manifold. A - Cornell Math
Math 205C - Topology Midterm 1.
Erin Pearse
a) State the definition of an n-dimensional topological (differentiable) manifold. An n-dimensional topological manifold is a topological space that is Hausdorff, has a countable basis at every point, and is locally Euclidean. That is, every point has a neighbourhood which is homeomorphic to an open set of Rn . An n-dimensional differentiable manifold M is an n-dimensional topological manifold with a differentiable structure. That is, there is a collection of coordinate charts f = {(Ui , ϕi )} which cover M such that i) ∀x ∈ M, ∃Ux ∈ f s.t. x ∈ Ux and Ux ∼ = V ⊆ Rn ii) For any two charts (U, ϕ) and (V, ψ) , U ∩V = 6 ∅ =⇒ are diffeomorphisms of ϕ (U ∩ V ) and ψ (U ∩ V ) in Rn .
ψ◦ϕ−1 and ϕ◦ψ −1
iii) U is maximal in the sense that any chart (U, ϕ) which compatible with f in the sense of (ii) is included in f. b) Describe all 1-dimensional manifolds. If a 1-dimensional manifold is compact, it is homeomorphic to S 1 . If a 1dimensional manifold is not compact, it is homeomorphic to R1 . Anything not falling into either category can readily be shown to be (i) not 1-dimensional, or (ii) not a topological manifold. c) Describe all closed orientable and non-orientable 2-dimensional manifolds. If a 2-dimensional closed manifold is orientable, then it is a sphere, a torus, or a connected sum of tori. That is, it is an n-genus torus (with a sphere corresponding to genus 0). If a 2-dimensional closed manifold is non-orientable, then it is a Klein bottle, projective plane, or connected sum of them. d) What are the fundamental groups of the manifolds in (b) and (c)? π1 (S 1 ) ∼ = Z and S 1 has universal covering space R1 . π1 (R1 ) ∼ = {1} and R1 is its own universal covering space. 2 ∼ π1 (S ) = {1} and S 2 has universal covering space R2 . π1 (T 2 ) ∼ = Z × Z and T 2 has universal covering space R2 . g g π1 (#i=1 T 2 ) ∼ = Z2g and #i=1 T 2 has universal covering space R2 . F (a,b) π1 (K) ∼ and K has universal covering space R2 . = ¡ 2 ¢ {aba−1 b} π1 RP ∼ = Z2 and RP2 has universal covering space S 2 .
2
Math 205C - Topology Midterm
Erin Pearse
2. Explain why each of the following figures either is or is not a topological manifold.
a)
This object is clearly 1-dimensional. Take x to be an intersection point with open nbd U as shown. Assuming this space has the natural (subspace) topology, U would have to be homeomorphic to an open set V ⊂ R1 . Since U − {x} consists of three connected components, and for V − {f (x)} consists of only two, f : U → V cannot be a homeomorphism. This is explained in a little better detail in (v), below. ¥
b)
By the same argument as above, any open nbd of the intersection point x cannot be homemorphic to an open set of R1 . ¥
c)
Though not a differentiable manifold, this space is clearly homeomorphic to (a, b) ⊂ R1 and is thus a 1-dimensional topological manifold.¥
d)
This space is not Hausdorff, as any open set containing x also contains y. Since it is not Hausdorff, it fails to be a topological manifold. ¥
e)
Any point of this space has a nbd homeomorphic to an open subset V ⊂ R , unless it lies in the intersection of the two planes. In this case, an open nbd U of x looks like two intersecting plates. Let l be the line that forms the intersection, and suppose f : U → V ⊂ R2 is a homeomorphism. But U \l consists of 4 distinct connected components, and f (U \l) can consist of at most two connected components. To see this, note that for f (l) to separate R2 into more components, f |l would not be injective. Thus, this space is not a topological manifold. ¥ 2
Math 205C - Topology Midterm
3.
Erin Pearse
3
a) Prove that R1 cannot be homeomorphic to R2 , for n > 2. Let f : R1 → Rn , and let U = R1 − {x} where x ∈ R1 . Then if f were a ∼ homeomorphism, f |R−{x} : R − {x} −−−−→ Rn would also be a homeomorphism. Then, R − {x} is disconnected ⇒ f (R − {x}) is disconnected. < But for n > 2, Rn will always be connected! . Therefore, such a homeomorphism cannot exist. ¥ b) Try to prove that Rn cannot be homeomorphic to Rm , for n 6= m. ⊂
Begin by taking n < m. Then we imbed f : Rn −−−−→ Rm by inclusion. Note that any open set of Rm must have dimension m. f (Rn ) = Rn has dimension n, so clearly Rn cannot be an open subset of Rm . Thus, f cannot be a homeomorphism, because it takes an open set to a non-open set.1 ¥
4.
a) Give an example of a closed 3-dimensional manifold whose fundamental group is (Z, +). i) S 2 × S 1 b) Give two examples of a closed simply-connected 4-dimensional manifold. i) S 2 × S 2 ii) S 4 c) Give two examples of a closed, simply-connected 3-dimensional manifold. i) S 3 ii) Umm ...
