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HW 5
Differential Geometry II, Spring 2014
HOMEWORK ASSIGNMENT 5
MICHAEL EICHMAIR
1. Bookkeeping questions
If you would like feedback on your work on some or all of these questions, please hand your written solutions
to your course assistant or put them into their mailbox in HG F 28 by Friday March 28.
Exercise 1.1 (Submanifolds). Let N be a manifold and S ⊂ N . We say that S is a submanifold of
N if there is a smooth structure on S compatible with the subset topology that turns the inclusion map
i : S → N into an embedding. Explain why there can only be one smooth structure on S with this property.
Solution: Denote the subspace topology on S by τ and suppose that the topological manifold (S, τ ) admits
two smooth atlases A1 and A2 that turn the inclusion map i : S → N into an embedding. We show that the
transition maps of A1 ∪ A2 are smooth, which implies that there can only be one smooth structure on (S, τ ).
Let ϕ1 : U1 → O1 be a chart of (S, τ, A1 ) and let ϕ2 : U2 → O2 be a chart of (S, τ, A2 ) such that O1 ∩ O2 6= ∅.
Consider a point p ∈ O1 ∩ O2 and let ϕ : U → O be a chart of N with p ∈ O. Since i : S → N is an
−1 ◦ ϕ : ϕ−1 (O) → U are immersions
embedding, it follows that the maps ϕ−1 ◦ ϕ1 : ϕ−1
2
2
1 (O) → U and ϕ
−1
−1
and homeomorphisms with their images ϕ (O1 ) and ϕ (O2 ). The technical lemma then implies that the
map
−1
−1
−1
◦ ϕ1
ϕ−1
◦ ϕ−1 ◦ ϕ2 : ϕ−1
1 ◦ ϕ2 = ϕ
2 (O ∩ O1 ) → ϕ1 (O ∩ O2 )
−1
−1
is smooth. We obtain that the transition map ϕ−1
1 ◦ ϕ2 : ϕ2 (O1 ) → ϕ1 (O2 ) is smooth. Analogously we
−1
−1
show that ϕ−1
2 ◦ ϕ1 : ϕ1 (O2 ) → ϕ2 (O1 ) is also smooth. We conclude that the atlases A1 and A2 are
equivalent.
Exercise 1.2 (Real projective space). Consider the equivalence relation on Rm+1 \ {0} where
p1 ∼ p2
if, and only if,
p1 = λp2
for some λ 6= 0.
Explain why the topological quotient of Rm+1 \ {0} with respect to this equivalence relation is a second
countable Hausdorff topological space. Show that the maps
Rm+1 \ {0}
given by
(x1 , . . . , xm ) 7→ [(x1 , . . . , xi−1 , 1, xi , . . . , xm )]
∼
are homeomorphisms with their images for every i = 1, . . . , m + 1 and that they give rise to a smooth m
m+1
dimensional atlas of R ∼\{0} . The resulting manifold is denoted by RPm and called the m dimensional real
projective space. Discuss the following questions:
ϕi : R m →
Date: August 21, 2014.
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1
HW 5
(1)
(2)
(3)
(4)
Differential Geometry II, Spring 2014
Describe the transition maps of the atlas for RPm described above explicitly.
Is RPm orientable?
Explain why RPm is called the space of lines in Rm+1 .
Explain why RP1 is diffeomorphic to the circle
S1 = {(x, y) ∈ R2 : x2 + y 2 = 1}.
Hint: Understand this geometrically first.
(5) We may think of RPm as the m dimensional sphere
Sm = {p ∈ Rm : |p| = 1}
after identifying antipodal points. Is RP2 diffeomorphic to S2 ?
Solution: Note that the quotient map π : Rm+1 \ {0} → RPm is an open map. Suppose we are given some
open set U ⊂ Rm+1 \{0}. The set CU := {tx : t ∈ R\{0}, x ∈ U } is open (the C is for “cone”). Furthermore,
we have that π −1 (π(U )) = CU is open. By the definition of the quotient topology, this is equivalent to π(U )
being open. It is a general fact that the quotient of a second countable space is second countable, provided
that the quotient map is open. Indeed, the image of a countable basis for the topology in Rm+1 \ {0} under
π is a countable basis for the quotient topology on RPm .
To see that RPm is Hausdorff, let p, q ∈ RPm be two points, p 6= q. Consider π −1 (p) = {tx : t 6= 0},
π −1 (q) = {sy : s 6= 0}. By assumption, we have tx 6= y for all t ∈ R. We can therefore find open sets U , V
containing π −1 (p) and π −1 (q), respectively, such that U , V are invariant under dilation and such that the
intersection of U and V in Rm+1 \ {0} is empty.1 This implies that π(U ) ∩ π(V ) = ∅, and since π is open,
that π(U ) and π(V ) are disjoint open neighborhoods of p, q. We conclude that RPm is Hausdorff.
m+1
(1) The map ϕi : Rm → R ∼\{0} given by (x1 , . . . , xm ) 7→ [(x1 , . . . , xi−1 , 1, xi , . . . , xm )] is the composition
of the embedding ι : Rm → Rm+1 , given by (x1 , . . . , xm ) 7→ (x1 , . . . , xi−1 , 1, xi , . . . , xm ) with the quotient
map π : Rm+1 → RPm . It follows from this that ϕi is continuous.
