Real Projective Space Carlos Henrique Silva Alcantara

Home , Real projective space, Topological manifold

n This text is an exercise of manifold theory and we are going show RP is a topological manifold, that is, it is topological space with Hausdorﬀ and second-countable topological properties and it is also locally euclidean.

n Definition 1. First definition of Projective Space. For some natural n ≥ 1, let RP := {V ⊂ n+1 n R : V vector space, dim(V ) = 1}. We introduce a topology on RP as follow: define n+1 n n −1 n+1 π : R \{0} → RP , π(x) = span{x}, we say U ⊂ RP is open if π (U) is open in R \{0}, n+1 n with induced topology of R , in addition, denote τ subset of parts of RP such that it is the n n collection of open sets of RP , it is easy to check (RP , τ) is a topological space. n Definition 2. Second definition of Projective Space. For some natural n ≥ 1, consider S := n+1 n {x ∈ R : kxk = 1}, in S we define x ∼ y iff x = y or A(x) = y, that is, if y is image of n n x by antipodal map (A(x) = −x, x ∈ S ). Obviously ∼ is a equivalent relation in S , the map n n n n q : S → S / ∼, q(x) = [x] = {−x, x} is called of projection map, we set S / ∼:= P , analogous n −1 n to previous definition, we have U open in P if q (U) is open in S , it is equipped with induced n+1 topology of R . n n Proposition 1. The RP is homeomorphic to P n n Proof. Let ϕ : P → RP , ϕ({−x, x}) = span{−x, x}, we are going show ϕ is a homeomorphism n n between RP and P . Claim. ϕ is a bijection.

We have ϕ({−x, x}) = ϕ({−y, y}) ⇐⇒ {λx : λ ∈ R} = {µy : µ ∈ R}, if t ∈ {λx : λ ∈ λ R} = {µy : µ ∈ R}, then t = λx = µy, λ, µ ∈ R, hence ktk = |λ| = |µ| =⇒ | µ | = 1, that is, λ = ±µ, therefore {−x, x} = {−y, y}, it shows ϕ is injective. For surjection we have just n n consider x ∈ r ∩ S , where r ∈ RP , then ϕ({−x, x}) = r. Claim. ϕ is an open map. n −1 −1 −1 It is true π|Sn is onto and continuous. Let U ⊂ P open, π |Sn (ϕ(U)) = (ϕ ◦ π|Sn ) (U), −1 n −1 where ϕ ({λx : λ ∈ R}) = {λx : λ ∈ R} ∩ S , it is easy to see ϕ ◦ π|Sn = q, hence −1 −1 −1 −1 π |Sn (ϕ(U)) = (ϕ ◦ π|Sn ) (U) = q (U) is open, therefore ϕ(U) is open. By two previous claims we have ϕ homeomorphism.

n Proposition 2. P is Hausdorﬀ. n+1 kx−yk Proof. Let x, y ∈ S , x 6= y, clearly y or −y is in same hemisphere of x, so 2 ≤ 1 or kx−(−y)k 1 2 ≤ 1, if < 2 min{kx − yk, kx + yk}, we must have B(, x) ∩ B(, y) = ∅, where B(·, ·) is a open ball, otherwise if z ∈ B(, x)∩B(, y) by triangle inequality kx−yk ≤ kx−zk+kz −yk < 2 < min{kx−yk, kx+yk} ≤ kx−yk, contradiction. Analogous we show B(, x)∩−B(, y) = ∅.

n n Denote U := B(, x) ∩ S and V := B(, y) ∩ S , by preceding statement we obtain U ∩ V = U ∩ −V = −U ∩ V = −U ∩ −V = ∅. The condition U ∩ −U = V ∩ −V = ∅ follow of < 1, otherwise, we have z ∈ {λx : λ ∈ R} and z ∈ B(, x) ∩ −B(, x), on the one hand kz −xk+kz −(−x)k = 2, on the other hand kz −xk, kz +xk < < 1 =⇒ kz −xk+kz +xk < 2, contradiction. Follow U, V, −U, −V are pairwise disjoint.

