solutions of exercises in - university of minnesotakim00657/note/warner_sol.pdfsolutions of...

Solutions of Exercises in”Foundations of Differentiable Manifolds and Lie Groups”

by Frank W. Warner

Dongkwan Kim

Department of Mathematical Sciences

Seoul National University

Contents

1 Chapter 1 1

2 Chapter 2 6

3 Chapter 3 13

4 Chapter 4 18

5 Chapter 5 21

6 Chapter 6 26

1 Chapter 1

1. Let x = (x1, ..., xn+1). Then for x ∈ Sn − n, the map pn : Sd − n → Rn is defined as pn(x) =1

1−xn+1(x1, ..., xn). Similarly, if x ∈ Sn − s, then pn(x) = 1

1+xn+1(x1, ..., xn). Therefore each function is

continuous on its domain, respectively. Also on ps(Sn − n − p), pn ◦ p−1s : ps(Sn − n − p) → pn(Sn −

n − p) ⊂ Rn is defined as pn ◦ p−1s (y1, ..., yn) = 1

k (y1, ..., yn)(k = ∑ni=1 y2

i ), which is smooth since(0, ..., 0) /∈ ps(Sn − n− p). Similarly, ps ◦ p−1

n is also smooth. Therefore, the maximal collection includingthese coordinate systems is indeed a differentiable structure on Sd.

2. Let f (t) = t3. Then idR ◦ f−1(t) = t13 , which is not differentiable at t = 0, and it means F 6= F1.

However, if we set ϕ : (R, F )→ (R, F1) : t 7→ t13 , then idR ◦ ϕ ◦ f−1(t) = t, which is a diffeomorphism.

Therefore, (R, F ) and (R, F1) are diffeomorphic.

3. By Theorem 1.11, given an open cover {Uα}, there exists a partition of unity {ϕα} subordinate to the cover{Uα}. If we set Vα = ϕ−1

α (R \ {0}), then it is also an open cover since for every m ∈ M, ∑α ϕα(m) = 1,which means ϕα(m) 6= 0 for some α, and m ∈ Vα. Also, by definition of support, Vα ⊂ Uα. Therefore{Vα} is the refinement satisfying the condition.

4. Suppose two closed subsets C, D ⊂ M are given. Then by regularity, ∀c ∈ C, ∃open subset Uc, Vc ⊂M s.t. c ∈ Uc, D ⊂ Vc, andUc ∩ Vc = ∅. Then {Uc}c∈C ∪ M \ C is an open cover of M. Hence, byparacompactness, there exists a refinement of {Uc}c∈C ∪ M \ C which is locally finite. By discardingopen sets which does not intersect C, one can obtain open cover of C which is locally finite, say {Uβ}.Then, ∀d ∈ D, ∃Vd s.t. Vd ∩ Uβ 6= ∅ for only finite β’s, which we can label them as β1, ..., βn. Since {Uβ}is a refinement, one can find ci ∈ C for each βi s.t. Uβ ⊂ Uci . If we set Wd = Vd ∩

⋂ni=1 Vci , then one can

easily see that ∀β, Wd ∩ Uβ = ∅ and Wd is an open neighborhood of d. If we set U =⋃

Uβ, W =⋃

Wd,then each open set is a neighborhood of C and D respectively which does not intersect each other. SinceC and D are arbitrary, M is normal.

5. (a) If~a = (a1, ..., ad) and ~v = (v1, ..., vd), ψ ◦ ϕ−1(~a,~v) = (ψ ◦ ϕ−1(~a),J (ψ ◦ ϕ−1)(~v), which is C∞.(b) ϕ : π−1(U) → R2d is a bijection between π−1(U) and ϕ(U)×Rd, which is open in R2d. Also, for

any (V, ψ), ψ−1(W ′) ∩ π−1(U) is of the form ϕ−1(W), which is ϕ−1(ϕ ◦ ψ−1(ψ(U ∩ V)×Rd ∩W ′)).Therefore, on the defined topology, ϕ is a homeomorphism and that T(M) is a 2d-dimensional locallyEuclidean space directly follows. For second countability, since M is second countable, we can choosecountable such (U, ϕ) that cover T(M). Since R2d is also second countable, it follows that T(M) issecond countable.

(c) By (a) and (b), it directly follows.

6. By Theorem 1.30, Corollary (a), ψ is a diffeomorphism if and only if dψ is surjective everywhere. Supposethat dψ is not surjective at m ∈ M. Since it is nonsingular, dimM = p < dimN = d. Pick a coordinatesystem on N, (U, ϕ), such that ϕ(U) = Rd. Then ϕ ◦ ψ : ψ−1(U) → Rd is surjective. Also, by secondcountability of M, one can find countable coordinate systems of ψ−1(U), say {(Vα, fα)}α∈A. Then Rd =⋃

α∈A ϕ ◦ψ ◦ f−1α (Rp). Using Proposition 1.35, since ϕ ◦ψ ◦ f−1

α is an immersion, one can obtain countablecover {Uβ} of Rp such that ϕ ◦ ψ ◦ f−1

α (Uβ) is a subset of a slice of a open set in Rd, which is nowheredense. Therefore, by Baire category theorem,

⋃α∈A ϕ ◦ ψ ◦ f−1

α (Rp) cannot be all of Rd, contradicting

1

assumption. Therefore, ψ is a diffeomorphism.

7. (a) For two differentiable structures on A, F and F ′, idA : (A, F )→ (A, F ′) is continuous, hence C∞ byTheorem 1.32. Since its inverse is also C∞, it is a diffeomorphism: that is, two structures are the same.

(b) Let (A, ι′) be A equipped with a manifold structure possibly different from induced structure formM. Then idA : A → A is continuous and hence C∞ by Theorem 1.32. Also it is nonsingular sincedι′ = d(ι ◦ idA) is. Therefore, by Exercise 6, it is a diffeomorphism, which means that a manifoldstructure on A is equal to that on A.

8. Use notations of Theorem 1.37. Since { ∂ fi∂sj|(r0,s0)

} is nonsingular(i.e., invertible), by continuity, there

exists an open set (r0, s0) ∈ V0 ×W0 ⊂ U such that d f is surjective. Also, (r0, s0) ∈ f−1(0), hencenonempty. Therefore, by Theorem 1.38, P = f |−1

V0×W0(0) is an imbedded submanifold of V0 ×W0 of

dimension c-d. Now, let π1 : V0 ×W0 → V0 be the first projection. Then, π1 ◦ ι : P → V0 is C∞, andd(π1 ◦ ι) : P(r0,s0)

→ V0r0 is an isomorphism. (∵ P(r0,s0)can be identified with ker d f ⊂ V0 ×W0(r0,s0)

,and d f |W0s0

is an isomorphism, V0 ×W0(r0,s0)= ker d f ⊕Ws0 = V0r0 ⊕W0s0 . Hence ker d f → V0r0 is an

isomorphism also.) Therefore, by Theorem 1.30, one can find an open set (r0, s0) ∈ V ×W ⊂ V0 ×W0

such that π1 ◦ ι|−1P∩V×W : V → P ∩ V ×W is an one-to-one onto C∞ map. Therefore, if we set g =

π2 ◦ ι ◦ (π1 ◦ ι|−1P∩V×W) : V → W (where π2 is the second projection), then it is the map satisfying the

condition.

9. ∂ f∂x = 3x2 + y, ∂ f

∂y = x + 3y2. Therefore, ∂ f∂x = ∂ f

∂y = 0 if and only if (x, y) = (0, 0), (− 13 ,− 1

3 ). There-

fore, by Theorem 1.38, f−1( f ( 13 , 1

3 )) is an imbedded submanifold, since (0, 0), (− 13 ,− 1

3 ) /∈ f−1( f ( 13 , 1

3 )).f−1( f (0, 0)) is not a manifold since (0,0) is a node. If p = (− 1

3 ,− 13 ), then f (x, y) = f ( 1

3 , 13 ) iff x3 + xy +

y3 − 127 = 0 iff (x + y− 1

3 )(x2 + y2 + 19 +

x3 + y

3 − xy) = 0 iff x + y = 13 , since x2 + y2 + 1

9 +x3 + y

3 − xy =12 ((x + 1

3 )2 + (y + 1

3 )2 + (x− y)2). Therefore f−1( f (− 1

3 ,− 13 )) is a line, which is an imbedded submani-

fold.

10. Since f (M) is compact, ∃m ∈ M s.t. | f (m)| is the biggest. If d f : Mm → Rnf (m)

is nonsingular, then byTheorem 1.30, Corollary (a), ∃U ⊂ f (M) neighborhood of f (m), which contradicts the maximality of| f (m)|.

11. Not closed : (−π2 , π

2 ) ⊂ R, g(x) = tan xNot imbedded : Second Example of 1.28, g(x) = xTo prove the converse : if there is a disk D(x, r) ⊂ Rn, one can make a C∞ function which is nonzeroexactly on D(x, r) and zero elsewhere. Since M is locally Euclidean, there exists a basis {Bα} of M suchthat ∀Bα, ∃gα s.t. g−1

α (R \ {0}) = Bα. Then by the assumption there is a C∞ function fα on N such thatfα ◦ ψ = gα. Since f−1

α (R \ {0}) ∩ ψ(M) = ψ(Bα), ψ(Bα) is open in ψ(M) with relative topology, whichmeans ψ is an imbedding.Therefore we can say M ⊂ N. If M is not closed in N, then there is n ∈ M \M and (U, ϕ) a coordinatesystem of n, and there is a C∞ function g on M such that g(m) = 1

ϕ(m)−ϕ(n) on the neighborhood of non M. However, since n is a limit point of M, it cannot be extended to all of N. Therefore, M should beclosed in N.

12. (a) Since d f = ∑ 2ridri, it is surjective except at the origin. Therefore one can use Theorem 1.38. Also,

2

since it is a submanifold with the relative topology, it should be the same as 1.5(d) by 1.33(b).(b) By following the argument, we need only to show that dψI is surjective. Since ψ(A) = (∑k aikajk)ij,

dψA = (∑k(ajkdaik + aikdajk))ij. ∴ dψI = (daij + daji)ij. But, since Gl(n, R)I ∼= Mn×n(R), it issurjective onto Symn(R)I ∼= Symn(R).

13. (a) If we set X = ∑ ai∂

∂xi, Y = ∑ bi

∂∂xi

locally as Proposition 1.43(b), [X, Y] = ∑i ∑j(ai∂bj∂xi− bi

∂aj∂xi

) ∂∂xj

.Therefore by 1.43(b), it is a smooth vector field.

(b) Using the same notation above,

[ f X, gY] = ∑i

∑j( f ai

∂gbj

∂xi− gbi

∂ f aj

∂xi)

∂

∂xj

= ∑i

∑j( f gai

∂bj

∂xi− f gbi

∂aj

∂xi+ f aibj

∂g∂xi− gajbi

∂ f∂xi

)∂

∂xj

= f g[X, Y] + f (Xg)Y− g(Y f )X

(c) [X, Y] = ∑i ∑j(ai∂bj∂xi− bi

∂aj∂xi

) ∂∂xj

= −∑i ∑j(bi∂aj∂xi− ai

∂bj∂xi

) ∂∂xj

= −[Y, X]

(d) Let Z = ∑ ci∂

∂xi. Then,

[[X, Y], Z] = ∑i

∑j

∑k((ai

∂bj

∂xi− bi

∂aj

∂xi)

∂ck∂xj

−cj(∂ai∂xj

∂bk∂xi

+ ai∂2bk

∂xj∂xi− ∂bi

∂xj

∂ak∂xi− bi

∂2ak∂xj∂xi

))∂

∂xk

and by direct calculation, Jacobi Identity can be proved.

14. No. If Xx = exp(x) ∂∂r and γ0(t) is the integral curve at 0 ∈ R, then by definition, dγ0

dt (t) = exp(γ0(t))and γ0(0) = 0. By solving the differential equation, γ0(t) = − log(−t + 1), and it is not defined on(−∞, ∞).

15. Use the equation of Exercises 13(a) above.

16. Consider the first example of 1.31 and a curve along the horizontal part of the submanifold. It is clearthat on 0 ∈ R2, the tangent vector of the integral curve should be horizontal, whereas N0 consists ofvertical vectors.

17. By 1.48(f),⋃

t>0Dt =⋃

t<0Dt = M, thus by compactness of M, ∃t0 > 0 s.t. Dt0 = D−t0 = M. Thusby 1.48(g), Xt0 is an endomorphism, and ∀n ∈ N, Xn

t0is defined on all of M. Therefore by 1.48(h),

∀N > 0,DN = M. Likewise, ∀N < 0,DN = M. Since it is clear that D0 = M, M is complete.

18. If it is one-to-one, the image should be infinite. Therefore, there exist a, b, c ∈ im f s.t. a < b < c.By erasing b, a and c can be separated by open sets (−∞, b) ∩ im f and (b, ∞) ∩ im f , hence im f isdisconnected, whereas R2 \ f−1(b) is connected since f−1(b) consists of one point. This contradicts

3

continuity of f .