Note that ϕi is injective: If ϕi (x1 , . . . , xm ) = ϕi (y 1 , . . . , y m ), then
(x1 , . . . , xi−1 , 1, xi , . . . , xm ) = t · (y 1 , . . . , y i−1 , 1, y i , . . . , y m )
for some t 6= 0. Comparison of the i-th components, implies that t = 1, hence (x1 , . . . , xm ) = (y 1 , . . . , y m ).
To show that ϕi is a homeomorphism with its image, we will give a continuous inverse. Let Ui =
{(x0 , . . . , xm ) ∈ Rm+1 : xi 6= 0}, then im(ι) ⊂ Ui and im(ϕi ) ⊂ π(Ui ). Now consider the map
0
x
xi−1 xi+1
xm
m
η : Ui → R , (x0 , . . . , xm ) 7→
,..., i , i ,..., i
xi
x
x
x
Then on the one hand we clearly have η ◦ ι = idRm . On the other hand, the map η is continuous and
constant on equivalence classes for ∼. This allows us to define a map η : π(Ui ) → Rm , [x] 7→ η(x), which
obviously satisfies η = η ◦ π. We claim that η is continuous. Indeed, for O ⊂ Rm open we must show that
η −1 (O) ⊂ RPm is open. But this is equivalent to π −1 (η −1 (O)) = (η ◦ π)−1 (O) = η −1 (O) ⊂ Ui being open
and this follows from η being continuous.
1For this, we could choose tiny balls B (x) and B (y) around x and y, and let U = CB (x) = {tz : t ∈ R, z ∈ B (x)},
ε
ε
ε
ε
V = CBε (y) = {tz : t ∈ R, z ∈ Bε (y)}. For ε > 0 small enough, U and V have the required form.
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HW 5
Differential Geometry II, Spring 2014
In fact it holds in general that for a continuous function η : (X, τ ) → (Y, τ 0 ) that is constant on the
equivalence classes of an equivalence relation ∼ on X, there exists a unique continuous function η : X/ ∼
→ Y . This is the universal property of the quotient topology.
We claim that ϕ−1
i = η. Indeed, we have η ◦ ϕi = η ◦ π ◦ ι = η ◦ ι = idRm . Or more explicitly,
η ◦ ϕi (x1 , . . . , xm ) = η([(x1 , . . . , xi−1 , 1, xi , . . . , xm )]) = η(x1 , . . . , xi−1 , 1, xi , . . . , xm ) = (x1 , . . . , xm ).
On the other hand we see
x0
xi−1 xi+1
xm
,
.
.
.
,
,
,
.
.
.
,
xi
xi
xi
xi
x0
xi−1
xi+1
xm
ϕi ◦ η[(x , . . . , x )] = ϕi
=
,
.
.
.
,
,
1,
,
.
.
.
,
xi
xi
xi
xi
x0
xi−1
xi+1
xm
= [(x0 , . . . , xm )]
=
xi i , . . . , xi i , xi , xi i , . . . , xi i
x
x
x
x
0
m
for [(x0 , . . . , xm )] ∈ π(Ui ). Hence ϕi is a homeomorphism with its image π(Ui ). Note that for i < j,
1
m
m
i
m
ψji = ϕ−1
i ◦ ϕj : {(x , . . . , x ) ∈ R : x 6= 0} → R
is given by
x1
xi−1 xi+1
xj−1 1 xj
xm
(x , . . . , x ) 7→
,..., i , i ,..., i , i, i,..., i .
xi
x
x
x
x x
x
m
This is a smooth map and furthermore it is easy to see that every point in RP lies in the image of some
ϕi .
Since each ϕi is a homeomorphism onto its image, the transition maps ϕ−1
i ◦ ϕj are all smooth and their
m+1
m
m
images cover RP . We conclude that {ϕi }i=1 is an atlas for RP .
1
m
(2) RPm is orientable if m is odd, and non-orientable if m is even.
Proof: Suppose m is odd. Let e1 , . . . , em be the standard basis on Rm . Note
ψji with i < j, the directional derivative ∂k ψji is given by

1

e ,

xi k



− 1 (x1 , . . . , xi−1 , xi+1 , . . . , xj−1 , 1, xj , . . . , xm ),
(xi )2
∂k ψji =
1


e ,

xi k−1


1
e
xi k
that for each transition map
k < i,
k = i,
i < k < j,
k ≥ j.