1 n Claim. q(U), q(V ) are open in P . x ∈ q−1(q(U)) =⇒ q(x) ∈ q(U), but q(x) = {−x, x} so x ∈ U or x ∈ −U, then q−1(q(U)) ⊂ −U ∪ U, conversely x ∈ −U ∪ U =⇒ q(x) ∈ q(U) =⇒ x ∈ q−1(q(U)), hence −U ∪ U ⊂ −1 −1 n q (q(U)), therefore q (q(U)) = −U ∪ U open in S , analogous for q(V ). Claim. q(U) ∩ q(V ) = ∅. If there is {−x, x} ∈ q(U) ∩ q(V ), then one of U ∩ V, −U ∩ V,U ∩ −V or −U ∩ V is non empty, contradiction. n By the previous affirmations, follow P Hausdorff. Definition 3. We say a topological space M is locally euclidean (of dimension n) if for ev- n ery point p ∈ M, there is a open neighborhood U and a map ϕ : U → R such that ϕ is homeomorphism.

n Theorem 3. RP is locally euclidean. n+1 Proof. Denote Vi = {x ∈ R \{0} : xi 6= 0} and Ui = π(Vi). −1 Claim. Vi = π (π(Vi)). −1 −1 Clearly Vi ⊂ π (π(Vi)), conversely y ∈ π (π(Vi)) =⇒ π(y) ∈ π(Vi), so exist x ∈ Vi such that π(y) = π(x), note y ∈ π(y) = {µy : µ ∈ R} = {λx : λ ∈ R} = π(x), then y = λx, hence yi = λxi, if λ = 0, then y = 0, contraction, therefore yi 6= 0 because x ∈ Vi, yi 6= 0 =⇒ y ∈ Vi, that is, −1 Vi ⊂ π (π(Vi)) and follow the claim.

Claim. Vi is open. n+1 Note that Vi is complement of pre-image by projection on i−coordinate of {0} in R \{0}. n Claim. Ui is open in RP . −1 −1 Follow by Vi = π (π(Vi)) = π (Ui), Vi open. n x1 xi−1 xi+1 xn+1 Deﬁne ϕi : Ui → R , ϕi([x1, ··· , xn+1]) = , ··· , , , ··· , . xi xi xi xi

Each ϕi is continuous. n It suﬃcient show continuity of ϕi ◦ π : Vi → R , but it is x1 xi−1 xi+1 xn+1 (x1, ··· , xn+1) 7→ , ··· , , , ··· , xi xi xi xi clearly continuous. n −1 Claim. ψ : R → Ui, ψ(u1, ··· , un) = [u1, ··· , ui, 1, ui+1, ··· , un] is ϕi . First (ϕi ◦ ψ)(u1, ··· , un) = ϕi([u1, ··· , ui−1, 1, ui+1, ··· , un]) = (u1, ··· , un) reciprocally, x1 xi−1 xi+1 xn+1 (ψ ◦ ϕi)([x1, ··· , xn+1]) = ψ , ··· , , , ··· , xi xi xi xi x x x x = 1 , ··· , i−1 , 1, i+1 , ··· , n+1 xi xi xi xi = [x1, ··· , xn+1].

−1 Claim. ϕi is continuous.

2 −1 n n+1 We have ϕi (u1, ··· , un) = (π ◦ fi)(u1, ··· , un), where fi : R → R , f(u1, ··· , un) = −1 (u1, ··· , ui, 1, ui+1, ··· , un), from continuity of π and f follow ϕi continuous. n n+1 If v ∈ RP , then exist x ∈ R such that v = [x], x 6= 0, then exist i ∈ {1, ··· , n + 1} with Sn+1 n n xi 6= 0, hence v ∈ Ui, therefore i=1 Ui = RP . Finally, for each point in RP we have a local n n homeomorphism with R , that is, RP is locally euclidean. n Corollary 4. RP is second-countable. n n Proof. Each Ui, i ∈ {1, ··· , n + 1}, is second-countable (homeomorphic to R ), because RP = Sn+1 n i=1 Ui, we have RP is second-countable. n Corollary 5. RP is a topological manifold of dimension n. n Proof. By Theorem 1 we have RP locally euclidean and second-countable, by Proposition 2 n n RP is Hausdorﬀ, because P is Hausdorﬀ and they are homeomorphic.

References

[L] Lee, J., Introduction to Smooth Manifolds. Graduate Texts in Mathematics, Vol. 218, Springer, 2002.