2b-submanifolds.tex

With differentials of smooth functions at hand, we are ready to discuss submanifolds: smaller manifolds sitting inside larger ones.
We have already seen an example at the beginning of the course.
In Exercise~\ref{exe:subsetsmanifolds}, we proved that any open subset $U\subseteq M$ can be made into a smooth manifold with a differentiable structure induced by the one of $M$.
These, somehow trivial, submanifolds are called \emph{open submanifolds}.
But there are many other examples beyond these ones.
In fact, you may have already seen them in multivariable analysis.

In this chapter we will briefly explore what subsets of manifolds are still manifolds of their own rights, how their topologies are related and how their smooth structures are related.
We will conclude the chapter answering the question opened at the beginning of the previous chapter: do tangent spaces really coincide with the intuitive understanding of a tangent hyperplane to a point on the manifold?

\section{Inverse function theorem}

Before getting at the core of the discussion, it is useful to
recall some results from multivariable analysis.

A function $f:\R^m \to \R^n$ between euclidean spaces has rank $k$ at $x\in\R^m$ if its ($n\times m$) Jacobian matrix $Df(x)$ has rank $k$.
The function has \emph{maximal rank}\footnote{Alternatively, it is of \emph{full rank}.} at $x$ if $k = \min(n,m)$.
When $n=m$, $f$ has maximal rank at $x$ if and only if the square matrix $Df(x)$ is an invertible matrix.

As for many local properties, this definition carries over to manifolds rather ``smoothly''.
%
\begin{definition}
    A smooth map $F:M\to N$ has \emph{rank $k$} at a point $p$ if its differential $dF_p$ has rank $k$, that is, if the linear subspace $dF_p(T_pM)$ has dimension $k$ inside $T_{F(p)}N$.
\end{definition}
%
The same applies to the inverse function theorem:
compare the following statements.
%
\begin{theorem}[Inverse function theorem for $\R^n$]\label{thm:ift}
  Let $U\subset\R^n$ open and $f:U \to \R^n$ be a smooth map.
  Assume that $f$ has maximal rank at some $x\in U$, then there exists an open neighbourhood $\Omega\subset U$ of $x$ such that  $f\big|_\Omega : \Omega \to f(\Omega)$ is a diffeomorphism.
\end{theorem}

\begin{theorem}[Inverse function theorem for manifolds]\label{thm:iftm}
  Let $F:M\to N$ be a smooth function between smooth manifolds without boundary of the same dimension $n$.
  Let $p\in M$ such that $F$ has maximal rank $n$ at $p$.
  Then there exists an open neighbourhood $V$ of $p$ such that the restriction $F|_V : V\to F(V)$ is a diffeomorphism.
\end{theorem}
% \begin{proof}
%   This is a purely local statement.
%   Let $(U,\varphi)$ be a chart about $p\in M$ and let $(V, \psi)$ be a chart about $F(p)\in N$.
%   Since both $\varphi$ and $\psi$ are diffeomorphisms (and thus have maximal rank), the derivative of
%   \begin{equation}
%     \psi \circ F \circ \varphi^{-1}: \varphi(U\cap F^{-1}(V)) \to \psi(F(U)\cap V)
%   \end{equation}
%   has rank $n$ at $\varphi(p)$.
%   Thus the inverse function theorem, Theorem~\ref{thm:ift}, there exists $\Omega \subset \varphi(U\cap F^{-1}(V))$ such that $\psi \circ F \circ \varphi^{-1}\big|_{\Omega}$ is a diffeomorphism.
%   Therefore, once again since both $\varphi$ and $\psi$ are diffeomorphisms, $F\big|_W : W \to F(W)$ is also a diffeomorphism for $W := \varphi^{-1}(\Omega)$.
% \end{proof}
\begin{exercise}
  Use the euclidean inverse function theorem (Theorem~\ref{thm:ift}) on $\R^n$ to prove Theorem~\ref{thm:iftm}.
\end{exercise}

Note that this theorem can fail for manifold with boundary.
A counterexample\footnote{Exercise: why?} is given by the inclusion map $\cH^n \hookrightarrow \R^n$.