19. (1.60⇐ 1.61) Follow the remark of 1.61. Then,

[Ya, Yb] =d−c

∑j=1

(∂ f jb ◦ γ

∂ya−

∂ f ja ◦ γ

∂yb

)∂

∂yc+j+

d−c

∑j=1

d−c

∑k=1

(∂ f jb ◦ γ

∂yc+kfka −

∂ f ja ◦ γ

∂yc+kfkb

)∂

∂yc+j

Since D is involutive, [Ya, Yb] ∈ D. But the expression above does not include ∂∂yi

terms for 1 ≤ i ≤ c,it must be zero. Therefore, the assumption 1.61(2) is satisfied. Then we obtain a map α as 1.61(3).Since a(r0, s) = s, (id, α)−1 is C∞ local homeomorphism, so shrinking the neighborhood if necessary,we obtain a local chart ϕ = (id, α)−1 ◦ γ : W → U × V and (z1, ..., zd) = (id, α)−1 ◦ (y1, ..., yd) iscoordinate functions. Also, if 1 ≤ i ≤ c, ∂

∂zi= ∑d

j=1∂yj∂zi

∂∂yj

= ∂∂yi

+ ∑d−cj=1 f ji ◦ γ ∂

∂yc+j= Yi by 1.61(4) and

(6). Therefore, if we let I be a submanifold defined by zj=constant for c + 1 ≤ j ≤ d, it is an integralmanifold of D.(1.60⇒ 1.61) Let Xi =

∂∂ri

+ ∑nj=1 bji

∂∂sj

for 1 ≤ i ≤ n and D be a distribution generated by Xi’s. Thenby similar calculation above, [Xa, Xb] = 0 ∈ D by assumption 1.61(2). Thus by 1.60, there exists a cubiccoordinate system (W, ϕ) with coordinate functions (x1, ..., xm, y1, ..., yn) defined as the theorem. Letus define α on the neighborhood of (0, 0) s.t. yj(r, αs(r)) = yj(r0, s)∀1 ≤ j ≤ n by implicit functiontheorem. We can define this because X1, ..., Xm, ∂

∂s1, ..., ∂

∂s1is a basis of tangent space and Xi(yj) = 0

since yj’s are constant on integral manifolds of D, which means { ∂yj∂sk} should be nonsingular. Then

clearly αs(r0) = s and∂yj∂ri

+ ∑nk=1

∂yj∂sk

∂αs,k∂ri

= 0. Also, Xi(yj) = 0 means∂yj∂ri

+ ∑nk=1 bki

∂yj∂sk

= 0. Since { ∂yj∂sk}

is nonsingular, condition (4) is satisfied.

20. Let M = R, N = (−π2 , π

2 ) and consider N as a submanifold imbedded in M and ϕ as an inclusion. IfX(t) = tan(t) ∂

∂r , then it satisfies the assumption since ϕ is one-to-one, but it cannot be extended to anyvector field on M.

21. It is clearly one-to-one, C∞, and nonsingular. To simplify the calculation, let us think the torus asT = R2�Z×Z and ϕ : R → T : x 7→ (x, αx). Then imϕ ∩ {(0, y) ∈ T} is dense in {(0, y) ∈ T}, sinceα is irrational. Also, given x0, imϕ ∩ {(x0, y) ∈ T} is dense in {(x0, y) ∈ T} since it is the transition ofimϕ ∩ {(0, y) ∈ T}. Since x0 is arbitrary, the image of ϕ is dense in T.

22. Let t0 be a point satisfying γ(t0) = 0. Then, the constant map whose image is γ(t0) is also an integralcurve. Therefore, by Theorem 1.48(c), γ should be a constant map.

23. Let {Uα} be an open cover of M such that ∀α, (Uα, ψα) is a chart. By Theorem 1.11, there exists a partitionof unity {ϕα} subordinate to the cover. Also, on each chart, there exists a positive definite inner productinduced by the standard inner product on the Euclidean space, say < , >α, and it is smooth on eachchart. If we set < , >m= ∑α < , >α,m ϕα(m), then it is the inner product which makes M becomeRiemannian manifold.

24. (a) (⇒) Obvious.(⇐) ∀(U, ϕ), ∀(V, ψ) charts of M and N respectively, (U × V, ϕ × ψ) is a chart of M × N. Then(ϕ× ψ) ◦ α = (ϕ ◦ π1 ◦ α, ψ ◦ π2 ◦ α) is C∞ if π1 ◦ α, π2 ◦ α are C∞.

(b) Let m ∈ U, n ∈ V, and (x1, ..., xs), (y1, ..., yt) be local coordinates of U, V respectively. Then (M ×

4

2 Chapter 2

1. (a) Let l : V×W → U be a bilinear map. Then by the universal property of free module, l : F (V×W)→U is well defined, and by the bilinearity, generators of 2.1(1) are in the kernel of l. Therefore, the mapl : V ⊗W → U is well defined. Also, it satisfies l ◦ ϕ = l by the definition of the map. Since l(v⊗w) isdetermined by the value of l and elements of the form v⊗ w generate V ⊗W, such l must be unique.The rest argument is directly followed by the properties of universal objects.

(b) Since V ×W ∼= W ×V, it is true because of the universal property.(c) For v ∈ V, let fv : W ×U → (V ×W)×U → (V ⊗W)×U → (V ⊗W)⊗U be the map (w, u) 7→

((v, w), u) 7→ (v⊗w, u) 7→ (v⊗w)⊗ u). Since it is bilinear, one can define fv : W⊗U → (V⊗W)⊗U.Then, if we set g : V × (W ⊗U)→ (V ⊗W)⊗U : (v, x) 7→ fv(x) for all x ∈ W ⊗U, it is also bilinear,hence one can define g : V ⊗ (W ⊗U) → (V ⊗W)⊗U, and g(v⊗ (w⊗ u)) = (v⊗ w)⊗ u). By thesame method, one can also define h : (V ⊗W)⊗U → V ⊗ (W ⊗U), and it is easy to see that g and hare inverses to each other. Therefore, V ⊗ (W ⊗U) ∼= (V ⊗W)⊗U.

(d) Let f : V∗ ×W → Hom(V, W) be a function such that f (ϕ, w)(v) = ϕ(v) · w. Since it is bilinear,one can define a map α : V∗ ⊗W → Hom(V, W). (1) Let {vi}, {wj} be each basis of V, W,respectively, and {v∗i } be a dual basis corresponding to {vi}, i.e., v∗a(vb) = δab. Then elements ofV∗⊗W can be represented as ∑i,j aijv∗i ⊗wj. If α(∑i,j aijv∗i ⊗wj) = 0, then ∀v ∈ V, ∑i,j aijv∗i (v) ·wj =

0. Especially, if one substitutes vi for v, then ∑j aijwj = 0∀i. Therefore, by linear independence ofbasis, aij = 0∀i, j, and it means α is injective.

(2) Let ϕ ∈ Hom(V, W). Then if ϕ(vi) = ∑j aijwj, set ψ = α(∑i,j aij ⊗ wj). Then ψ and ϕ has the samevalue at vi, so it must be the same linear map. Therefore, α is surjective.

Therefore, α is an isomorphism. Hence, dim V∗ ⊗W = dim Hom(V, W) = (dimV)(dimW), anddim V ⊗W = dim V∗∗ ⊗W = (dimV∗)(dimW) = (dimV)(dimW).

(e) By arguments similar to the above, {ei ⊗ f j} span V ⊗W. Since the number of {ei ⊗ f j} is the same asthe dimension of V ⊗W by (d), it must be a basis.

2. (a) Let V = R2, and consider e1⊗ e2− e2⊗ e1 ∈ V⊗V. If it is decomposable, say (ae1 + be2)⊗ (ce1 + de2),then ac = bd = 0, ad = 1, bc = −1, which is impossible.

(b) If the dimension is 1 or 2, then it is trivially true. Let dimV = 3, V =< v1, v2, v3 >. Then we only needto check that all elements of Λ2(V) are decomposable. However for any element x = av1 ∧ v2 + bv2 ∧v3 + cv3 ∧ v1 ∈ Λ2(V), if a = 0, then x = (bv2− cv1)∧ v3, and a 6= 0, then x = (v1− b

a v3)∧ (av2− cv3).Therefore x is always decomposable.

(c) If V = R4, then e1 ∧ e2 + e3 ∧ e4 is indecomposable. If it is decomposable, say (a1e1 + a2e2 + a3e3 +

a4e4) ∧ (b1e1 + b2e2 + b3e3 + b4e4), then a1b2 − b1a2 = a3b4 − b3a4 = 1, a1b3 − b1a3 = a1b4 − b1a4 =

a2b3 − b2a3 = a2b4 − b2a4 = 0. If a1 = 0, b1a2 = −1, b1a3 = b1a4 = 0, so a3 = a4 = 0, which contradictsa3b4 − b3a4 = 1. If a1 6= 0, b3 = b1a3

a1, b4 = b1a4

a1, which also contradicts a3b4 − b3a4 = 1.

(d) No. Let α = e1 ∧ e2 + e3 ∧ e4 ∈ Λ2(R4). Then α ∧ α = 2e1 ∧ e2 ∧ e3 ∧ e4 ∈ Λ4(R

4), which is not zero.

3. (a) That u ∧ v ∈ λk+l(V) is a direct consequence of associativity of tensor products. Since the lastequation is bilinear, we only need to show if u and v are decomposable. However, if u = u1 ∧ ... ∧ uk

and v = v1 ∧ ... ∧ vl , v ∧ u = v1 ∧ ... ∧ vl ∧ u1 ∧ ... ∧ uk = (−1)lu1 ∧ v1 ∧ ... ∧ vl ∧ u2 ∧ ... ∧ uk = ... =(−1)klu1 ∧ ...∧ uk ∧ v1 ∧ ...∧ vl = (−1)klu ∧ v.

(b) Let us follow the argument on the text. Since an arbitrary element of I(V) has zero determinant,e1 ⊗ ...⊗ en cannot be in I(V), i.e., e1 ∧ ... ∧ en 6= 0. That {eΦ} span Λ(V) is obvious. If ∑Φ aΦeΦ = 0,

6

since homogeneous parts should be zero, ∀0 ≤ k ≤ n, ∑|Φ|=k aΦeΦ = 0. However, for some Φ0, ifwe wedge eΦc

0on the equation, then we get aΦ0 e1 ∧ ... ∧ en = 0, hence aΦ0 = 0. Since Φ0 is arbitrary,

it means {eΦ} is linearly independent. Therefore, {eΦ} is a basis. The rest arguments are triviallyfollowed.

(c) h can be defined because of the universal property of tensor products and I(V) should be in the kernelof the induced map since h is alternating. Its uniqueness is also followed by the universal propertyof tensor products. If W = R, since h ∈ Λk(V)∗ and h ∈ Ak(V), ϕ∗ : Λk(V)∗ → Ak(V) gives anisomorphism.

4. Since (2), (3), and (4) are bilinear forms, we only need to consider when f and g are both induced bydecomposable elements. Therefore if ϕ : Λ(V)∗ → A(V) is the isomorphism, and f = ϕ ◦ α(w∗1 ∧ ... ∧w∗p), g = ϕ ◦ α(w∗p+1 ∧ ...∧ w∗p+q). Then

f ∧α g(v1, ..., vp+q) = w∗1 ∧ ...∧ w∗p ∧ w∗p+1 ∧ ...∧ w∗p+q(v1 ∧ ...∧ vp+q)

= det(w∗i vj)1≤i,j≤p+q

= ∑σ∈Sp+q

sgn(σ)p+q

∏i=1

w∗i vσ(i)

= ∑π:p,qshu f f les

∑σ1∈Sp

∑σ2∈Sq

sgn(π)sgn(σ1)sgn(σ2)p

∏i1=1

w∗i1 vπ(σ1(i1))

q

∏i2=1

w∗p+i2 vπ(p+σ2(i2))


sgn(π)det(w∗i vπ(i))1≤i≤p det(w∗j vπ(j))p+1≤j≤p+q


sgn(π) f (vπ(1), ..., vπ(p))g(vπ(p+1), ..., vπ(p+q))

On the other hand, let f = ϕ ◦ β(w∗1 ∧ ...∧ w∗p), g = ϕ ◦ β(w∗p+1 ∧ ...∧ w∗p+q). Then

f ∧β g(v1, ..., vp+q) = w∗1 ∧ ...∧ w∗p ∧ w∗p+1 ∧ ...∧ w∗p+q(v1 ∧ ...∧ vp+q)

=1

(p + q)!det(w∗i vj)1≤i,j≤p+q

=1

(p + q)! ∑σ∈Sp+q

sgn(σ)p+q

∏i=1

w∗i vσ(i)

=1

(p + q)! ∑π:p,qshu f f les

∑σ1∈Sp

∑σ2∈Sq

sgn(π)sgn(σ1)sgn(σ2)p

∏i1=1

w∗i1 vπ(σ1(i1))

q

∏i2=1

w∗p+i2 vπ(p+σ2(i2))

=1

(p + q)! ∑π:p,qshu f f les


=p!q!

(p + q)! ∑π∈Sp+q


=1

(p + q)! ∑π∈Sp+q


7

Therefore,

f ∧α g(v1, ..., vp+q) = ∑π:p,qshu f f les


=1

p!q! ∑π∈Sp+q


=(p + q)!

p!q!f ∧β g

5. (where γ is an integral curve of X)

LX( f )|m = limt→0

δXt( fXt(m))− fm

t= lim

t→0

fXt(m) − f (m)

t

= limt→0

f (γm(t))− f (m)

t= dγ(

ddt|0)( f )

= Xm( f )

6. (Since√

t is defined on [0, ε), the limit should be calculated when t → 0+.) We will use L’Hopital’s rule.