1I
understand why this argument is wrong - the result is due to a confusion of the topologies on Rn relative to the homeomorphism and to Rm . However, this problem asks us to try to prove the theorem, and this was q the best I could come up with, short of re-deriving Brouwer’s Invariance of Domain from scratch. ° `
4
Math 205C - Topology Midterm
Erin Pearse
5. Let M be a topological (differentiable) manifold with an infinite fundamental group. ˜ is a non-compact topological (differentiable) manProve that its universal covering M ifold. ˜ is a universal covering, it is path-connected, so for any point m0 ∈ M , we Since M can use the lifting correspondence to establish a bijection π1 (M ) Φ : ³ ´ −−−→ p−1 (m0 ) ˜ π1 M (cf. [Munk] Thm. 54.6b). Then we immediately have that ¯ ¯ ¯ ¯ ¯ −1 ¯ ¯ π1 (M ) ¯ ¯p (m0 )¯ = ¯ ³ ´ ¯ by bijection ¯ ˜ ¯¯ ¯ π1 M |π1 (M )| |{1}| = |π1 (M )| =∞ =
M is simply connected by hypothesis
Now let U be an open map, F nbd of m0 so that m0 ∈ U ⊂ M . Because p is a covering ∼ we have p−1 (U ) = i∈I Vi where I is an infinite indexing set and p|Vi : Vi −−−−→ U is a homeomorphism. −1 Let {xi }∞ (m0 ). Then {xi } is an infinite i=1 by any countable sequence of points in p ˜ sequence in M , and xi ∈ Vi ∀i. Since the definition of covering map guarantees that all the Vi are disjoint, {xi } clearly cannot have any limit points. Thus, by the ˜ equivalence of compactness and limit point compactness in metrizable spaces, M must not be compact. ¥
Math 205C - Topology Midterm
Erin Pearse
5
6. Define the following terms: a) rank of a differentiable mapping The rank of a differentiable mapping F at p is the rank at a = ϕ(p) of the Jacobian matrix ∂f1 ∂f1 · · · ∂x1 ∂xn .. .. f∗ |a = ... . . ∂fm ∂fm · · · ∂xn a ∂x1 b) immersion An immersion is a differentiable mapping f : M → N such that dim M = rank f at every point of M . In this case, f∗ : Tp (M ) → Tϕ(p) (M ) is injective. c) submersion An submersion is a differentiable mapping f : M → N such that dim N = rank f at every point of N . In this case, f∗ : Tp (M ) → Tϕ(p) (M ) is surjective. d) imbedding An imbedding is an injective immersion f : M → N which is homeomorphic to its image f (M ) ⊂ N . In this case, the topology on f (M ) induced by f and the subspace topology on f (M ) induced by N are identical. e) submanifold (i.e., immersed submanifold) A submanifold or immersed submanifold is just the image of an immersion, that is, the image of a manifold M under an immersion. In this case, the topology on f (M ) is taken as being induced by the topology on M , via the immersive map. f) imbedded submanifold An imbedded submanifold is the image of an imbedding, that is, the image of a manifold under an immersion.
6
Math 205C - Topology Midterm
7.
Erin Pearse
a) Is an injective immersion an imbedding? If not, give two examples. i) Define the figure-eight map G : R → R2 as ¡ ¡ ¢ ¡ ¢¢ G (t) = 2 cos π2 + 2 arctan t , sin 2 π2 + 2 arctan t G−1 (t) will not be continuous at (0, 0) for reasons similar to those discussed in #2(b,e). The essential ideal is that any sufficiently small neighbourhood of (0, 0) which is open in the subspace topology of R2 , will be mapped by G−1 (t) to a disjoint union of three subsets of R. ¤ ii) Define a variant of the topologist’s sine curve as ¡1 ¢ t > 21 t , sin πt 0 6 t 6 21 (smooth) F (t) = t 0 on M ii) {supp (fα )} forms a locally finite covering of M P iii) α fα (x) = 1, ∀x ∈ M b) Let p 6= q be two distinct points on a differentiable manifold M. Prove that there exists a C ∞ function f : M → R such that f (p) = 100 and f (q) = 1001. Let g be the imbedding of M into Rn , as guaranteed by the Whitney Imbedding Theorem. Then let bc, ε (x) : Rn → R be the “bump function” centered at c with . support {x ∈ Rn .. |x − c| 6 ε}, as defined in class. Then set ε = 1 kg(p) − g(q)k and define b as follows:
2
b (x) = 100bp,ε (x) + 1001bq,ε (x) Finally, we define f = b◦g. Because C ∞ (Rn ) is an algebra, f is clearly C ∞ . ¥ Alternative proof: As a manifold, M is Hausdorff, so we can find two disjoint open sets U and V such that p ∈ U and q ∈ V . Then since M is locally compact, we can find a neighbourhood C of x such that C¯ is compact and C¯ ⊂ U , and we can find a ¯ is compact and D ¯ ⊂ V . By ([Boot] III.3.4, neighbourhood D of y such that D ∞ P.67) there exist C functions 0, x ∈ M \U x ∈ M \V 0, bC (x) = 0 < α < 1, x ∈ U \C and bD (x) = 0 < β < 1, x ∈ V \D . 1, 1, x∈C x∈D Now define f (x) = 100bC (x) + 1001bD (x) . ¥
Math 205C - Topology Midterm
13.