It follows that
det[∂ψij ] =
=
(−1)
det(e1 , . . . , ei−1 , ej−1 , ei , . . . , ej−2 , ej , . . . , em )
(xi )m+1
(−1)j−i
(−1)j−i
det(e
,
.
.
.
,
e
)
=
.
1
m
(xi )m+1
(xi )m+1
j−i
(−1)
j−i . Thus, the orientations induced by ϕ and ϕ are compatible
If m is odd, then the sign of (x
i
j
i )m+1 is (−1)
m
m
if (j − i) is even, and not compatible if (j − i) is odd. Let τ : R → R be given by τ (x1 , . . . , xm ) =
(−x1 , x2 , . . . , xm ). Consider the atlas A0 consisting of the charts ϕi , for i even, and ϕj ◦ τ , for j odd, where
0 ≤ i, j ≤ m. Since det[∂τ ] = −1 < 0, it follows from the computation above that the differentials of all
transition maps of this atlas A0 have positive determinant. Thus, A0 is an oriented atlas for RPm .
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HW 5
Differential Geometry II, Spring 2014
Now suppose that m is even. We claim that if m is even, then RPm is not orientable. Suppose RPm had
an oriented atlas A. Each one of the charts ϕi from the standard atlas {ϕi }m+1
i=1 constructed above has
connected domain. It follows that each ϕi either has an orientation compatible with A, or else ϕi ◦ τ has an
orientation compatible with A. Consider the two charts ϕ1 and ϕ2 . We have seen above that the transition
−1
1
map ψ21 = ϕ−1
1 ◦ ϕ2 satisfies det[∂ψ21 ] = (x1 )m+1 . Since m is even, this is positive for x < 0 and negative
for x1 > 0. But this would be incompatible with each one of the possibilities
(i)
(ii)
(iii)
(iv)
ϕ1 and ϕ2 are compatible with the orientation of RPm ,
ϕ1 and ϕ2 ◦ τ are compatible with the orientation of RPm ,
ϕ1 ◦ τ and ϕ2 are compatible with the orientation of RPm ,
ϕ1 ◦ τ and ϕ2 ◦ τ are compatible with the orientation of RPm .
We therefore must conclude that RPm is not orientable for m even.
(3) The point [x] ∈ RPm corresponds to the line {tx : t ∈ R} in Rm+1 . This correspondence between points
in RPm and one-dimensional linear subspaces of Rm+1 is bijective.
(4) We can consider RP1 as the quotient RP1 ∼
= S1 / ∼, where p ∼ q ⇐⇒ p = ±q. We can identify S1 with
the unit circle in C. Consider the map ϑ : S1 → S1 , z 7→ z 2 . Note that this map is constant on fibers of
π : S1 → RP1 . It follows from the universal property of quotient spaces, that ϑ induces a continuous map
ϑ : RP1 → S1 satisfying ϑ(π(z)) = ϑ(z). We claim that ϑ is a diffeomorphism.
ϑ is a homeomorphism: By the universal property of quotient spaces, ϑ is continuous. Since RP1 is compact
as the continuous image of a compact space, and since S1 is Hausdorff, it follows that ϑ is a closed map.
ϑ is surjective, because ϑ : S1 → S1 is surjective. To see that ϑ is injective, suppose ϑ([x]) = ϑ([y]) for
[x], [y] ∈ RP1 , where [x] = π(x), [y] = π(y) for x, y ∈ S1 . It follows that ϑ(x) = ϑ(π(x)) = ϑ(π(y)) = ϑ(y),
i.e. x2 = y 2 . This is only the case if x = ±y, i.e. only if [x] = π(x) = π(y) = [y]. Thus, ϑ : RP1 → S1 is a
continuous, closed and bijective map. This shows that ϑ is a homeomorphism.
ϑ is a diffeomorphism: It remains to prove that the map ϑ : RP1 → S1 and its inverse are smooth. We note
that the maps π : S1 → RP1 and ϑ : S1 → S1 are surjective. Using the standard charts (a, b) → S1 , t 7→
(cos(t), sin(t)) with a < b ≤ a + 2π for S1 and the charts for RP1 described above, we verify that the maps
π and ϑ are smooth and that their derivatives have full rank everywhere. Let p ∈ RP1 and q ∈ S 1 with
p = π(q). By the inverse function theorem, there exist open neighborhoods U in RP1 containing p and V in S1
containing q such that π : V → U is a diffeomorphism. It follows that ϑ|U = (ϑ ◦ π −1 )|U : U → S1 is smooth.
−1
Thus, the map ϑ : RP1 → S1 is smooth and in a similar vein we argue that its inverse ϑ : S1 → RP1 is
also smooth.
