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
1− xn+1 ( x1 , ..., xn ). Similarly, if x ∈ Sn − s, then pn ( x ) = 1+ xn+1 ( x1 , ..., xn ). Therefore each function is
1 : p (S − n − p) → p (S −
continuous on its domain, respectively. Also on ps (Sn − n − p), pn ◦ p−
s n
n n
s
n
1
n
−
1
2
n − p) ⊂ R is defined as pn ◦ ps (y1 , ..., yn ) = k (y1 , ..., yn )(k = ∑i=1 yi ), which is smooth since
1
(0, ..., 0) ∈
/ ps (Sn − n − p). Similarly, ps ◦ p−
n is also smooth. Therefore, the maximal collection including
these coordinate systems is indeed a differentiable structure on Sd .
1
2. Let f (t) = t3 . Then idR ◦ f −1 (t) = t 3 , which is not differentiable at t = 0, and it means F 6= F1 .
1
However, if we set ϕ : (R, F ) → (R, F1 ) : t 7→ t 3 , 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
1
{Uα }. If we set Vα = ϕ−
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, by
paracompactness, there exists a refinement of {Uc }c∈C ∪ M \ C which is locally finite. By discarding
fβ }.
open sets which does not intersect C, one can obtain open cover of C which is locally finite, say {U
fd s.t. V
fd ∩ U
fβ 6= ∅ for only finite β’s, which we can label them as β 1 , ..., β n . Since {U
fβ }
Then, ∀d ∈ D, ∃V
Tn
f
f
is a refinement, one can find ci ∈ C for each β i s.t. Uβ ⊂ Uci . If we set Wd = Vd ∩ i=1 Vci , then one can
fβ = ∅ and Wd is an open neighborhood of d. If we set U = S U
fβ , W = S Wd ,
easily see that ∀ β, Wd ∩ U
then each open set is a neighborhood of C and D respectively which does not intersect each other. Since
C and D are arbitrary, M is normal.
5. (a) If ~a = ( a1 , ..., ad ) and ~v = (v1 , ..., vd ), ψ̃ ◦ ϕ̃−1 (~a, ~v) = (ψ ◦ ϕ−1 (~a), J (ψ ◦ ϕ−1 )(~v), which is C ∞ .
(b) ϕ̃ : 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), ψ̃−1 (W 0 ) ∩ π −1 (U ) is of the form ϕ̃−1 (W ), which is ϕ̃−1 ( ϕ̃ ◦ ψ̃−1 (ψ(U ∩ V ) × Rd ∩ W 0 )).
Therefore, on the defined topology, ϕ̃ is a homeomorphism and that T ( M ) is a 2d-dimensional locally
Euclidean space directly follows. For second countability, since M is second countable, we can choose
countable such (U, ϕ) that cover T ( M ). Since R2d is also second countable, it follows that T ( M) is
second 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. Suppose
that dψ is not surjective at m ∈ M. Since it is nonsingular, dimM = p < dimN = d. Pick a coordinate
system on N, (U, ϕ), such that ϕ(U ) = Rd . Then ϕ ◦ ψ : ψ−1 (U ) → Rd is surjective. Also, by second
countability of M, one can find countable coordinate systems of ψ−1 (U ), say {(Vα , f α )}α∈ A . Then Rd =
S
−1
p
−1
α∈ A ϕ ◦ ψ ◦ f α (R ). Using Proposition 1.35, since ϕ ◦ ψ ◦ f α is an immersion, one can obtain countable
cover {Uβ } of R p such that ϕ ◦ ψ ◦ f α−1 (Uβ ) is a subset of a slice of a open set in Rd , which is nowhere
S
dense. Therefore, by Baire category theorem, α∈ A ϕ ◦ ψ ◦ f α−1 (R p ) cannot be all of Rd , contradicting
1
assumption. Therefore, ψ is a diffeomorphism.
7. (a) For two differentiable structures on A, F and F 0 , id A : ( A, F ) → ( A, F 0 ) is continuous, hence C ∞ by
Theorem 1.32. Since its inverse is also C ∞ , it is a diffeomorphism: that is, two structures are the same.
(b) Let ( Ã, ι0 ) be A equipped with a manifold structure possibly different from induced structure form
M. Then id A : Ã → A is continuous and hence C ∞ by Theorem 1.32. Also it is nonsingular since
dι0 = d(ι ◦ id A ) is. Therefore, by Exercise 6, it is a diffeomorphism, which means that a manifold
structure on à is equal to that on A.
∂f
8. Use notations of Theorem 1.37. Since { ∂s i |(r0 ,s0 ) } is nonsingular(i.e., invertible), by continuity, there
j
exists an open set (r0 , s0 ) ∈ V0 × W0 ⊂ U such that d f is surjective. Also, (r0 , s0 ) ∈ f −1 (0), hence
−1
(0) is an imbedded submanifold of V0 × W0 of
nonempty. Therefore, by Theorem 1.38, P = f |V
0 ×W0
dimension c-d. Now, let π1 : V0 × W0 → V0 be the first projection. Then, π1 ◦ ι : P → V0 is C ∞ , and
d(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 |W0s is an isomorphism, V0 × W0(r0 ,s0 ) = ker d f ⊕ Ws0 = V0r0 ⊕ W0s0 . Hence ker d f → V0r0 is an
0
isomorphism also.) Therefore, by Theorem 1.30, one can find an open set (r0 , s0 ) ∈ V × W ⊂ V0 × W0
1
∞ map. Therefore, if we set g =
such that π1 ◦ ι|−
P∩V ×W : V → P ∩ V × W is an one-to-one onto C
−1
π2 ◦ ι ◦ (π1 ◦ ι| P∩V ×W ) : V → W (where π2 is the second projection), then it is the map satisfying the
condition.
9.
∂f
∂x
= 3x2 + y, ∂∂yf = x + 3y2 . Therefore,
∂f
∂x
=
∂f
∂y
= 0 if and only if ( x, y) = (0, 0), (− 13 , − 13 ). There-
/ f −1 ( f ( 31 , 13 )).
fore, by Theorem 1.38, f −1 ( f ( 13 , 13 )) is an imbedded submanifold, since (0, 0), (− 13 , − 13 ) ∈
f −1 ( f (0, 0)) is not a manifold since (0,0) is a node. If p = (− 31 , − 13 ), then f ( x, y) = f ( 31 , 13 ) iff x3 + xy +
1
y3 − 27
= 0 iff ( x + y − 31 )( x2 + y2 + 19 + 3x + 3y − xy) = 0 iff x + y = 13 , since x2 + y2 + 19 + 3x + y3 − xy =
1
1 2
1 2
2
−1 ( f (− 1 , − 1 )) is a line, which is an imbedded submani2 (( x + 3 ) + ( y + 3 ) + ( x − y ) ). Therefore f
3
3
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 by
Theorem 1.30, Corollary (a), ∃U ⊂ f ( M ) neighborhood of f (m), which contradicts the maximality of
| f (m)|.
11. Not closed : (− π2 , π2 ) ⊂ R, g( x ) = tan x
Not imbedded : Second Example of 1.28, g( x ) = x
To prove the converse : if there is a disk D ( x, r ) ⊂ Rn , one can make a C ∞ function which is nonzero
exactly on D ( x, r ) and zero elsewhere. Since M is locally Euclidean, there exists a basis { Bα } of M such
that ∀ Bα , ∃ gα s.t. gα−1 (R \ {0}) = Bα . Then by the assumption there is a C ∞ function f α on N such that
f α ◦ ψ = gα . Since f α−1 (R \ {0}) ∩ ψ( M) = ψ( Bα ), ψ( Bα ) is open in ψ( M ) with relative topology, which
means 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 coordinate
1
system of n, and there is a C ∞ function g on M such that g(m) = ϕ(m)−
on the neighborhood of n
ϕ(n)
on M. However, since n is a limit point of M, it cannot be extended to all of N. Therefore, M should be
closed in N.
12. (a) Since d f = ∑ 2ri dri , 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 dψ I is surjective. Since ψ( A) = (∑k aik a jk )ij ,
dψ A = (∑k ( a jk daik + aik da jk ))ij . ∴ dψ I = (daij + da ji )ij . But, since Gl (n, R) I ∼
= Mn×n (R), it is
∼
surjective onto Symn (R) I = Symn (R).
∂b
∂a
13. (a) If we set X = ∑ ai ∂x∂ , Y = ∑ bi ∂x∂ locally as Proposition 1.43(b), [ X, Y ] = ∑i ∑ j ( ai ∂xj − bi ∂xj ) ∂x∂ .
i
i
i
i
j
Therefore by 1.43(b), it is a smooth vector field.
(b) Using the same notation above,
[ f X, gY ] =
∑ ∑ ( f ai
i
=
j
∂b j
∂g
∂f
∂
j
f g[ X, Y ] + f ( Xg)Y − g(Y f ) X
∂a
∂b
∂a j
∑ ∑( f gai ∂xi − f gbi ∂xi + f ai bj ∂xi − ga j bi ∂xi ) ∂x j
i
=
∂ f aj ∂
∂gb j
− gbi
)
∂xi
∂xi ∂x j
∂a
∂b
(c) [ X, Y ] = ∑i ∑ j ( ai ∂xj − bi ∂xj ) ∂x∂ = − ∑i ∑ j (bi ∂xj − ai ∂xj ) ∂x∂ = −[Y, X ]
i
i
j
i
i
j
(d) Let Z = ∑ ci ∂x∂ . Then,
i
[[ X, Y ], Z ] =
∂b j
∂a j ∂c
∑ ∑ ∑((ai ∂xi − bi ∂xi ) ∂xkj
i
j
−c j (
k
∂ai ∂bk
∂ 2 bk
∂b ∂a
∂2 a k
∂
+ ai
− i k − bi
))
∂x j ∂xi
∂x j ∂xi
∂x j ∂xi
∂x j ∂xi ∂xk
and by direct calculation, Jacobi Identity can be proved.