An important observation at this point is that the rank of the mapping
is a crucial property in the inverse function theorem and it is really a
property of its differential.
If we map our manifold via some function $F$ or via its charts into a
euclidean space in such a way that the rank of the mapping remains fixed,
and thus all the tangent spaces will be mapped to euclidean spaces of same dimensions,
we may be able to use the shape of the mapping itself to describe the
shape of the manifold.

In fact, if we restrict our attention to constant rank maps, that is,
maps whose rank is the same at all points on the manifold, we can go quite a
long way and show that any manifold locally looks like a projection or an inclusion.
The tool to get there is the following, we will see more clearly the link with
projections and inclusions in the next subsection.

\begin{theorem}[Rank theorem]\label{thm:rank}
  Let $F : M^m \to N^n$ be a smooth function between smooth manifolds without boundary\footnote{The theorem can be extended to allow $M$ with boundary and $N$ without boundary assuming $\ker dF_p \not\subseteq T_p\partial M$ but we will omit this case here to keep the discussion more contained and avoid unnecessary technicalities.}.
  Assume that $F$ is of rank $k$ at all points $p\in M$.
  Then, for all $p\in M$ there exist smooth charts $(U, \varphi)$ centred at $p$ and $(V, \psi)$ centred at $F(p)$ with $F(U)\subseteq V$, such that the coordinate representation of $F$ with respect to the charts $\varphi$ and $\psi$ has the form
  \begin{align}
    \psi \circ F \circ \varphi^{-1} : \varphi(U) \subseteq \R^m &\to \psi(F(U)) \subseteq \R^n \\
    (x^1, \ldots, x^k, x^{k+1}, \ldots, x^m) &\mapsto (x^1, \ldots, x^k, \LaTeXunderbrace{0, \ldots, 0}_{n-k}).
  \end{align}
\end{theorem}

We will present here an adaptation of the general proof from \cite[Theorem 4.12]{book:lee}.
If some detail seems unclear, I would recommend you to go through the proof step by step by setting $n=m=2$ and $k=1$, and writing everything down explicitly.

\begin{proof}
  We start with two important observations.

  \newthought{First}. The statement is local on charts: without loss of generality,
  we can fix local coordinates on $M$ and $N$ respectively centred around $p$ and $F(p)$
  and replace them altogether by two open subsets $U\subseteq \R^m$ and $V \subseteq \R^n$.
  So, from now on, $F: U \subseteq \R^m \to V \subseteq \R^n$, $p=(0,\ldots,0)$
  and $F(p) = F(0,\ldots,0) = (0,\ldots,0)$.

  \newthought{Second}. The fact that $F$, and thus $dF$, is of rank $k$, translates to
  the euclidean setting to the fact that the matrix $DF$ is a $m\times n$ matrix of rank $k$
  and thus with an invertible $k\times k$ submatrix.
  Rearranging the coordinates we can assume, again without loss of generality, that this
  $k\times k$ minor is is the upper left block of the matrix $DF$, that is, the block
  $\left( \frac{\partial F^i}{\partial x^j} \right)$, $i,j = 1,\ldots, k$.

  In what follows, we will denote the coordinates as follows:
  $(x,y) = (x^1, \ldots, x^k, y^1, \ldots, y^{m-k})\in\R^m$
  and $(u,v) = (u^1, \ldots, u^k, v^1, \ldots, v^{n-k})\in\R^n$.
  \medskip

  \newthought{Right hand side}.
  Writing $F(x,y) = (Q(x,y), R(x,y))$ for some smooth maps $Q: U \to \R^k$ and $R: U \to \R^{n-k}$,
  our choice of coordinates after the above observations implies that the gradient of $Q$ with respect to $x$ is regular, that is,
  $\det \left( \frac{\partial Q^i}{\partial x^j} \right) \neq 0$ at $(x,y) = (0,0)$.