Erin Pearse
13
a) State the Whitney Imbedding Theorem The “easy” Whitney Imbedding Theorem states: Any differentiable manifold M of dimension n may be imbedded differentiably as a closed submanifold of R2n+1 . The “hard” Whitney Imbedding Theorem states: For n > 0, every paracompact Hausdorff n-manifold can be imbedded into R2n . Furthermore, it may be immersed in R2n−1 if n > 1. b) Let M be a compact differentiable manifold of dimension n. Prove that there exists an imbedding f : M → Rn for sufficiently large n, without using (a). Take n = dim M and let {Aα } be any open covering of M . By ([Boot] V.4.1), there exists a countable, locally finite refinement {Ui }∞ i=1 consisting of coordinate neighbourhoods (Ui , ϕi ) with i) ϕi (Ui ) = B3n (0) ∀i, and n ii) for Vi = ϕ−1 i (B1 (0)) ⊂ Ui , M =
S∞ i=1
Vi
Since M is compact, we need consider only a finite set of coordinate charts {(Ui , Vi , ϕi )}ki=1 . Now by ([Boot] V.4.4), we can take a subordinate C ∞ partition of unity {gi }ki=1 such that gi = 1 on Vi . Define the C ∞ maps fi : M → Rn by ( g i ϕ i , x ∈ Ui fi (x) = 0, otherwise so that each fi is immersive on Vi . Now for RN = Rn+1 × · · · × Rn+1 = R(n+1)k and Fi = (fi , gi ), define F : M → RN by F (p) = (F1 (p) , . . . , Fk (p)) F is componentwise C ∞ and thus C ∞ . F is injective. Pick x 6= y, with y ∈ Vi . case i) x ∈ Vi . Then fi |Vi = ϕi |Vi =⇒ F (x) 6= F (y)
(ϕi is injective)
/ Vi . Then gi (y) = 1 6= gi (x) =⇒ F (x) 6= F (y) case ii) x ∈ F is an immersion. For x ∈ M , x ∈ Vi for some i. Then =⇒ gi , and hence g, is immersive at x, as in the proof of #16. Now we can apply #7(d)iii to obtain the desired result. (See [Boot], p.196) ¥
14
Math 205C - Topology Midterm
Erin Pearse
14. Let M be a differentiable manifold. Prove that there exists a proper differentiable map f : M → Rn , for any positive integer n. Pick some point x0 ∈ M and define f (x) = d(x0 , x), where d is the Riemannian metric. This is justified, because it is possible to define a C ∞ Riemannian metric on every C ∞ manifold (cf. [Boot] V.4.5, p.195), and because a Riemannian manifold (that is, manifold on which a Riemannian metric Φ is defined) is a complete metric space in which the metric topology and manifold topology coincide (cf. [Boot] V.3.1, p.189). f is given explicitly by (Z µ µ ) ¶¶1/2 v dp dp .. Φ dt . c (u) = x0 , c (v) = x, u 6 t 6 v f (x) = inf 1 , c(t)∈D dt dt u Then f : M → R+ is a metric and hence differentiable. For any compact interval [a, b] ⊂ R+ , its preimage will be an annulus about x0 (i.e., f −1 ([a, b]) = . {x ∈ R .. a 6 |x − x0 | 6 b} ). This is enough to show that the preimage of any compact set is compact. Since f : M → R is proper and differentiable, we can define the imbedding ⊂ g : R −−−−→ Rn by inclusion. Then g◦f will work, for any n.