∂
0
14. No. If Xx = exp( x ) ∂r
and γ0 (t) is the integral curve at 0 ∈ R, then by definition, dγ
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 clear
that on 0 ∈ R2 , the tangent vector of the integral curve should be horizontal, whereas N0 consists of
vertical vectors.
17. By 1.48(f), t>0 Dt = t<0 Dt = M, thus by compactness of M, ∃t0 > 0 s.t. Dt0 = D−t0 = M. Thus
by 1.48(g), Xt0 is an endomorphism, and ∀n ∈ N, Xtn0 is defined on all of M. Therefore by 1.48(h),
∀ N > 0, D N = M. Likewise, ∀ N < 0, D N = M. Since it is clear that D0 = M, M is complete.
S
S
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 is
disconnected, 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 ∂ f
∑
j =1
◦ γ ∂ f ja ◦ γ
−
∂y a
∂yb
jb
d−c d−c
∂
+∑ ∑
∂yc+ j j=1 k=1
∂ f jb ◦ γ
∂ f ja ◦ γ
f ka −
f
∂yc+k
∂yc+k kb
∂
∂yc+ j
Since D is involutive, [Ya , Yb ] ∈ D . But the expression above does not include ∂y∂ terms for 1 ≤ i ≤ c,
i
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 ) is
∂y
coordinate functions. Also, if 1 ≤ i ≤ c, ∂z∂ = ∑dj=1 ∂z j ∂y∂ = ∂y∂ + ∑dj=−1c f ji ◦ γ ∂y∂ = Yi by 1.61(4) and
i
i
j
i
c+ j
(6). Therefore, if we let I be a submanifold defined by z j =constant for c + 1 ≤ j ≤ d, it is an integral
manifold of D .
(1.60 ⇒ 1.61) Let Xi = ∂r∂ + ∑nj=1 b ji ∂s∂ for 1 ≤ i ≤ n and D be a distribution generated by Xi ’s. Then
i
j
by similar calculation above, [ Xa , Xb ] = 0 ∈ D by assumption 1.61(2). Thus by 1.60, there exists a cubic
coordinate system (W, ϕ) with coordinate functions ( x1 , ..., xm , y1 , ..., yn ) defined as the theorem. Let
us define α on the neighborhood of (0, 0) s.t. y j (r, αs (r )) = y j (r0 , s)∀1 ≤ j ≤ n by implicit function
theorem. We can define this because X1 , ..., Xm , ∂s∂ , ..., ∂s∂ is a basis of tangent space and Xi (y j ) = 0
1
1
∂y
since y j ’s are constant on integral manifolds of D , which means { ∂s j } should be nonsingular. Then
k
∂y
∂y ∂α
clearly αs (r0 ) = s and ∂r j + ∑nk=1 ∂s j ∂rs,k = 0. Also, Xi (y j ) = 0 means
i
i
k
is nonsingular, condition (4) is satisfied.
∂y j
∂ri
∂y
∂y
+ ∑nk=1 bki ∂s j = 0. Since { ∂s j }
k
k
20. Let M = R, N = (− π2 , π2 ) and consider N as a submanifold imbedded in M and ϕ as an inclusion. If
∂
X (t) = tan(t) ∂r
, then it satisfies the assumption since ϕ is one-to-one, but it cannot be extended to any
vector field on M.
21. It is clearly one-to-one, C ∞ , and nonsingular. To simplify the calculation, let us think the torus as
T = R2 Z × Z and ϕ : R → T : x 7→ ( x, αx ). Then imϕ ∩ {(0, y) ∈ T } is dense in {(0, y) ∈ T }, since
α is irrational. Also, given x0 , imϕ ∩ {( x0 , y) ∈ T } is dense in {( x0 , y) ∈ T } since it is the transition of
imϕ ∩ {(0, y) ∈ T }. Since x0 is arbitrary, the image of ϕ is dense in T.
22. Let t0 be a point satisfying γ̇(t0 ) = 0. Then, the constant map whose image is γ(t0 ) is also an integral
curve. 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 partition
of unity { ϕα } subordinate to the cover. Also, on each chart, there exists a positive definite inner product
induced by the standard inner product on the Euclidean space, say < , >α , and it is smooth on each
chart. If we set < , >m = ∑α < , >α,m ϕα (m), then it is the inner product which makes M become
Riemannian 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
N )(m,n) = ⊕i R ∂x∂ |(m,n) ⊕ j R ∂y∂ |(m,n) ∼
= Mm ⊕ Nn .
i
j
(c) On U × V, X̃ = ∑i ai ( x1 , ..., xs ) ∂x∂ and Ỹ = ∑ j b j (y1 , ..., yt ) ∂y∂ . Therefore, by direct calculation,
i
j
[ X̃, Ỹ ] = 0.
(d) Let v = ∑i ci ∂x∂ |(m0 ,n0 ) + ∑ j d j ∂y∂ |(m0 ,n0 ) ∈ ( M × N )(m0 ,n0 ) . Then dπ1 (v) = ∑i ci ∂x∂ |m0 , dπ2 (v) =
i
j
i
∑ j d j ∂y∂ j |n0 . Also,
∂ f ◦ i n0
∂xi |m0
=
∂ f ◦ i m0
∂f
∂xi |(m0 ,n0 ) , ∂y j |n0
∂f
∑ j d j ∂y j |(m0 ,n0 ) = v1 ( f ◦ in0 ) + v2 ( f ◦ im0 ).
5
=
∂f
∂y j |(m0 ,n0 ) .
∂f
Summing up, v( f ) = ∑i ci ∂x |(m0 ,n0 ) +
i
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 ) →
¯ Therefore, the map
U is well defined, and by the bilinearity, generators of 2.1(1) are in the kernel of l.
˜l : V ⊗ W → U is well defined. Also, it satisfies l˜ ◦ ϕ = l by the definition of the map. Since l˜(v ⊗ w) is
determined 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 f v : 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 f˜v : W ⊗ U → (V ⊗ W ) ⊗ U.
Then, if we set g : V × (W ⊗ U ) → (V ⊗ W ) ⊗ U : (v, x ) 7→ f˜v ( x ) for all x ∈ W ⊗ U, it is also bilinear,
hence one can define g̃ : V ⊗ (W ⊗ U ) → (V ⊗ W ) ⊗ U, and g̃(v ⊗ (w ⊗ u)) = (v ⊗ w) ⊗ u). By the
same method, one can also define h̃ : (V ⊗ W ) ⊗ U → V ⊗ (W ⊗ U ), and it is easy to see that g̃ and h̃
∼ (V ⊗ W ) ⊗ U.
are inverses to each other. Therefore, 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 }, {w j } be each basis of V, W,
respectively, and {vi∗ } be a dual basis corresponding to {vi }, i.e., v∗a (vb ) = δab . Then elements of
V ∗ ⊗ W can be represented as ∑i,j aij vi∗ ⊗ w j . If α(∑i,j aij vi∗ ⊗ w j ) = 0, then ∀v ∈ V, ∑i,j aij vi∗ (v) · w j =
0. Especially, if one substitutes vi for v, then ∑ j aij w j = 0∀i. Therefore, by linear independence of
basis, aij = 0∀i, j, and it means α is injective.
(2) Let ϕ ∈ Hom(V, W ). Then if ϕ(vi ) = ∑ j aij w j , set ψ = α(∑i,j aij ⊗ w j ). Then ψ and ϕ has the same
value 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 ), and
dim 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 as
the 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 need
to 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 − ba v3 ) ∧ ( av2 − cv3 ).
Therefore x is always decomposable.
(c) If V = R4 , then e1 ∧ e2 + e3 ∧ e4 is indecomposable. If it is decomposable, say ( a1 e1 + a2 e2 + a3 e3 +
a4 e4 ) ∧ (b1 e1 + b2 e2 + b3 e3 + b4 e4 ), then a1 b2 − b1 a2 = a3 b4 − b3 a4 = 1, a1 b3 − b1 a3 = a1 b4 − b1 a4 =
a2 b3 − b2 a3 = a2 b4 − b2 a4 = 0. If a1 = 0, b1 a2 = −1, b1 a3 = b1 a4 = 0, so a3 = a4 = 0, which contradicts
a3 b4 − b3 a4 = 1. If a1 6= 0, b3 = b1aa3 , b4 = b1aa4 , which also contradicts a3 b4 − b3 a4 = 1.
1
1
(d) No. Let α = e1 ∧ e2 + e3 ∧ e4 ∈ Λ2 (R4 ). Then α ∧ α = 2e1 ∧ e2 ∧ e3 ∧ e4 ∈ Λ4 (R4 ), which is not zero.
3. (a) That u ∧ v ∈ λk+l (V ) is a direct consequence of associativity of tensor products. Since the last
equation 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)l u1 ∧ v1 ∧ ... ∧ vl ∧ u2 ∧ ... ∧ uk = ... =
(−1)kl u1 ∧ ... ∧ uk ∧ v1 ∧ ... ∧ vl = (−1)kl u ∧ 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 aΦ eΦ = 0,
6
since homogeneous parts should be zero, ∀0 ≤ k ≤ n, ∑|Φ|=k aΦ eΦ = 0. However, for some Φ0 , if
we wedge eΦ0c on 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 trivially
followed.
(c) h̃ can be defined because of the universal property of tensor products and I (V ) should be in the kernel
of the induced map since h is alternating. Its uniqueness is also followed by the universal property
of tensor products. If W = R, since h̃ ∈ Λk (V )∗ and h ∈ Ak (V ), ϕ∗ : Λk (V )∗ → Ak (V ) gives an
isomorphism.