8

Let g(x, y, z, w) = f (Y−xX−yYzXw)(m). Then,

∂g∂x

(x, y, z, w) = −YY−xX−yYzXw(m)( f )

∂g∂y

(x, y, z, w) = −XX−yYzXw(m)( f ◦Y−x)

∂g∂z

(x, y, z, w) = YYzXw(m)( f ◦Y−x ◦ X−y)

∂g∂w

(x, y, z, w) = XXw(m)( f ◦Y−x ◦ X−y ◦Yz)

∂2g∂x2 (x, y, z, w) = YY−xX−yYzXw(m)(Y( f ))

∂2g∂y2 (x, y, z, w) = XX−yYzXw(m)(X( f ◦Y−x))

∂2g∂z2 (x, y, z, w) = YYzXw(m)(Y( f ◦Y−x ◦ X−y))

∂2g∂w2 (x, y, z, w) = XXw(m)(X( f ◦Y−x ◦ X−y ◦Yz))

∂2g∂x∂y

(x, y, z, w) = XX−yYzXw(m)(Y( f ) ◦Y−x)

∂2g∂x∂z

(x, y, z, w) = −YYzXw(m)(Y( f ) ◦Y−x ◦ X−y)

∂2g∂x∂w

(x, y, z, w) = −XXw(m)(Y( f ) ◦Y−x ◦ X−y ◦Yz)

∂2g∂y∂z

(x, y, z, w) = −YYzXw(m)(X( f ◦Y−x) ◦ X−y)

∂2g∂y∂w

(x, y, z, w) = −XXw(m)(X( f ◦Y−x) ◦ X−y ◦Yz)

∂2g∂z∂w

(x, y, z, w) = XXw(m)(Y( f ◦Y−x ◦ X−y) ◦Yz)

9

Therefore,

limt→0+

f (β(t))− f (β(0))t

= limt→0+

f (β(t2))− f (β(0))t2

= limt→0+

g(t, t, t, t)− g(0, 0, 0, 0)t2

= limt→0+

dg(t, t, t, t)/dt2t

= limt→0+

d2g(t, t, t, t)/dt2

2

(∵ dg(t, t, t, t)/dt|t=0 = −Ym( f )− Xm( f ) + Ym( f ) + Xm( f ) = 0)

=12(Ym(Y( f )) + Xm(X( f )) + Ym(Y( f )) + Xm(X( f )))

+Xm(Y( f ))−Ym(Y( f ))− Xm(Y( f ))−Ym(X( f ))

−Xm(X( f )) + Xm(Y( f ))

= Xm(Y( f ))−Ym(X( f )) = [X, Y]|m( f )

7. For m ∈ M, let x1, ..., xn be local coordinates of m. Since the equation is linear with respect to ω, it isenough to consider the case when ω is decomposable, i.e., ω = gdxm1 ∧ ...∧ dxmp . Then,

LY0(ω(Y1, ..., Yp)) = Y0(g det(Y1(xmi ), ..., Yp(xmi ))1≤i≤p)

= Y0(g)det(Y1(xmi ), ..., Yp(xmi ))1≤i≤p

+p

∑j=1

g det(Y1(xmi ), ..., Y0(Yj(xmi )), ..., Yp(xmi ))1≤i≤p

LY0(ω)(Y1, ..., Yp) = Y0(g)det(Y1(xmi ), ..., Yp(xmi ))1≤i≤p

+p

∑j=1

g det(Y1(xmi ), ..., Yj(Y0(xmi )), ..., Yp(xmi ))1≤i≤p

ω(Y1, ..., LY0Yj, ..., Yp) = det(Y1, ..., Y0(Yj(xmi )), ..., Yp)1≤i≤p − det(Y1, ..., Yj(Y0(xmi )), ..., Yp)1≤i≤p

(e) follows by combining these equations.

8. (⇒) ∀(m, n) ∈ M × N, ∀Y2, ..., Yp ∈ (M × N)(m,n), i(X)ω(m,n)(Y2, ..., Yp) = i(X)δπ(α)(m,n)(Y2, ..., Yp) =

δπ(α)(m,n)(X(m,n), Y2, ..., Yp) = αm(dπ(X(m,n)), dπ(Y2), ..., dπ(Yp)) = 0. Also, i(X)dω(m,n)(Y2, ..., Yp) =

i(X)d(δπ(α))(m,n)(Y2, ..., Yp) = d(δπ(α))(m,n)(X(m,n), Y2, ..., Yp) = δπ(dα)(m,n)(X(m,n), Y2, ..., Yp) =

dαm(dπ(X(m,n)), dπ(Y2), ..., dπ(Yp)) = 0. Since LX = i(X) ◦ d + d ◦ i(X), LXω = 0.(⇐) Let (m, n) ∈ M × N, and choose local coordinates of m ∈ M, n ∈ N, say {x1, ..., xs}, {y1, ..., yt},such that local coordinates of (m, n) ∈ M× N consist of {x1, ..., xs, y1, ..., yt}. Then, ω can be describedas ∑|Φ|+|Ψ|=p gΦ,ΨdxΦ ∧ dyΨ. And we can take a vector field X on M × N as ∂

∂xi× (C∞ function which

is zero outside the local chart and 1 at (m, n)). Then X(m,n) = ∂∂xi|(m,n) and dπ(X) = 0 Therefore, by

assumption, LXω = 0. But, LXω = ∑|Φ|+|Ψ|=p∂gΦ,Ψ

∂xidxΦ ∧ dyΨ. Therefore ∀xi,

∂gΦ,Ψ∂xi

= 0. If we letιm′ : N → M× N : n′ 7→ (m′, n′), (διmω)n = ∑|Ψ|=p g∅,Ψ(m, n)dyΨ. Therefore, by the result above, on

10

the small neighborhood of m ∈ M and for all m′ in the neighborhood, (δι′mω)n is the same as (διmω)n.Since (m, n) is arbitrary, we can say that διmω is locally the same regardless of the choice of m. Also, sinceM is connected, we conclude that διmω is always the same regardless of the choice of m ∈ M. Let α bethis p-form on N. Then, ω = δπ(α). Indeed, for all (m, n) ∈ M× N, and X1, ..., Xp ∈ (M× N)(m,n),

(ω− δπ(α))(m,n)(X1, ..., Xp) = ω(m,n)(X1, ..., Xp)− δπ(διmω)(m,n)(X1, ..., Xp)

= ω(m,n)(X1, ..., Xp)−ω(m,n)(d(ιm ◦ π)(X1), ..., d(ιm ◦ π)(Xp))

= ω(m,n)(X1 − d(ιm ◦ π)(X1), ..., Xp − d(ιm ◦ π)(Xp))

The last term is zero since if we set X as a vector field of M× N such that X(m,n) = X1 − d(ιm ◦ π)(X1)

and dπX = 0, (it is possible since dπ(X1 − d(ιm ◦ π)(X1)) = 0, the last term is (i(X)ω)(m,n)(X2 − d(ιm ◦π)(X2), ..., Xp − d(ιm ◦ π)(Xp)) which should be zero by the assumption.

9. (⇐) If a1v1 + ... + arvr = 0 and some of ai’s are nonzero, then without loss of generality, one can assumea1 6= 0. Dividing the equation by a1, one can also assume a1 = 1. Then v1 ∧ ...∧ vn = −a2v2− ...− arvr ∧v2 ∧ ...∧ vr = 0.(⇒) Suppose v1, ..., vp are linearly independent, then there is a linear map ϕ : V → kr : vi 7→ ei. Byuniversal property, it induces a map Λr(ϕ) : Λr(V)→ Λr(kr). Since ϕ(v1 ∧ ...∧ vp) = e1 ∧ ...∧ ep 6= 0 by2.6, v1 ∧ ...∧ vp 6= 0.

10. (⇒) Let vi = ∑j aijwj. Then by direct calculation, v1 ∧ ...∧ vr = (det A)w1 ∧ ...∧ wr.(⇐) If w ∈< w1, ..., wr > \ < v1, ...vr >, v1 ∧ ... ∧ vr ∧ w 6= 0 by Exercise 9, which contradicts w1 ∧ ... ∧

wr ∧ w = 0. ∴< w1, ..., wr >⊂< v1, ...vr > and vice versa.

11. (Condition⇒ (a)) Since dωi ∈ I by definition of differential ideals.((a)⇒ (b)) dω = dω1 ∧ ...∧ωr + ... + (−1)r+1ω1 ∧ ...∧ dωr = α ∧ω by using (a).((b)⇒ (a)) dω = dω1 ∧ ... ∧ ωr + ... + (−1)r+1ω1 ∧ ... ∧ dωr = α ∧ ω. Since ω ∧ ωi = 0, by wedging ωi

on the equation, we get dωi ∧ω = 0. Therefore, dωi ∈< ω1, ..., ωr >.((a)⇒ condition) d < ω1, ..., ωr >⊂< dω1, ..., dωr, ω1, ..., ωr > by property of derivations, and the resultdirectly follows.

12. For another basis w1, ..., wn, let wi = ∑j aijvj. Then Aw1 ∧ ... ∧ Awn = ∑j a1j Avj ∧ ... ∧ ∑j anj Avj =

det(aij)Av1 ∧ ... ∧ Avn. Therefore by Exercise 10, det A does not depend on the choice of basis. Ifwe choose standard basis e1, ..., en, the equation of det A follows directly. Also, for two matrices A, B,Bv1 ∧ ...∧ Bvn = (det B)v1 ∧ ...∧ vn, so ABv1 ∧ ...∧ ABvn = (det B)Av1 ∧ ...∧ Avn = (det B)(det A)v1 ∧...∧ vn, which proves det A det B = det AB.

13. The orthonormality comes from direct calculations.(5) Since ∗∗ is linear, we only need to show this property for the basis elements. If Φ = {i1, ..., ip} ⊂{1, ..., n} and let f be the element which satisfies eΦ ∧ f = e1 ∧ ...∧ en, then f ∧ eΦ = (−1)p(n−p)e1 ∧ ...∧en. ∴ ∗ ∗ eΦ = ± ∗ f = (−1)p(n−p)eΦ.(6) Since the equation is bilinear to v and w, we only need to show this for the basis elements. Thereforewe can assume that v = eΦ1 , w = eΦ2 . If v 6= w, < v, w >= ∗(w ∧ ∗v) = ∗(v ∧ ∗w) = 0 by definition ofstar and inner product. If v = w, < v, w >= ∗(w ∧ ∗v) = ∗(v ∧ ∗w) = 1 also by definition. Hence (6) is

11

proved.

14. It suffices to show that for all v ∈ Λp+1(V) and w ∈ Λp(V), < γ(v), w >=< v, ξ ∧ w >. But it isequivalent to(using Exercise 13):

(−1)np < w, ∗(ξ ∧ ∗v) >=< ξ ∧ w, v >

⇔ (−1)np ∗ (w ∧ (−1)p(n−p)ξ ∧ ∗v) =< ξ ∧ w, v >

⇔ (−1)p ∗ (w ∧ ξ ∧ ∗v) =< ξ ∧ w, v >

⇔ ∗(ξ ∧ w ∧ ∗v) =< ξ ∧ w, v >

which is true by Exercise 13.

15. (It is true only if ξ 6= 0) Obviously (ξ∧)2 = 0. Conversely, if ξ ∧ w = 0, then w = ξ ∧ γ( w<ξ,ξ> ).

To prove this, it is enough to show that for all v ∈ Λp+1(V), < ξ ∧ γ(w), v >=< ξ, ξ >< w, v >.But < ξ ∧ γ(w), v >=< γ(w), γ(v) >=< (−1)np ∗ (ξ ∧ ∗w), (−1)np ∗ (ξ ∧ ∗v) >= ∗(∗(ξ ∧ ∗w) ∧(−1)p(n−p)ξ ∧ ∗v) = ∗(ξ ∧ ∗v ∧ ∗(ξ ∧ ∗w)) =< ξ ∧ ∗v, ξ ∧ ∗w >. If we consider basis {vi} of Vwhich contains ξ and an expression of w in terms of the basis derived by {vi}, say ∑|I|=p+1 aIvI whereI = {vi1 , ..., vip+1} ⊂ {vi} and vI = vi1 ∧ ... ∧ vip+1 , (to clarify the definition, it is assumed that ordersof elements of all finite subset of {vi} are determined, possibly by using Axiom of Choice if necessary.)Since ξ ∧w = 0, it follows that all I which aI 6= 0 should contain ξ. Therefore, considering an expressionof ∗w derived by w, we see that there is no ξ in the expression of ∗w. It means that if calculating< ξ ∧ ∗v, ξ ∧ ∗w > by splitting matrices and getting determinants, the first column of matrices shouldbe (< ξ, ξ >, 0, ..., 0). It means that < ξ ∧∗v, ξ ∧∗w > should be < ξ, ξ >< ∗v, ∗w >=< ξ, ξ >< v, w >.

16. ∑i θi ∧ ωi = 0. If we wedge ω1 ∧ ... ∧ ωi ∧ ... ∧ ωp, θi ∧ ω1 ∧ ... ∧ ωp = 0. ∴ θi ∈< ω1, ..., ωp >. So it ispossible to write θi = ∑j Aijωj, where Aij are C∞ functions. One can derive Aij = Aji by just puttingthese expressions into the given equation and noting that ωi ∧ωj’s are linearly independent.

12

3 Chapter 3

1. Let e ∈ U ⊂ G be a locally Euclidean open neighborhood of e. Then by Theorem 3.18, G =⋃∞

n=1 Un.Now, fix a countable basis of U, say {Bi}. Then for all open set V ⊂ Un and for all points g ∈ V, g can beexpressed as g = g1...gn by definition. Also, since multiplication on G is continuous, we can find openneighborhoods of gk, say Vk, such that V1...Vn ⊂ V. Then, we can find Bik ’s such that gk ∈ Bik ⊂ Vk, andas a result g = g1...gn ∈ Bi1 ...Bin ∈ V. Therefore it follows from this that {Bn

i } becomes a basis of Un andUn is second countable, and so is G.