  In particular, this implies that we can extend the mapping with
  the identity on the rest of the coordinates to obtain a regular map
  on the whole neighbourhood.
  Let $\varphi : U \to \R^m$ be defined by $\varphi(x,y) = (Q(x,y), y)$. Then,
  \begin{equation}
    D\varphi(0,0) =
    \begin{pmatrix}
      \frac{\partial Q^i}{\partial x^j}(0,0) & \frac{\partial Q^i}{\partial y^j}(0,0) \\
      0 & \id_{\R^{m-k}}
    \end{pmatrix}
  \end{equation}
  has nonvanishing determinant by hypothesis.
  The Inverse Function Theorem then implies that there are open neighborhoods $U_0$ of $(0,0)$
  and $W_0$ of $\varphi(0,0)$ such that $\varphi : U_0 \to W_0$ is a diffeomorphism.
  Up to shrinking both domains appropriately, we will assume that $W_0$ is an open cube.

  Denoting $\varphi^{-1}(x,y) = (A(x,y), B(x,y))$ for some smooth functions
  $A : W_0 \to \R^k$ and $B : W_0 \to \R^{m-k}$, we get
  \begin{equation}
    (x,y) = \varphi(A(x,y), B(x,y)) = (Q(A(x,y), B(x,y)), B(x,y)).
  \end{equation}
  That is, $B(x,y) = y$ and therefore $\varphi^{-1}(x,y) = (A(x,y), y)$.
  Moreover, $\varphi \circ \varphi^{-1} = \id$ and thus $Q(A(x,y),y) = x$.

  This leaves us with
  \begin{equation}
    F\circ\varphi^{-1}(x,y) = (x, \widetilde{R}(x,y)),
  \end{equation}
  where $\widetilde{R} : W_0 \to\R^{m-k}$ is defined by $\widetilde{R}(x,y) = R(A(x,y), y)$.
  Therefore, for $(x,y)\in W_0$ we have
  \begin{equation}\label{eq:diff_DFp-1}
    D(F\circ\varphi^{-1})(x,y) = \begin{pmatrix}
      \id_{\R^k} & 0 \\
      \frac{\partial\widetilde{R}^i}{\partial x^j}(x,y) & \frac{\partial\widetilde{R}^i}{\partial y^j}(x,y)
    \end{pmatrix}.
  \end{equation}
  The composition with a diffeomorphism does not change the rank of the function $F$ and
  thus the map $F\circ\varphi^{-1}$ is of rank $k$ on $W_0$.
  You can see it for instance using the chain rule
  $D(F\circ\varphi^{-1})(x,y) = DF(\varphi^{-1}(x,y))\, D\varphi^{-1}(x,y)$ and
  observing that this is the product of an invertible square matrix with $DF$.
  The claim then follows from a classical theorem in linear algebra.

  Since $\id_{\R^k}$ makes the first $k$ columns of the matrix \eqref{eq:diff_DFp-1}
  linearly independent, the matrix can have rank $k$ only if the derivatives
  $\frac{\partial \widetilde{R}^i}{\partial y^j}$ vanish identically on $W_0$.
  That means that $\widetilde{R}$ is independent of the corresponding variables
  $(y^1, \ldots, y^{m-k})$.

  Defining $S(x) = \widetilde{R}(x, 0)$, we have
  \begin{equation}
    F\circ\varphi^{-1}(x,y) = (x, S(x)).
  \end{equation}

  \newthought{Left hand side}.
  If we can find a smooth chart $\psi : V_0 \to \R^n$ in some neighbourhood $V_0$ of $(0,0)$,
  we will have concluded the proof.
  Define
  \begin{equation}
    \psi(u, v) = (u, v - S(u))
    \quad\mbox{on}\quad
    V_0 = \{(u,v) \in V \,\mid\, (u,0)\in W_0\}\subset V.
  \end{equation}
  The map is invertible, with an explicit inverse
  \begin{equation}
    \psi^{-1}(s,t) = (s, t + S(s)).
  \end{equation}
  Therefore, it defines a diffeomorphism onto its image and thus a smooth chart.
  We only need to check that $F\circ\varphi^{-1}(W_0)\subseteq V_0$, since that would
  imply $F(U_0) \subseteq V_0$, and that $V_0$  is a neighbourhood of $(0,0)$.
  The latter claim follows immediately from the fact that
  both $V$ and $W_0$ are neighbourhoods of $(0,0)$.
  The first claim follows by definition of $V_0$ since that requires $(x,0)\in W_0$,
  which is clearly the case by setting $y=0$.