4. Since (2), (3), and (4) are bilinear forms, we only need to consider when f and g are both induced by
decomposable elements. Therefore if ϕ : Λ(V )∗ → A(V ) is the isomorphism, and f = ϕ ◦ α(w1∗ ∧ ... ∧
w∗p ), g = ϕ ◦ α(w∗p+1 ∧ ... ∧ w∗p+q ). Then
f
∧α
g(v1 , ..., v p+q ) = w1∗ ∧ ... ∧ w∗p ∧ w∗p+1 ∧ ... ∧ w∗p+q (v1 ∧ ... ∧ v p+q )
=
det(wi∗ v j )1≤i,j≤ p+q
=
∑
p+q
σ ∈S p+q
sgn(σ ) ∏ wi∗ vσ(i)
i =1
∑
=
∑ ∑
sgn(π )sgn(σ1 )sgn(σ2 )
π:p,qshu f f les σ1 ∈S p σ2 ∈Sq
=
p
q
i1 =1
i2 =1
∏ wi∗1 vπ(σ1 (i1 )) ∏ w∗p+i2 vπ( p+σ2 (i2 ))
∑
sgn(π ) det(wi∗ 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) )
π:p,qshu f f les
=
π:p,qshu f f les
On the other hand, let f = ϕ ◦ β(w1∗ ∧ ... ∧ w∗p ), g = ϕ ◦ β(w∗p+1 ∧ ... ∧ w∗p+q ). Then
f
∧β
g(v1 , ..., v p+q ) = w1∗ ∧ ... ∧ w∗p ∧ w∗p+1 ∧ ... ∧ w∗p+q (v1 ∧ ... ∧ v p+q )
=
1
det(wi∗ v j )1≤i,j≤ p+q
( p + q)!
=
1
( p + q)!
=
1
( p + q)!
∑
σ ∈S p+q
p+q
sgn(σ) ∏ wi∗ vσ(i)
∑
i =1
∑ ∑
sgn(π )sgn(σ1 )sgn(σ2 )
π:p,qshu f f les σ1 ∈S p σ2 ∈Sq
p
q
i1 =1
i2 =1
∏ wi∗1 vπ(σ1 (i1 )) ∏ w∗p+i2 vπ( p+σ2 (i2 ))
=
1
∑ sgn(π ) det(wi∗ vπ(i) )1≤i≤ p det(w∗j vπ( j) ) p+1≤ j≤ p+q
( p + q)! π:p,qshu
f f les
=
p!q!
( p + q)!
=
1
( p + q)!
∑
sgn(π ) det(wi∗ 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) )
π ∈S p+q
π ∈S p+q
7
Therefore,
f ∧α g(v1 , ..., v p+q )
∑
=
sgn(π ) f (vπ (1) , ..., vπ ( p) ) g(vπ ( p+1) , ..., vπ ( p+q) )
π:p,qshu f f les
1
p!q!
=
∑
sgn(π ) f (vπ (1) , ..., vπ ( p) ) g(vπ ( p+1) , ..., vπ ( p+q) )
π ∈S p+q
( p + q)!
f ∧β g
p!q!
=
5. (where γ is an integral curve of X)
L X ( f )|m
= lim
δXt ( f Xt (m) ) − f m
t
t →0
= lim
t →0
= lim
t →0
f Xt ( m ) − f ( m )
t
f (γm (t)) − f (m)
d
= dγ( |0 )( f )
t
dt
= Xm ( f )
6. (Since
√
t is defined on [0, ε), the limit should be calculated when t → 0+ .) We will use L’Hôpital’s rule.
8
Let g( x, y, z, w) = f (Y− x X−y Yz Xw )(m). Then,
∂g
( x, y, z, w)
∂x
∂g
( x, y, z, w)
∂y
∂g
( x, y, z, w)
∂z
∂g
( x, y, z, w)
∂w
= −YY−x X−y Yz Xw (m) ( f )
= − XX−y Yz Xw (m) ( f ◦ Y− x )
= YYz Xw (m) ( f ◦ Y− x ◦ X−y )
= XXw (m) ( f ◦ Y− x ◦ X−y ◦ Yz )
∂2 g
( x, y, z, w)
∂x2
= YY−x X−y Yz Xw (m) (Y ( f ))
∂2 g
( x, y, z, w)
∂y2
= XX−y Yz Xw (m) ( X ( f ◦ Y− x ))
∂2 g
( x, y, z, w)
∂z2
= YYz Xw (m) (Y ( f ◦ Y− x ◦ X−y ))
∂2 g
( x, y, z, w)
∂w2
= XXw (m) ( X ( f ◦ Y− x ◦ X−y ◦ Yz ))
∂2 g
( x, y, z, w)
∂x∂y
= XX−y Yz Xw (m) (Y ( f ) ◦ Y− x )
∂2 g
( x, y, z, w)
∂x∂z
= −YYz Xw (m) (Y ( f ) ◦ Y− x ◦ X−y )
∂2 g
( x, y, z, w)
∂x∂w
= − XXw (m) (Y ( f ) ◦ Y− x ◦ X−y ◦ Yz )
∂2 g
( x, y, z, w)
∂y∂z
= −YYz Xw (m) ( X ( f ◦ Y− x ) ◦ X−y )
∂2 g
( x, y, z, w)
∂y∂w
= − XXw (m) ( X ( f ◦ Y− x ) ◦ X−y ◦ Yz )
∂2 g
( x, y, z, w)
∂z∂w
= XXw (m) (Y ( f ◦ Y− x ◦ X−y ) ◦ Yz )
9
Therefore,
lim
t →0+
f ( β(t)) − f ( β(0))
t
=
=
=
=
(∵ dg(t, t, t, t)/dt|t=0
lim
f ( β(t2 )) − f ( β(0))
t2
lim
g(t, t, t, t) − g(0, 0, 0, 0)
t2
lim
dg(t, t, t, t)/dt
2t
lim
d2 g(t, t, t, t)/dt2
2
t →0+
t →0+
t →0+
t →0+
= −Ym ( f ) − Xm ( f ) + Ym ( f ) + Xm ( f ) = 0)
=
1
(Ym (Y ( f )) + Xm ( X ( f )) + Ym (Y ( f )) + Xm ( X ( f )))
2
+ 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 is
enough to consider the case when ω is decomposable, i.e., ω = gdxm1 ∧ ... ∧ dxm p . 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
+ ∑ g det(Y1 ( xmi ), ..., Y0 (Yj ( xmi )), ..., Yp ( xmi ))1≤i≤ p
j =1
LY0 (ω )(Y1 , ..., Yp )
= Y0 ( g) det(Y1 ( xmi ), ..., Yp ( xmi ))1≤i≤ p
p
+ ∑ g det(Y1 ( xmi ), ..., Yj (Y0 ( xmi )), ..., Yp ( xmi ))1≤i≤ p
j =1
ω (Y1 , ..., LY0 Yj , ..., 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 L X = i ( X ) ◦ d + d ◦ i ( X ), L X ω = 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 described
as ∑|Φ|+|Ψ|= p gΦ,Ψ dxΦ ∧ dyΨ . And we can take a vector field X on M × N as ∂x∂ × (C ∞ function which
i
is zero outside the local chart and 1 at (m, n)). Then X(m,n) = ∂x∂ |(m,n) and dπ ( X ) = 0 Therefore, by
i
assumption, L X ω = 0. But, L X ω = ∑|Φ|+|Ψ|= p ∂xΦ,Ψ dxΦ ∧ dyΨ . Therefore ∀ xi , ∂xΦ,Ψ = 0. If we let
i
i
ι m0 : N → M × N : n0 7→ (m0 , n0 ), (δι m ω )n = ∑|Ψ|= p g∅,Ψ (m, n)dyΨ . Therefore, by the result above, on
∂g
10
∂g
the small neighborhood of m ∈ M and for all m0 in the neighborhood, (δι0m ω )n is the same as (δι m ω )n .
Since (m, n) is arbitrary, we can say that δι m ω is locally the same regardless of the choice of m. Also, since
M is connected, we conclude that δι m ω is always the same regardless of the choice of m ∈ M. Let α be
this p-form on N. Then, ω = δπ (α). Indeed, for all (m, n) ∈ M × N, and X1 , ..., X p ∈ ( M × N )(m,n) ,
(ω − δπ (α))(m,n) ( X1 , ..., X p ) = ω(m,n) ( X1 , ..., X p ) − δπ (δι m ω )(m,n) ( X1 , ..., X p )
= ω(m,n) ( X1 , ..., X p ) − ω(m,n) (d(ι m ◦ π )( X1 ), ..., d(ι m ◦ π )( X p ))
= ω(m,n) ( X1 − d(ι m ◦ π )( X1 ), ..., X p − d(ι m ◦ π )( X p ))
The last term is zero since if we set X as a vector field of M × N such that X(m,n) = X1 − d(ι m ◦ π )( X1 )
and dπX = 0, (it is possible since dπ ( X1 − d(ι m ◦ π )( X1 )) = 0, the last term is (i ( X )ω )(m,n) ( X2 − d(ι m ◦
π )( X2 ), ..., X p − d(ι m ◦ π )( X p )) which should be zero by the assumption.
9. (⇐) If a1 v1 + ... + ar vr = 0 and some of ai ’s are nonzero, then without loss of generality, one can assume
a1 6= 0. Dividing the equation by a1 , one can also assume a1 = 1. Then v1 ∧ ... ∧ vn = − a2 v2 − ... − ar vr ∧
v2 ∧ ... ∧ vr = 0.
(⇒) Suppose v1 , ..., v p are linearly independent, then there is a linear map ϕ : V → kr : vi 7→ ei . By
universal property, it induces a map Λr ( ϕ) : Λr (V ) → Λr (kr ). Since ϕ(v1 ∧ ... ∧ v p ) = e1 ∧ ... ∧ e p 6= 0 by
2.6, v1 ∧ ... ∧ v p 6= 0.
10. (⇒) Let vi = ∑ j aij w j . 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+1 ω1 ∧ ... ∧ dωr = α ∧ ω by using (a).
((b) ⇒ (a)) dω = dω1 ∧ ... ∧ ωr + ... + (−1)r+1 ω1 ∧ ... ∧ 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 result
directly follows.
