# 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 ${\displaystyle {\mathcal {C}}^{n}(M)}$ functions.

## Definition of a submanifold

Definition 4.1:

Let ${\displaystyle M}$ be a ${\displaystyle d}$-dimensional manifold of class ${\displaystyle {\mathcal {C}}^{n}}$, and let ${\displaystyle A}$ be it's maximal atlas. If ${\displaystyle m\in \{1,\ldots ,d\}}$, we call a subset ${\displaystyle N\subseteq M}$ a submanifold of dimension ${\displaystyle m}$ iff ${\displaystyle m}$ is the largest number in ${\displaystyle \{1,\ldots ,d\}}$ such that for each ${\displaystyle p\in N}$ there exists ${\displaystyle (O,\phi )\in A}$ such that ${\displaystyle p\in O}$ and

${\displaystyle \phi (O\cap N)\subseteq \{(x_{1},\ldots ,x_{d})\in \mathbb {R} ^{d}|x_{d-m+1}=x_{d-m+2}=\ldots =x_{d}=0\}}$

## How to obtain a submanifold out of a set of certain functions

Lemma 4.2: Let ${\displaystyle M}$ be a ${\displaystyle d}$-dimensional manifold of class ${\displaystyle {\mathcal {C}}^{n}}$ with atlas ${\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$, let ${\displaystyle A}$ be it's maximal atlas, let ${\displaystyle (O,\phi )\in A}$, and let ${\displaystyle V\subseteq O}$ be an open subset of ${\displaystyle O}$. Then ${\displaystyle (V,\phi |_{V})\in A}$.

Proof:

1. We show that ${\displaystyle \phi |_{V}:V\to \phi (V)}$ is a chart.

It is a homeomorphism since the restriction of a homeomorphism is a homeomorphism, and if ${\displaystyle V\subseteq O}$ is open, then ${\displaystyle \phi (V)}$ is open in ${\displaystyle \phi (O)}$ since ${\displaystyle \phi }$ is a homeomorphism, and further, due to the definition of the subspace topology and since ${\displaystyle \phi (V)}$ is open in ${\displaystyle \phi (O)}$, we have ${\displaystyle \phi (V)=W\cap \phi (O)}$ for an open set ${\displaystyle W\subseteq \mathbb {R} ^{d}}$, and hence ${\displaystyle \phi (V)}$ is open in ${\displaystyle \mathbb {R} ^{d}}$ as the intersection of two open sets.

2. We show that ${\displaystyle \phi |_{V}}$ is compatible with all ${\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$.

Let ${\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$.

We have:

${\displaystyle \theta |_{V\cap U}\circ (\phi |_{V})|_{V\cap U}^{-1}=(\theta |_{O\cap U}\circ \phi |_{O\cap U}^{-1})|_{\phi (V\cap U)}}$

and

${\displaystyle (\phi |_{V})|_{V\cap U}\circ \theta ^{-1}|_{V\cap U}=(\phi |_{O\cap U}\circ \theta |_{O\cap U}^{-1})|_{\theta (V\cap U)}}$

, which can be verified by direct calculation. But these are ${\displaystyle n}$-times differentiable (or continuous if ${\displaystyle n=0}$), since they are restrictions of ${\displaystyle n}$-times differentiable (or continuous if ${\displaystyle n=0}$) functions; this is since ${\displaystyle \theta }$ and ${\displaystyle \phi }$ are compatible. Due to the definitions of ${\displaystyle {\mathcal {C}}}$ and ${\displaystyle {\mathcal {C}}}$ respectively, the lemma is proved.${\displaystyle \Box }$

Lemma 4.3: Let ${\displaystyle M}$ be a ${\displaystyle d}$-dimensional manifold of class ${\displaystyle {\mathcal {C}}^{n}}$ with atlas ${\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$, let ${\displaystyle A}$ be it's maximal atlas, let ${\displaystyle (O,\phi )\in A}$, and let ${\displaystyle \Phi :\phi (O)\to \mathbb {R} ^{d}}$ be a diffeomorphism of class ${\displaystyle {\mathcal {C}}^{n}}$. Then we have: ${\displaystyle (O,\Phi \circ \phi )\in A}$.

Proof:

1. We show that ${\displaystyle \Phi \circ \phi }$ is a chart.

By invariance of domain, and since ${\displaystyle \phi (O)}$ is open in ${\displaystyle \mathbb {R} ^{d}}$ since ${\displaystyle \phi }$ is a chart, ${\displaystyle (\Phi \circ \phi )(O)}$ is open in ${\displaystyle \mathbb {R} ^{d}}$. Furthermore, ${\displaystyle \Phi }$ and ${\displaystyle \phi }$ are homeomorphisms (${\displaystyle \Phi }$ is a homeomorphism because every diffeomorphism is a homeomorphism), and therefore ${\displaystyle \Phi \circ \phi }$ is a homeomorphism as well. Thus, ${\displaystyle \Phi \circ \phi }$ is a chart.

2. We show that ${\displaystyle \Phi \circ \phi }$ is compatible with all ${\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$.

Let ${\displaystyle (U,\theta )\in \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$.