  Putting all pieces together, we get
  \begin{equation}
    \psi \circ F \circ \varphi^{-1} (x,y) = \psi(x, S(x)) = (x, S(x) - S(x)) = (x,0),
  \end{equation}
  concluding the proof.
\end{proof}

\begin{exercise}
  Formulate and prove a version of the Rank theorem for a map $F : M^m \to N^n$ of constant rank $k$,
  where $M$ is a smooth manifold with boundary, $N$ is a smooth manifold without boundary
  and $\ker dF_p \not\subseteq T_p\partial M$.
\end{exercise}

\section{Embeddings, submersions and immersions}

Looking at the statement of the Rank Theorem, one can already see that there can be different possibilities depending on the relation between, $m$, $n$ and $k$. This warrants a definition.

\begin{marginfigure}
  \includegraphics{2_8_1-immersion-embedding}
\end{marginfigure}
\begin{definition}
  Let $M^m$ and $N^n$ be differentiable manifolds and $F:M\to N$ a smooth function.
  \begin{itemize}
    \item $F$ is a \emph{immersion} if $dF_p$ is injective for all $p\in M$ ($\Rightarrow\; m\leq n$);
    \item $F$ is a \emph{submersion} if $dF_p$ is surjective for all $p\in M$ ($\Rightarrow\; m\geq n$);
    \item $F$ is a \emph{embedding}\footnote{This is a particular case of a more general concept, the topological embedding, which is defined as an injective continuous map that is a homeomorphism onto its image.} if $F$ is an injective immersion that is also a diffeomorphism onto its image $F(M)\subset N$, where the topology on $F(M)$ is the subspace topology as a subset of $N$.
  \end{itemize}
\end{definition}

\begin{example}
  \begin{enumerate}
    \item The prototype of an immersion is the inclusion of $\R^m$ in a higher-dimensional $\R^n$:
          \begin{align}
            i & : \R^m \hookrightarrow \R^n,                                                                                 \\
            i & : \left(x^1, \ldots, x^m\right) \mapsto \left(x^1, \ldots, x^m, \LaTeXunderbrace{0, \ldots, 0}_{n-m}\right).
          \end{align}

          Indeed, the $n\times m$ matrix
          \begin{equation}
            di_x = Di(x)
            = \begin{pmatrix}
              1      & 0      & \cdots & 0      \\
              0      & 1      & \cdots & 0      \\
              \vdots &        & \ddots & \vdots \\
              0      & \cdots & \cdots & 1      \\
              0      & \cdots & \cdots & 0      \\
              \vdots &        &        & \vdots \\
              0      & \cdots & \cdots & 0
            \end{pmatrix}
          \end{equation}
          has full rank (equal to $m$) and is therefore injective.
          Moreover, the map $i$ is injective and continuously invertible on its range, so it is also an embedding.

    \item The prototype for a submersion is the projection of $\R^m$ onto a lower-dimensional $\R^n$: $\pi\left(x^1,\ldots,x^n,x^{n+1},\ldots,x^m\right) = \left(x^1,\ldots,x^n\right)$.

          Indeed, the $n\times m$ matrix
          \begin{equation}
            d\pi_x = D\pi(x)
            = \begin{pmatrix}
              1      & 0      & \cdots & 0      & 0      & \cdots & 0      \\
              0      & 1      & \cdots & 0      & \vdots &        & \vdots \\
              \vdots &        & \ddots & \vdots & \vdots &        & \vdots \\
              0      & \cdots & \cdots & 1      & 0      & \cdots & 0
            \end{pmatrix}
          \end{equation}
          has full rank (equal to $n$) and is therefore surjective.
          Hence, $\pi$ is a submersion.