12. For another basis w1 , ..., wn , let wi = ∑ j aij v j . Then Aw1 ∧ ... ∧ Awn = ∑ j a1j Av j ∧ ... ∧ ∑ j anj Av j =
det( aij ) Av1 ∧ ... ∧ Avn . Therefore by Exercise 10, det A does not depend on the choice of basis. If
we 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 , ..., i p } ⊂
{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. Therefore
we can assume that v = eΦ1 , w = eΦ2 . If v 6= w, < v, w >= ∗(w ∧ ∗v) = ∗(v ∧ ∗w) = 0 by definition of
star 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 ∈ Λ p+1 (V ) and w ∈ Λ p (V ), < γ(v), w >=< v, ξ ∧ w >. But it is
equivalent 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.
w
15. (It is true only if ξ 6= 0) Obviously (ξ ∧)2 = 0. Conversely, if ξ ∧ w = 0, then w = ξ ∧ γ( <ξ,ξ
> ).
To prove this, it is enough to show that for all v ∈ Λ p+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 V
which contains ξ and an expression of w in terms of the basis derived by {vi }, say ∑| I |= p+1 a I v I where
I = {vi1 , ..., vi p+1 } ⊂ {vi } and v I = vi1 ∧ ... ∧ vi p+1 , (to clarify the definition, it is assumed that orders
of 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 a I 6= 0 should contain ξ. Therefore, considering an expression
of ∗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 should
be (< ξ, ξ >, 0, ..., 0). It means that < ξ ∧ ∗v, ξ ∧ ∗w > should be < ξ, ξ >< ∗v, ∗w >=< ξ, ξ >< v, w >.
ci ∧ ... ∧ ω p , θi ∧ ω1 ∧ ... ∧ ω p = 0. ∴ θi ∈< ω1 , ..., ω p >. So it is
16. ∑i θi ∧ ωi = 0. If we wedge ω1 ∧ ... ∧ ω
possible to write θi = ∑ j Aij ω j , where Aij are C ∞ functions. One can derive Aij = A ji by just putting
these expressions into the given equation and noting that ωi ∧ ω j ’s are linearly independent.
12
3
Chapter 3
n
1. Let e ∈ U ⊂ G be a locally Euclidean open neighborhood of e. Then by Theorem 3.18, G = ∞
n =1 U .
n
Now, fix a countable basis of U, say { Bi }. Then for all open set V ⊂ U and for all points g ∈ V, g can be
expressed as g = g1 ...gn by definition. Also, since multiplication on G is continuous, we can find open
neighborhoods of gk , say Vk , such that V1 ...Vn ⊂ V. Then, we can find Bik ’s such that gk ∈ Bik ⊂ Vk , and
as a result g = g1 ...gn ∈ Bi1 ...Bin ∈ V. Therefore it follows from this that { Bin } becomes a basis of U n and
U n is second countable, and so is G.
S
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 determinant 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 = ( ss , − sts 1 + t) is continuous.
1
1
(i) It is also a manifold, and ( A, v)( A1 , v1 )−1 = ( AA1−1 , − AA1−1 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. But
ω ( X1 , ..., Xr )(σ) = ωσ ( X1σ , ..., Xrσ ) = (δlσ−1 ωe )( X1σ , ..., Xrσ ) = ωe (dlσ−1 X1σ , ..., dlσ−1 Xrσ ), so we need
to check that ωe (dlσ−1 X1σ , ..., dlσ−1 Xrσ ) is smooth. Now let ω̃ be a smooth form such that ω̃e = ωe .
Then for ϕ : G × G → G : ( a, b) 7→ ab, δϕω̃ 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, δϕω̃ ([0, X1 ], ..., [0, Xr ])(ι(σ )) is smooth
with respect to σ. But, δϕω̃ ([0, X1 ], ..., [0, Xr ])(ι(σ )) = δϕω̃(σ−1 ,σ) ([0, X1 ](σ−1 ,σ) , ..., [0, Xr ](σ−1 ,σ) ) =
ω̃e (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 ) = dla Xib ( f ). Therefore ω̃e (dϕ[0, X1 ](σ−1 ,σ) , ..., dϕ[0, Xr ](σ−1 ,σ) ) =
ωe (dlσ−1 X1σ , ..., dlσ−1 Xrσ ), and the result follows.
∗ (G) = T
∗
(b) δlσ is an algebra endomorphism and Elinv
σ ∈ G { δlσ − invariant elements}, so Elinv ( G ) is a
∗ ( G ) → Λ ( G ∗ ) : ω 7 → ω ( e ) is a homomorphism. And since ω ( e ) = ω 0 ( e ) ⇒
subalgebra. Also, e : Elinv
e
ωσ = δlσ−1 ωe = δlσ−1 ωe0 = ωσ0 ⇒ ω = ω 0 , e is injective. Furthermore, for all χ ∈ Λ( Ge∗ ), if we set
∗ ( G ). Also,
ω ( a) = δla−1 χ, then (δlτ ω )(σ ) = δlτ (ω (τσ)) = δlτ δlσ−1 τ −1 χ = δlσ−1 χ = ω (σ), so ω ∈ Elinv
e(ω ) = χ, so e 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 constant
on G. Also, if α∗ : Ge∗ → g∗ is the dual map of α defined on Proposition 3.7, then α∗ (e(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,
2 ( G ) and dω ( X , X ) = − ω [ X , X ], so we can write dω =
dωi ∈ Elinv
∑ j<k c jki ωk ∧ ω j + ω 0 where
i
j
i
j
i
k
k
2 ( G ) → Λ ( G ) ∼ g ∧ G is an isomorphism, ω 0 must be
ω 0 ( X, Y ) = 0∀ X, Y ∈ ( g). But, since e : Elinv
e =
2
zero. 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 : X̃ → X is a covering and X is simply connected, then by (a), there is a unique continuous map
α̃ : X → X̃ s.t. π ◦ α̃ = id X . Therefore, π ◦ (α̃ ◦ π ) = π. Then by uniqueness, α̃ ◦ π = id X̃ also.
Therefore, α̃ 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 homeomorphic to open ball in Euclidean space. For all U, U 0 ∈ U , U ∩ U 0 has at most countably many components,
each of which is path connected. Let X be a countable set containing one point from each component of
U ∩ U 0 (including U = U 0 ). For each U ∈ U and each x, x 0 ∈ X s.t. x, x 0 ∈ U, let pU
x,x 0 be a path joining x
0
and x and contained in U.
Since the fundamental groups based at any two points in the same component of M are isomorphic, and
X 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 0 .
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 represented 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 are
finitely many numbers 0 = a0 < a1 < ... < ak = 1 such that [ ai−1 , ai ] ⊂ f −1 (U ) for some U ∈ U . For
each i, let f i be the restriction of f to the interval [ ai01 , ai ], reparametrized so that its domain is [0, 1], and
let Ui ∈ U be a coordinate ball containing the image of f i . For each i, we have f ( ai ) ∈ Bi ∩ Bi+1 , and
there 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 constant
path cq based at q. Then, because gi−1 · gi is path homotopic to a constant path,
f
∼
f 1 · ... · f k
∼
g0 · f 1 · g1−1 · g1 · f 2 · g2−1 · ... · gk−−11 · gk−1 · f k · gk−1
∼
f˜1 · f˜2 · ... · f˜n
where f˜i = gi−1 · f i · gi−1 . For each i, f˜i is a path in Ui from xi−1 to xi . Since Bi is simply connected, f˜i is
14
U
path homotopic to p xii−1 ,xi . It follows that f is path homotopic to a special loop, as claimed.
9. (⇒) part is easy since δlσ δϕω = δ( ϕ ◦ lσ )ω = δ(l ϕ(σ) ◦ ϕ)ω = δϕω. For the converse, let I be an
ideal generated by δπ1 δϕ(ωi ) − δπ2 (ωi ) for projections π1 : G × H → G, π2 : G × H → H. Then by
direct calculation similar to 3.15, (6), it is a differential ideal. Therefore, there exists I, maximal connected
integral manifold of I through (e, e) But since δπ1 , δπ2 , and δϕ pull left invariant forms back to left
invariant 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(following
2.33), it is an integral manifold of I . Since G is connected, g( G ) ⊂ I by maximality. Now, let π1 | I : I → G
be a restriction map which is a homomorphism. Then, dπ1 : I(e,e) → Ge is an isomorphism since for
( X, Y ) ∈ I(e,e) ⇔ ∀i, [δπ1 δϕ(ωi ) − δπ2 (ωi )]( X, Y ) = 0 ⇔ Y = dϕ( X ). Therefore by Theorem 3.26, π1 | I is
a 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
−1
−1
(`−
G , ϕ (` G )) · ` is also a path in I which joins ( e, e ) and ( e, ϕ ( σ ) · τ ) since g ( G ) ⊂ I. However, since ”Gcoordinate” of this path is constantly e and there is a local isomorphism from a neighborhood of (e, e) ∈ I
to a neighborhood of e ∈ G, it should be a constant path, which means ϕ(σ−1 ) · τ = e. Therefore, since I
is a Lie subgroup, (σ, τ ) ∈ I ⇒ (σ−1 , τ −1 ) ∈ I ⇒ ϕ(σ ) = τ ⇒ I ⊂ g( G ). Thus I = g( G ) and it means
that 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. −02 −01 = e A . Then regarding A as an element of gl(2, C), one can find a
−1
matrix L = ac db ∈ gl(2, C) s.t. LAL−1 is upper triangular. Then L −02 −01 L−1 = Le A L−1 = e LAL is
−1
−2 0
1
−2ad+bc
ab
=
also upper triangular. L −02 −01 L−1 = ad−
bc
−cd − ad+2bc , thus cd = 0. If c = 0, L 0 −1 L
−1
a
−
1
−
−2 db
−2 0
1
1
−2ad ab
ab
c
. If d = 0, L 0 −1 L
= −bc bc
=
. Therefore, diagonal
ad
0 2bc =
0 − ad
0 −2
−1
0 −1
elements of e LAL should 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 upper
−1
triangular, diagonal elements of e LAL 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. Evaluating
∂
|t=0 , we obtain the given result.
both sides at ∂t
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 is
discrete, 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 : G̃ → G is the universal covering. Then since
π is a local homeomorphism, ker π is discrete, so ker π ∈ Z ( G̃ ), which says that ker π is abelian. But
by lifting property, the fundamental group of G can be identified with a subgroup of ker π, so it is also
abelian.