2. (a) It is definitely a differential manifold. Also, subtraction is continuous under standard topology.(b) It is also almost obvious.(c) Since it is a subgroup of C∗, and zero set of |z| − 1, which is nonsingular.(d) It is also a manifold, and operation is continuous by definition of product topology.(e) It directly follows from (c) and (d).(f) Since matrix multiplication is defined by a polynomial, and inverse by a polynomial divided by de-terminant which is continuous.(g) Since it is the zero set of elements below the diagonal of Gl(n, R).(h) It surely is a manifold. Also, (s, t)(s1, t1)

−1 = ( ss1

,− st1s1

+ t) is continuous.(i) It is also a manifold, and (A, v)(A1, v1)

−1 = (AA−11 ,−AA−1

1 v1 + v) is continuous.

3. (a) It is a vector space, and Lie bracket operation satisfies given conditions by Proposition 1.45.(b) Trivial Lie bracket trivially satisfies the conditions.(c) [B, A] = BA− AB = −[A, B], and [[A, B], C] + [[B, C], A] + [[C, A], B] = ABC− BAC−CAB+CBA+

BCA− CBA− ABC + ACB + CAB− ACB− BCA + BAC = 0.(d) [ax + by, cx + dy] = (ad− bc)y. It is a Lie algebra by direct calculation.(e) Cross product also satisfies the conditions. Note that (a× b)× c = b(a · c)− a(b · c).

4. (a) It suffices to check that ∀Xi(1 ≤ i ≤ r) smooth vector fields on G, ω(X1, ..., Xr) is smooth. Butω(X1, ..., Xr)(σ) = ωσ(X1σ, ..., Xrσ) = (δlσ−1 ωe)(X1σ, ..., Xrσ) = ωe(dlσ−1 X1σ, ..., dlσ−1 Xrσ), so we needto check that ωe(dlσ−1 X1σ, ..., dlσ−1 Xrσ) is smooth. Now let ω be a smooth form such that ωe = ωe.Then for ϕ : G× G → G : (a, b) 7→ ab, δϕω is smooth. Also, [0, Xi] is a smooth vector field on G× G,and ι : G → G × G : τ 7→ (τ−1, τ) is smooth. Therefore, δϕω([0, X1], ..., [0, Xr])(ι(σ)) is smoothwith respect to σ. But, δϕω([0, X1], ..., [0, Xr])(ι(σ)) = δϕω(σ−1,σ)([0, X1](σ−1,σ), ..., [0, Xr](σ−1,σ)) =

ωe(dϕ[0, X1](σ−1,σ), ..., dϕ[0, Xr](σ−1,σ)). Furthermore, dϕ[0, Xi](a,b)( f ) = [0, Xi](a,b)( f ◦ ϕ) = 0a( f ◦ϕ ◦ ι1b) + Xib( f ◦ ϕ ◦ ι2a) = Xib( f ◦ la) = dlaXib( f ). Therefore ωe(dϕ[0, X1](σ−1,σ), ..., dϕ[0, Xr](σ−1,σ)) =

ωe(dlσ−1 X1σ, ..., dlσ−1 Xrσ), and the result follows.(b) δlσ is an algebra endomorphism and E∗linv(G) =

⋂σ∈G{δlσ − invariant elements}, so E∗linv(G) is a

subalgebra. Also, ε : E∗linv(G) → Λ(G∗e ) : ω 7→ ω(e) is a homomorphism. And since ω(e) = ω′(e) ⇒ωσ = δlσ−1 ωe = δlσ−1 ω′e = ω′σ ⇒ ω = ω′, ε is injective. Furthermore, for all χ ∈ Λ(G∗e ), if we setω(a) = δla−1 χ, then (δlτω)(σ) = δlτ(ω(τσ)) = δlτδlσ−1τ−1 χ = δlσ−1 χ = ω(σ), so ω ∈ E∗linv(G). Also,ε(ω) = χ, so ε is surjective.

(c) ω(X)(σ) = ωσ(Xσ) = (δlσ−1 ω)σ(Xσ) = ωe(dlσ−1 Xσ) = ωe(Xe) = ω(X)(e), so ω(X) is constanton G. Also, if α∗ : G∗e → g∗ is the dual map of α defined on Proposition 3.7, then α∗(ε(ω))(X) =

13

α(ω(e))(X) = ω(e)(X(e)) = ω(X)(e).(d) dω(X, Y) = X(ω(Y))−Y(ω(X))−ω([X, Y]) = −ω([X, Y]) since ω(X), ω(y) is constant.(e) Since g is closed under Lie bracket, such cijk’s exist, and the relation (3) is directly followed by

properties of Lie bracket, i.e., anti-symmetry and Jacobi identity. Also, since δlσ commutes with d,dωi ∈ E2

linv(G) and dωi(Xj, Xk) = −ωi[Xj, Xk], so we can write dωi = ∑j<k cjkiωk ∧ ωj + ω′ whereω′(X, Y) = 0∀X, Y ∈ (g). But, since ε : E2

linv(G) → Λ2(Ge) ∼= g ∧G is an isomorphism, ω′ must bezero. Hence the Maurer-Cartan equations are derived.

5. Consider id, π2 : Z×R→ Z×R.

6. (⇐) part is obvious since dω(X, Y) = −ω[X, Y]. For (⇒) part, dω(X, Y) = −ω[X, Y] = 0 by assumption,so by using Maurer-Cartan equation, dω should be in I .

7. (a) See [1], pp. 61-62, of follow the sketch given in the textbook.(b) See [1], pp. 64-65, of follow the sketch given in the textbook.(c) If π : X → X is a covering and X is simply connected, then by (a), there is a unique continuous map

α : X → X s.t. π ◦ α = idX . Therefore, π ◦ (α ◦ π) = π. Then by uniqueness, α ◦ π = idX also.Therefore, α and π are inverses to each other, which means π is a homeomorphism.

8. (Following the proof of [4], Prop. 1.9.) There is a countable open cover U which elements are homeomor-phic to open ball in Euclidean space. For all U, U′ ∈ U , U ∩U′ has at most countably many components,each of which is path connected. Let X be a countable set containing one point from each component ofU ∩U′(including U = U′). For each U ∈ U and each x, x′ ∈ X s.t. x, x′ ∈ U, let pU

x,x′ be a path joining xand x′ and contained in U.Since the fundamental groups based at any two points in the same component of M are isomorphic, andX contains at least one point in each component of M, we may as well choose a point q ∈ X as base point.Define a ”special loop” to be a loop based at q that is equal to a finite product of paths of the form pU

x,x′ .Clearly, the set of special loops is countable, and each special loop determines an element of π1(M, q).To show that π1(M, q) is countable, therefore, it suffices to show that every element of π1(M, q) is repre-sented by a special loop.Suppose f : [0, 1] → M is any loop based at q. The collection of components of sets of the form f−1(U)

as U ranges over U is an open cover of [0, 1], so by compactness it has a finite subcover. Thus there arefinitely many numbers 0 = a0 < a1 < ... < ak = 1 such that [ai−1, ai] ⊂ f−1(U) for some U ∈ U . Foreach i, let fi be the restriction of f to the interval [ai01, ai], reparametrized so that its domain is [0, 1], andlet Ui ∈ U be a coordinate ball containing the image of fi. For each i, we have f (ai) ∈ Bi ∩ Bi+1, andthere is some xi ∈ X that lies in the same component of Bi ∩ Bi+1 as f (ai). Let gi be a path in Bi ∩ Bi+1

from xi to f (ai), with the understanding that x0 = xk = q, and g0 and gk are both equal to the constantpath cq based at q. Then, because g−1

i · gi is path homotopic to a constant path,

f ∼ f1 · ... · fk

∼ g0 · f1 · g−11 · g1 · f2 · g−1

2 · ... · g−1k−1 · gk−1 · fk · g−1

k

∼ f1 · f2 · ... · fn

where fi = gi−1 · fi · g−1i . For each i, fi is a path in Ui from xi−1 to xi. Since Bi is simply connected, fi is

14

path homotopic to pUixi−1,xi . It follows that f is path homotopic to a special loop, as claimed.

9. (⇒) part is easy since δlσδϕω = δ(ϕ ◦ lσ)ω = δ(lϕ(σ) ◦ ϕ)ω = δϕω. For the converse, let I be anideal generated by δπ1δϕ(ωi) − δπ2(ωi) for projections π1 : G × H → G, π2 : G × H → H. Then bydirect calculation similar to 3.15, (6), it is a differential ideal. Therefore, there exists I, maximal connectedintegral manifold of I through (e, e) But since δπ1 , δπ2 , and δϕ pull left invariant forms back to leftinvariant forms, by Thm 3.19, Corollary (c), I is a Lie subgroup of G× H.Meanwhile, let (G, g) be a graph of ϕ where g(σ) = (σ, ϕ(σ)). Then also by direct calculation(following2.33), it is an integral manifold of I . Since G is connected, g(G) ⊂ I by maximality. Now, let π1|I : I → Gbe a restriction map which is a homomorphism. Then, dπ1 : I(e,e) → Ge is an isomorphism since for(X, Y) ∈ I(e,e) ⇔ ∀i, [δπ1δϕ(ωi)− δπ2(ωi)](X, Y) = 0⇔ Y = dϕ(X). Therefore by Theorem 3.26, π1|I isa covering map.If (σ, τ) ∈ I, then by path-connectedness, there is a path ` = (`G, `H) from (e, e) to (σ, τ) in I. Then(`−1

G , ϕ(`−1G )) · ` is also a path in I which joins (e, e) and (e, ϕ(σ−1) · τ) since g(G) ⊂ I. However, since ”G-

coordinate” of this path is constantly e and there is a local isomorphism from a neighborhood of (e, e) ∈ Ito a neighborhood of e ∈ G, it should be a constant path, which means ϕ(σ−1) · τ = e. Therefore, since Iis a Lie subgroup, (σ, τ) ∈ I ⇒ (σ−1, τ−1) ∈ I ⇒ ϕ(σ) = τ ⇒ I ⊂ g(G). Thus I = g(G) and it meansthat ϕ is a homomorphism.Connectedness is necessary: Consider G = Z, H = R, ϕ : Z→ R : x 7→ x2.

10. If there exist A ∈ gl(2, R) s.t.( −2 0

0 −1)= eA. Then regarding A as an element of gl(2, C), one can find a

matrix L =(

a bc d

)∈ gl(2, C) s.t. LAL−1 is upper triangular. Then L

( −2 00 −1

)L−1 = LeAL−1 = eLAL−1

is

also upper triangular. L( −2 0

0 −1)

L−1 = 1ad−bc

(−2ad+bc ab−cd −ad+2bc

), thus cd = 0. If c = 0, L

( −2 00 −1

)L−1 =

1ad

(−2ad ab

0 −ad

)=(−2 b

d0 −1

). If d = 0, L

( −2 00 −1

)L−1 = 1

−bc(

bc ab0 2bc

)=(−1 − a

c0 −2

). Therefore, diagonal

elements of eLAL−1should be -2 and -1. Meanwhile, if we let λ1, λ2 be eigenvalues of A(and LAL−1),

since A is real, two eigenvalues are either both real or complex conjugates. But since LAL−1 is uppertriangular, diagonal elements of eLAL−1

are eλ1 and eλ2 , which should not be -2 and -1 in either cases.Therefore, there is no A satisfying the condition.

11. For η defined as the text, using Exercise 1.24(d), dη(X, Y) = dηdι1e (X) + dηdι2e (Y) = X + Y for ι1e : G →G× G : σ 7→ (σ, e) and ι2e : G → G× G : σ 7→ (e, σ). Therefore, dα = dη(dσ, dβ) = dσ + dβ. Evaluatingboth sides at ∂

∂t |t=0, we obtain the given result.

12. For any σ ∈ G, a path ` from e to σ, and τ ∈ ker ϕ, ` · τ · `−1 is a curve in ker ϕ also. Since ker ϕ isdiscrete, the curve should be constant, which means σ · τ · σ−1 = τ. ∴ ker ϕ ⊂ Z(G).For the second statement, for any Lie group G, let π : G → G is the universal covering. Then sinceπ is a local homeomorphism, ker π is discrete, so ker π ∈ Z(G), which says that ker π is abelian. Butby lifting property, the fundamental group of G can be identified with a subgroup of ker π, so it is alsoabelian.

13. Let the Lie algebra of 3.5(d) be L = (R2, [·, ·]). For any non-abelian Lie algebra L = (R2, [·, ·]) whichis defined by [e1, e2] = se1 + te2, let f : L → L : ae1 + be2 7→ ((at − bs)e1 + (as + bt)e2). It is bydefinition a linear isomorphism, and f ([ae1 + be2, ce1 + de2]) = (ad− bc) f (se1 + te2) = (ad− bc)(s2 +

t2)e2, [ f (ae1 + be2), f (ce1 + de2)] = [(at− bs)e1 + (as + bt)e2, (ct− ds)e1 + (cs + dt)e2] = ((at− bs)(cs +

15

dt)− (as + bt)(ct− ds))e2 = (ad− bc)(s2 + t2)e2. Therefore f is a Lie algebra isomorphism.For the statement below, think of (R>0, ·)× (R,+) instead of 3.3(h). Then it can be easily checked that{x ∂

∂x , x ∂∂y} is left invariant tangent fields that satisfies [x ∂

∂x , x ∂∂y ] = x ∂

∂y . Therefore, by Theorem 3.27,Corollary, it is (up to isomorphism) unique simply connected 2-dimensional non-abelian Lie group.