    \item Let $m=1$, $n > 1$ and $\gamma:\R\to\R^n$ a smooth curve.
          The map $\gamma$ is an immersion if and only if its velocity vector satisfies $\gamma'(t)\neq0$ for all $t\in\R$.
          If the curve intersects itself, e.g $\gamma(t_1) = \gamma(t_2)$ for some $t_1\neq t_2$, then $f$ is not an embedding.
  \end{enumerate}
\end{example}

\begin{remark}
  It is important to keep stressing that surjectivity of submersions or injectivity of immersions are properties of the differentials, not of the maps themselves.
  %
  For example, if $U\subset M$ open, the inclusion $i: U \to M$ is both an immersion and a submersion.
\end{remark}

\begin{definition}
  Let $M$ and $N$ be smooth manifolds such that $M\subset N$ as a set.
  We say that $M$ is an \emph{embedded (or regular) submanifold} of $N$ if the inclusion $M\hookrightarrow N$ is an embedding. If the inclusion is just an immersion, we say that $M$ is an \emph{immersed submanifold}.
\end{definition}

It is probably no surprise at this point that smooth maps are going to be useful in providing ways to nicely include a manifold into the another and in giving new ways to construct manifolds in the first place.
In the rest of this chapter we will try to give an answer to the following questions:
\begin{itemize}
  \item if $F$ is an immersion, what can we say about its image $F(M)$ as a subset of $N$?
  \item if $F$ is a submersion, what can we say about its levelsets $F^{-1}(q) \subset M$?
\end{itemize}
And what can we say about the corresponding tangent spaces?


To begin with, since all our result so-far have been local in nature, we can use locality to our advantage.
Say that we have an immersion $F$, then we can restrict its domain to obtain a local embedding, that is
a function that is locally an embedding onto its image.
The proof of this fact will show that a $k$-dimensional submanifold is also a $k$-dimensional manifold
whose charts are the ones obtained from the Rank Theorem after we drop the final $n-k$ components.

\begin{marginfigure}
  \includegraphics{2_8_9-mapping.pdf}
  \caption{Theorem~\ref{thm:rank}, case of Proposition~\ref{prop:local_embedding}, in a picture.}
\end{marginfigure}
\begin{proposition}\label{prop:local_embedding}
  Let $M^m$ and $N^n$ be smooth manifolds without boundary\footnote{This is not necessary: the result holds also on manifolds with boundary but we need a modified version of the Rank Theorem in that case.} and $F:M\to N$ a smooth function.
  Then $F$ is an immersion if and only if $F$ is a \emph{local embedding}, that is, for any $p\in M$, there exists a neighbourhood $U$ of $p$ such that $F\big|_U : U \to N$ is an embedding.
\end{proposition}
\begin{exercise}
  Prove Proposition~\ref{prop:local_embedding}. \\
  \textit{\small Hint: for the nontrivial direction use the Rank Theorem~\ref{thm:rank} and construct appropriate charts}
\end{exercise}

In fact we can say more if the manifold is compact.

\begin{exercise}
  Prove the following statements.
  \begin{enumerate}
    \item If $M$ is compact, an injective immersion $F:M\to N$ is always an embedding.
    \item This is not necessarily the case in the non-compact case, give a counterexample.
  \end{enumerate}
\end{exercise}

% \begin{lemma}
%   Assume that around any point $p\in M$ there is a chart $(V, (y^i))$ of the form
%   \begin{equation}
%     M\cap V = \left\{ q \in V \;\mid\; y^{m+1}(q)=\cdots=y^n(q)=0\right\} \subset N.
%   \end{equation}
%   Then, if we endow $M$ with the subspace topology on $N$, $M$ is a topological manifold of dimension $m$.
%   Furthermore, it has a smooth structure that makes it into an embedded submanifold of $N$.
% \end{lemma}
% \begin{proof}[Sketch]
%   Let $\pi: \R^n\to\R^m$ be the projection as in the examples above.
%   Let $p\in M$ and let $(V,\psi)$ be a chart with coordinates $(y^i)$ of the form above.
%   If we endow $M$ with the subspace topology, then $\sigma:= \pi \circ \psi\big|_{M\cap V}$ is a homeomorphism.
%   Repeating this at any point we end up with a collection of maps satisfying the hypotheses of Lemma~\ref{lem:manifold_chart}.
%   Thus $M$ is a smooth manifold of dimension $m$ and its topology coincides with the subspace topology.
%
%   Finally, with the inclusion $i:M\hookrightarrow N$ one has that $\psi \circ i\circ \sigma^{-1} (p^1,\ldots,p^m) = (p^1,\ldots,p^m,0,\ldots,0)$ which is smooth.
% \end{proof}
%
% A non-trivial consequence of the previous lemma is the following proposition\footnote{Refer to \cite[Proposition 5.8 and Proposition 5.31]{book:lee}.}.
%
% \marginnote[1em]{In Proposition~\ref{prop:uniqdiffeoinclusion} it is not enough to ask that $\iota$ is smooth! As counterexample consider the two manifolds $(\R, \cA_1)$ with $\cA_1 := \{(\R, \id_\R)\}$ and $(\R, \cA_2)$ with $\cA_2 := \{(\R, x\mapsto x^3)\}$. The inclusion of open sets in $\R$ is smooth in both cases but is a diffeomorphism only in one.}
% \begin{proposition}\label{prop:uniqdiffeoinclusion}
%   Let $M$ be a smooth manifold and $U\subset M$ an open set.
%   Then $U$ has a unique differentiable structure such that the inclusion $\iota:U\hookrightarrow M$ is a diffeomorphism.
% \end{proposition}