(5) To see this identification, consider the restriction of the quotient map π : Rm+1 \{0} → RPm to Sm . The
fibers of the map π|Sm : Sm → RPm are precisely given by the equivalence classes of the equivalence relation
m
p ∼ q ⇐⇒ p = ±q. It follows that the map π|Sm : Sm → RPm induces a homeomorphism S∼ → RPm .
We could have defined RPm to be the quotient space of Sm by this equivalence relation ∼. Considering
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HW 5
Differential Geometry II, Spring 2014
the atlas used to define a smooth structure on RPm , one checks that π|Sm : Sm → RPm is in fact a local
diffeomorphism.
RP2 cannot be diffeomorphic to S2 . We have seen in (2) that RP2 is not orientable. On the other hand, S2
is orientable.
2. Project light
The goal of this week’s project light is to gain some familiarity with the abstract construction of the tangent
space of a smooth manifold. You should think about all the claims made below before your exercise class,
discuss with your classmates, and contribute actively when you go over this material with your assistant.
This material (with full proofs) is examinable for this course.
2.1. Tangent space of domains. To begin with, make sure that you understand the example below.
Example 2.1. Let M be an m dimensional domain with standard smooth structure [{idM : M → M }]. A
chart ϕ : U → O of M is a diffeomorphism of m dimensional domains U and O where O ⊂ M is open. A
concrete realization of the tangent bundle
π
T M −→ M
is determined as follows. We have that T M = M × Rm with the standard smooth structure of a 2m
dimensional domain and that π : M × Rm → M is projection to the first factor. The vector space structure
of
(T M )p = {p} × Rm = π −1 ({p})
is that of Rm ignoring the base point. The chart
ϕ∗ : U × Rm → O × Rm = π −1 (O)
of T M associated with ϕ : U → O is the usual tangent map
(x, w) 7→ (ϕ(x), ∂ϕ|x w).
Exercise 2.1 (Tangent maps and short curves). Let p be an interior point of a manifold M . A short curve
of M at p is a smooth map γ : (−ε, ε) → M with γ(0) = p for some ε > 0. Two short curves γ1 , γ2 of M at
p are said to be equivalent if their tangent maps agree at the origin. Show carefully that
d (T M )p = γ∗
:
γ
is
a
short
curve
of
M
at
p
.
dt t=0
Here,
d
|t=0 = (0, 1) ∈ T(−ε, ε) = (−ε, ε) × R.
dt
d
Let g ∈ C ∞ (M ) and v = γ∗ ( dt
|t=0 ) ∈ (T M )p . Show that
v(g) = (g ◦ γ)0 (0).
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HW 5
Differential Geometry II, Spring 2014
This is a useful interpretation of the directional derivative. Let f : M → N be a smooth map. Show that f
takes equivalent short curves of M at p to equivalent short curves of N at f (p). The tangent map
(T M )p → (T N )f (p)
of f at p can be thought of as follows. Let v ∈ (T M )p and let γ be a short curve of M at p with tangent
map v at the origin. Then f∗ (v) ∈ (T N )f (p) is the tangent map of the short curve f ◦ γ of N at f (p). How
does this discussion change if p is a boundary point?
Solution: Choose a chart ϕ : U → O, where U ⊂ Rm is an interior chart domain and O ⊂ M is an open
neighborhood of p. Let v ∈ Tp M be given. We show that there is ε > 0 and a smooth curve γ : (−ε, ε) → M
d
∼ m
with γ(0) = p, such that γ∗ ( dt
|t=0 ) = v. Let w = ϕ−1
∗ v ∈ Tϕ−1 (p) U = R . Since U is open, there exists
ε > 0 such that η(t) := ϕ−1 (p) + tw ∈ U for t ∈ (−ε, ε). Here w is identified with an element of Rm under
d
|t=0 ) = w and η(0) = ϕ−1 (p). We define a
the canonical identification Tϕ−1 (p) U ∼
= Rm . Note that η∗ ( dt
smooth curve γ : (−ε, ε) → M by γ(t) = ϕ ◦ η(t). Then
This proves that Tp M ⊂ {γ∗
inclusion is trivial.
d
d
γ∗ ( |t=0 ) = ϕ∗ ◦ η∗ ( |t=0 ) = ϕ∗ w = v.
dt
dt
d
dt t=0 : γ : (−ε, ε) → M smooth for some ε > 0 and γ(0) = p}. The other
d
If p ∈ ∂M , then {γ∗ dt
: γ : (−ε, ε) → M smooth for some ε > 0 and γ(0) = p} = Tp ∂M . The same
t=0
argument as above and choosing a boundary chart ϕ : U → O, implies the inclusion ⊃.