13. Let the Lie algebra of 3.5(d) be L = (R2 , [·, ·]). For any non-abelian Lie algebra L̄ = (R2 , [·, ·]) which
is defined by [e1 , e2 ] = se1 + te2 , let f : L̄ → L : ae1 + be2 7→ (( at − bs)e1 + ( as + bt)e2 ). It is by
definition 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 ∂y
} is left invariant tangent fields that satisfies [ x ∂x
, x ∂y
] = x ∂y
. Therefore, by Theorem 3.27,
{ x ∂x
Corollary, it is (up to isomorphism) unique simply connected 2-dimensional non-abelian Lie group.
14. Let A =
01
00
,B=
00
01
. Then e A e B =
1e
0e
, e A+ B =
1 e −1
0 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 which
blocks consist of exponent of original blocks. For A an elementary Jordan matrix, let A = λI · N, where
( I − N )K
diagonal entries of N are 1’s. Since ( I − N )k = 0 for k > n, log N = − ∑∞
exists and elog N = N
k =1
k
by calculation dealing with formal series. Also, λI = elog λI for any log λ regardless of choosing the
branch of log. Since log λI is in the center, elog λI +log N = elog λI elog 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 an
isomorphism. Also, for X a left invariant vector field on G, drσ dϕ( X ) = dϕdlσ−1 ( X ) = dϕ( X ), so
dϕ( X ) ∈ h. Also, for η : G × G → G : (σ, τ ) → στ, η (idG , ϕ) is a constant function to e. Thus
0 = 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 group
up to isomorphism. Then, since π : Rn → G is the universal cover of G, G ' Rn / ker π. Also, the kernel
of a covering homomorphism is discrete. Then following the argument in the text, ker π = ⊕ik=1 Zvi for
some linearly independent elements v1 , ..., vk . Therefore, G ' Rn / ⊕ik=1 Zvi ' Rn−k × T k .
19. Clearly A B (V ) is an abstract subgroup of Aut(V ) and dB is a subalgebra of End(V ). According to
T
Theorem 3.42, A B (V ) is a closed Lie subgroup of Aut(V ) since A B (V ) = v,w∈V ker(α 7→ (α(v), α(w)) −
(v, w)) is closed.
Let a be the Lie algebra of A B (V ). If l ∈ a, exp tl ∈ A B (V ). ∴ ((exp tl )(v), (exp tl )(w)) = (v, w). Taking
their derivatives at t = 0, (lv, w) + (v, lw) = 0. ∴ l ∈ dB . Conversely, suppose that l ∈ dB . Then
B(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. But
since et(l ⊗1+1⊗l ) = et(l ⊗1) et(1⊗l ) = (etl ⊗ 1)(1 ⊗ etl ) = etl ⊗ etl , exp tl ∈ A B (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 subgroup
of A( G̃ ) consisting of those automorphisms of G̃ which map D onto D. However, if g : G → G ∈ A( G ),
˜ 1 = id
then by lifting property, there is a unique map g̃ : G̃ → G̃ s.t. π ◦ g̃ = g ◦ π, and since g̃ ◦ g−
G̃
c
c
by uniqueness, g̃ ∈ A( G̃ ). By definition g̃( D ) ⊂ D and g̃( D ) ⊂ D , thus g̃( D ) = D since it is an
automorphism. Conversely, if h ∈ A( G̃ ) and h( D ) = D, there exists a map ĥ : G → G s.t. ĥ ◦ π = π ◦ h
16
and it is also an automorphism since ker(π ◦ h) = ker π. Then it is easy to see that g̃ˆ = g and ĥ˜ = h.
21. ϕ : U (n) → S1 × SU (n) : A 7→ (det A, detA A ) is an diffeomorphism.
22. (a) First, find an eigenvalue λ and a corresponding unit eigenvector v. Since λ||v|| = λv̄t v = v̄t Av =
v̄t Āt v = λ̄v̄t v = λ̄||v||, λ should be real. Then if vt w̄ = 0, < v, Aw >= vt Aw = vt Āw̄ = vt At w̄ =
λvt w̄ = 0. Therefore < v > and < v >⊥ is both A-stable. Then we can use induction on the rank
of A to find orthonormal basis which consists of {vi } eigenvectors of A. Set B = (v1 v2 ...vn )−1 For
the real symmetric case, first find an eigenvalue λ(which is real) and a corresponding eigenvector v
which can be complex valued. Since Av̄ = Av = λv = λv̄, v̄ 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 e A =
−
−
B−1 e BAB 1 B. Since eigenvalues of A is real, e BAB 1 is a diagonal matrix whose diagonal entries are all
positive. Therefore, since B ∈ U (n), e A is also positive definite hermitian. Meanwhile, for a positive
definite hermitian matrix H, we can find B ∈ U (n) as above. Then since eigenvalues of H are all
−1
positive, we can find a real diagonal matrix A s.t. e A = BHB−1 . Then e B AB = H, and B−1 AB is a
0
hermitian matrix. If e A = e A for two hermitian matrices A and A0 , find B ∈ U (n) s.t. BAB−1 is real
−1
0 −1
diagonal, and note that e BAB = e BA B is positive definite diagonal. Also, there exists C ∈ U (n) s.t.
−1 −1
0 −1 −1
CBA0 B−1 C −1 is real diagonal, and eCBAB C = eCBA B C is also positive definite diagonal. But
since CBAB−1 C −1 is also real diagonal by direct calculation, it means CBAB−1 C −1 = CBA0 B−1 C −1 ,
which says A = A0 . Therefore exp is well-defined one-to-one onto.
For a real symmetric A, e A is positive definite hermitian by above and real, so positive definite
symmetric. Injectivity is explained above. For a positive definite symmetric matrix H, we can
find B ∈ O(n) as (a), and there exists a real diagonal matrix A s.t. e A = BHB−1 . Then note that
−1
e B AB = H and B−1 AB is real symmetric.
23. If σ = PR = P1 R1 , then P2 = PRRt Pt = P1 R1 R1t P1t = P12 . Find A, B real symmetric matrices satisfying
e A = P and e B = P1 , and we see that e2A = (e A )2 = P2 = P12 = (e B )2 = e2B , so by injectivity, 2A = 2B
or A = B. It follows that P = P1 . Therefore, R = R1 also.
24. σ σ̄t is positive definite hermitian, so ∃ B ∈ U (n) s.t. Bσ σ̄t B−1 is positive definite diagonal. Set P =
1
B−1 ( Bσ σ̄t B−1 ) 2 B, which is positive definite hermitian, and R = P−1 σ a unitary matrix.
25. U (n) is connected, and there exists a path tI + (1 − t) P joining I and P a positive definite hermitian
matrix. 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 0 ( x ) = δ f x (i ( N ( x ))ω f ( x) ) for a smooth
nowhere-vanishing normal vector field N along ( X, f ). Then it is non-vanishing d-form on X, since for
x ∈ X and X1 , ..., Xd ∈ Xx which are linearly independent, ωx0 ( X1 , ..., Xd ) 6= 0.
(⇒) Let ω 0 be a nowhere-vanishing d-form on X. For x ∈ X choose X1 , ..., Xd ∈ Xx which satisfies
f1 ∧ ... ∧ X
fd = ωx . Then we can pick N ( x ) ∈ (Rd+1 ) f ( x) s.t. N ( x ) is orthogonal to d f ( Xx ) and
X
ω (d f ( X1 ), ..., d f ( Xd ), N ( x )) = 1 where ω = dr1 ∧ ... ∧ drd+1 . N ( x ) is independent of the choice of
fd = ωx , one can express Yi = ∑ j aij X j ,
X1 , ..., Xd ∈ Xx , since for Y1 , ..., Yd which also satisfies Ye1 ∧ ... ∧ Y
then using Exercise 2.10, we see that det aij = 1. Thus ω (d f (Y1 ), ..., d f (Yd ), N ( x )) = 1 also. Then since
d f ( Xx )⊥ is 1-dimensional, N ( x ) which satisfies above condition is uniquely determined. Also, it is by
definition nowhere-vanishing. It remains to show that N is smooth, but locally on the neighborhood of
f1 ∧ ... ∧ X
fd = ω 0 on the neighborx ∈ X we can choose smooth vector field X1 , ..., Xd which satisfies X
hood, and d f ( X1 ), ..., d f ( Xd ) is also smooth on the neighborhood of f ( x ) over f ( X ). Therefore locally N
is 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 such
that it is preserved by antipodal map. Now regard Sn as imbedded to Rn+1 , and let N be the outward
unit normal vector field on Sn and ω = dr1 ∧ ... ∧ drn+1 . Then ω 0 = i ( N )ω is nowhere-vanishing n-form
on Sn . If ϕ : Rn+1 → Rn+1 : x 7→ − x, ϕ|Sn is the antipodal map, and δϕ x ωx0 = (−1)n i ( N ( x ))ω− x =
0 . Therefore, if n is odd, ϕ is orientation-preserving, therefore Pn is
(−1)n+1 i ( N (− x )ω− x ) = (−1)n+1 ω−
x
oriented. But if n is even, ϕ is orientation-reversing, so we cannot take a nowhere-vanishing n-form on
Sn 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 , X j,m )}i,j where g is given Riemannian metric. Then near p ∈ X, it is a smooth function to the set
of 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 = f 1 e1 + ... + f n en . Then Ṽ = f 1 ω1 + ... + f n ωn , so ∗Ṽ =
f 1 ω2 ∧ ... ∧ ωn − ... + (−1)n−1 f n ω1 ∧ ... ∧ ωn−1 . Therefore on ∂D, ∗Ṽ = f 1 ω2 ∧ ... ∧ ωn since ω1 |∂D = 0.
R
R
On the other hand, hV, ~ni = f 1 and ω2 ∧ ... ∧ ωn is the volume form on ∂D. Thus ∂D ∗Ṽ = ∂D f 1 . Since
R
R
R
R
D divV = D ∗ d ∗ Ṽ = D d ∗ Ṽ = ∂D ∗Ṽ by Stokes’ theorem, the result follows.