%\newthought{Up to this point}, the first manifold either had the same dimension or was smaller than the second one.
%What if it is larger?

Let's now focus on the opposite situation, submersions.
For this case we will need to be able to identify some special points depending on the rank of the mapping.

\begin{marginfigure}
  \includegraphics{2_8-crit_pts.pdf}
  \caption{Beware of the subtleties here. The map $F=\pi_x\circ i$ for the inclusion $i:\bT^2\hookrightarrow\R^3$ and the projection $\pi_x(x,y,z)=x$.
    So $dF_p = d (\pi_x)_{i(p)} \circ d i_p$. The latter is zero if the image of $T_p\bT^2$ by $d i_p: T_p\bT^2\hookrightarrow T_p\R^3$ is contained in the $yz$-plane (the reason will be clear by the end of the chapter): the critical points depicted here are exactly those points for which the tangent plane is the $yz$-plane.}
  \label{fig:2_8-crit_pts}
\end{marginfigure}

\begin{definition}
  Let $F:M^m \to N^n$, $m\geq n$, be a smooth map between smooth manifolds.
  A point $p\in M$ is said to be a \emph{regular point} of $F$ if $dF$ has maximal rank $n$ at $p$, while it is called a \emph{critical point} otherwise.

  Similarly, a point $q\in N$ is called a \emph{regular value} if every point in $F^{-1}(q)$ is a regular point, and \emph{critical value} otherwise. If $q\not\in F(M)$, then $q$ is considered a regular value (in the sense that there is nothing to check in its preimage by $F$).
  Cf. Figure~\ref{fig:2_8-crit_pts}.
\end{definition}

With this definition at hand, we are ready to state one of the most important theorems in this lecture.
Differently from the previous ones, the statement is not local.

\begin{theorem}[Regular levelset theorem]\label{thm:impl_fun}
  Let $m\geq n$ and let $F: M^m \to N^n$ be a smooth map between smooth manifolds.
  If $q\in N$ is a regular value of $F$ and $P := F^{-1}(q)$ is not empty, then $P$ is a topological manifold of dimension $m-n$.
  Moreover, there exists a smooth structure on $P$ which makes it into a smooth embedded submanifold of $M$.
\end{theorem}

If you think about it carefully, this is an analogue of the implicit function theorem for manifolds, where some coordinate functions are implicitly defined as solution of a levelset equation.
While the proof of Theorem~\ref{thm:impl_fun} is not particularly complicated and is also a direct application of the implicit function theorem, we will not pursue it here. You can find it in \cite[Theorem 5.12]{book:lee} or \cite[Theorem 9.9]{book:tu}.

In the previous chapter, all manifolds were introduced as a set on top of which we had to explicitly provide a differentiable structure. With this theorem we can finally get it all for free as long as we provide an appropriate submersion $F$.

\begin{remark}
  If $F:M\to N$ is a submersion, Theorem~\ref{thm:impl_fun} implies that any $p\in M$ belongs to the $(m-n)$-dimensional embedded submanifold $F^{-1}(F(p))$.
\end{remark}

