Integration and Manifolds Solutions Fall 2007

Michael Stoll, Nikita Selinger (1) Zero-dimensional Submanifolds. Show that the set

 1 T = n : n ∈ {1, 2,... } ∪ {0} is not a zero-dimensional submanifold of R. More generally, show that a subset S ⊂ Rn is a zero-dimensional Cq-submanifold of Rn if and only if S is discrete. Solution. If S is discrete then for any point p ∈ S there exists a neighborhood n 0 p ∈ U ⊂ R such that U ∩ S = {p}. Then the map fp : {p} → R is a chart of S. Clearly all such charts cover S and are compatible because their domains do not intersect. If S is not discrete then there exists a point p ∈ S such that U ∩ S 6= {p} for any neighborhood p ∈ U. Therefore we can not have a zero-dimensional chart covering p (its preimage must be a neighborhood of p consisting of a single point).

(2) Some Curves in the Plane. For the following functions f : R2 → R, sketch the set defined by f(x, y) = 0 and decide (with proof) in which cases the set is a 1-dimensional C1-submanifold of R2: (a) f(x, y) = x2 − y2 − 1 (b) f(x, y) = (x2 + y2)x (c) f(x, y) = xy Solution. (a) f(x, y) = 0 is the zero set of a smooth function whose derivative has full rank at each point in the zero set. (b) f(x, y) = 0 is the line x = 0 which is clearly a manifold. (c) f(x, y) = 0 is not a manifold because it cannot be written as a graph near the point (0, 0).

(3) The M¨obius Strip. 3
Let f :]−1, 1[ × R → R ,(x, t) 7→ (1 + x cos t) cos 2t, (1 + x cos t) sin 2t, x sin t . Show that the image of f is a 2-dimensional C∞-submanifold of R3. Solution. First of all one can check that the rank of Df is always equal to 2.

This is straightforward. Note that f(x, t) is 2π periodic in t and f(x1, t1)=f(x2, t2) k if and only if x1 = (−1) x2, t1 − t2 = πk for some k ∈ Z. After that we can  −π π  consider two charts with domains f(]−1, 1[ × 2 , 2 ) and f(]−1, 1[ × ]0, π[). The map f −1 maps both subsets diffeomorphically into R2. (4) Products.

q n1 q Let M1 be a k1-dimensional C -submanifold of R , M2 a k2-dimensional C - n2 n1 n2 n1+n2 submanifold of R . Show that M1 × M2 ⊂ R × R = R is a (k1 + k2)- dimensional Cq-submanifold of Rn1+n2 . Solution. This is obvious from the very first definition of Cq-submanifold of Rn. q ni ni−ki Locally Mi is the zero set of some C function fi : R → R of rank ni − ki (i = 1, 2). Hence M1 × M2 is the zero set of

n1 n2 n1+n2 n1−k1 n2−k2 n1+n2−k1−k2 f1 × f2 : R × R = R → R × R = R

and the rank of f1 × f2 is n1 + n2 − k1 − k2. (It is equally simple using the other definitions.)

(5) Products of Abstract Manifolds Let M and M 0 be k-, respectively k0-dimensional abstract Cq-manifolds. Show that the product M × M 0 can be made into a (k + k0)-dimensional Cq-manifold in such a way that both projections M × M 0 → M and M × M 0 → M 0 are Cq-maps. (20 points) Solution. For any two charts φ : U → V and φ0 : U 0 → V 0 of M and M 0 respectively, we can consider the chart φ × φ0 : U × U 0 → V × V 0. One can verify that any two such charts are Cq-compatible. These charts endow M × M 0 with the structure of a (k + k0)-dimensional manifold. The projections maps M × M 0 → M and M × M 0 → M 0 look like actual coordinate projections in our charts, hence they are differentiable.