R
R
R
R
R
R
R
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 hd f , dgi
R
R
R
R
R
R
R
fi −
∴ D h grad f , gradgi − D f ∆g = D hdff , dg
D f ∆g = D d ( f ∗ dg ) = ∂D f ∗ dg = ∂D ∗( f dg )
∂g
Using the proof of Exercise 4, ∗( f dg) = h g
f dg, ~ni = f ∂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 space
of 1-forms is 1-dimensional, we only need to show that they have the same nonzero value for some
arbitrary n-tuples of vectors. If we put X1 , ..., Xn , the left side becomes ω ( X1 , ..., Xn )2 , which is the same
18
as X̃1 ∧ ... ∧ X˜n ( X1 , ..., Xn ) = det{h Xi , X j i} since ω is the volume form.
7. Since λ̄(στ ) = λ̄(σ )λ̄(τ ), we only need to show that it is C ∞ at e. On the local chart of e with coordinate functions x1 , ..., xn centered at e, ω can be expressed as f ( x1 , ..., xn )dx1 ∧ ... ∧ dxn where f is C ∞ .
Since λ̄(σ )ωe = δrσ (ωσ ), λ̄(σ ) f (0, ..., 0)dx1 ∧ ... ∧ dxn = f ( x1 (σ), ..., xn (σ ))d( x1 ◦ rσ ) ∧ ... ∧ d( xn ◦ rσ ) =
f ( x1 (σ ), ..., xn (σ ))
det{hdxi ◦rσ ,dx j i}
dx1
det{hdxi ,dx j i}
∧ ... ∧ dxn , so λ̄(σ) =
f ( x1 (σ ),...,xn (σ )) det{hdxi ◦rσ ,dx j i}
f (0,...,0)
det{hdxi ,dx j i}
which is C ∞ .
8. (This is one of characteristics of Haar measure on a unimodular Lie group. Here is the sketch of the
proof.) Since δrσ δα(ω ) = δαδlσ−1 (ω ) = δα(ω ), it is right invariant. Let ω 0 = δα(ω ) and think of µ, µ0
R
measures on G corresponding to ω and ω 0 respectively. Then we can define f 7→ G f µ0 which is wellR
defined right-invariant integral on G. Now think of G×G f (τσ)µ(σ)µ0 (τ ); if we first integrate for µ it
R
R
R
gives ( G f ω )( G ω 0 ). and if we integrate for µ0 first, we get G f ω 0 . By Fubini’s theorem, they must be
R
R
R
R
R
R
R
the same. Also, G ω 0 = G δα(ω ) = α(G) ω = 1, so G f ω = G f ω 0 . Since G f ω 0 = α(G) ( f ◦ α)ω, the
result follows.
9. (⇒) Since hdlσ X, dlσ Y i = hdrτ X, drτ Y i = h X, Y i, Ad( G ) in Aut( g) is actually in O( g). Since O( g) is
compact, 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, so
all elements of Ad( G ) preserve the inner product. Now for σ ∈ G and X, Y ∈ Gσ , define h X, Y iσ =
hdlσ−1 X, dlσ−1 Y ie . Then by definition it is left-invariant, and since hdrτ X, drτ Y iστ =
hdlτ −1 σ−1 drτ X, dlτ −1 σ−1 drτ Y ie = hdAdτ −1 dlσ−1 X, dAdτ −1 dlσ−1 Y ie = hdlσ−1 X, dlσ−1 Y ie = h X, Y iσ , it is also
right-invariant.
10. (a) Since Rn ∼
= D (0, 1) ⊂ Rn , singular homology groups of the two must be the same. Then by de Rham
theorem, it is also the same as de Rham cohomology groups. By Poincaré Lemma, it must be zero for
given conditions.
(b) By de Rham theorem, it suffices to show that the 0-th singular homology group is R. But since it is
isomorphic 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)k α ∧ dβ = 0. If β = dγ, then d(α ∧ γ) = dα ∧ γ + (−1)k α ∧ β, so α ∧ β is exact.
13. By Stokes’ theorem,
R
z
α=
R
int z
dα = −8
R
int z
dx ∧ dy = −8.
14. It is closed by direct calculation. Also d(sin xy + x2 ) = α, it is exact. Therefore,
R
z
α=
R
int z
dd f = 0.
R
R xdy−ydx R dx∧dy
15. It is closed by direct calculation. Also, S1 α = S1 2π
= D π = 1. If δi (α) is exact, then integral
1
1
on S must be zero since S 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
R
(c) S2 σ = 3Vol (S2 ), so it is not zero. If it is exact it must be zero since S2 is a compact manifold without
boundary.
c ∧...∧dr
∑n (−1)i−1 r dr ∧...∧dr
n
i 1
i
. It is closed by similar calculation to (b).
(d) ∗α = i=1
n
(∑in=1 ri2 ) 2
R
(e) Sn−1 ∗α = nVol (Sn−1 ), so it is not zero, and ∗α is not exact.
1
1
17. HdeRham
(S2 ; R) = 0 6= R2 = HdeRham
( 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 (−e, 1 + e) on M × (−e, 1 + e). For a k-form ω =
∂
ω1 + dt ∧ ω2 on M × (−e, 1 + e), i ( ∂t
)ω = ω2 is a (k-1)-form, and its expression does not contain dt
R1
R1 ∂
term. Therefore, we can define 0 i ( ∂t )ωdt = 0 ω2 dt which is a (k-1)-form on M. Denote it Dk (α). For
convenience of the notation, denote d M as a differential operator on M, and d = d M + dt as a differential
R1 ∂
R1
operator on M × (−e, 1 + e). Then Dk d(ω ) = 0 i ( ∂t
)(dω1 − dt ∧ dω2 )dt = 0 L ∂ (ω1 ) − d M ω2 dt, and
∂t
R1
R1
dDk (ω ) = d M 0 ω2 dt = 0 d M ω2 dt. Therefore dDk (ω ) + Dk d(ω ) = i1∗ ω1 − i0∗ ω1 = i1∗ ω − i0∗ ω by
definition of Lie derivative and i0 , i1 (i0∗ (dt ∧ ω2 ) = i1∗ (dt ∧ ω2 ) = 0). It means i0∗ and i1∗ are the same
map on the cohomology level. Since f = F ◦ i0 and g = F ◦ i1 , f ∗ = i0∗ ◦ F ∗ and g∗ = i1∗ ◦ F ∗ also induce
the 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 that
ω = δ f (i (~n)(dr1 ∧ ... ∧ drn+1 )) satisfies ω (v1 , ..., vn ) = 1 for each point m ∈ M and v1 , ..., vn ∈ Mm which
is an oriented orthonormal basis of Mm . Since ω (v1 , ..., vn ) = dr1 ∧ ... ∧ drn+1 (~n, d f (v1 ), ..., d f (vn )) = 1
by assumption, the result follows.
∂
∂
∂
− f y ∂y
+ ∂z
). Therefore,
(b) From vector calculus we obtain ~n = q 2 1 2 (− f x ∂x
f x + f y +1
ω
= δϕ(i (~n)(dx ∧ dy ∧ dz))
= δϕ( q
=
=
1
f x2
+ f y2 + 1
1
q
f x2 + f y2 + 1
1
q
f x2 + f y2 + 1
(− f x dy ∧ dz + f y dx ∧ dz + dx ∧ dy))
(− f x dy ∧ d f + f y dx ∧ d f + dx ∧ dy)
( f x2 + f y2 + 1)dx ∧ dy =
20
q
f x2 + f y2 + 1 dx ∧ dy
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 containing
f ( 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 homeomorphism
of 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 a
homeomorphism, and f |π ( L) is its inverse, it is also continuous. Since x ∈ X is arbitrary, f is continuous
on 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 a
homeomorphism 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 ). Also
f ( 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 is
a 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 ) is
nonempty. 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 a
section) by Exercise 1, the second statement follows.
4. It is because f x = 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 f 0 6= 0.
5. It is because projection of sheaves are local homeomorphisms and a sheaf mapping is locally equal to the
composition 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 = π1−1 (π2 ( ϕ(U ) ∩ ψ(U ))), which is an open neighborhood of x
since π1 (U ) ⊃ π2 ( ϕ(U ) ∩ ψ(U )), ϕ(y) = ( ϕ ◦ (π1−1 ◦ π2 ◦ ψ))(y) = (( ϕ ◦ π1−1 ◦ 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 projection
map, 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, then
R ∩ π −1 (U ) + f (U ) is open in S .
Proof) Let p x ∈ R ∩ π −1 (U ) + f (U ) and π ( p x ) = x. By definition, there exist q x ∈ R and r x ∈ f (U )
21
s.t. p x = q x + r x . Let U be an open neighborhood of p x which is locally homeomorphic to π (U ). Since
addition is continuous, there exist V ⊂ R ∩ π −1 (U ) and W ⊂ f (U ) neighborhoods of q x and r x respectively, s.t. π (V ) = π (W ), V + W ⊂ U, and π |V : V → π (V ), π |W : W → π (W ) are homeomorphisms. Let Y = π (V ) = π (W ). Then V + W ⊂ U ∩ π −1 (Y ). Also, y ∈ U ∩ π −1 (Y ), then
there exists aπ (y) ∈ V ∩ Sπ (y) and bπ (y) ∈ W ∩ Sπ (y) . Then aπ (y) + bπ (y) ∈ U ∩ Sπ (y) . But since
π |U : U → π (U ) is 1-1, y = aπ (y) + bπ (y) ∈ V + W. Therefore V + W = U ∩ π −1 (Y ) is open. But
it means p x ∈ V + W ⊂ R ∩ π −1 (U ) + f (U ), so R ∩ π −1 (U ) + f (U ) is open.
Now come back to the question. First, we need to show that π̄ : 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. Let
f : π (U ) → U be the inverse of π |U , which is a section. Since ϕ−1 ( ϕ(U )) = U + R ∩ π −1 (π (U )) and
it is open in S by Lemma, ϕ(U ) is open. Also if W ⊂ ϕ(U ), ϕ−1 (W ) = f −1 (π̄ (W )) + R ∩ π −1 (π̄ (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 )
−1
is open. It means ϕ−1 is also continuous. Therefore, ϕ|U is a homeomorphism, and π̄ | ϕ(U ) = π ◦ ϕ|U
is
a homeomorphism also. Clearly x ∈ ϕ(U ). Therefore π̄ : R → M is a local homeomorphism.
