In this chapter, we will show what submanifolds are, and how we can obtain, under a condition, a submanifold out of some
functions.
Definition 4.1:
Let
be a
-dimensional manifold of class
, and let
be it's maximal atlas. If
, we call a subset
a submanifold of dimension
iff
is the largest number in
such that for each
there exists
such that
and

How to obtain a submanifold out of a set of certain functions
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Lemma 4.2: Let
be a
-dimensional manifold of class
with atlas
, let
be it's maximal atlas, let
, and let
be an open subset of
. Then
.
Proof:
1. We show that
is a chart.
It is a homeomorphism since the restriction of a homeomorphism is a homeomorphism, and if
is open, then
is open in
since
is a homeomorphism, and further, due to the definition of the subspace topology and since
is open in
, we have
for an open set
, and hence
is open in
as the intersection of two open sets.
2. We show that
is compatible with all
.
Let
.
We have:

and

, which can be verified by direct calculation. But these are
-times differentiable (or continuous if
), since they are restrictions of
-times differentiable (or continuous if
) functions; this is since
and
are compatible. Due to the definitions of
and
respectively, the lemma is proved.
Lemma 4.3: Let
be a
-dimensional manifold of class
with atlas
, let
be it's maximal atlas, let
, and let
be a diffeomorphism of class
. Then we have:
.
Proof:
1. We show that
is a chart.
By invariance of domain, and since
is open in
since
is a chart,
is open in
. Furthermore,
and
are homeomorphisms (
is a homeomorphism because every diffeomorphism is a homeomorphism), and therefore
is a homeomorphism as well. Thus,
is a chart.
2. We show that
is compatible with all
.
Let
.
We have:

And also:

These functions are
-times differentiable (or continuous if
), because they are compositions of functions, which are
-times differentiable (or continuous if
); this is since
and
are compatible. By definition of
and
respectively, we are finished with the proof of this lemma.
Theorem 4.4:
Let
be a
-dimensional manifold of class
, where
must be
here, with maximal atlas
, let
and let
(remember def. 1.5). If for each
there exists
such that
and the matrix

has rank
, then the set
is a submanifold of dimension
of
.
Proof:
Since the matrix

has rank
, it has
linearly independent columns (this is a theorem from linear algebra). Therefore there exists a permutation
such that the last
columns of thee matrix

Hence, the
matrix

is invertible (one can prove the invertibility of the transpose by induction and Laplace's formula). But the latter matrix is the Jacobian matrix of the function
given by

at
. By the inverse function theorem, there exists an open set
such that
and
is a diffeomorphism.
Since
is a homeomorphism, and in particular is continuous,
is an open subset of
. Due to lemma 4.2,
. Due to lemma 4.3,
. But it also holds that for
such that
:

Hence,
is a submanifold of dimension
.
- Torres del Castillo, Gerardo (2012). Differentiable Manifolds. Boston: Birkhäuser. ISBN 978-0-8176-8271-2.