(6) The Torus as a Submanifold of R3 Let T ⊂ R3 be the set consisting of all points in R3 having distance r > 0 from a fixed circle of radius R > 0. Under which conditions on R and r is this a submanifold of R3? Show that in this case, T is C∞-diffeomorphic to the product S1 × S1. Solution. We may assume that the circle is parameterized as (R cos φ, R sin φ, 0). Then T can be parameterized as f(φ, ψ) = (R cos φ + r cos φ cos ψ, R sin φ + r sin φ cos ψ, r sin ψ) . √ If r ≥ R then near the point (0, 0, r2 − R2), T is not a graph, and so T is not a manifold. If r < R then T is a manifold: the map

1 1

F : S × S → T, (x1, y1), (x2, y2) 7→ x1(R + rx2), y1(R + rx2), ry2 induced by f is a diffeomorphism (it is bijective, and the derivative at every point is invertible). (7) Transitivity of Compatibility of Charts q Let A be a C -atlas on M and let φ, φ0 be two charts on M. Show that if both φ q q and φ0 are C -compatible with all charts in A, then φ and φ0 are C -compatible with each other.

Solution. Take any p which is in the intersections of the domains of φ and φ0. −1 −1 −1 Then p ∈ A where (a : A → B) ∈ A is a chart. Then φ◦φ0 = (φ◦a )◦(a◦φ0 ) is a composition of two Cq mappings in some neighborhood of p.

(8) Real Projective Space Let n ≥ 0. Show that the quotient manifold Sn/{±1} (see Example 1.14 in the notes) and the manifold of all lines through the origin in Rn+1 (Example 1.16) are C∞-diffeomorphic. Solution. Let f be a map sending every point of the sphere to the line passing through this point and the origin. Two different points are sent to the same line if and only if they lie on the same line with the origin, hence are antipodal. So f

is a well defined mapping from the first manifold M1 to the second M2. Clearly n+1 −1 f is a bijection. One can define a map sx : M1 → R as a local branch of π in some small enough neighborhood of x ∈ M1, where π is the projection map

from the sphere to M1, which is clearly differentiable. On the other hand, we 0 n+1 have the map π : R \{0} → M2 that sends a point to the line spanned by it; this map is differentiable, as can be checked using the charts given in Ex. 1.16. 0 Then f = π ◦ sx is differentiable. We get a local inverse of f as π ◦ t`, where t` n maps a line near ` ∈ M2 to its point of intersection with S near a fixed point in n −1 ` ∩ S . Again, it is easily seen that t` is differentiable, hence f is differentiable as well.

(9) A Differentiable Map Let T be the image of the map

2
3 F : R 3 (ϕ, ψ) 7−→ cos ϕ(2 + cos ψ), sin ϕ(2 + cos ψ), sin ψ ∈ R ; this is the torus from Homework Problem (2.2) when R = 2 and r = 1. Define f : T → S2 (the “outer unit normal vector”) as 1 ∂F ∂F fF (ϕ, ψ)
= v , where v = (ϕ, ψ) × (ϕ, ψ) . kvk ∂ϕ ∂ψ ∞ Show that f is a well-defined C -map, and for all p ∈ T , determine rk Dfp. Solution. One computes f(F (φ, ψ)) = (cos φ cos ψ, sin φ cos ψ, sin ψ). This map is clearly C∞ as a map from R2 to R3 and its rank (or rank of its derivative) at every point is 2 except for the points with ψ = π/2 + πk where rank is 1. We know that F is a local diffeomorphism (see problem 2.2), hence the considered map is differentiable and its rank is 2 except for the two circles in T given by (2 cos φ, 2 sin φ, ±1). (10) Line Bundles Show that a line bundle π : E → M is trivial if and only if it has a nonvanishing

section. This is a differentiable map s : M → E such that π◦s = idM (“section”)

with s(p) 6= 0 ∈ Ep for all p ∈ M (“nonvanishing”). ∼ Solution. If E = M ×R then take the map s(m) = (m, 1). This will clearly de- fine a non-vanishing section. On the other hand, if s is a non-vanishing section, then v = a(v)s(π(v)) for any v ∈ E where a(v) ∈ R. a : E → R is differen- tiable because it is linear on every fiber and s is differentiable. This defines a diffeomorphism (π, a): E → M × R.