¥
Alternative proof: To avoid the use of Riemannian metrics, this proof can be adapted as follows: use the Whitney embedding theorem to embed M in some RN , then replace Φ with the standard n-dimensional Euclidean metric. The rest of the proof remains essentially the same. ¥
Math 205C - Topology Midterm
Erin Pearse
15
15. Prove that there exists no immersion f : S n → Rn . S n is compact and f is continuous, so f (S n ) is compact. Since Rn is Hausdorff, f (S n ) is also closed in Rn . Following ([Munk] p.102), we define the boundary of a set A ⊂ X by ∂A = A ∩ (X − A) Then we can pick a point x ∈ ∂f (S n ) and then also a point y ∈ f −1 (x) ⊂ f −1 (∂f (S n )). Now, if f really were an immersion, the Df would be nonsingular everywhere. In particular, Df would be nonsingular at x. Then we can find an open nbd U ⊂ S n of x such that V = f (U ) is open in Rn by the Inverse Function Theorem. Yet f (U ) is clearly not open, because x ∈ f (U ) and any open nbd of x is not contained in f (U ) by the choice x ∈ ∂f (S n ). To see this, note that the definition of boundary makes x a limit point of (Rn − f (S n )). ¥ Alternative proof: Note that S n and Rn are manifolds of the same dimension, and S n is compact. Then we know by #11 that any immersion f : S n → Rn would be a surjection and that f is an open map. Let {Uα }a∈A be an open covering of Rn . Then {f −1 (Uα )}a∈A n is an open covering of S n , so it must have a finite subcover {f −1 (Uαi )}i=1 . Thus, n {f (f −1 (Uαi ))}i=1 = {Uαi }ni=1 (equality because f is surjective) shows that any open < covering of Rn has a finite subcovering. . (Rn is not compact) ¥
16
Math 205C - Topology Midterm
Erin Pearse
16. Let f : M → Rn be an injective immersion. Prove that there exists an imbedding F : M → Rn+1 such that f (M ) is a closed subset of Rn+1 . We have f (x) = (f1 (x) , f2 (x) , . . . , fn (x)), and we know from the proof of #14 that we can find a proper differentiable map g : M → R. Define F (x) = (f1 (x) , f2 (x) , . . . , fn (x) , g (x)) So that F : M → Rn+1 . We will show that F is a proper injective immersion. Then the result will follow by #8(b). F is proper. ∞ Let {xi }∞ i=1 be any divergent sequence in M . Then {g (xi )}i=1 will be a divergent sequence because g is proper. Then it must also be the case that ∞ {F (xi )}∞ i=1 = {(f1 (xi ) , . . . , fn (xi ) , g (xi ))}i=1
is a divergent sequence. This shows that F is proper.
¤
F is injective. Let m1 6= m2 be distinct points of M . Then f is injective, so f (m1 ) 6= f (m2 ). This implies F (m1 ) = (f (m1 ) , g (m1 )) 6= (f (m2 ) , g (m2 )) = F (m2 ) . ¤ F is an immersion. First we note that f : M → N is an immersion ⇐⇒ Df : Tp (M ) → Tf (p) (N ) is injective. Pick v ∈ Tp (M ) such that DF (v) = 0. Then ∂f1 ∂f1 · · · ∂x ∂x1 m .. .. v1 ... . . .. = DF (v) = ∂fn . ∂fn ∂x · · · ∂x m 1 vm ∂g ∂g · · · ∂xm ∂x1 This can only be true if
(?)
0 .. .. . . = ∂fn 0 · v dx ∂g 0 · v dx ∂f1 dx
·v
0 .. .. Df (v) = = . . ∂fn 0 ·v dx ∂f1 dx
·v
And since f is an immersion, we know by (?) that Df is injective. Hence, this can only happen when v = [0, . . . , 0]. This shows that DF (v) = 0
⇒
v = 0,
i.e., that DF is injective. Then by (?) again, F is an immersion.
¤
Math 205C - Topology Midterm
Erin Pearse
17
Proof of (?) f : M → N is an immersion ⇐⇒ Df : Tp (M ) → Tf (p) (N ) is injective. Note that dim M = dim Tp (M ) and dim V = rank A + dim (ker A) , for any linear mapping A defined on a vector space V . Since Df : Tp (M ) → Tf (p) (N ) is linear, this gives dim Tp (M ) = rank Df + dim (ker Df ) Now we have f : M → N is an immersion ⇐⇒ dim M = rank Df ⇐⇒ dim (ker Df ) = 0 ⇐⇒ ker Df = {0} ⇐⇒ Df is injective. ¥
18
Math 205C - Topology Midterm
Erin Pearse
References [Bred]
Bredon, Glen E. (1993) Topology and Geometry. Springer-Verlag.
[Boot]
Boothby, William M. (1986) An Introduction to Differentiable Manifolds and Riemannian Geometry (Second Edition). Academic Press.
[Hirs]
Hirsch, Morris W. (1976) Differential Topology. Springer-Verlag.
[Mass]
Massey, William S. (1967) Algebraic Topology: An Introduction. Springer-Verlag.
[Miln]
Milnor, John W. (1965) Topology from the Differentiable Viewpoint. Princeton University Press.
[Munk]
Munkres, James R. (2000) Topology (Second Edition). Prentice Hall.