We have:

${\displaystyle \theta |_{U\cap O}\circ (\Phi \circ \phi )|_{U\cap O}^{-1}=\theta |_{U\cap O}\circ \phi |_{U\cap O}^{-1}\circ \Phi |_{\phi (O\cap U)}^{-1}}$

And also:

${\displaystyle (\Phi \circ \phi )|_{U\cap O}\circ \theta |_{U\cap O}^{-1}=\Phi |_{\phi (O\cap U)}\circ \phi |_{O\cup U}\circ \theta |_{U\cap O}^{-1}}$

These functions are ${\displaystyle n}$-times differentiable (or continuous if ${\displaystyle n=0}$), because they are compositions of functions, which are ${\displaystyle n}$-times differentiable (or continuous if ${\displaystyle n=0}$); this is since ${\displaystyle \phi }$ and ${\displaystyle \theta }$ are compatible. By definition of ${\displaystyle {\mathcal {C}}^{n}((\Phi \circ \phi )(O),\mathbb {R} ^{d})}$ and ${\displaystyle {\mathcal {C}}^{n}(\psi (O),\mathbb {R} ^{d})}$ respectively, we are finished with the proof of this lemma.${\displaystyle \Box }$

Theorem 4.4:

Let ${\displaystyle M}$ be a ${\displaystyle d}$-dimensional manifold of class ${\displaystyle {\mathcal {C}}^{n}}$, where ${\displaystyle n}$ must be ${\displaystyle \geq 1}$ here, with maximal atlas ${\displaystyle A}$, let ${\displaystyle m\in \{1,\ldots ,d\}}$ and let ${\displaystyle f_{1},\ldots ,f_{m}\in {\mathcal {C}}^{n}(M)}$ (remember def. 1.5). If for each ${\displaystyle p\in N}$ there exists ${\displaystyle (O,\phi )\in A}$ such that ${\displaystyle p\in O}$ and the matrix

${\displaystyle {\begin{pmatrix}\left(\partial _{x_{1}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{d}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))\\\vdots &\ddots &\vdots \\\left(\partial _{x_{1}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{d}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))\end{pmatrix}}}$

has rank ${\displaystyle m}$, then the set ${\displaystyle \left\{q\in M|f_{1}(q)=\ldots =f_{m}(q)=0\right\}}$ is a submanifold of dimension ${\displaystyle d-m}$ of ${\displaystyle M}$.

Proof:

Since the matrix

${\displaystyle {\begin{pmatrix}\left(\partial _{x_{1}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{d}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))\\\vdots &\ddots &\vdots \\\left(\partial _{x_{1}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{d}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))\end{pmatrix}}}$

has rank ${\displaystyle m}$, it has ${\displaystyle m}$ linearly independent columns (this is a theorem from linear algebra). Therefore there exists a permutation ${\displaystyle \sigma :\{1,\ldots ,d\}\to \{1,\ldots ,d\}}$ such that the last ${\displaystyle m}$ columns of thee matrix

${\displaystyle {\begin{pmatrix}\left(\partial _{x_{\sigma (1)}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{\sigma (d)}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))\\\vdots &\ddots &\vdots \\\left(\partial _{x_{\sigma (1)}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))&\cdots &\left(\partial _{x_{\sigma (d)}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))\end{pmatrix}}}$

Hence, the ${\displaystyle d\times d}$ matrix

${\displaystyle {\begin{pmatrix}1&0&\cdots &0&0&0&\cdots &0\\0&1&\cdots &0&0&&&\\\vdots &\ddots &\ddots &\ddots &\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &1&0&&&\\0&0&\cdots &0&1&0&\cdots &0\\\left(\partial _{x_{\sigma (1)}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))&&\cdots &&&&\cdots &\left(\partial _{x_{\sigma (d)}}(f_{1}\circ \phi ^{-1})\right)(\phi (p))\\\vdots &&&&\ddots &&&\vdots \\\left(\partial _{x_{\sigma (1)}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))&&\cdots &&&&\cdots &\left(\partial _{x_{\sigma (d)}}(f_{m}\circ \phi ^{-1})\right)(\phi (p))\end{pmatrix}}}$

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 ${\displaystyle \Phi :\mathbb {R} ^{d}\to \mathbb {R} ^{d}}$ given by

${\displaystyle \Phi (x_{1},\ldots ,x_{d})={\begin{pmatrix}x_{1}\\\vdots \\x_{d-m}\\(f_{1}\circ \phi ^{-1})(x_{\sigma (1)},\ldots ,x_{\sigma (d)}\\\vdots \\(f_{m}\circ \phi ^{-1})(x_{\sigma (1)},\ldots ,x_{\sigma (d)})\end{pmatrix}}}$

at ${\displaystyle \phi (p)}$. By the inverse function theorem, there exists an open set ${\displaystyle V\subseteq \mathbb {R} ^{d}}$ such that ${\displaystyle \phi (p)\in V}$ and ${\displaystyle \Phi |_{V}}$ is a diffeomorphism.

Since ${\displaystyle \phi }$ is a homeomorphism, and in particular is continuous, ${\displaystyle \phi ^{-1}(V)}$ is an open subset of ${\displaystyle O}$. Due to lemma 4.2, ${\displaystyle (\phi ^{-1}(V),\phi |_{\phi ^{-1}(V)})\in A}$. Due to lemma 4.3, ${\displaystyle (\phi ^{-1}(V),\Phi \circ \phi |_{\phi ^{-1}(V)})\in A}$. But it also holds that for ${\displaystyle q}$ such that ${\displaystyle f_{1}(q)=\ldots =f_{m}(q)=0}$:

${\displaystyle \Phi \circ \phi |_{\phi ^{-1}(V)}(q)=(\phi _{1}(q),\ldots ,\phi _{d-m}(q),f_{1}(q),\ldots ,f_{m}(q))=(\phi _{1}(q),\ldots ,\phi _{d-m}(q),0,\ldots ,0)}$

Hence, ${\displaystyle \left\{q\in M|f_{1}(q)=\ldots =f_{m}(q)=0\right\}}$ is a submanifold of dimension ${\displaystyle d-m}$.${\displaystyle \Box }$

## Sources

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