(11) Pulling Back Vector Bundles Let π : E → M be a vector bundle of rank n, and let f : M 0 → M be differen- tiable. Show how to construct a rank n vector bundle π0 : E0 → M 0 such that f extends to a map F : E0 → E, which gives isomorphisms on the fibers (i.e., the 0 0 induced map Fp : Ep → Ef(p) is an isomorphism for all p ∈ M ). Solution. Define E0 = {(x, e) ∈ M 0 × E : f(x) = π(e)} ⊂ M 0 × E, with 0 0 0 n projection map π : E → M ,(x, e) 7→ x. If φ : E|U → U × R is a local 0 0 −1 n trivialization of E, we define φ : Ef −1(U) → f (U) × R by (x, e) 7→ φ(e) 0 (note that e ∈ Ef(x) by definition of E ). One checks easily that these local trivializations are compatible. This gives E0 the structure of a rank n vector bundle over M 0. Finally, the map F : E0 → E,(x, e) 7→ e is a differentiable 0 0 0 extension of f to E . Clearly F acts on fibers by isomorphisms (F |Ex : Ex → Ef(x),(x, e) 7→ e is essentially the identity map).

(12) A Non-Trivial Line Bundle Let f : S1 → S1, z 7→ z2 (viewing S1 ⊂ C). Show that the pull-back of the line bundle E0 from Example 3.2 under f is trivial, but E0 itself is not trivial. Solution. Let us show that E0 is not trivial. Use problem 3.3. Suppose there exists a non-vanishing continuous section s : z → (z, a(z)) where 0 6= √ a(z) ∈ R z. We can normalize (divide by |a(z)|) and assume that |a(z)| = 1 and also (change to −a(z) if necessary) that a(1) = 1. Then we must have a(eit) = eit/2 for 0 ≤ t ≤ 2π, which gives a contradiction (there is no continuous √ branch of z on S1). If we pull back E0 under f : z → z2 we obtain a bundle f ∗E0 = {(z, y) ∈ S1 × C : y ∈ Rz}. This bundle is trivial, for instance {(z, z)} is obviously a non-vanishing section.

(13) TM is Orientable Prove that the tangent bundle TM of a manifold M is an orientable manifold. Solution. Take any two charts of TM intersecting at p. The derivative of the transition map consists of two equal blocks: one for the coordinates in M and

one for the coordinates in TpM. Therefore the determinant of this matrix is a square of a nonzero real number, hence positive. (14) Orientable Double Cover For a connected differentiable manifold M define M ∗ to be the set of pairs (p, O), where p ∈ M and O is an orientation of M near x. Then for every open, oriented and connected subset U of M that is the domain of a chart φ : U → V ⊂ Rk , ∗ ± ± we define charts φ± = φ ◦ pr1 : U × {O } → V , where O are the two possible orientations of U. Show that M ∗ has a natural orientation, and is connected if and only if M is not orientable. Solution. For any chart φ : U × {O} of M ∗ from the definition above define orientation to be the same as the corresponding orientation O± of M. Then if two charts U, U 0 intersect at some point (p, O±) that means that the orientation on both coincides. Therefore any two intersecting charts are coherent (i.e. the transition map is orientation-preserving). If M has an orientation O, then M ∗ is the disjoint union of M × {O} and M × {O0} (where O0 is the opposite orientation), so M ∗ is not connected. On the other hand, if M ∗ is not connected, then we can restrict π : M ∗ → M to one of the connected components; this gives a diffeomorphism between M and this component of M ∗. Since M ∗ (and therefore its components) is orientable, this implies that M is also orientable. (To see the claim about the diffeomorphism, let M 0 ⊂ M ∗ be a connected component. Consider the function s : M → {0, 1, 2}, p 7→ #(π−1(p) ∩ M 0). If U is a sufficiently small connected neighborhood of p, then π−1(U) is a disjoint union of two diffeomorphic copies of U, and each of them is either contained in M 0 or disjoint to it. So the function s is locally constant, hence constant. Since M 0 6= ∅,M ∗, s must be constant 1.)