% We can gather this observation and the previous results (the inverse and the implicit function theorems) into the following proposition (of which we are also omitting the proof).
%
% \begin{proposition}\label{prop:submanifolds_and_R}
%   The following assertions are equivalent.
%   \begin{enumerate}[(i)]
%     \item $P^n\subset M^m$ is a $n$-dimensional submanifold, $n \leq m$.
%     \item $P$ is locally the image of an embedding of a subset of $\R^n$.
%           That is, for every $p\in P$ there exists $V\subset P$ open neighbourhood of $p$, an open set $U\subset\R^n$ and an embedding
%           \begin{equation}
%             \varphi : U \to M \quad\mbox{such that}\quad \varphi(U)=V.
%           \end{equation}
%     \item $P$ is locally a level set of a submersion into $\R^{m-n}$.
%           That is, for every $p\in P$ there exists $V\subset P$ open neighbourhood of $p$ and a submersion $\psi: V \to\R^{m-n}$ such that
%           \begin{equation}
%             M\cap V = \{q\in V \;\mid\; \psi(q) = 0\}.
%           \end{equation}
%   \end{enumerate}
% \end{proposition}

\begin{remark}\label{rmk:WhitneyET}
  Whitney Embedding Theorem states that any smooth $n$-dimensional manifold can be smoothly embedded into $\R^{2n}$.
  Thus any abstract manifold is diffeomorphic to a submanifold of $\R^m$ (for some $m$).
\end{remark}

\begin{remark}
  The concepts expressed in this section are extremely relevant in the contexts of mechanics and topology.
  Some good keywords to know more, here, could be Morse Theory, Floer Homology or Arnold's conjecture.
  We will not get into this, but I will refer you to a nice article from Quanta Magazine that touches upon these topics \cite{article:quanta:floer}.
\end{remark}

\begin{example}\label{ex:s2}
  The sphere $\bS^2 = \{x\in\R^3 \mid \|x\| = 1\}$ is a $2$-dimensional embedded submanifold of $N=\R^3$.
  This is an immediate consequence of Theorem~\ref{thm:impl_fun}: let $F(x) = \|x\|^2 -1 : \R^3 \to \R$, then $F$ is smooth, $\bS^2 = \{x\in\R^3\mid F(x)=0\}$ and, denoting $t$ the coordinate on $\R$, $dF_x(v)= v^i \frac{\partial F}{\partial x^i}|_x \frac{\partial}{\partial t}|_0 = (2x\cdot v) \frac{\partial}{\partial t}|_0$, that is, as a 1x3 matrix $dF_x = 2(x^1\; x^2\; x^3)$ so it is of maximal rank $1$ for all $x\in\bS^2$.
\end{example}

\begin{example}
  Let $N = \R^2$ and $P = \{ x\in N \;\mid\; x^2 = |x^1| \}$.
  Then $P$ is \emph{not} a submanifold, but it can be equipped with a manifold structure.
  For example with the global atlas $\{(P,\; (x^1,x^2)\mapsto x^1)\}$, $P$ is a manifold diffeomorphic to $\R$.
\end{example}

\begin{exercise}
  A real-valued function $f:M\to\R$ on a manifold has a local maximum at $p\in M$ if there is a neighbourhood $U\subset M$ of $p$ such that $f(p) \geq f(q)$ for all $q\in U$.
  \begin{enumerate}
    \item Show that if a differentiable function $f:(a,b)\to\R$, has a local maximum at $x\in (a,b)$, then $f'(x) = 0$.
    \item Prove that a local maximum of a function $f\in C^\infty(M)$ is a critical point of $f$.\\
          \textit{\small Hint: choose $X_p\in T_pM$ and let $\gamma(t)$ be a curve in $M$ starting at $p$ with initial velocity $X_p$. The $f\circ \gamma$ is a real-valued function with local maximum at $0$...}
  \end{enumerate}
\end{exercise}

\newthought{We still have a question pending} since the beginning of the previous chapter.
Is the tangent space to a sphere the one that we naively imagine (see Figure~\ref{fig:tan-embedded-sphere})?
To finally answer the question, we will prove one last proposition.

\begin{proposition}
  Let $F:M^m\to N^n$ be a smooth map between smooth manifolds.
  Let $q\in N$ be a regular value of $F$ such that $P:=F^{-1}(q)\neq\emptyset$ and let $i:P\hookrightarrow M$ denote the inclusion.
  Then, for all $p\in P$, one has
  \begin{equation}
    d i_p(T_p P) = \ker dF_p.
  \end{equation}
\end{proposition}
\begin{proof}
  Both $d i_p(T_p P)\subset T_p M$ and $\ker dF_p \subset T_p M$ are linear subspaces of the same dimension $m-n$, therefore we only need to show that one contains the other, e.g. $d i_p(T_p P) \subset \ker dF_p$.