Erin Pearse
a) State the definition of an n-dimensional topological (differentiable) manifold. An n-dimensional topological manifold is a topological space that is Hausdorff, has a countable basis at every point, and is locally Euclidean. That is, every point has a neighbourhood which is homeomorphic to an open set of Rn . An n-dimensional differentiable manifold M is an n-dimensional topological manifold with a differentiable structure. That is, there is a collection of coordinate charts f = {(Ui , ϕi )} which cover M such that i) ∀x ∈ M, ∃Ux ∈ f s.t. x ∈ Ux and Ux ∼ = V ⊆ Rn ii) For any two charts (U, ϕ) and (V, ψ) , U ∩V = 6 ∅ =⇒ are diffeomorphisms of ϕ (U ∩ V ) and ψ (U ∩ V ) in Rn .
ψ◦ϕ−1 and ϕ◦ψ −1
iii) U is maximal in the sense that any chart (U, ϕ) which compatible with f in the sense of (ii) is included in f. b) Describe all 1-dimensional manifolds. If a 1-dimensional manifold is compact, it is homeomorphic to S 1 . If a 1dimensional manifold is not compact, it is homeomorphic to R1 . Anything not falling into either category can readily be shown to be (i) not 1-dimensional, or (ii) not a topological manifold. c) Describe all closed orientable and non-orientable 2-dimensional manifolds. If a 2-dimensional closed manifold is orientable, then it is a sphere, a torus, or a connected sum of tori. That is, it is an n-genus torus (with a sphere corresponding to genus 0). If a 2-dimensional closed manifold is non-orientable, then it is a Klein bottle, projective plane, or connected sum of them. d) What are the fundamental groups of the manifolds in (b) and (c)? π1 (S 1 ) ∼ = Z and S 1 has universal covering space R1 . π1 (R1 ) ∼ = {1} and R1 is its own universal covering space. 2 ∼ π1 (S ) = {1} and S 2 has universal covering space R2 . π1 (T 2 ) ∼ = Z × Z and T 2 has universal covering space R2 . g g π1 (#i=1 T 2 ) ∼ = Z2g and #i=1 T 2 has universal covering space R2 . F (a,b) π1 (K) ∼ and K has universal covering space R2 . = ¡ 2 ¢ {aba−1 b} π1 RP ∼ = Z2 and RP2 has universal covering space S 2 .
2
Math 205C - Topology Midterm
Erin Pearse
2. Explain why each of the following figures either is or is not a topological manifold.
a)
This object is clearly 1-dimensional. Take x to be an intersection point with open nbd U as shown. Assuming this space has the natural (subspace) topology, U would have to be homeomorphic to an open set V ⊂ R1 . Since U − {x} consists of three connected components, and for V − {f (x)} consists of only two, f : U → V cannot be a homeomorphism. This is explained in a little better detail in (v), below. ¥
b)
By the same argument as above, any open nbd of the intersection point x cannot be homemorphic to an open set of R1 . ¥
c)
Though not a differentiable manifold, this space is clearly homeomorphic to (a, b) ⊂ R1 and is thus a 1-dimensional topological manifold.¥
d)
This space is not Hausdorff, as any open set containing x also contains y. Since it is not Hausdorff, it fails to be a topological manifold. ¥
e)
Any point of this space has a nbd homeomorphic to an open subset V ⊂ R , unless it lies in the intersection of the two planes. In this case, an open nbd U of x looks like two intersecting plates. Let l be the line that forms the intersection, and suppose f : U → V ⊂ R2 is a homeomorphism. But U \l consists of 4 distinct connected components, and f (U \l) can consist of at most two connected components. To see this, note that for f (l) to separate R2 into more components, f |l would not be injective. Thus, this space is not a topological manifold. ¥ 2
Math 205C - Topology Midterm
3.
Erin Pearse
3
a) Prove that R1 cannot be homeomorphic to R2 , for n > 2. Let f : R1 → Rn , and let U = R1 − {x} where x ∈ R1 . Then if f were a ∼ homeomorphism, f |R−{x} : R − {x} −−−−→ Rn would also be a homeomorphism. Then, R − {x} is disconnected ⇒ f (R − {x}) is disconnected. < But for n > 2, Rn will always be connected! . Therefore, such a homeomorphism cannot exist. ¥ b) Try to prove that Rn cannot be homeomorphic to Rm , for n 6= m. ⊂
Begin by taking n < m. Then we imbed f : Rn −−−−→ Rm by inclusion. Note that any open set of Rm must have dimension m. f (Rn ) = Rn has dimension n, so clearly Rn cannot be an open subset of Rm . Thus, f cannot be a homeomorphism, because it takes an open set to a non-open set.1 ¥
4.
a) Give an example of a closed 3-dimensional manifold whose fundamental group is (Z, +). i) S 2 × S 1 b) Give two examples of a closed simply-connected 4-dimensional manifold. i) S 2 × S 2 ii) S 4 c) Give two examples of a closed, simply-connected 3-dimensional manifold. i) S 3 ii) Umm ...