For the other inclusion, let γ : (−ε, ε) → M be a smooth curve with γ(0) = p. By shrinking ε if necessary,
we may assume that the trace of γ lies in the neighborhood O of p. Consider the curve η : (−ε, ε) → U given
1
by η = ϕ−1 ◦ γ. Write η(t) = (η 1 (t), . . . , η m (t)). Since η 1 (0) = 0 and η 1 (t) ≤ 0, it follows that dηdt(0) |t=0 = 0.
d
d
d
Therefore η∗ ( dt
|t=0 ) ∈ Tϕ−1 (p) ∂U and γ∗ ( dt
|t=0 ) = ϕ∗ ◦ η∗ ( dt
|t=0 ) ∈ Tp ∂M . Therefore the same definition
in terms of curves will not give us back the entire tangent space at boundary points. If we considered short
curves instead, then we would get a half-space at boundary points. More precisely, we have that the set
d
{γ∗ dt
: γ : [0, ε) → M smooth for some ε > 0 and γ(0) = p} consists of all inward pointing tangent
t=0
vectors in Tp M . The proof is practically the same as for interior points of M . The only difference is that
we have to consider a boundary chart ϕ : U → O instead and all curves with domain (−ε, ε) have to be
exchanged for short curves with domain [0, ε). The fact that these short curves are inward-pointing follows
from the corresponding statement on U in the closed left half-space.
Exercise 2.2 (Classification of compact one dimensional manifolds). Let Γ ⊂ Rn be a connected compact
one-dimensional submanifold with boundary. Show with all bells and whistles that there is a unit speed
embedding γ : [a, b] → Rn with trace Γ. Deduce from Whitney’s embedding theorem that every connected
compact one-dimensional manifold with boundary is diffeomorphic to a closed interval. Adapt the statement
of this question to the case of connected one-dimensional manifolds that are closed, and prove it.
Solution: We assume that Γ ⊂ Rn is a compact, connected 1-manifold with (non-empty) boundary.
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HW 5
Differential Geometry II, Spring 2014
Since Γ is compact, we can find a finite atlas A = {ϕk : Ik → Ok : 1 ≤ k ≤ N }. In fact, by inductively
altering our charts (and maybe splitting charts in half, throwing out unnecessary charts etc.), we can obtain
an atlas that satisfies the following additional conditions (for claim (iv) see below):
(i) Each Ik is a connected interval,
(ii) Ik is of the form Ik = [ak , bk ), or Ik = (ak , bk ] (for a boundary chart) or of the form Ik = (ak , bk ) (for
an interior chart), for 1 ≤ k ≤ N ,
(iii) each ϕk is parametrized by arc-length,
(iv) the intersection Ok ∩ O` has at most one component for all k, `,
(v) we have Ok 6⊂ O` for all k 6= `,
(vi) all charts have compatible orientation.
Concerning (iv), assume that we have two charts ϕj : Ij → Oj of a 1-dimensional manifold Γ (with or
without boundary), j = 1, 2, for which (i)-(ii) hold. We claim that in this case O1 ∩ O2 has at most two
connected components. Observe that O1 ∩ O2 ⊂ Γ is relatively open in Γ and hence in O1 . But as ϕ1 is a
homeomorphism, it follows that ϕ−1 (O1 ∩ O2 ) ⊂ I1 is an open subset. Let I ⊂ I1 be a connected component
of this set, then we distinguish the following cases:
• If I is of the form I = [c1 , d1 ] it follows from I ⊂ I1 open, I1 connected, that I = I1 and hence
O1 ∩ O2 = O1 is connected.
• If I is of the form I = [c1 , d1 ) we find that I1 = [c1 , b1 ) or I1 = [c1 , b1 ] as otherwise c1 would be
an interior point of I1 . A similar result holds for I = (c1 , d1 ]. In any case we find that one of the
endpoints of I (i.e. sup I or inf I) agrees with the corresponding endpoint of I1 .
• If I is of the form I = (c1 , d1 ) then either I = I1 , finishing the proof as in the first case, or one
of c1 , d1 is contained in I1 . Assume c1 ∈ I1 and consider the map ψ12 = ϕ−1
2 ◦ ϕ1 : I → I2 . Since
it is continuous and injective, it follows that it is monotone. Then as I2 is bounded, the limit
y = limx&c1 ψ12 (x) exists in R. But it cannot be contained in I2 , because otherwise ϕ1 (c1 ) = ϕ2 (y)
by continuity of ψ12 and hence c1 ∈ ϕ−1
1 (O1 ∩ O2 ), a contradiction. Thus, for I2 it either holds that
y = inf I2 or that y = sup I2 . Applying the analogous argument to the open interval ψ12 (I) ⊂ I2 , we
have that one of c1 , d1 is either the supremum or the infimum of I1 .
Combining the arguments above, one sees that every component of ϕ−1
1 (O1 ∩ O2 ) shares either its supremum
or its infimum with I1 . Hence there can be at most two of them, completing the argument. Then we can
modify a finite atlas A satisfying (i) - (iii) to satisfy (iv) as well by separating charts in at most two pieces
successively. Let us denote the obtained atlas by A again.