2) π̄ −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 map
ϕ ◦ ϕ : S ◦ S → T ◦ T . Let ψ : S ◦ S → S be a composition map. Then ψ̄ : T ◦ T → T is a composition
map on T . To check that it is continuous, we need to verify that ψ̄−1 (U ) is open in T ◦ T for an open
set U ⊂ T . But it is by definition equal to whether ( ϕ ◦ ϕ)−1 (ψ̄−1 (U ))) is open, and it is open since
∗ ϕ ◦ ϕ)−1 (ψ̄−1 (U )) = ψ−1 ( ϕ−1 (U )). Therefore the composition map is continuous.
Therefore π̄ : T → M is a sheaf. It remains to show that ϕ is a sheaf homomorphism, but it follows from
the proof of 1).
8. Let ϕ : β(α(S)) → S : ρ p,U f → f ( p).
1) If ρ p,U f = ρ0p0 ,U 0 f 0 , p = p0 , and ∃V ⊂ U ∩ U 0 and p ∈ V s.t. f V = f V0 . Therefore f ( p) = f 0 ( p0 ) and ϕ is
well-defined.
2) Let s ∈ S and s ∈ V s.t. π |V : V → π (V ) be a homeomorphism. If ϕ(ρ p,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 means
ρ p,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→ ρ p,U f . (Since π is a local homomorphism, we can find
−1
a section f = π |U
and p ∈ M s.t. f ( p) = x ∀ x ∈ S .)
0
0
1) If f ( p) = f ( p ) for f ∈ Γ(S , U ) and f 0 ∈ Γ(S , U 0 ), p = p0 ∈ U ∩ U 0 , and ∃V ⊂ U ∩ U 0 s.t. f |V = f 0 |V
by Exercise 3. Therefore ρ p,U f = ρ p0 ,U 0 f 0 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 = limm∈U (α(S) ⊗ α(T ))U = limm∈U α(S)U ⊗ α(T )U = limm∈U Γ(S , U ) ⊗ Γ(T , U )
−→
−→
−→
= lim
Γ(S , U ) ⊗ limm∈U Γ(T , U ) = Sm ⊗ Tm , since direct limit and tensor product commute. The last
m
∈
U
−→
−→
equality follows from β(α(S)) ' S by Exercise 8.
10. Let C p and Cq be two skyscraper sheaves with p 6= q. Then C p ⊗ Cq is a zero sheaf, but Γ(C p , 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-section
as 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 and
U1 , U2 , U3 ∈ Φ. Using the process of Theorem 5.12, we can find s ∈ SU3 s.t. t|U3 is the image of s. Then
c
s|U3 \U1 has a zero image, so it is in RU3 \U1 . Now consider a cover U2 , U1 . Since R is fine, there exist an
c
c
endomorphism ϕ s.t. supp( ϕ) ⊂ U1 and ϕ = 1 on U2c . Also, supp( ϕ ◦ s|U3 \U1 ) ⊂ U1 , so we can extend
it to all of U3 which is valued zero on U1 . Denote it s0 ∈ SU3 . Then the image of s − s0 is also t|U3 and
by construction s − s0 |U3 \U c = 0, so supp(s − s0 ) ⊂ U2 ⊂ U3 . Thus one can extend it to all of M which
2
is valued zero on U3c , and s − s0 ∈ ΓΦ (S). Also, t is the image of s − s0 . The statement is proved.
13.
...
/ H q −1 ( E ∗ )
∂
/ H q (C ∗ )
f
/ H q ( D∗ )
g
/ H q ( E∗ )
∂
0
/ C q −1
/ D q −1
/ E q −1
/ 0
0
/ Cq
/ Dq
/ Eq
/ 0
0
/ C q +1
/ D q +1
/ E q +1
/ 0
/ H q +1 ( C ∗ )
/ ...
1) f ∂ = 0. Let [e] ∈ Eq−1 for e ∈ Z ( Eq−1 ) and d ∈ D q−1 , d0 ∈ D q , c ∈ C q s.t. d 7→ e, d 7→ d0 , c 7→ d0 . Then
by definition of ∂, f ∂[e] = f [c] = [d0 ] = 0 ∈ H q ( D ∗ ).
2) im∂ ⊃ ker f . Let c ∈ Z (C q ) and d0 ∈ D q s.t. c 7→ d0 . If f [c] = [d0 ] = 0, ∃d ∈ D q−1 s.t. d 7→ d0 . Let
e ∈ Eq−1 s.t. d 7→ e. Then e 7→ 0 ∈ Eq since c 7→ d0 7→ 0 ∈ Eq . Therefore e ∈ Z ( Eq−1 ), and ∂[e] = [c] by
definition of ∂.
3) g f = 0. Since the original sequence is exact.
4) im f ⊃ ker g. Let d ∈ Z ( D q ) and d 7→ e ∈ Eq . If g[d] = [e] = 0, ∃e0 ∈ Eq−1 s.t. e0 7→ e. Let d0 ∈ D q−1
s.t. d0 7→ e0 and d˜ ∈ D q s.t. d0 7→ d.˜ Then d − d˜ 7→ 0 ∈ Eq , so ∃c ∈ C q s.t. c 7→ d − d.˜ Now let c0 ∈ C q+1
s.t. c 7→ c0 . Then c0 7→ 0 ∈ D q+1 since d − d˜ 7→ 0 ∈ D q+1 , so c0 = 0 since C q+1 → D q+1 is injective.
Therefore c ∈ Z (C q ), and f [c] = [d − d˜] = [d].
5) ∂g = 0. Let d ∈ Z ( D q ), e ∈ Eq s.t. d 7→ e. Then d 7→ 0 ∈ D q+1 , so e 7→ 0 ∈ Eq+1 , which means
e ∈ Z ( Eq ). Also, ∂g[d] = ∂[e] = 0 by definition of ∂, since 0 ∈ C q+1 7→ 0 ∈ D q+1 .
6) img ⊃ ker ∂. Let e ∈ Z ( Eq ), d ∈ D q , d0 ∈ D q+1 , c0 ∈ C q+1 s.t. d 7→ e, d 7→ d0 , c0 7→ d0 . Then
∂[e] = [c0 ] = 0 means ∃c ∈ C q s.t. c 7→ c0 . Let d˜ ∈ D q s.t. c 7→ d.˜ Then d − d˜ 7→ 0 ∈ D q+1 , so
23
d − d˜ ∈ Z ( D q ). Also, g[d − d˜] = [e].
14.
0
/ Γ (K ⊗ S )
0
/ Γ (K ⊗ S )
/ Γ(C0 ⊗ S )
/ Γ(C1 ⊗ S )
ϕ0
id
ϕ1
/ Γ(C˜0 ⊗ S )
/ Γ(C˜1 ⊗ S )
/ Γ(C2 ⊗ S )
/ ...
ϕ2
/ Γ(C˜2 ⊗ S )
/ ...
(a) Since Z (Γ(C0 ⊗ S )) = Γ(K ⊗ S ) = Z (Γ(C˜0 ⊗ 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
/ C˜q ⊗ T
(c) By using 5.17.
15. If U = i Vi , f i ∈ A p (Vi s.t. f i = f j on Vi ∩ Vj , then set f ∈ A p (U ) s.t. f ( x1 , ..., x p+1 ) = f i ( x1 , ..., x p+1 ) if
x1 , ..., x p+1 ∈ Vi and otherwise f = 0. Then f is well-defined, and f |Vi = f i .
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.
S
16. 0
0
/ Z/2Z
×2
π
/ Z/4Z
/ Z/2Z ⊗ Z/2Z
/ Z/2Z
/ 0 is exact, but
×2⊗id=0
/ Z/4Z ⊗ Z/2Z
π ⊗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 which
is fine.
18. Since exactness is local statement, we only need to think of module cases. However, since stalks of
torsionless sheaves are flat, the result follows. Or a long exact sequence can be divided into short exact
sequences, and use Proposition 5.15.
p
p
19. Naturally f : M → N induces f ∗ : H∆ ( N; G ) → H∆ ( M; G ) s.t. f ∗ ([ ϕ])(σ ) = ϕ( f (σ )) which is welldefined. Also that it satisfies functorial properties is straightforward. Therefore the last statement
follows from a basic property of a functor.
p
20. Denote the given map ϕ : H∆ ∞ ( M; R) → ∞ H p ( M; R)∗ . Let [ f ] ∈ ker ϕ. Then f (σ ) = 0 for any cycle σ.
Now set g ∈ S p−1 ( M, R) s.t. for (p-1)-boundary ∂τ, g(∂τ ) = f (τ ) and otherwise 0. It is well-defined
since if ∂τ = ∂τ 0 then τ − τ 0 is a cycle, so f (τ − τ 0 ) = 0. Then dg = f , so [ f ] = 0. Therefore ϕ is
injective.
By the way, if g ∈ ∞ H p ( M; R)∗ , set f ∈ S p ( 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. (Maybe
the 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 E p ( M ), then ∆ = δd + dδ = (−1)n( p+2)+1 ∗ d ∗ d + (−1)n( p+1)+1 d ∗ d∗. (Be careful about
signs.) Therefore ∗∆ = (−1)n( p+2)+1+(n− p) p d ∗ 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 2
and 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 ∈ ( H p )⊥ 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 ∈ ( H p )⊥ , ∆Gα = α, ∆Gβ = β By definition. Thus, h Gα, βi = h Gα, ∆Gβi = h∆Gα, Gβi =
hα, Gβi.
(c) Suppose {αn } is a bounded sequence. Then { Gαn } is also bounded by (a), and {∆Gαn } is also
bounded because ||∆Gαn || = ||αn − H (αn )|| ≤ ||αn ||. Then use 6.6 to obtain the result.