(15) Derivations Let M be a manifold. A derivation on C∞(M) is an R-linear map D : C∞(M) → C∞(M) satisfying the Leibniz Rule D(fg) = (Df)g + f(Dg). Show that every vector field X : M → TM gives rise to a derivation

DX : f 7−→ p 7→ Xp(f)

(considering Xp = X(p) as a derivation at p), and that every derivation arises in this way.

Solution. Clearly DX maps differentiable functions on M to functions on M. Linearity and the Leibniz Rule are satisfied at every point p by definition of ∞ Xp : C (p) → R, therefore they are satisfied in general. The only thing to check ∞ ∞ here is that DX (f) is C for any C function f. But in any chart φ we have −1 ∞ −1 that DX (f)(p) = Dp(f ◦ φ ) · φ∗Xp which is a C function because f ◦ φ ∞ and φ∗Xp are C . The other direction goes exactly the same way. If D is a ∞ derivation on C (M) then Dp : f → D(f)(p) is a derivation at point p, i.e. a tangent vector. To see that the vector field defined this way is C∞, it suffices to k ∞ work on R (use charts), in which case it follows from the fact that D(xj) is C for the coordinate functions. (16) Riemannian Manifolds A Riemannian manifold (M, g) is a manifold M equipped with a Riemannian metric g. This associates to each p ∈ M an inner product on the tangent

space TpM that varies smoothly with p. In other words, for any chart with

domain U ⊂ M, the inner product gp (which can be identified with a positive ∼ k definite symmetric matrix, by the identification of TM|U = U × R ) is a differ- entiable function of p. Prove that any manifold M admits a Riemannian metric. Solution. Clearly any subset of Rn carries a Riemannian structure. Cover M by charts {φi : Ui → Vi} and consider the Riemannian structure gi on each n Ui pulled back from R by the chart maps. Take a partition of unity {φi} P subordinate to {Ui}. Then i giφi is a Riemannian structure on the whole M. Note that a linear combination with positive coefficients of positive definite symmetric matrices is again a positive definite symmetric matrix.

(17) Embeddings Let M be a compact k-dimensional manifold.

(a) Show that there is a finite partition of unity (ψj)1≤j≤n on M such that the

support of each ψj is contained in the domain of a chart φj. (b) Show that the map

nk+n
F : M −→ R , x 7−→ ψ1(x)φ1(x), . . . , ψn(x)φn(x), ψ1(x), . . . , ψn(x) nk+n embeds M as a submanifold of R . (The maps ψjφj are extended by zero outside the domain of definition of φj.) I.e., F (M) is a submanifold M 0 of Rnk+n, and F : M → M 0 is a diffeomorphism. Solution. (a) Take a collection of charts covering M. We can find a finite subcollection which still covers M because M is compact. Take a partition of unity subordinate to this finite collection. (b) Clearly F is differentiable because each component of F is a product of

differentiable functions (φi need not be even continuous in our definition but

after we multiply it by ψi we ”smoothen” this out because φi is differentiable on

supp(ψi)). It is also injective because if F (m1) = F (m2) then ψi(m1) = ψi(m2) for all i and their sum is 1 so at least one of them is non-zero. Then the

corresponding φi(m1) = φi(m2) hence m1 = m2 because φi is a bijection. The only thing left is to prove that the derivative of F is invertible. If we consider F

on the domain of φi : Ui → Vi and pull it down to Vi, then part of it is given by −1 −1
x 7→ ψi(φi (x))x, ψi(φi )(x) . One can check that the derivative of this has −1 rank k if ψi(φi )(x) 6= 0. Since the supports of the ψi cover M, this proves that DF is invertible everywhere on M.