Note that conditions (i) and (iii) together with the fact that A is oriented, imply that each transition map
ψk` = ϕ−1
` ◦ ϕk is a translation, i.e. for all k, ` there exists ck` ∈ R such that ψk` (t) = t + ck` .
We use A to construct the desired embedding γ : I → Rn with trace Γ. Our proof is by induction on the
number N of charts in the atlas A.
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7
HW 5
Differential Geometry II, Spring 2014
Note that the case N = 1 can never occur. This is because Γ is compact, whereas the image of a chart is
never compact. Therefore, no single chart covers Γ.
We start our induction with the case N = 2. In this case we have two charts ϕ1 : I1 → O1 , ϕ2 : I2 → O2 .
Note that O1 and O2 are both open in Γ and that Γ is connected. It follows that O1 ∩ O2 6= ∅. Note that for
N = 2, conditions (i), (ii), and (iv) together with the fact that the transition map is a translation imply that
the intervals I1 , I2 are of the form “[a1 , b1 ) and (a2 , b2 ]”, or “(a1 , b1 ] and [a2 , b2 )”, respectively. Reindexing
if necessary, we may assume the former.
Let ψ12 (t) = t + c12 be the transition map. Let I = [a1 , b1 ) ∪ (a2 − c12 , b2 − c12 ] = [a1 , b2 − c12 ]. We define a
unit speed embedding γ : I → Rn , by setting

ϕ (t),
t ∈ [a1 , b1 ),
1
γ(t) =
ϕ2 (t + c12 ), t + c12 ∈ (a2 , b2 ].
Note that since ϕ1 (t) = ϕ2 (ψ12 (t)) = ϕ2 (t + c12 ) on the overlap of the respective domains, this is a welldefined smooth function. Note further that I is compact, Rn is Hausdorff and γ is injective. Thus γ is a
homeomorphism with its image. Finally the inverse function γ −1 : γ(I) → I is smooth (check on the atlas
A). Since we assume N = 2 and γ(I) = O1 ∪ O2 = Γ, it follows that γ is the required diffeomorphism of
the closed interval I with Γ.
The induction step is similar: Let N > 2 be a given integer. Suppose that A is an oriented atlas for a
compact connected 1-manifold Γ ⊂ Rn satisfying the conditions (i)-(vi) above. We assume that A consists
of precisely N charts. We prove that it is possible to find an oriented atlas A0 for Γ, still satisfying (i)-(vi),
but with fewer than N charts. Using the induction hypothesis, the classification of compact connected
1-manifolds with boundary then follows.
The idea is to merge two of our charts to produce a slightly larger chart: Consider a chart ϕk for the
boundary of Γ such that Ik = [ak , bk ) (if such a chart should not exist, then we can start out by choosing
a chart with domain Ik = (ak , bk ] and adapt the rest of our argument in the obvious way. Note that this
distinction is necessary because of the orientation). Reindexing if necessary, we may assume that k = 1.
Since Γ is compact, whereas ϕ1 (I1 ) = O1 is not, O1 cannot be all of Γ. Since Γ is connected, there must
exist some 1 < k ≤ N such that O1 ∩ Ok 6= ∅. For otherwise, we could write Γ as the disjoint union of the
S
relatively open sets O1 and N
k=2 Ok . Reindexing if necessary, we may assume that O1 ∩ O2 6= ∅.
There are three cases: Either I2 = (a2 , b2 ) or I2 = [a2 , b2 ) or I2 = (a2 , b2 ].
We can rule out the second case: This is due to the fact that on the one hand, our transition maps are
given by translation, which would imply ϕ1 (a1 ) = ϕ2 (a2 ). On the other hand we also have that I1 6⊂ I2 and
I2 6⊂ I1 by condition (v). These two requirements are incompatible with each other.
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HW 5
Differential Geometry II, Spring 2014
In the third case we can argue as follows: The same construction used in the proof of the case N = 2 yields a
smooth embedding γ from a closed interval I onto the union O1 ∪ O2 ⊂ Γ. But then γ(I) = O1 ∪ O2 is both
compact as a continuous image of a compact set I, and open in Γ as the union of the two open sets O1 and
O2 . This implies that γ(I) ⊂ Γ is both a closed and open subset. By connectedness, the only non-empty
subset of Γ which is both open and closed, is Γ itself. Therefore γ(I) = Γ. It follows that γ : I → Γ is a
diffeomorphism.
