood U ∩ ∂M (=φ (V ∩ ∂R+)) of x ∈ ∂M is homeomorphic n n to an open subset of ∂R+. The latter is defined as {x ∈ R | xn = 0}, i.e. isomorphic to Rn−1. 2.3 Examples of differentiable manifolds

(a) Suppose Mj are smooth mj-manifolds, for j = 1, 2. Show that there is a smooth structure on M1 × M2 such that M1 × M2 is a smooth (m1 + m2)-manifold, and the canonical projections πj : M1 × M2 → Mj are smooth, for j = 1, 2. (b) Show that any open subset of a smooth manifold is again a smooth manifold of the same dimension. (c) A Lie group is a C∞-manifold G having a group structure such that the multiplication map µ : G × G → G and the inverse map ι : G → G, ι(c) = x−1

are both C∞. Show that GL(n, R) is a Lie group and compute its dimension.

Solution:

(a) Let {(Ui, ϕi)}i∈I , {(Vj, ψj)}j∈J be an atlas of M1 and M2, respectively. We claim that {(Ui×Vj, (ϕi, ψj))}(i,j)∈I×J is an atlas for the product manifold, where (ϕi, ψj)(p1, p2) := m1 m2 ∼ m1+m2 (ϕi(p1), ψj(p2)) ∈ R × R = R . Note that M1×M2 is second countable and Hausdorﬀ since M1 and M2 are. Moreover, the (ϕi, ψj) are homeomorphisms since their components are and the sets {Ui × Vj}i∈I,j∈J cover M × N. Thus M1 × M2 is a topological manifold of dimension (m1 + m2). The transition functions

−1 −1 −1 (ϕi, ψj) ◦ (ϕk, ψl) = (ϕi ◦ ϕk , ψj ◦ ψl )

are smooth on (ϕk, ψl)((Ui × Vj) ∩ (Uk × Vl)) = ϕk(Ui ∩ Uk) × ψl(Vj ∩ Vl) because M and N are smooth manifolds. Furthermore,

−1 −1 ϕi ◦ π1 ◦ (ϕk, ψl) = ϕi ◦ ϕk

is smooth on ϕk(Ui ∩ Uk) × ψl(Vl), therefore π1 is a smooth function and by a similar argument, the same follows for π2.

(b) Let {(Ui, ϕi)}i∈I be an atlas for the smooth manifold M and V ⊂ M open. Then

{(Ui ∩ V, ϕi|Ui∩V )}i∈I is an atlas for V and the transition functions of the charts restricted to V are still smooth. This makes V into a smooth manifold of the same dimension as M. 2 (c) We identify the n × n matrices with Rn , since both spaces are isomorphic. By deﬁnition, n×n GL(n, R) = {A ∈ R | det(A) 6= 0}. −1 Hence, for det : Rn×n → R, we have that GL(n, R) = det (R \{0}), which is open since det is continuous. By part (c) of this exercise, it is a smooth manifold of dimension n2. Let A, B ∈ Rn×n. As

n X µ(A, B)ij = AikBkj, k=1 multiplication is smooth. By Cramer’s rule, 1 ι(A) = (−1)i+j((j, i)−minor of A). ij det(A)

As det(A) 6= 0, inversion is also smooth and consequently GL(n, R) is a Lie group.