1I
understand why this argument is wrong - the result is due to a confusion of the topologies on Rn relative to the homeomorphism and to Rm . However, this problem asks us to try to prove the theorem, and this was q the best I could come up with, short of re-deriving Brouwer’s Invariance of Domain from scratch. ° `
4
Math 205C - Topology Midterm
Erin Pearse
5. Let M be a topological (differentiable) manifold with an infinite fundamental group. ˜ is a non-compact topological (differentiable) manProve that its universal covering M ifold. ˜ is a universal covering, it is path-connected, so for any point m0 ∈ M , we Since M can use the lifting correspondence to establish a bijection π1 (M ) Φ : ³ ´ −−−→ p−1 (m0 ) ˜ π1 M (cf. [Munk] Thm. 54.6b). Then we immediately have that ¯ ¯ ¯ ¯ ¯ −1 ¯ ¯ π1 (M ) ¯ ¯p (m0 )¯ = ¯ ³ ´ ¯ by bijection ¯ ˜ ¯¯ ¯ π1 M |π1 (M )| |{1}| = |π1 (M )| =∞ =
M is simply connected by hypothesis
Now let U be an open map, F nbd of m0 so that m0 ∈ U ⊂ M . Because p is a covering ∼ we have p−1 (U ) = i∈I Vi where I is an infinite indexing set and p|Vi : Vi −−−−→ U is a homeomorphism. −1 Let {xi }∞ (m0 ). Then {xi } is an infinite i=1 by any countable sequence of points in p ˜ sequence in M , and xi ∈ Vi ∀i. Since the definition of covering map guarantees that all the Vi are disjoint, {xi } clearly cannot have any limit points. Thus, by the ˜ equivalence of compactness and limit point compactness in metrizable spaces, M must not be compact. ¥
Math 205C - Topology Midterm
Erin Pearse
5
6. Define the following terms: a) rank of a differentiable mapping The rank of a differentiable mapping F at p is the rank at a = ϕ(p) of the Jacobian matrix ∂f1 ∂f1 · · · ∂x1 ∂xn .. .. f∗ |a = ... . . ∂fm ∂fm · · · ∂xn a ∂x1 b) immersion An immersion is a differentiable mapping f : M → N such that dim M = rank f at every point of M . In this case, f∗ : Tp (M ) → Tϕ(p) (M ) is injective. c) submersion An submersion is a differentiable mapping f : M → N such that dim N = rank f at every point of N . In this case, f∗ : Tp (M ) → Tϕ(p) (M ) is surjective. d) imbedding An imbedding is an injective immersion f : M → N which is homeomorphic to its image f (M ) ⊂ N . In this case, the topology on f (M ) induced by f and the subspace topology on f (M ) induced by N are identical. e) submanifold (i.e., immersed submanifold) A submanifold or immersed submanifold is just the image of an immersion, that is, the image of a manifold M under an immersion. In this case, the topology on f (M ) is taken as being induced by the topology on M , via the immersive map. f) imbedded submanifold An imbedded submanifold is the image of an imbedding, that is, the image of a manifold under an immersion.
6
Math 205C - Topology Midterm
7.
Erin Pearse
a) Is an injective immersion an imbedding? If not, give two examples. i) Define the figure-eight map G : R → R2 as ¡ ¡ ¢ ¡ ¢¢ G (t) = 2 cos π2 + 2 arctan t , sin 2 π2 + 2 arctan t G−1 (t) will not be continuous at (0, 0) for reasons similar to those discussed in #2(b,e). The essential ideal is that any sufficiently small neighbourhood of (0, 0) which is open in the subspace topology of R2 , will be mapped by G−1 (t) to a disjoint union of three subsets of R. ¤ ii) Define a variant of the topologist’s sine curve as ¡1 ¢ t > 21 t , sin πt 0 6 t 6 21 (smooth) F (t) = t 0 on M ii) {supp (fα )} forms a locally finite covering of M P iii) α fα (x) = 1, ∀x ∈ M b) Let p 6= q be two distinct points on a differentiable manifold M. Prove that there exists a C ∞ function f : M → R such that f (p) = 100 and f (q) = 1001. Let g be the imbedding of M into Rn , as guaranteed by the Whitney Imbedding Theorem. Then let bc, ε (x) : Rn → R be the “bump function” centered at c with . support {x ∈ Rn .. |x − c| 6 ε}, as defined in class. Then set ε = 1 kg(p) − g(q)k and define b as follows:
2
b (x) = 100bp,ε (x) + 1001bq,ε (x) Finally, we define f = b◦g. Because C ∞ (Rn ) is an algebra, f is clearly C ∞ . ¥ Alternative proof: As a manifold, M is Hausdorff, so we can find two disjoint open sets U and V such that p ∈ U and q ∈ V . Then since M is locally compact, we can find a neighbourhood C of x such that C¯ is compact and C¯ ⊂ U , and we can find a ¯ is compact and D ¯ ⊂ V . By ([Boot] III.3.4, neighbourhood D of y such that D ∞ P.67) there exist C functions 0, x ∈ M \U x ∈ M \V 0, bC (x) = 0 < α < 1, x ∈ U \C and bD (x) = 0 < β < 1, x ∈ V \D . 1, 1, x∈C x∈D Now define f (x) = 100bC (x) + 1001bD (x) . ¥
Math 205C - Topology Midterm
13.