The last case to consider is the case where I2 = (a2 , b2 ): Let ψ12 (t) = t + c12 be the transition map. Let
I˜ = [a1 , b1 ) ∪ (a2 − c12 , b2 − c12 ) = [a1 , b2 − c12 ). Let Õ = O1 ∪ O2 . We define a chart ϕ̃ : I˜ → Õ, parametrized
by arclength, by setting

ϕ (t),
t ∈ [a1 , b1 ),
1
ϕ̃(t) =
ϕ2 (t + c12 ), t + c12 ∈ (a2 , b2 ).
Note that since ϕ1 (t) = ϕ2 (ψ12 (t)) = ϕ2 (t+c12 ) on the overlap of the respective domains, this is well-defined.
Let à = {ϕ̃ : I˜ → Õ} ∪ {ϕk : Ik → Ok : 3 ≤ k ≤ N }. It is readily checked that à is oriented and satisfies
conditions (i)-(iv) and (vi). Note that à has N − 1 charts. Getting rid of redundant charts in Ã, we obtain
an atlas A0 for Γ, with strictly fewer than N charts, satisfying conditions (i)-(vi). By induction hypothesis,
it follows that Γ is diffeomorphic to a closed interval. This finishes the proof for ∂Γ 6= ∅.
Now suppose that Γ is a compact connected manifold without boundary. Let ϕ : I → O be a coordinate
chart for Γ, parametrized by arclength with I an open interval. Choose a non-empty, open interval K ⊂ I,
such that the closure K of K in R is still contained in I. The manifold Γ̃ = Γ \ ϕ(K) is a 1-manifold with at
most 2 components, each of which is a compact connected 1-dimensional manifold with nonempty boundary.
Think carefully how to modify an atlas of Γ to obtain an atlas of Γ \ ϕ(K) and why the boundary ∂K ⊂ I is
mapped by ϕ into the boundary of Γ \ ϕ(K). By the classification of connected 1-manifolds with boundary,
it follows that each component is diffeomorphic to a closed interval.
If Γ̃ had two components, both of which diffeomorphic to a closed interval, then it would follow that ∂ Γ̃
contains 4 points. Only two of these points lie in γ(I). This would imply that Γ must have non-empty
boundary itself (the remaining two points). This is contrary to our assumption ∂Γ = ∅.
Therefore Γ̃ can only have one component, i.e. Γ̃ is diffeomorphic to a closed interval. Let ϕ̃ : I˜ → Õ be a
diffeomorphism of an open interval I˜ onto the interior Õ = Γ̃ \ ∂ Γ̃ of Γ̃. We may choose ϕ̃ to be parametrized
by arclength.
Note that A = {ϕ : I → O, ϕ̃ : I˜ → Õ} is an atlas for Γ, and that O ∩ Õ = O \ γ(K) has precisely two
(open) components C1 , C2 . Write I = (a, b), I˜ = (ã, b̃). Since ϕ, ϕ̃ are parametrized by arclength, and after
˜ ϕ−1 (C1 ) = (0, c1 ) for some
modifying our charts ϕ, ϕ̃ if necessary, we may assume that I = (0, `), I˜ = (0, `),
˜ for some c̃1 ∈ (0, `).
˜
c1 ∈ (0, `), and ϕ̃−1 (C1 ) = (c̃1 , `)
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HW 5
Differential Geometry II, Spring 2014
˜ → (0, c1 ) is then given by
The transition map ϕ−1 ◦ ϕ̃ : (c̃1 , `)
(ϕ−1 ◦ ϕ̃)(t) = t − c̃1 .
On the other component C2 , we can similarly write ϕ−1 (C2 ) = (c2 , `), ϕ̃−1 (C2 ) = (0, c̃2 ), with c1 < c2 , and
c̃2 < c̃1 . The transition map ϕ−1 ◦ ϕ̃ : (0, c̃2 ) → (c2 , `) must be given by
(ϕ−1 ◦ ϕ̃)(t) = t + c2 .
We define a smooth map γ : (0, c̃1 + `) → Γ by setting

ϕ̃(t),
˜
t ∈ (0, `),
γ(t) =
ϕ(t − c̃1 ), t ∈ (c̃1 , c̃1 + `).
Set τ := ` + c̃1 − c̃2 . Note that ` − c2 = c̃2 and `˜− c̃1 = c1 . It follows that for all sufficiently small t ∈ (0, c̃2 ),
we have t + τ = t + c̃1 + (` − c̃2 ) ∈ (c̃1 + ` − c̃2 , c̃1 + `), and
γ(t + τ ) = ϕ(t + τ − c̃1 ) = ϕ(t + ` − c̃2 ) = ϕ(t + c2 ) = ϕ((ϕ−1 ◦ ϕ̃)(t)) = ϕ̃(t) = γ(t).
Thus, γ can be extended to a smooth τ -periodic curve with image Γ, which we can interpret as a smooth
function γ : S1 → Γ. From the construction it is clear that γ is a local diffeomorphism. The fact that γ has
no smaller periods than τ is easily checked. Hence, γ : S1 → Γ is a diffeomorphism.