14. Let A =(

0 10 0), B =

(0 00 1). Then eAeB =

(1 e0 e), eA+B =

( 1 e−10 e

)15. If suffices to check that all Jordan canonical forms are in the image, since basis change and exponential

function commute, and the exponent of block diagonal matrices is block diagonal matrices of whichblocks consist of exponent of original blocks. For A an elementary Jordan matrix, let A = λI · N, wherediagonal entries of N are 1’s. Since (I−N)k = 0 for k > n, log N = −∑∞

k=1(I−N)K

k exists and elog N = Nby calculation dealing with formal series. Also, λI = elog λI for any log λ regardless of choosing thebranch of log. Since log λI is in the center, elog λI+log N = elog λIelog N = A.

16. It is a Lie algebra because of Proposition 1.55 and if we denote it by h, h → Ge : X 7→ X(e) is anisomorphism. Also, for X a left invariant vector field on G, drσdϕ(X) = dϕdlσ−1(X) = dϕ(X), sodϕ(X) ∈ h. Also, for η : G × G → G : (σ, τ) → στ, η(idG, ϕ) is a constant function to e. Thus0 = dη(didG, dϕ) = didG + dϕ and dϕ(X)(e) = −X(e). Also, by the same argument, dϕ(h) ⊂ g. Then,since ϕ is an involution, dϕdϕ = id, so g and h are isomorphic via dϕ.

17. Consider (Q, discrete topology) ↪→ (R, standard topology).

18. By Theorem 3.50, Corollary (b) and Theorem 3.28, Rn is the unique simply connected abelian Lie groupup to isomorphism. Then, since π : Rn → G is the universal cover of G, G ' Rn/ ker π. Also, the kernelof a covering homomorphism is discrete. Then following the argument in the text, ker π = ⊕k

i=1Zvi forsome linearly independent elements v1, ..., vk. Therefore, G ' Rn/⊕k

i=1 Zvi ' Rn−k × Tk.

19. Clearly AB(V) is an abstract subgroup of Aut(V) and dB is a subalgebra of End(V). According toTheorem 3.42, AB(V) is a closed Lie subgroup of Aut(V) since AB(V) =

⋂v,w∈V ker(α 7→ (α(v), α(w))−

(v, w)) is closed.Let a be the Lie algebra of AB(V). If l ∈ a, exp tl ∈ AB(V). ∴ ((exp tl)(v), (exp tl)(w)) = (v, w). Takingtheir derivatives at t = 0, (lv, w) + (v, lw) = 0. ∴ l ∈ dB. Conversely, suppose that l ∈ dB. ThenB(l ⊗ 1 + 1⊗ l) = 0 by definition. Thus, B(l ⊗ 1 + 1⊗ l)n = 0 for n ≥ 1 and B ◦ et(l⊗1+1⊗l) = B. Butsince et(l⊗1+1⊗l) = et(l⊗1)et(1⊗l) = (etl ⊗ 1)(1⊗ etl) = etl ⊗ etl , exp tl ∈ AB(V), and l ∈ a. Therefore,a = dB.

20. Thanks to the outline, we only need to prove that A(G) is naturally isomorphic with the closed subgroupof A(G) consisting of those automorphisms of G which map D onto D. However, if g : G → G ∈ A(G),then by lifting property, there is a unique map g : G → G s.t. π ◦ g = g ◦ π, and since g ◦ ˜g−1 = idGby uniqueness, g ∈ A(G). By definition g(D) ⊂ D and g(Dc) ⊂ Dc, thus g(D) = D since it is anautomorphism. Conversely, if h ∈ A(G) and h(D) = D, there exists a map h : G → G s.t. h ◦ π = π ◦ h

16

and it is also an automorphism since ker(π ◦ h) = ker π. Then it is easy to see that ˆg = g and ˜h = h.

21. ϕ : U(n)→ S1 × SU(n) : A 7→ (det A, Adet A ) is an diffeomorphism.

22. (a) First, find an eigenvalue λ and a corresponding unit eigenvector v. Since λ||v|| = λvtv = vt Av =

vt Atv = λvtv = λ||v||, λ should be real. Then if vtw = 0, < v, Aw >= vt Aw = vt Aw = vt Atw =

λvtw = 0. Therefore < v > and < v >⊥ is both A-stable. Then we can use induction on the rankof A to find orthonormal basis which consists of {vi} eigenvectors of A. Set B = (v1v2...vn)−1 Forthe real symmetric case, first find an eigenvalue λ(which is real) and a corresponding eigenvector vwhich can be complex valued. Since Av = Av = λv = λv, v is also an eigenvector of λ. Therefore,Re(v) and/or Im(v) is an real eigenvector of λ and after scaling, we can find an unit real eigenvector.Now use the same induction as above.

(b) For a hermitian matrix A, by (a) we can find B ∈ U(n) s.t. BAB−1 is diagonal. Then eA =

B−1eBAB−1B. Since eigenvalues of A is real, eBAB−1 is a diagonal matrix whose diagonal entries are allpositive. Therefore, since B ∈ U(n), eA is also positive definite hermitian. Meanwhile, for a positivedefinite hermitian matrix H, we can find B ∈ U(n) as above. Then since eigenvalues of H are allpositive, we can find a real diagonal matrix A s.t. eA = BHB−1. Then eB−1 AB = H, and B−1 AB is ahermitian matrix. If eA = eA′ for two hermitian matrices A and A′, find B ∈ U(n) s.t. BAB−1 is realdiagonal, and note that eBAB−1

= eBA′B−1is positive definite diagonal. Also, there exists C ∈ U(n) s.t.

CBA′B−1C−1 is real diagonal, and eCBAB−1C−1= eCBA′B−1C−1

is also positive definite diagonal. Butsince CBAB−1C−1 is also real diagonal by direct calculation, it means CBAB−1C−1 = CBA′B−1C−1,which says A = A′. Therefore exp is well-defined one-to-one onto.For a real symmetric A, eA is positive definite hermitian by above and real, so positive definitesymmetric. Injectivity is explained above. For a positive definite symmetric matrix H, we canfind B ∈ O(n) as (a), and there exists a real diagonal matrix A s.t. eA = BHB−1. Then note thateB−1 AB = H and B−1 AB is real symmetric.

23. If σ = PR = P1R1, then P2 = PRRtPt = P1R1Rt1Pt

1 = P21 . Find A, B real symmetric matrices satisfying

eA = P and eB = P1, and we see that e2A = (eA)2 = P2 = P21 = (eB)2 = e2B, so by injectivity, 2A = 2B

or A = B. It follows that P = P1. Therefore, R = R1 also.

24. σσt is positive definite hermitian, so ∃B ∈ U(n) s.t. BσσtB−1 is positive definite diagonal. Set P =

B−1(BσσtB−1)12 B, which is positive definite hermitian, and R = P−1σ a unitary matrix.

25. U(n) is connected, and there exists a path tI + (1− t)P joining I and P a positive definite hermitianmatrix. Thus by similar argument of Theorem 3.68 and using Exercise 24, Gl(n, C) is connected.

26. Follow the argument and use actions SU(n)× X → X for (b) and SO(n)×Pn−1 → Pn−1 for (c).

17

4 Chapter 4

1. (⇐) Let ω = dr1 ∧ ... ∧ drd+1 be the volume form on Rd+1 and ω′(x) = δ fx(i(N(x))ω f (x)) for a smoothnowhere-vanishing normal vector field N along (X, f ). Then it is non-vanishing d-form on X, since forx ∈ X and X1, ..., Xd ∈ Xx which are linearly independent, ω′x(X1, ..., Xd) 6= 0.(⇒) Let ω′ be a nowhere-vanishing d-form on X. For x ∈ X choose X1, ..., Xd ∈ Xx which satisfiesX1 ∧ ... ∧ Xd = ωx. Then we can pick N(x) ∈ (Rd+1) f (x) s.t. N(x) is orthogonal to d f (Xx) andω(d f (X1), ..., d f (Xd), N(x)) = 1 where ω = dr1 ∧ ... ∧ drd+1. N(x) is independent of the choice ofX1, ..., Xd ∈ Xx, since for Y1, ..., Yd which also satisfies Y1 ∧ ... ∧ Yd = ωx, one can express Yi = ∑j aijXj,then using Exercise 2.10, we see that det aij = 1. Thus ω(d f (Y1), ..., d f (Yd), N(x)) = 1 also. Then sinced f (Xx)⊥ is 1-dimensional, N(x) which satisfies above condition is uniquely determined. Also, it is bydefinition nowhere-vanishing. It remains to show that N is smooth, but locally on the neighborhood ofx ∈ X we can choose smooth vector field X1, ..., Xd which satisfies X1 ∧ ... ∧ Xd = ω′ on the neighbor-hood, and d f (X1), ..., d f (Xd) is also smooth on the neighborhood of f (x) over f (X). Therefore locally Nis also smooth(locally you can use Gram-Schmidt process, and scaling if necessary, then the result N(x)would be a smooth function of the local coordinates of x).

2. Taking a nowhere-vanishing n-form on Pn is equivalent to take a nowhere-vanishing n-form on Sn suchthat it is preserved by antipodal map. Now regard Sn as imbedded to Rn+1, and let N be the outwardunit normal vector field on Sn and ω = dr1 ∧ ...∧ drn+1. Then ω′ = i(N)ω is nowhere-vanishing n-formon Sn. If ϕ : Rn+1 → Rn+1 : x 7→ −x, ϕ|Sn is the antipodal map, and δϕxω′x = (−1)ni(N(x))ω−x =

(−1)n+1i(N(−x)ω−x) = (−1)n+1ω′−x. Therefore, if n is odd, ϕ is orientation-preserving, therefore Pn isoriented. But if n is even, ϕ is orientation-reversing, so we cannot take a nowhere-vanishing n-form onSn which is preserved by antipodal map, which means Pn is not oriented.

3. At first find local vector fields X1, ..., Xn near p ∈ X s.t. X1,p, ..., Xn,p is orthonormal, and let L(m) =

{g(Xi,m, Xj,m)}i,j where g is given Riemannian metric. Then near p ∈ X, it is a smooth function to the setof n× n matrix. Also L(p) = I. Therefore locally L is positive definite and also symmetric by definition.Thus there is R s.t. R2 = L, which is also smooth and symmetric. Define (Y1, ..., Yn)t = R−1(X1, ..., Xn)t.Then we see that g(Yi, Yj) = δij, which means that they are local orthonormal frame fields.

4. Use the notation on the problem. Let V = f1e1 + ... + fnen. Then V = f1ω1 + ... + fnωn, so ∗V =

f1ω2 ∧ ... ∧ ωn − ... + (−1)n−1 fnω1 ∧ ... ∧ ωn−1. Therefore on ∂D, ∗V = f1ω2 ∧ ... ∧ ωn since ω1|∂D = 0.On the other hand, 〈V,~n〉 = f1 and ω2 ∧ ...∧ωn is the volume form on ∂D. Thus

∫∂D ∗V =

∫∂D f1. Since∫

D divV =∫

D ∗d ∗ V =∫

D d ∗ V =∫

∂D ∗V by Stokes’ theorem, the result follows.

5. −∫

D f ∆g =∫

D f (∗d ∗ dg) =∫

D f (d ∗ dg) =∫

D d( f ∗ dg)−∫

D d f ∧ ∗dg =∫

D d( f ∗ dg)−∫

D〈d f , dg〉∴∫

D〈grad f , gradg〉 −∫

D f ∆g =∫

D〈d f , dg〉 −∫

D f ∆g =∫

D d( f ∗ dg) =∫

∂D f ∗ dg =∫

∂D ∗( f dg)Using the proof of Exercise 4, ∗( f dg) = 〈 f dg,~n〉 = f ∂g

∂n on ∂D. Therefore Green’s 1st equation follows.For the 2nd equation, interchange f and g and subtract from the former one.

6. If we prove the latter equation, the former one directly follows. For the latter one, since the vector spaceof 1-forms is 1-dimensional, we only need to show that they have the same nonzero value for somearbitrary n-tuples of vectors. If we put X1, ..., Xn, the left side becomes ω(X1, ..., Xn)2, which is the same

18

as X1 ∧ ...∧ Xn(X1, ..., Xn) = det{〈Xi, Xj〉} since ω is the volume form.

7. Since λ(στ) = λ(σ)λ(τ), we only need to show that it is C∞ at e. On the local chart of e with coor-dinate functions x1, ..., xn centered at e, ω can be expressed as f (x1, ..., xn)dx1 ∧ ... ∧ dxn where f is C∞.Since λ(σ)ωe = δrσ(ωσ), λ(σ) f (0, ..., 0)dx1 ∧ ... ∧ dxn = f (x1(σ), ..., xn(σ))d(x1 ◦ rσ) ∧ ... ∧ d(xn ◦ rσ) =

f (x1(σ), ..., xn(σ))det{〈dxi◦rσ ,dxj〉}

det{〈dxi ,dxj〉}dx1 ∧ ...∧ dxn, so λ(σ) = f (x1(σ),...,xn(σ))

f (0,...,0)det{〈dxi◦rσ ,dxj〉}

det{〈dxi ,dxj〉}which is C∞.