Erin Pearse
13
a) State the Whitney Imbedding Theorem The “easy” Whitney Imbedding Theorem states: Any differentiable manifold M of dimension n may be imbedded differentiably as a closed submanifold of R2n+1 . The “hard” Whitney Imbedding Theorem states: For n > 0, every paracompact Hausdorff n-manifold can be imbedded into R2n . Furthermore, it may be immersed in R2n−1 if n > 1. b) Let M be a compact differentiable manifold of dimension n. Prove that there exists an imbedding f : M → Rn for sufficiently large n, without using (a). Take n = dim M and let {Aα } be any open covering of M . By ([Boot] V.4.1), there exists a countable, locally finite refinement {Ui }∞ i=1 consisting of coordinate neighbourhoods (Ui , ϕi ) with i) ϕi (Ui ) = B3n (0) ∀i, and n ii) for Vi = ϕ−1 i (B1 (0)) ⊂ Ui , M =
S∞ i=1
Vi
Since M is compact, we need consider only a finite set of coordinate charts {(Ui , Vi , ϕi )}ki=1 . Now by ([Boot] V.4.4), we can take a subordinate C ∞ partition of unity {gi }ki=1 such that gi = 1 on Vi . Define the C ∞ maps fi : M → Rn by ( g i ϕ i , x ∈ Ui fi (x) = 0, otherwise so that each fi is immersive on Vi . Now for RN = Rn+1 × · · · × Rn+1 = R(n+1)k and Fi = (fi , gi ), define F : M → RN by F (p) = (F1 (p) , . . . , Fk (p)) F is componentwise C ∞ and thus C ∞ . F is injective. Pick x 6= y, with y ∈ Vi . case i) x ∈ Vi . Then fi |Vi = ϕi |Vi =⇒ F (x) 6= F (y)
(ϕi is injective)
/ Vi . Then gi (y) = 1 6= gi (x) =⇒ F (x) 6= F (y) case ii) x ∈ F is an immersion. For x ∈ M , x ∈ Vi for some i. Then =⇒ gi , and hence g, is immersive at x, as in the proof of #16. Now we can apply #7(d)iii to obtain the desired result. (See [Boot], p.196) ¥
14
Math 205C - Topology Midterm
Erin Pearse
14. Let M be a differentiable manifold. Prove that there exists a proper differentiable map f : M → Rn , for any positive integer n. Pick some point x0 ∈ M and define f (x) = d(x0 , x), where d is the Riemannian metric. This is justified, because it is possible to define a C ∞ Riemannian metric on every C ∞ manifold (cf. [Boot] V.4.5, p.195), and because a Riemannian manifold (that is, manifold on which a Riemannian metric Φ is defined) is a complete metric space in which the metric topology and manifold topology coincide (cf. [Boot] V.3.1, p.189). f is given explicitly by (Z µ µ ) ¶¶1/2 v dp dp .. Φ dt . c (u) = x0 , c (v) = x, u 6 t 6 v f (x) = inf 1 , c(t)∈D dt dt u Then f : M → R+ is a metric and hence differentiable. For any compact interval [a, b] ⊂ R+ , its preimage will be an annulus about x0 (i.e., f −1 ([a, b]) = . {x ∈ R .. a 6 |x − x0 | 6 b} ). This is enough to show that the preimage of any compact set is compact. Since f : M → R is proper and differentiable, we can define the imbedding ⊂ g : R −−−−→ Rn by inclusion. Then g◦f will work, for any n.