This completes the classification of connected compact 1-dimensional manifolds.
2.2. Linear algebra. We will need the linear algebra facts presented below when constructing new tensor
bundles from old ones. This is a good opportunity for you to review tensor notation. It is important that
you understand the final statement well.
Let E1 , . . . , Ek be finite dimensional vector spaces.
Recall that the tensor product E1∗ ⊗ . . . ⊗ Ek∗ is identified with the vector space
{multilinear maps E1 × . . . × Ek → R}
together with the tensor product map
⊗ : E1∗ × . . . × Ek∗ → E1∗ ⊗ . . . ⊗ Ek∗
that takes (e1 , . . . , ek ) ∈ E1∗ × . . . × Ek∗ to the multilinear map e1 ⊗ . . . ⊗ ek given by
(e1 , . . . , ek ) 7→ e1 (e1 ) · · · ek (ek ).
Let B1 , . . . , Bk be bases of E1 , . . . , Ek respectively. Then
{b1 ⊗ . . . ⊗ bk : b1 ∈ B1 , . . . , bk ∈ Bk }
is a basis of
{multilinear maps E1 × . . . × Ek → R}.
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10
HW 5
Differential Geometry II, Spring 2014
A linear map L : E → F has a unique linear extension
L∗ : E
. . ⊗ E} → F
. . ⊗ F}
| ⊗ .{z
| ⊗ .{z
k times
k times
to k fold contravariant tensor products such that
L∗ (e1 ⊗ . . . ⊗ ek ) = (Le1 ) ⊗ . . . (Lek )
for all e1 , . . . , ek ∈ E.
The adjoint L∗ : F ∗ → E ∗ of a linear map L : E → F has a unique linear extension
∗
∗
. . ⊗ E }∗
L∗ : F
. . ⊗ F }∗ → E
| ⊗ .{z
| ⊗ .{z
` times
` times
to ` fold covariant tensor products such that
L∗ (f 1 ⊗ . . . ⊗ f ` ) = (L∗ f 1 ) ⊗ . . . ⊗ (L∗ f ` )
for all f 1 , . . . , f ` ∈ F ∗ .
If L : E → F is an isomorphism, then so is L∗ : F ∗ → E ∗ , and
L∗−1 = L−1∗ .
An isomorphism L : E → F between vector spaces has a unique linear extension
∗
. . ⊗ E }∗ → F
. . ⊗ F} ⊗ |F ∗ ⊗ .{z
. . ⊗ F }∗
L∗ : E
. . ⊗ E} ⊗ E
| ⊗ .{z
| ⊗ .{z
| ⊗ .{z
k times
` times
k times
` times
to tensor products of mixed type (`, k) such that
L∗ (e1 , . . . , ek , e1 , . . . , e` ) = (Le1 ) ⊗ . . . ⊗ (Lek ) ⊗ (L−1∗ e1 ) ⊗ . . . ⊗ (L−1∗ e` )
for all e1 , . . . , ek ∈ E and e1 , . . . , e` ∈ E ∗ .
Assume now that E and F are finite dimensional.
Identifying tensor products with multilinear maps as above, we see that
∼
=⊗k E
∼
=⊗k F
}|
{
z
}|
{
z
∗
∗
∗
∗
→
R}
→
→
R}
L∗ : {multilinear maps E
×
.
.
.
×
E
{multilinear
maps
F
×
.
.
.
×
F
|
{z
}
|
{z
}
k times
is pre-composition with the adjoint
L∗
:
F∗
k times
→
E∗
in each slot.
Similarly,
∼
=⊗` F ∗
∼
=⊗` E ∗
}|
{
}|
{
z
z
L∗ : {multilinear maps |F × .{z
. . × F} → R} → {multilinear maps E
×
.
.
.
×
E
→
R}
|
{z
}
` times
` times
is pre-composition with L : E → F in each slot.
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11
HW 5
Differential Geometry II, Spring 2014
If L : E → F is an isomorphism, then
(⊗k E)⊗(⊗` E ∗ )
}|
{
z
∗
∗
×
.
.
.
E
L∗ : {multilinear maps E
×
E
×
.
.
.
×
E
→
R}
|
{z
} |
{z
}
k times
` times
→
(⊗k F )⊗(⊗` F ∗ )
z
}|
{
×
.
.
.
×
F
→
R}
{multilinear maps |F ∗ ×{z
. . . F }∗ × F
|
{z
}
k times
is pre-composition with the adjoint
the last ` slots.
L∗
:
F∗
→
E∗
` times
in the first k slots and with the inverse L−1 : F → E in
Fix bases of E and F and corresponding dual bases of E ∗ and F ∗ . We see that the matrices representing the
extensions of L as above are polynomials in the entries of the matrices representing the adjoint L : E → F
and the inverse L−1 : F → E.
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12