R
3. It suffices to show that Rn (1 + x12 + ... + xn2 )−k dx1 ...dxn converges for k ≥ [ n2 ] + 1. By change of
R
variables yi = √ xi 2 for 1 ≤ i ≤ n − 1 and yn = xn , it is equal to Rn (1 + y2n )−k (1 + y21 + ... +
1+ x n
R
R
n −1
n −1
y2n−1 )−k (1 + y2n ) 2 dy1 ...dyn = Rn (1 + y21 + ... + y2n−1 )−k (1 + y2n ) 2 −k dy1 ...dyn = Rn−1 (1 + y21 + ... +
R
n −1
n −1
∂( x ,...,x )
y2n−1 )−k dy1 ...dyn−1 R (1 + y2n ) 2 −k dyn (∵ ∂(y1 ,...,yn ) = (1 + y2n ) 2 ). The former part converges by inducn
1
tion on n, and the latter one converges when 2k − (n − 1) > 1, or k > n2 . It is equivalent to k ≥ [ n2 ] + 1
because k is an integer.
4. ∑t[α]=0 || D α ϕ||2 = ∑t[α]=0 ∑ξ ξ 2α | 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 ξ 2α ≤ (1 + |ξ |2 )t for all ξ. The second inequality is clear because
each term in ∑t[α]=0 ξ 2α 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 ◦0 ψ = ϕ ◦ Aψ. Also it is smooth because the volume form on M is smooth.
6. I will show only when α = f dx1 ∧ ... ∧ dx p . The general case is similar. By definition, dα =
∂f
(−1) p ∑in= p+1 ∂x
dx1 ∧ ... ∧ dx p ∧ dxi , and ∗α = f dx p+1 ∧ ... ∧ dxn . (assuming ∗1 = dx1 ∧ ... ∧ dxn ) Also
i
n
(
p
+
1
)+
1 ∗ d ∗ α = (−1)n( p+1)+1 p (−1) p(i −1)+( p−1)( p−i ) ∂ f dx ∧ ... ∧ dx
ci ∧ ... ∧ dx p . Furtherδα = (−1)
∑
1
more, by tedious calculation, we get ∆α =
i =1
2
− ∑in=1 ∂∂x2f dx1
i
∂2 ϕ
∂xi
∧ ... ∧ dx p .
∂2 ϕ
7. By 6.16(14), − ∂x∂y = ∑(a,b)∈Z2 abϕ(a,b) ei(ax+by) . Thus || ∂x∂y ||2 = ∑(a,b)∈Z2 a2 b2 | ϕ(a,b) |2 by Parseval identity. Similarly, we get ||∆ϕ||2 = ∑(a,b)∈Z2 ( a2 + b2 )2 | ϕ(a,b) |2 . Thus the given inequality follows from
a2 b2 ≤
( a2 + b2 )2
.
4
8. If { f n } is bounded in C1 , then it is uniformly bounded and equicontinuous by assumption, thus by
Arzela-Ascoli theorem it contains a convergent subsequence.
9. Differential operators given in the following 4 problems are all elliptic. Therefore we can deduce that a
solution 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) = −u0
(b) Adjoint: D ∗ (u) = −u0 − u
(c) Adjoint: D ∗ (u) = u00
(d) Adjoint: D ∗ (u) = u00 + 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 hδα, δαi = hdδα, αi = 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 uξ eix◦ξ as a formal sum is
uniquely determined. It remains to show that in fact u ∈ H−∞ . By 6.22(2), we know that |l ( ϕ)| ≤
const ∑[α]≤k || D α ϕ||∞ ≤ const|| ϕ|| N for some N 0. In other words, | ∑ξ uξ ϕξ | ≤ const|| ϕ|| N . By
squaring both side, it becomes | ∑ξ uξ ϕξ |2 ≤ const|| ϕ||2N . Since multiplying eiθ (θ ∈ R) does not
change absolute values of ϕξ , we may assume that uξ k 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 ∈
N+M
H− N . Now set | ϕξ | = (1 + |ξ |2 )− 2 for some M > [ n2 ] + 1. Then ϕ ∈ H− N by 6.16(7), and the inequality 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)| = |hu, ϕi0 | ≤ ||u||s || ϕ||−s
s
≤ const
∑ || Dα ϕ||2
(6.18(e))
(6.16(15))
[α]≤−s
≤ const
s
∑
|| D α ϕ||2∞ ≤ const
[α]≤−s
∑
|| D α ϕ||∞
[α]≤−s
as desired.
R
12. It suffices to show that if M α = 0, then it is an exact form. En = imd ⊕ Hn because imδ = 0. Let
R
α = dγ + e with e harmonic. Since [α] 7→ M α is the natural isomorphism, [α] = [e] = 0. Hence by
6.11, e = 0. In other words, α is exact.

13. Let ϕn = ∑ a∈Z una eiax , where una = 
1
a2
if | a| < n
0
otherwise
. Then || ϕn ||2 = ∑ a∈Z |una |2 < ∑ a∈Z
1
a4
and
||∆ϕn ||2 = ∑ a∈Z a2 |una |2 < ∑ a∈Z a12 are bounded. Also, ϕn is a finite sum, so ϕn ∈ P . However, its limit
ϕ = ∑ a∈Z a12 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 is
that ∆ is an elliptic operator. Also, by ”Reduction to the Periodic Case” argument, we only need to
show this locally.
15. It is elliptic because x + iy 6= 0 for ( x, y) 6= (0, 0). Hence every holomorphic function is C ∞ . Also every
∂2
∂2
∂
∂
∂
∂
holomorphic function is harmonic because ∆ = ∂x
2 + ∂y2 = ( ∂x − i ∂y )( ∂x + i ∂y ). If a holomorphic
R
R
R
∂f
function f has compact support, then by Green’s 1st identity, ∂D f ∂n = D h grad f , grad f i − D f ∆ f for
R
every disc D. Since ∆ f = 0, if we take a sufficiently large disc D, then it becomes D h grad f , grad f i = 0.
27
R
By letting the radius of D goes to infinity, we see that C h grad f , grad f i = 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, λhu, ui = hλu, ui = h∆u, ui = h(dδ + δd)u, ui = hdu, dui + hδu, δui ≥ 0. Thus
λ ≥ 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 = λ0 v, then λhu, vi = h∆u, vi = hu, ∆vi = λ0 hu, vi. Thus hu, vi = 0.
(e) hψj , Gψj i = h∆Gψj , Gψj i = hdGψj , dGψj i + hδGψj , δGψj i ≥ 0. For the latter part, we need to
show that l ((∆ − η1 )∗ α) = 0. It is equivalent to lim j→∞ η h Gϕ j , (∆ − η1 )∗ αi = 0 ⇔ lim j→∞ η h(∆ −
1
η ) Gϕ j , α i = 0 ⇔ lim j→∞ h η ϕ j − Gϕ j , α i = 0, which is true. Also it is not trivial; take N such that
|| Gϕ N − η ϕ N || < ε and || Gϕi − Gϕ N || < ε for i > N. Since || ϕ N || = 1 and by the inequality
|hv, wi| ≥ ||v|| − |hv, v − wi| ≥ ||v||(||v|| − ||v − w||), we see that |h Gϕi , η ϕ N i| ≥ η (η − 2ε) > 0 for
sufficiently small ε > 0. Thus l ( ϕ N ) 6= 0.
(f) It is the same as (e).
(g) α − ∑ik=1 hα, ui iui ∈ ( H p )⊥ , so there exists β ∈ ( H p )⊥ . If k + 1 ≤ j, hα, u j i = hα − ∑ik=1 hα, ui iui , u j i =
h Gβ, u j i = h β, Gu j i = λ1j h β, u j i. Thus hα, u j iu j = λ1j h β, u j iu j = G (h β, u j iu j ). 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, E p ( M ) = im∆ ⊕ H p = im∆2 ⊕ V ⊕ H p with V = ker∆2 ∩ im∆. If ∆α ∈ V,
h∆α, ∆αi = h∆2 α, αi = 0, so ∆α = 0. Thus V = 0 and E p ( M) = im∆2 ⊕ H p . It means ∆2 α = β is
solvable if and only if β ⊥ H p .
18. We can assume the matrix which consists of coefficients of highest order operators is symmetric because
∂
∂
n
t
∂xi and ∂x j commutes. Also, L is elliptic at x if for all v ∈ R \ {0}, v ( ai j ( x ))ij v > 0, which means that
( ai j( x ))ij is definite. But for , the corresponding matrix 10 −01 is not definite, so it is not elliptic. For
example, if we let

x3 + 3xy2
if x ≥ |y|


 3x y + y3
if − t ≤ x ≤ t
u( x, y) = 

y
3
 −3x − y
if t ≤ x ≤ −t

− x3 − 3xy2 if x ≤ |t|
it is easy to see that u = 0, but u ∈
/ C 3.
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 ∈ E p ( M). (In fact, the minimum eigenvalue
of ∆ is 0.)
20. The first assertion is true because lσ are isometries, thus they commute with d and δ. Then the second
assertion 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, and
we 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, by
∂ϕ
∂ϕ
6.16(15), const|| ϕn ||21 ≤ || ϕn ||2 + || ∂xn ||2 + || ∂yn ||2 . Since { ϕn } is monotone increasing, || ϕn ||2 →
R
∂ϕn
1 2
x
r ≤ 21 (log log r ) dxdy < ∞ by Monotone convergence theorem. Also, ∂x = r ( 1 +r ) log( 1 +r ) and we
n
n
R
∂ϕ
x
2 dxdy = π . Similarly, || ∂ϕn ||2 → π .
can use similar argument to see that || ∂xn ||2 → r≤ 1 ( r2 log
)
∂x
log 2
log 2
r
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 may
take 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 the
volume form), we conclude that ϕγ also preserves imL∗ . Thus u ◦ γ ∈ imL∗ . Let u ◦ γ = L∗ v0 . Then
LL∗ v = LL∗ v0 ⇒ LL∗ (v − v0 ) = 0 ⇒ 0 = h LL∗ (v − v0 ), v − v0 i = h L∗ (v − v0 ), L∗ (v − v0 )i ⇒ L∗ v = L∗ v0 ,
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.
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