8. (This is one of characteristics of Haar measure on a unimodular Lie group. Here is the sketch of theproof.) Since δrσδα(ω) = δαδlσ−1(ω) = δα(ω), it is right invariant. Let ω′ = δα(ω) and think of µ, µ′

measures on G corresponding to ω and ω′ respectively. Then we can define f 7→∫

G f µ′ which is well-defined right-invariant integral on G. Now think of

∫G×G f (τσ)µ(σ)µ′(τ); if we first integrate for µ it

gives (∫

G f ω)(∫

G ω′). and if we integrate for µ′ first, we get∫

G f ω′. By Fubini’s theorem, they must bethe same. Also,

∫G ω′ =

∫G δα(ω) =

∫α(G) ω = 1, so

∫G f ω =

∫G f ω′. Since

∫G f ω′ =

∫α(G)( f ◦ α)ω, the

result follows.

9. (⇒) Since 〈dlσX, dlσY〉 = 〈drτX, drτY〉 = 〈X, Y〉, Ad(G) in Aut(g) is actually in O(g). Since O(g) iscompact, the closure of Ad(G) is also compact.(⇐) Give to g a non-degenerate inner product. Let `inAd(G). Since Ad(G) is a subgroup of Aut(g)and has compact closure, `n should be bounded for all n ∈ Z. It is satisfied only when ||`|| = 1, soall elements of Ad(G) preserve the inner product. Now for σ ∈ G and X, Y ∈ Gσ, define 〈X, Y〉σ =

〈dlσ−1 X, dlσ−1Y〉e. Then by definition it is left-invariant, and since 〈drτX, drτY〉στ =

〈dlτ−1σ−1 drτX, dlτ−1σ−1 drτY〉e = 〈dAdτ−1 dlσ−1 X, dAdτ−1 dlσ−1Y〉e = 〈dlσ−1 X, dlσ−1Y〉e = 〈X, Y〉σ, it is alsoright-invariant.

10. (a) Since Rn ∼= D(0, 1) ⊂ Rn, singular homology groups of the two must be the same. Then by de Rhamtheorem, it is also the same as de Rham cohomology groups. By Poincare Lemma, it must be zero forgiven conditions.(b) By de Rham theorem, it suffices to show that the 0-th singular homology group is R. But since it isisomorphic to the free R-module generated by path-components, the result follows.

11. By de Rham theorem, it suffices to calculate singular homology groups. But it can be easily seen that

∞ H0(M; R) = R, ∞ H1(M; R) = R, ∞Hn(M; R) = 0for n ≥ 2.

12. d(α ∧ β) = dα ∧ β + (−1)kα ∧ dβ = 0. If β = dγ, then d(α ∧ γ) = dα ∧ γ + (−1)kα ∧ β, so α ∧ β is exact.

13. By Stokes’ theorem,∫

z α =∫

int z dα = −8∫

int z dx ∧ dy = −8.

14. It is closed by direct calculation. Also d(sin xy + x2) = α, it is exact. Therefore,∫

z α =∫

int z dd f = 0.

15. It is closed by direct calculation. Also,∫

S1 α =∫

S1xdy−ydx

2π =∫

Ddx∧dy

π = 1. If δi(α) is exact, then integralon S1 must be zero since S1 is a compact manifold without boundary. Therefore δi(α) is not exact. Also,if α is exact, then δi(α) is also exact since δ and d commute. Therefore, α is also not exact.

16. (a) It is because ∞H1(S2; R) = 0, and by de Rham theorem.(b) By direct calculation.

19

(c)∫

S2 σ = 3Vol(S2), so it is not zero. If it is exact it must be zero since S2 is a compact manifold withoutboundary.

(d) ∗α = ∑ni=1(−1)i−1ridr1∧...∧dri∧...∧drn

(∑ni=1 r2

i )n2

. It is closed by similar calculation to (b).

(e)∫

Sn−1 ∗α = nVol(Sn−1), so it is not zero, and ∗α is not exact.

17. H1deRham(S

2; R) = 0 6= R2 = H1deRham(T

2; R), using singular homology theory and de Rham theorem.

18. (a) Since H1(M; R) = 0.(b) σ in Exercise 16(b) satisfies the condition.(c) Since H2(D; R) = 0 and 0 6= [σ] ∈ H2(M; R).

19. Let t be a variable corresponding to the interval (−ε, 1 + ε) on M × (−ε, 1 + ε). For a k-form ω =

ω1 + dt ∧ ω2 on M × (−ε, 1 + ε), i( ∂∂t )ω = ω2 is a (k-1)-form, and its expression does not contain dt

term. Therefore, we can define∫ 1

0 i( ∂∂t )ωdt =

∫ 10 ω2dt which is a (k-1)-form on M. Denote it Dk(α). For

convenience of the notation, denote dM as a differential operator on M, and d = dM + dt as a differentialoperator on M× (−ε, 1 + ε). Then Dkd(ω) =

∫ 10 i( ∂

∂t )(dω1 − dt ∧ dω2)dt =∫ 1

0 L ∂∂t(ω1)− dMω2dt, and

dDk(ω) = dM∫ 1

0 ω2dt =∫ 1

0 dMω2dt. Therefore dDk(ω) + Dkd(ω) = i∗1ω1 − i∗0ω1 = i∗1ω − i∗0ω bydefinition of Lie derivative and i0, i1 (i∗0(dt ∧ ω2) = i∗1(dt ∧ ω2) = 0). It means i∗0 and i∗1 are the samemap on the cohomology level. Since f = F ◦ i0 and g = F ◦ i1, f ∗ = i∗0 ◦ F∗ and g∗ = i∗1 ◦ F∗ also inducethe same cohomology maps.

20. (a) Since the vector space of n-form on M is 1-dimensional on each fiber, we only need to show thatω = δ f (i(~n)(dr1 ∧ ...∧ drn+1)) satisfies ω(v1, ..., vn) = 1 for each point m ∈ M and v1, ..., vn ∈ Mm whichis an oriented orthonormal basis of Mm. Since ω(v1, ..., vn) = dr1 ∧ ... ∧ drn+1(~n, d f (v1), ..., d f (vn)) = 1by assumption, the result follows.(b) From vector calculus we obtain~n = 1√

f 2x+ f 2

y+1(− fx

∂∂x − fy

∂∂y + ∂

∂z ). Therefore,

ω = δϕ(i(~n)(dx ∧ dy ∧ dz))

= δϕ(1√

f 2x + f 2

y + 1(− fxdy ∧ dz + fydx ∧ dz + dx ∧ dy))

=1√

f 2x + f 2

y + 1(− fxdy ∧ d f + fydx ∧ d f + dx ∧ dy)

=1√

f 2x + f 2

y + 1( f 2

x + f 2y + 1)dx ∧ dy =

(√f 2x + f 2

y + 1)

dx ∧ dy

20

5 Chapter 5

1. (Following Lecture 1, Proposition 1 of [6]) Let f be the 0-section, x ∈ X and let G be an open set containingf (x) = 0x. Then there is an open set G1 such that 0x ∈ G1 ⊂ G and π|G1 is a homeomorphism of G1

onto open π(G1). Since 0x + 0x = 0x, and addition is continuous, there exist open sets H, K with 0x ∈ H,0x ∈ K such that H + K ⊂ G1. Let L = G1 ∩ H ∩ K, then L is open, 0x ∈ L and π|L is a homeomorphismof L onto open π(L). Clearly x = π(0x) ∈ π(L). If y ∈ π(L) there exists q ∈ L with π(q) = y.Then q ∈ H ∩ K ∩ Sy, so q + q ∈ G1 ∩ Sy. However, since q ∈ G1 and π(G1) is 1-1, {q} = Sy ∩ G1.Therefore q + q = q, which means q = 0y. Thus if y ∈ π(L), f (y) = 0y ∈ L. Since π|L : L → π(L) is ahomeomorphism, and f |π(L) is its inverse, it is also continuous. Since x ∈ X is arbitrary, f is continuouson X.

2. If f : U → S is a section and x ∈ U, then there exist an open set f (x) ∈ V s.t. π|V : V → π(V) is ahomeomorphism and π(V) ⊂ U. Since f is continuous, f−1(V) is open, and x ∈ f−1(V) ⊂ π(V) ⊂U. Let W = π−1 f−1(V) ∩ V. Then it is an open neighborhood of f (x), and π(W) ⊂ f−1(V). Alsof ( f−1(V)) ⊂ W. But since π ◦ f | f−1(V) = id f−1(V), π(W) = f−1(V). Also, π|W : W → π(W) = f−1(V)

is a homeomorphism, f ( f−1(V)) = W. Since f | f−1(V) is continuous with a continuous inverse π|W , it isa homeomorphism. Therefore f is a local homeomorphism and hence an open map.

3. Let f , g be two sections and f (x) = g(x). Then there exist an open set x ∈ U and x ∈ V s.t. f : U → f (U)

and g : V → g(V) are local homeomorphisms. Since f (x) = g(x) ∈ f (U) ∩ g(V), f (U) ∩ g(V) isnonempty. Also, on W = π( f (U)∩ g(V)), which is an open neighborhood of x, f (y) = ( f ◦ (π ◦ g))(y) =(( f ◦ π) ◦ g)(y) = g(y) if y ∈ W. Therefore f = g on W. Since 0-section is continuous (and indeed asection) by Exercise 1, the second statement follows.

4. It is because fx = 0 does not mean only f (x) = 0, but ” f is zero on the neighborhood of x”. If f : R →R : x 7→ x, f (0) = 0 but f0 6= 0.

5. It is because projection of sheaves are local homeomorphisms and a sheaf mapping is locally equal to thecomposition of a projection and the inverse of another projection.

6. Let π1 : S1 → X and π2 : S2 → X be two sheaves on X and ϕ, ψ : S1 → S2 be two sheaf mappings.Suppose ϕ(x) = ψ(x). There exist an open set x ∈ U that π1 : U → π1(U) is a local homeomorphism.Also, contracting U if necessary, we can assume that ϕ : U → ϕ(U), ψ : U → ψ(U), π2 : ϕ(U) →π2(ϕ(U)), π2 : ψ(U) → π2(ϕ(U)) are local homeomorphisms. Since ϕ(x) = ψ(x) ∈ ϕ(U) ∩ ψ(U),ϕ(U) ∩ ψ(U) is nonempty. Also, on W = π−1

1 (π2(ϕ(U) ∩ ψ(U))), which is an open neighborhood of xsince π1(U) ⊃ π2(ϕ(U) ∩ ψ(U)), ϕ(y) = (ϕ ◦ (π−1

1 ◦ π2 ◦ ψ))(y) = ((ϕ ◦ π−11 ◦ π2) ◦ ψ)(y) = ψ(y) if

y ∈W. Therefore ϕ = ψ on W.By the way, if ψ is a ”0-sheaf-mapping”, then it is the composition of the 0-section and the projectionmap, hence continuous. Therefore, it really is a sheaf mapping and the second statement follows.

7. First, a lemma:Lemma Let π : S → X be a sheaf and R ⊂ S be a subsheaf. If f : U → S is a section, thenR∩ π−1(U) + f (U) is open in S .Proof) Let px ∈ R ∩ π−1(U) + f (U) and π(px) = x. By definition, there exist qx ∈ R and rx ∈ f (U)

21

s.t. px = qx + rx. Let U be an open neighborhood of px which is locally homeomorphic to π(U). Sinceaddition is continuous, there exist V ⊂ R ∩ π−1(U) and W ⊂ f (U) neighborhoods of qx and rx re-spectively, s.t. π(V) = π(W), V + W ⊂ U, and π|V : V → π(V), π|W : W → π(W) are home-omorphisms. Let Y = π(V) = π(W). Then V + W ⊂ U ∩ π−1(Y). Also, y ∈ U ∩ π−1(Y), thenthere exists aπ(y) ∈ V ∩ Sπ(y) and bπ(y) ∈ W ∩ Sπ(y). Then aπ(y) + bπ(y) ∈ U ∩ Sπ(y). But sinceπ|U : U → π(U) is 1-1, y = aπ(y) + bπ(y) ∈ V + W. Therefore V + W = U ∩ π−1(Y) is open. Butit means px ∈ V + W ⊂ R∩ π−1(U) + f (U), soR∩ π−1(U) + f (U) is open.

Now come back to the question. First, we need to show that π : T → M is a sheaf.1) For x ∈ T , there exists a y ∈ S s.t. ϕ(y) = x and y ∈ U s.t. π|U : U → π(U) is a homeomorphism. Letf : π(U) → U be the inverse of π|U , which is a section. Since ϕ−1(ϕ(U)) = U +R ∩ π−1(π(U)) andit is open in S by Lemma, ϕ(U) is open. Also if W ⊂ ϕ(U), ϕ−1(W) = f−1(π(W)) +R∩ π−1(π(W))

is also open in U by Lemma. Therefore ϕ : U → ϕ(U) is continuous. It is clearly onto, and 1-1 since ϕ

preserves fibers. Also, if V ⊂ U is open, ϕ−1(ϕ(V)) = V +R∩ π−1(π(V)) is open by Lemma, so ϕ(V)

is open. It means ϕ−1 is also continuous. Therefore, ϕ|U is a homeomorphism, and π|ϕ(U) = π ◦ ϕ|−1U is

a homeomorphism also. Clearly x ∈ ϕ(U). Therefore π : R → M is a local homeomorphism.2) π−1(m) is clearly a K-module.3) ϕ × ϕ : S × S → T × T is also a quotient map, and restricting to S ◦ S , we obtain a quotient mapϕ ◦ ϕ : S ◦ S → T ◦ T . Let ψ : S ◦ S → S be a composition map. Then ψ : T ◦ T → T is a compositionmap on T . To check that it is continuous, we need to verify that ψ−1(U) is open in T ◦ T for an openset U ⊂ T . But it is by definition equal to whether (ϕ ◦ ϕ)−1(ψ−1(U))) is open, and it is open since∗ϕ ◦ ϕ)−1(ψ−1(U)) = ψ−1(ϕ−1(U)). Therefore the composition map is continuous.Therefore π : T → M is a sheaf. It remains to show that ϕ is a sheaf homomorphism, but it follows fromthe proof of 1).

