1Manifolds
III Differential Geometry
1.4 Submanifolds
You have a manifold, and a subset of it is a manifold, so you call it a submanifold.
Definition
(Embedded submanifold)
.
Let
M
be a manifold with
dim M
=
n
,
and
S
be a submanifold of
M
. We say
S
is an embedded submanifold if for all
p ∈ S
, there are coordinates
x
1
, · · · , x
n
on some chart
U ⊆ M
containing
p
such
that
S ∩ U = {x
k+1
= x
k+2
= · · · = x
n
= 0}
for some k. Such coordinates are known as slice coordinates for S.
This is a rather technical condition, rather than “a subset that is also a
manifold under the inherited smooth structure”. The two definitions are indeed
equivalent, but picking this formulation makes it easier to prove things about it.
Lemma.
If
S
is an embedded submanifold of
M
, then there exists a unique
differential structure on
S
such that the inclusion map
ι
:
S → M
is smooth and
S inherits the subspace topology.
Proof.
Basically if
x
1
, · · · , x
n
is a slice chart for
S
in
M
, then
x
1
, · · · , x
k
will be
coordinates on S.
More precisely, let π : R
n
→ R
k
be the projection map
π(x
1
, · · · , x
n
) = (x
1
, · · · , x
k
).
Given a slice chart (
U, ϕ
) for
S
in
M
, consider
˜ϕ
:
S ∩U → R
k
by
˜ϕ
=
π ◦ϕ
. This
is smooth and bijective, and is so a chart on
S
. These cover
S
by assumption.
So we only have to check that the transition functions are smooth.
Given another slice chart (
V, ξ
) for
S
in
M
, we let
˜
ξ
=
π ◦ ξ
, and check that
˜
ξ ◦ ˜ϕ
−1
= π ◦ ξ ◦ ϕ
−1
◦ j,
where j : R
k
→ R
n
is given by j(x
1
, · · · , x
k
) = (x
1
, · · · , x
k
, 0, · · · , 0).
From this characterization, by looking at local charts, it is clear that
S
has
the subspace topology. It is then easy to see that the embedded submanifold is
Hausdorff and second-countable, since these properties are preserved by taking
subspaces.
We can also check easily that
ι
:
S → M
is smooth, and this is the only
differential structure with this property.
It is also obvious from the slice charts that:
Proposition.
Let
S
be an embedded submanifold. Then the derivative of the
inclusion map ι : S → M is injective.
Sometimes, we like to think of a subobject not as a subset, but as the inclusion
map
ι
:
S → M
instead. However, when we are doing topology, there is this
funny problem that a continuous bijection need not be a homeomorphism. So if
we define submanifolds via inclusions maps, we get a weaker notion known as an
immersed submanifold.
Definition
(Immersed submanifold)
.
Let
S, M
be manifolds, and
ι
:
S → M
be a smooth injective map with D
ι|
p
:
T
p
S → T
p
M
injective for all
p ∈ S
. Then
we call (
ι, S
) an immersed submanifold. By abuse of notation, we identify
S
and
ι(S).
Example.
If we map
R
into
R
2
via the following figure of eight (where the arrow
heads denote the “end points” of
R
), then this gives an immersed submanifold
that is not an embedded submanifold.
Example.
Consider the line
R
, and define the map
f
:
R → T
2
=
R
2
/Z
2
by
f
(
x
) =
αx
, where
α
is some irrational number. Then this map gives an immersed
submanifold of
T
2
, but is not an embedded submanifold, since
R
certainly does
not have the subspace topology from T
2
.
How do we construct submanifolds? The definition is rather difficult to work
with. It is not immediately clear whether
S
n
= {x ∈ R
n+1
: |x| ≤ 1} ⊆ R
n+1
is an embedded submanifold, even though it feels like it should be.
More generally, if
M, N
are manifolds,
F ∈ C
∞
(
M, N
) and
c ∈ N
, under
what circumstances will
F
−1
(
c
) be an embedded submanifold of
M
? The answer
is that c has to be a regular value.
Definition
(Regular value)
.
Let
F ∈ C
∞
(
M, N
) and
c ∈ N
. Let
S
=
F
−1
(
c
).
We say
c
is a regular value if for all
p ∈ S
, the map D
F |
p
:
T
p
M → T
c
N
is
surjective.
Proposition.
Let
F ∈ C
∞
(
M, N
), and let
c ∈ N
. Suppose
c
is a regular value.
Then S = F
−1
(c) is an embedded submanifold of dimension dim M − dim N.
Proof.
We let
n
=
dim M
and
m
=
dim N
. Notice that for the map D
F
to be
surjective, we must have n ≥ m.
Let
p ∈ S
, so
F
(
p
) =
c
. We want to find a slice coordinate for
S
near
p
.
Since the problem is local, by restricting to local coordinate charts, we may wlog
assume N = R
m
, M = R
n
and c = p = 0.
Thus, we have a smooth map
F
:
R
n
→ R
m
with surjective derivative at 0.
Then the derivative is
∂F
j
∂x
i
0
i=1,...,n; j=1,...,m
,
which by assumption has rank
m
. We reorder the
x
i
so that the first
m
columns
are independent. Then the m × m matrix
R =
∂F
j
∂x
i
0
i,j=1,...,m
is non-singular. We consider the map
α(x
1
, · · · , x
n
) = (F
1
, · · · , F
m
, x
m+1
, · · · , x
n
).
We then obtain
Dα|
0
=
R ∗
0 I
,
and this is non-singular. By the inverse function theorem,
α
is a local diffeomor-
phism. So there is an open
W ⊆ R
n
containing 0 such that
α|
W
:
W → α
(
W
) is
smooth with smooth inverse. We claim that α is a slice chart of S in R
n
.
Since it is a smooth diffeomorphism, it is certainly a chart. Moreover, by
construction, the points in
S
are exactly those whose image under
F
have the
first m coordinates vanish. So this is the desired slice chart.
Example.
We want to show that
S
n
is a manifold. Let
F
:
R
n+1
→ R
be
defined by
F (x
0
, · · · , x
n
) =
X
x
2
i
.
Then F
−1
(1) = S
n
. We find that
DF |
p
= 2(x
0
, · · · , x
n
) 6= 0
when p ∈ S
n
. So S
n
is a manifold.
Example.
Consider the orthogonal group. We let
M
n
∼
=
R
n
2
be the space of
all n × n matrices with the obvious smooth structure. We define
N = {A ∈ M
n
: A
T
= A}.
Since this is a linear subspace, it is also a manifold. We define
F : M
n
→ N
A 7→ AA
T
.
Then we have
O(n) = F
−1
(I) = {A : AA
T
= I}.
We compute the derivative by looking at
F (A + H) = (A + H)(A + H)
T
= AA
T
+ HA
T
+ AH
T
+ HH
T
.
So we have
DF |
A
(H) = HA
T
+ AH
T
.
Now if A ∈ O(n), then we have
DF |
A
(HA) = HAA
T
+ AA
T
H
T
= H + H
T
for any
H
. Since every symmetric matrix is of the form
H
+
H
T
, we know
DF |
A
: T
A
M
n
→ T
F (A)
N is surjective. So O(n) is a manifold.