# Differentiable Manifolds/Submanifolds

In this chapter, we will show what submanifolds are, and how we can obtain, under a condition, a submanifold out of some functions.

## Definition of a submanifold[edit | edit source]

**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[edit | edit source]

**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 .

## Sources[edit | edit source]

- Torres del Castillo, Gerardo (2012).
*Differentiable Manifolds*. Boston: Birkhäuser. ISBN 978-0-8176-8271-2.