8. Let ϕ : β(α(S))→ S : ρp,U f → f (p).1) If ρp,U f = ρ′p′ ,U′ f

′, p = p′, and ∃V ⊂ U ∩U′ and p ∈ V s.t. fV = f ′V . Therefore f (p) = f ′(p′) and ϕ iswell-defined.2) Let s ∈ S and s ∈ V s.t. π|V : V → π(V) be a homeomorphism. If ϕ(ρp,U f ) ∈ V, f (p) ∈ V,therefore f (p) ∈ f (U) ∩ V 6= ∅, which is open. Also π( f (U) ∩ V) ⊂ U ∩ π(V) is open. Then it meansρp,U f ∈ {ρq,π( f (U)∩V) f |π( f (U)∩V)|q ∈ π( f (U) ∩V)} ⊂ ϕ−1(V). Thus ϕ is continuous.3) It clearly preserves fibers and composition.On the other hand, let ψ : S → β(α(S)) : f (p) 7→ ρp,U f . (Since π is a local homomorphism, we can finda section f = π|−1

U and p ∈ M s.t. f (p) = x ∀x ∈ S .)1) If f (p) = f ′(p′) for f ∈ Γ(S , U) and f ′ ∈ Γ(S , U′), p = p′ ∈ U ∩U′, and ∃V ⊂ U ∩U′ s.t. f |V = f ′|Vby Exercise 3. Therefore ρp,U f = ρp′ ,U′ f ′ and ψ is well-defined.2) For O f = {ρp, U f |p ∈ U}, ψ−1(O f ) = f (U), which is open. Therefore ψ is continuous.3) It clearly preserves fibers and composition.Then it is certain that ϕ and ψ are inverses to each other. Therefore β(α(S)) is isomorphic to S .

9. (S ⊗ T )m = lim−→m∈U(α(S)⊗ α(T ))U = lim−→m∈Uα(S)U ⊗ α(T )U = lim−→m∈UΓ(S , U)⊗ Γ(T , U)

= lim−→m∈UΓ(S , U)⊗ lim−→m∈UΓ(T , U) = Sm ⊗Tm, since direct limit and tensor product commute. The lastequality follows from β(α(S)) ' S by Exercise 8.

10. Let Cp and Cq be two skyscraper sheaves with p 6= q. Then Cp ⊗ Cq is a zero sheaf, but Γ(Cp, M)⊗

22

Γ(Cq, M) ' C⊗C, which is not zero.

11. Let M = R, and ϕ : M → R : x 7→ x. Denote the given mapping of the exercise as Φ and the 0-sectionas f . Then f (M) ⊂ C ∞(M) is open, but Φ−1( f (M)) = f (M) ∩ C ∞(M)0, which is not open.

12. Let t ∈ ΓΦ(T ). Since Supp(t) ∈ Φ, ∃U1, U2, U3 s.t. supp(t) ⊂ U1 ⊂ U1 ⊂ U2 ⊂ U2 ⊂ U3 ⊂ U3 andU1, U2, U3 ∈ Φ. Using the process of Theorem 5.12, we can find s ∈ SU3 s.t. t|U3 is the image of s. Thens|U3\U1

has a zero image, so it is inRU3\U1. Now consider a cover U2, U1

c. SinceR is fine, there exist an

endomorphism ϕ s.t. supp(ϕ) ⊂ U1c and ϕ = 1 on Uc

2. Also, supp(ϕ ◦ s|U3\U1) ⊂ U1

c, so we can extendit to all of U3 which is valued zero on U1. Denote it s′ ∈ SU3 . Then the image of s− s′ is also t|U3 andby construction s− s′|U3\Uc

2= 0, so supp(s− s′) ⊂ U2 ⊂ U3. Thus one can extend it to all of M which

is valued zero on Uc3, and s− s′ ∈ ΓΦ(S). Also, t is the image of s− s′. The statement is proved.

13.

... // Hq−1(E∗)∂// Hq(C∗)

f// Hq(D∗)

g// Hq(E∗)

∂// Hq+1(C∗) // ...

0 // Cq−1 //

��

Dq−1 //

��

Eq−1 //

��

0

0 // Cq //

��

Dq //

��

Eq //

��

0

0 // Cq+1 // Dq+1 // Eq+1 // 0

1) f ∂ = 0. Let [e] ∈ Eq−1 for e ∈ Z(Eq−1) and d ∈ Dq−1, d′ ∈ Dq, c ∈ Cq s.t. d 7→ e, d 7→ d′, c 7→ d′. Thenby definition of ∂, f ∂[e] = f [c] = [d′] = 0 ∈ Hq(D∗).2) im∂ ⊃ ker f . Let c ∈ Z(Cq) and d′ ∈ Dq s.t. c 7→ d′. If f [c] = [d′] = 0, ∃d ∈ Dq−1 s.t. d 7→ d′. Lete ∈ Eq−1 s.t. d 7→ e. Then e 7→ 0 ∈ Eq since c 7→ d′ 7→ 0 ∈ Eq. Therefore e ∈ Z(Eq−1), and ∂[e] = [c] bydefinition of ∂.3) g f = 0. Since the original sequence is exact.4) im f ⊃ ker g. Let d ∈ Z(Dq) and d 7→ e ∈ Eq. If g[d] = [e] = 0, ∃e′ ∈ Eq−1 s.t. e′ 7→ e. Let d′ ∈ Dq−1

s.t. d′ 7→ e′ and d ∈ Dq s.t. d′ 7→ d. Then d− d 7→ 0 ∈ Eq, so ∃c ∈ Cq s.t. c 7→ d− d. Now let c′ ∈ Cq+1

s.t. c 7→ c′. Then c′ 7→ 0 ∈ Dq+1 since d− d 7→ 0 ∈ Dq+1, so c′ = 0 since Cq+1 → Dq+1 is injective.Therefore c ∈ Z(Cq), and f [c] = [d− d] = [d].5) ∂g = 0. Let d ∈ Z(Dq), e ∈ Eq s.t. d 7→ e. Then d 7→ 0 ∈ Dq+1, so e 7→ 0 ∈ Eq+1, which meanse ∈ Z(Eq). Also, ∂g[d] = ∂[e] = 0 by definition of ∂, since 0 ∈ Cq+1 7→ 0 ∈ Dq+1.6) img ⊃ ker ∂. Let e ∈ Z(Eq), d ∈ Dq, d′ ∈ Dq+1, c′ ∈ Cq+1 s.t. d 7→ e, d 7→ d′, c′ 7→ d′. Then∂[e] = [c′] = 0 means ∃c ∈ Cq s.t. c 7→ c′. Let d ∈ Dq s.t. c 7→ d. Then d − d 7→ 0 ∈ Dq+1, so

23

d− d ∈ Z(Dq). Also, g[d− d] = [e].

14.0 // Γ(K ⊗S ) //

id��

Γ(C0 ⊗S ) //

ϕ0��

Γ(C1 ⊗S ) //

ϕ1��

Γ(C2 ⊗S ) //

ϕ2��

...

0 // Γ(K ⊗S ) // Γ(C0 ⊗S ) // Γ(C1 ⊗S ) // Γ(C2 ⊗S ) // ...

(a) Since Z(Γ(C0 ⊗S )) = Γ(K ⊗S ) = Z(Γ(C0 ⊗S )), ϕ|Z(Γ(C0⊗S )) = idΓ(K ⊗S ). Hence condition(a) is satisfied.(b)Since the following diagram commutes:

Cq ⊗S

��

// Cq ⊗T

��

Cq ⊗S // Cq ⊗T

(c) By using 5.17.

15. If U =⋃

i Vi, fi ∈ Ap(Vi s.t. fi = f j on Vi ∩Vj, then set f ∈ Ap(U) s.t. f (x1, ..., xp+1) = fi(x1, ..., xp+1) ifx1, ..., xp+1 ∈ Vi and otherwise f = 0. Then f is well-defined, and f |Vi = fi.However, if U = R, V1 = (−∞, 1), V2 = (−1, ∞), and f ∈ A1(U) s.t. f (2,−2) = 1 and otherwise f = 0,then f |V1 = 0 and f |V2 = 0, but f 6= 0.

16. 0 // Z/2Z×2// Z/4Z

π// Z/2Z // 0 is exact, but

0 // Z/2Z⊗Z/2Z×2⊗id=0

// Z/4Z⊗Z/2Zπ⊗id

// Z/2Z⊗Z/2Z // 0 is not exact.

17. Choose any sheaf K which is not find and consider the sheaf of germs of discrete sections of K whichis fine.

18. Since exactness is local statement, we only need to think of module cases. However, since stalks oftorsionless sheaves are flat, the result follows. Or a long exact sequence can be divided into short exactsequences, and use Proposition 5.15.

19. Naturally f : M → N induces f∗ : Hp∆(N; G) → Hp

∆(M; G) s.t. f∗([ϕ])(σ) = ϕ( f (σ)) which is well-defined. Also that it satisfies functorial properties is straightforward. Therefore the last statementfollows from a basic property of a functor.

20. Denote the given map ϕ : Hp∆∞(M; R) → ∞Hp(M; R)∗. Let [ f ] ∈ ker ϕ. Then f (σ) = 0 for any cycle σ.

Now set g ∈ Sp−1(M, R) s.t. for (p-1)-boundary ∂τ, g(∂τ) = f (τ) and otherwise 0. It is well-definedsince if ∂τ = ∂τ′ then τ − τ′ is a cycle, so f (τ − τ′) = 0. Then dg = f , so [ f ] = 0. Therefore ϕ isinjective.By the way, if g ∈ ∞ Hp(M; R)∗, set f ∈ Sp(M, R) s.t. f (σ) = g([σ]). Then it is well-defined, and

24

d f (τ) = f (∂τ) = g([∂τ]) = 0, so f is a cocycle. Also, ϕ( f ) = g by the setting. Hence ϕ is surjective.

21. If periods of σ and τ are all integer-valued, they must be zero since {σ}({az}) = a{σ}({z}) for all a ∈ R

and p-cycle z, and similarly for τ. Therefore, σ ∧ τ is exact and has integer periods, i.e. zero. (Maybethe purpose of this exercise is to think of cohomology theory over Z, but I am not sure.)

25

6 Chapter 6

1. If ∆ operates on Ep(M), then ∆ = δd + dδ = (−1)n(p+2)+1 ∗ d ∗ d + (−1)n(p+1)+1d ∗ d∗. (Be careful aboutsigns.) Therefore ∗∆ = (−1)n(p+2)+1+(n−p)pd ∗ d + (−1)n(p+1)+1 ∗ d ∗ d∗ and ∆∗ = (−1)n(n−p+2)+1 ∗ d ∗d ∗+(−1)n(n−p+1)+1+p(n−p)d ∗ d. Since n(p + 2) + 1 + (n− p)p ≡ n(n− p + 1) + 1 + p(n− p) mod 2and n(p + 1) + 1 ≡ n(n− p + 2) + 1 mod 2, ∗∆ = ∆∗ as desired.

2. (a) Obviously G is linear. To show that it is bounded, note that there exists a c > 0 such that ||β|| ≤c||∆β|| for all β ∈ (Hp)⊥ by 6.8(4). Hence if we let G(α) = ω, then ||G(α)|| = ||ω|| ≤ c||∆ω|| =c||α− H(α)|| ≤ c||α||. Thus G is bounded.

(b) For all α, β ∈ (Hp)⊥, ∆Gα = α, ∆Gβ = β By definition. Thus, 〈Gα, β〉 = 〈Gα, ∆Gβ〉 = 〈∆Gα, Gβ〉 =〈α, Gβ〉.

(c) Suppose {αn} is a bounded sequence. Then {Gαn} is also bounded by (a), and {∆Gαn} is alsobounded because ||∆Gαn|| = ||αn − H(αn)|| ≤ ||αn||. Then use 6.6 to obtain the result.

3. It suffices to show that∫

Rn(1 + x21 + ... + x2

n)−kdx1...dxn converges for k ≥ [ n

2 ] + 1. By change ofvariables yi = xi√

1+x2n

for 1 ≤ i ≤ n − 1 and yn = xn, it is equal to∫

Rn(1 + y2n)−k(1 + y2

1 + ... +

y2n−1)

−k(1 + y2n)

n−12 dy1...dyn =

∫Rn(1 + y2

1 + ... + y2n−1)

−k(1 + y2n)

n−12 −kdy1...dyn =

∫Rn−1(1 + y2

1 + ... +

y2n−1)

−kdy1...dyn−1∫

R(1+ y2

n)n−1

2 −kdyn (∵ ∂(x1,...,xn)∂(y1,...,yn)

= (1+ y2n)

n−12 ). The former part converges by induc-

tion on n, and the latter one converges when 2k− (n− 1) > 1, or k > n2 . It is equivalent to k ≥ [ n

2 ] + 1because k is an integer.