¥
Alternative proof: To avoid the use of Riemannian metrics, this proof can be adapted as follows: use the Whitney embedding theorem to embed M in some RN , then replace Φ with the standard n-dimensional Euclidean metric. The rest of the proof remains essentially the same. ¥
Math 205C - Topology Midterm
Erin Pearse
15
15. Prove that there exists no immersion f : S n → Rn . S n is compact and f is continuous, so f (S n ) is compact. Since Rn is Hausdorff, f (S n ) is also closed in Rn . Following ([Munk] p.102), we define the boundary of a set A ⊂ X by ∂A = A ∩ (X − A) Then we can pick a point x ∈ ∂f (S n ) and then also a point y ∈ f −1 (x) ⊂ f −1 (∂f (S n )). Now, if f really were an immersion, the Df would be nonsingular everywhere. In particular, Df would be nonsingular at x. Then we can find an open nbd U ⊂ S n of x such that V = f (U ) is open in Rn by the Inverse Function Theorem. Yet f (U ) is clearly not open, because x ∈ f (U ) and any open nbd of x is not contained in f (U ) by the choice x ∈ ∂f (S n ). To see this, note that the definition of boundary makes x a limit point of (Rn − f (S n )). ¥ Alternative proof: Note that S n and Rn are manifolds of the same dimension, and S n is compact. Then we know by #11 that any immersion f : S n → Rn would be a surjection and that f is an open map. Let {Uα }a∈A be an open covering of Rn . Then {f −1 (Uα )}a∈A n is an open covering of S n , so it must have a finite subcover {f −1 (Uαi )}i=1 . Thus, n {f (f −1 (Uαi ))}i=1 = {Uαi }ni=1 (equality because f is surjective) shows that any open < covering of Rn has a finite subcovering. . (Rn is not compact) ¥
16
Math 205C - Topology Midterm
Erin Pearse
16. Let f : M → Rn be an injective immersion. Prove that there exists an imbedding F : M → Rn+1 such that f (M ) is a closed subset of Rn+1 . We have f (x) = (f1 (x) , f2 (x) , . . . , fn (x)), and we know from the proof of #14 that we can find a proper differentiable map g : M → R. Define F (x) = (f1 (x) , f2 (x) , . . . , fn (x) , g (x)) So that F : M → Rn+1 . We will show that F is a proper injective immersion. Then the result will follow by #8(b). F is proper. ∞ Let {xi }∞ i=1 be any divergent sequence in M . Then {g (xi )}i=1 will be a divergent sequence because g is proper. Then it must also be the case that ∞ {F (xi )}∞ i=1 = {(f1 (xi ) , . . . , fn (xi ) , g (xi ))}i=1
is a divergent sequence. This shows that F is proper.
¤
F is injective. Let m1 6= m2 be distinct points of M . Then f is injective, so f (m1 ) 6= f (m2 ). This implies F (m1 ) = (f (m1 ) , g (m1 )) 6= (f (m2 ) , g (m2 )) = F (m2 ) . ¤ F is an immersion. First we note that f : M → N is an immersion ⇐⇒ Df : Tp (M ) → Tf (p) (N ) is injective. Pick v ∈ Tp (M ) such that DF (v) = 0. Then ∂f1 ∂f1 · · · ∂x ∂x1 m .. .. v1 ... . . .. = DF (v) = ∂fn . ∂fn ∂x · · · ∂x m 1 vm ∂g ∂g · · · ∂xm ∂x1 This can only be true if
(?)
0 .. .. . . = ∂fn 0 · v dx ∂g 0 · v dx ∂f1 dx
·v
0 .. .. Df (v) = = . . ∂fn 0 ·v dx ∂f1 dx
·v
And since f is an immersion, we know by (?) that Df is injective. Hence, this can only happen when v = [0, . . . , 0]. This shows that DF (v) = 0
⇒
v = 0,
i.e., that DF is injective. Then by (?) again, F is an immersion.
¤
Math 205C - Topology Midterm
Erin Pearse
17
Proof of (?) f : M → N is an immersion ⇐⇒ Df : Tp (M ) → Tf (p) (N ) is injective. Note that dim M = dim Tp (M ) and dim V = rank A + dim (ker A) , for any linear mapping A defined on a vector space V . Since Df : Tp (M ) → Tf (p) (N ) is linear, this gives dim Tp (M ) = rank Df + dim (ker Df ) Now we have f : M → N is an immersion ⇐⇒ dim M = rank Df ⇐⇒ dim (ker Df ) = 0 ⇐⇒ ker Df = {0} ⇐⇒ Df is injective. ¥
18
Math 205C - Topology Midterm
Erin Pearse
References [Bred]
Bredon, Glen E. (1993) Topology and Geometry. Springer-Verlag.
[Boot]
Boothby, William M. (1986) An Introduction to Differentiable Manifolds and Riemannian Geometry (Second Edition). Academic Press.
[Hirs]
Hirsch, Morris W. (1976) Differential Topology. Springer-Verlag.
[Mass]
Massey, William S. (1967) Algebraic Topology: An Introduction. Springer-Verlag.
[Miln]
Milnor, John W. (1965) Topology from the Differentiable Viewpoint. Princeton University Press.
[Munk]
Munkres, James R. (2000) Topology (Second Edition). Prentice Hall.