  Take $f\in C^\infty(N)$ and $v\in T_p P$. By the chain rule\footnote{Proposition~\ref{thm:chainrule_mfld}} we get
  \begin{align}
    (d F_p \circ d i_p)(v)(f) = d(F\circ i)_p(v)(f) = v(f\circ F\circ i).
  \end{align}
  Since $F\circ i\big|_{P} \equiv q$ constant, $f\circ F\circ i\in C^\infty(P)$ is the constant function $p \mapsto f(q)$ and by Corollary~\ref{cor:derzero} we have $v(f\circ F\circ i)=0$.
\end{proof}

\begin{example}
  We have seen in Example~\ref{ex:s2} that $\bS^2 = F^{-1}(0)$ is a smooth manifold of dimension $2$.
  Denoting the inclusion by $i:\bS^2 \hookrightarrow\R^3$, one has
  \begin{equation}\label{ex:tan_sph}
    di_p(T_p\bS^2) = \cT_p(p^\perp)
  \end{equation}
  where $\cT_p:\R^3\to T_p\R^3$ is the map defined in Exercise~\ref{ex:tg_curve_iso} and
  \begin{equation}
    p^\perp := \big\{q\in\R^3 \;\mid\; \left\langle p, q\right\rangle = 0\big\},
  \end{equation}
  where $\left\langle\cdot,\cdot\right\rangle$ is the usual euclidean dot product. The latter directly comes from computing $dF_p$ and its kernel, which we essentially already did in Example~\ref{ex:s2}.
  Take a long deep breath and unfold the definitions in~\eqref{ex:tan_sph}, here it may be useful to draw a picture\footnote{Which is generally always the case in geometry and topology, and most other mathematical fields.}.
  Equation~\eqref{ex:tan_sph} implies that the tangent space to $\bS^2$ at a point $p$ is the plane tangent to $\bS^2$ at $p$, as claimed in Figure~\ref{fig:tan-embedded-sphere}.
\end{example}

\begin{exercise}
  Show that the above reasoning holds verbatim for $\bS^n\subset\R^{n+1}$.
\end{exercise}

\begin{exercise}
  Let $U\subset\R^n$ open and $f:U\to\R$ smooth.
  Define $g:U\to\R^{n+1}$ by
  \begin{equation}
    g(x) = (x, f(x)).
  \end{equation}
  Show that $g$ is a smooth embedding and, therefore, that $g(U)$ is a smooth embedded $n$-dimensional submanifold\footnote{$g(U)$ is the the \emph{graph} of $f$!} of $\R^{n+1}$.
\end{exercise}

\begin{exercise}\label{exe:onsubmanifold}
  Show that the orthogonal matrices
  \begin{equation}
    O(n) := O(n, \R) = \{ Q\in \mathrm{Mat}(n, \R) \mid Q^TQ=\id \}
  \end{equation}
  form a $n(n-1)/2$-dimensional submanifold of the $n^2$-manifold $\mathrm{Mat}(n, \R)$ of $n\times n$-matrices.

  Show also that
  \begin{equation}
    T_Q O(n) = \left\lbrace B \in \mathrm{Mat}(n, \R) \mid (Q^{-1} B)^T = -Q^{-1}B \right\rbrace,
  \end{equation}
  and, thus, that $T_{\id} O(n)$ is the space of skew-symmetric matrices
  \begin{equation}
    T_{\id} O(n) = \left\{ B \in \mathrm{Mat}(n, \R) \mid B^T = -B \right\}.
  \end{equation}
  \textit{\small Hint: Find a suitable map $F: \mathrm{Mat}(n, \R) \to \mathrm{Sym}(n)$ such that $F^{-1}(\{p\}) = O(n)$ for some point $p$ in the image, e.g. $0$ or $\id_n$.
    Here $\mathrm{Sym}(n)$ denotes the space of symmetric matrices.}
\end{exercise}