4. ∑t[α]=0 ||D

α ϕ||2 = ∑t[α]=0 ∑ξ ξ2α|ϕξ |2. Thus we only need to show that there exists c > 0 which depends

on t and n such that c(1+ |ξ|2)t ≤ ∑t[α]=0 ξ2α ≤ (1+ |ξ|2)t for all ξ. The second inequality is clear because

each term in ∑t[α]=0 ξ2α is in the expansion of (1 + |ξ|2)t with coefficients ≥ 1. Also, if we set c = n−t,

then it is easy to see that the first inequality holds because of similar reasons.

5. Since both are defined by pointwise inner products, there exists a positive-definite Hermitian matrix Mx

for each x ∈ Rn such that ϕ ◦′ ψ = ϕ ◦ Aψ. Also it is smooth because the volume form on M is smooth.

6. I will show only when α = f dx1 ∧ ...∧ dxp. The general case is similar. By definition, dα =

(−1)p ∑ni=p+1

∂ f∂xi

dx1 ∧ ... ∧ dxp ∧ dxi, and ∗α = f dxp+1 ∧ ... ∧ dxn. (assuming ∗1 = dx1 ∧ ... ∧ dxn) Also

δα = (−1)n(p+1)+1 ∗ d ∗ α = (−1)n(p+1)+1 ∑pi=1(−1)p(i−1)+(p−1)(p−i) ∂ f

∂xidx1 ∧ ... ∧ dxi ∧ ... ∧ dxp. Further-

more, by tedious calculation, we get ∆α = −∑ni=1

∂2 f∂x2

idx1 ∧ ...∧ dxp.

7. By 6.16(14), − ∂2 ϕ∂x∂y = ∑(a,b)∈Z2 abϕ(a,b)ei(ax+by). Thus || ∂2 ϕ

∂x∂y ||2 = ∑(a,b)∈Z2 a2b2|ϕ(a,b)|2 by Parseval iden-

tity. Similarly, we get ||∆ϕ||2 = ∑(a,b)∈Z2(a2 + b2)2|ϕ(a,b)|2. Thus the given inequality follows from

a2b2 ≤ (a2+b2)2

4 .

8. If { fn} is bounded in C1, then it is uniformly bounded and equicontinuous by assumption, thus byArzela-Ascoli theorem it contains a convergent subsequence.

9. Differential operators given in the following 4 problems are all elliptic. Therefore we can deduce that asolution exists if and only if the given function is in the orthogonal subspace of the kernel of the formal

26

adjoint.(a) Adjoint: D∗(u) = −u′

(b) Adjoint: D∗(u) = −u′ − u(c) Adjoint: D∗(u) = u′′

(d) Adjoint: D∗(u) = u′′ + πu

10. By Hodge decomposition theorem, we may represent any form ω as ω = δα + dβ + γ with γ harmonic.If it is closed, then dω = dδα = 0. Since 〈δα, δα〉 = 〈dδα, α〉 = 0, we see that δα = 0. Hence ω = dβ + γ

and the result follows.

11. (⇒) Since P is a (infinite dimensional) vector space, the element u = ∑ξ uξeix◦ξ as a formal sum isuniquely determined. It remains to show that in fact u ∈ H−∞. By 6.22(2), we know that |l(ϕ)| ≤const ∑[α]≤k ||Dα ϕ||∞ ≤ const||ϕ||N for some N � 0. In other words, |∑ξ uξ ϕξ | ≤ const||ϕ||N . Bysquaring both side, it becomes |∑ξ uξ ϕξ |2 ≤ const||ϕ||2N . Since multiplying eiθ(θ ∈ R) does notchange absolute values of ϕξ , we may assume that uξ ‖ ϕξ . Then ∑ξ |uξ |2|ϕξ |2 ≤ (∑ξ |uξ ||ϕξ |)2 =

|∑ξ uξ ϕξ |2 ≤ const||ϕ||2N . Since P is dense in H−N , the inequality above is also satisfied when ϕ ∈H−N . Now set |ϕξ | = (1 + |ξ|2)− N+M

2 for some M > [ n2 ] + 1. Then ϕ ∈ H−N by 6.16(7), and the in-

equality says that ∑ξ |uξ |2|ϕξ |2 = ∑ξ |uξ |2(1 + |ξ|2)−N−M < ∞, that is to say u ∈ H−N−M.(⇐) If u ∈ H−∞, there exists s < 0 such that u ∈ Hs. Then

|l(ϕ)| = |〈u, ϕ〉0| ≤ ||u||s||ϕ||−s (6.18(e))

≤ const√

∑[α]≤−s

||Dα ϕ||2 (6.16(15))

≤ const√

∑[α]≤−s

||Dα ϕ||2∞ ≤ const ∑[α]≤−s

||Dα ϕ||∞

as desired.

12. It suffices to show that if∫

M α = 0, then it is an exact form. En = imd ⊕ Hn because imδ = 0. Letα = dγ + ε with ε harmonic. Since [α] 7→

∫M α is the natural isomorphism, [α] = [ε] = 0. Hence by

6.11, ε = 0. In other words, α is exact.

13. Let ϕn = ∑a∈Z una eiax, where un

a =

1a2 if |a| < n

0 otherwise. Then ||ϕn||2 = ∑a∈Z |un

a |2 < ∑a∈Z1a4 and

||∆ϕn||2 = ∑a∈Z a2|una |2 < ∑a∈Z

1a2 are bounded. Also, ϕn is a finite sum, so ϕn ∈ P . However, its limit

ϕ = ∑a∈Z1a2 eiax /∈ H3. Thus it is not in P .

14. The statement can be proved by following the proof of 6.32. The only characteristic of ∆ used here isthat ∆ is an elliptic operator. Also, by ”Reduction to the Periodic Case” argument, we only need toshow this locally.

15. It is elliptic because x + iy 6= 0 for (x, y) 6= (0, 0). Hence every holomorphic function is C∞. Also everyholomorphic function is harmonic because ∆ = ∂2

∂x2 + ∂2

∂y2 = ( ∂∂x − i ∂

∂y )(∂

∂x + i ∂∂y ). If a holomorphic

function f has compact support, then by Green’s 1st identity,∫

∂D f ∂ f∂n =

∫D〈grad f , grad f 〉 −

∫D f ∆ f for

every disc D. Since ∆ f = 0, if we take a sufficiently large disc D, then it becomes∫

D〈grad f , grad f 〉 = 0.

27

By letting the radius of D goes to infinity, we see that∫

C〈grad f , grad f 〉 = 0, and it means grad f = 0

everywhere, or f is constant. But f has support, so f should be zero everywhere.

16. (a) If ∆u = λu and u 6= 0, λ〈u, u〉 = 〈λu, u〉 = 〈∆u, u〉 = 〈(dδ + δd)u, u〉 = 〈du, du〉+ 〈δu, δu〉 ≥ 0. Thusλ ≥ 0.

(b) If not, it contradicts to 6.6 if we have an orthonormal sequence.(c) If not, it also contradicts to 6.6 if we have an orthonormal sequence.(d) If ∆u = λu and ∆v = λ′v, then λ〈u, v〉 = 〈∆u, v〉 = 〈u, ∆v〉 = λ′〈u, v〉. Thus 〈u, v〉 = 0.(e) 〈ψj, Gψj〉 = 〈∆Gψj, Gψj〉 = 〈dGψj, dGψj〉 + 〈δGψj, δGψj〉 ≥ 0. For the latter part, we need to

show that l((∆ − 1η )∗α) = 0. It is equivalent to limj→∞ η〈Gϕj, (∆ − 1

η )∗α〉 = 0 ⇔ limj→∞ η〈(∆ −

1η )Gϕj, α〉 = 0 ⇔ limj→∞〈ηϕj − Gϕj, α〉 = 0, which is true. Also it is not trivial; take N such that||GϕN − ηϕN || < ε and ||Gϕi − GϕN || < ε for i > N. Since ||ϕN || = 1 and by the inequality|〈v, w〉| ≥ ||v|| − |〈v, v− w〉| ≥ ||v||(||v|| − ||v− w||), we see that |〈Gϕi, ηϕN〉| ≥ η(η − 2ε) > 0 forsufficiently small ε > 0. Thus l(ϕN) 6= 0.

(f) It is the same as (e).(g) α−∑k

i=1〈α, ui〉ui ∈ (Hp)⊥, so there exists β ∈ (Hp)⊥. If k + 1 ≤ j, 〈α, uj〉 = 〈α−∑ki=1〈α, ui〉ui, uj〉 =

〈Gβ, uj〉 = 〈β, Guj〉 = 1λj〈β, uj〉. Thus 〈α, uj〉uj =

1λj〈β, uj〉uj = G(〈β, uj〉uj). Then it follows from it

and the definition of λ. λn → ∞ because of (c).(h) Locally it is true, and by compactness of M, we can find c and k such that it is still true globally.

Then it follows from (g).

17. By Hodge decomposition, Ep(M) = im∆⊕ Hp = im∆2 ⊕ V ⊕ Hp with V = ker∆2 ∩ im∆. If ∆α ∈ V,〈∆α, ∆α〉 = 〈∆2α, α〉 = 0, so ∆α = 0. Thus V = 0 and Ep(M) = im∆2 ⊕ Hp. It means ∆2α = β issolvable if and only if β ⊥ Hp.

18. We can assume the matrix which consists of coefficients of highest order operators is symmetric because∂

∂xiand ∂

∂xjcommutes. Also, L is elliptic at x if for all v ∈ Rn \ {0}, vt(ai j(x))ijv > 0, which means that

(ai j(x))ij is definite. But for �, the corresponding matrix( 1 0

0 −1)

is not definite, so it is not elliptic. Forexample, if we let

u(x, y) =

x3 + 3xy2 if x ≥ |y|

3xy + y3 if − t ≤ x ≤ t

−3xy − y3 if t ≤ x ≤ −t

−x3 − 3xy2 if x ≤ |t|

it is easy to see that �u = 0, but u /∈ C3.

19. Since ∆ + c is elliptic, if β ⊥ ker(∆ + c) then the equation can be solved. But (∆ + c)α = 0 ⇔ ∆α =

−cα ⇔ α = 0 by assumption, so it can be solved for all β ∈ Ep(M). (In fact, the minimum eigenvalueof ∆ is 0.)

20. The first assertion is true because lσ are isometries, thus they commute with d and δ. Then the secondassertion directly follows from it. The third assertion is that ϕi ◦ lσ is continuous as a function of σ,which is also true by properties of Lie group.

21. The formal adjoint exists by similar argument to 6.24 (integration by parts) and the fact that integration

28

of exact forms over M is zero. Also, the proof of 6.5 and 6.6 is still valid because it is valid locally, andwe can glue solution with respect to transition maps of vector bundles. Then (a) follows from 6.6, and(b) follows from similar arguments to 6.8.

22. For n sufficiently large, or n > 10, then ||ϕn||∞ = ϕ(0, 0) = log log n. On the other hand, by6.16(15), const||ϕn||21 ≤ ||ϕn||2 + || ∂ϕn

∂x ||2 + || ∂ϕn

∂y ||2. Since {ϕn} is monotone increasing, ||ϕn||2 →∫

r≤ 12(log log 1

r )2dxdy < ∞ by Monotone convergence theorem. Also, ∂ϕn

∂x = xr( 1

n +r) log( 1n +r)

and we

can use similar argument to see that || ∂ϕn∂x ||

2 →∫

r≤ 12( x

r2 log r )2dxdy = π

log 2 . Similarly, || ∂ϕn∂x ||

2 → πlog 2 .

Thus ||ϕn||1 is bounded and there is no constant c satisfying the given condition.

23. By assumption, there exists a solution u of Lu = f . Also, since C∞(M) = imL∗ ⊕ ker L, we maytake u = L∗v ∈ imL∗. Since f = f ◦ γ = Lu ◦ γ = L(u ◦ γ), u ◦ γ is also a solution. Meanwhile,L(β ◦ γ) = 0 ⇔ Lβ ◦ γ = 0 ⇔ Lβ = 0, ϕγ : C∞(M) → C∞(M) : β 7→ β ◦ γ preserves ker L. Also,ϕγ is an isomorphism (with the inverse ϕγ−1 ) and preserves inner product (because it preserves thevolume form), we conclude that ϕγ also preserves imL∗. Thus u ◦ γ ∈ imL∗. Let u ◦ γ = L∗v′. ThenLL∗v = LL∗v′ ⇒ LL∗(v− v′) = 0⇒ 0 = 〈LL∗(v− v′), v− v′〉 = 〈L∗(v− v′), L∗(v− v′)〉 ⇒ L∗v = L∗v′,we conclude that u = u ◦ γ, or u is invariant under γ.

29

References

[1] Allen Hatcher, ”Algebraic Topology”, Cambridge University Press, 2002.

[2] Daniel Murfet, ”Automorphisms of Power Series Rings”, in his blog(http://therisingsea.org), 2005.

[3] Frank W. Warner, ”Foundations of Differentiable Manifolds and Lie Groups”, Springer, 1983.

[4] John M. Lee, ”Introduction to Smooth Manifolds”, 2nd ed., Springer, 2010.

[5] Paul Loya, ”An introductory course in differential geometry and the Atiyah-Singer index theorem”,Binghamton University, 2005.

[6] C. H. Dowker, ”Lectures on Sheaf Theory”, Tata Institute of Fundamental Research, 1956.

30

solutions of exercises in - university of minnesotakim00657/note/warner_sol.pdfsolutions of...

Documents