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\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item The \emph{ellipsoid} $E \subset \RR^3$
        is $f^{-1}(0)$, for $f : \RR^3 \to \RR$
        \[ f(x, y, z) = \frac{x^2}{a^2} +
        \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1 \]
        $\forall p \in E = f^{-1}(0)$, $Df |_p
        \neq 0$, so $E$ is a smooth surface in
        $\RR^2$.
    \item Surfaces of revolution. \\
        Let $\gamma : [a, b] \to \RR^3$ be a
        smooth map with image in the $xz$-plane
        \[ \gamma(t) = (f(t), 0, g(t)) \]
        Assume $\gamma$ is injective, $\gamma'(t)
        \neq 0 \forall t$ and $f > 0$. Rotate this
        curve around the $z$-axis.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/df99520aa3b511ed.png}
        \end{center}
        The associated surface of revolution as
        \emph{allowable} parametrization
        \[ \sigma(u, v) = (f(u) \cos v, f(u) \sin
        v, g(u)) \]
        \[ (u, v) \subseteq (a, b) \times (\theta,
        \theta + 2\pi) \]
        $\theta \in [0, 2\pi)$ fixed. $\sigma$
        homemorphic onto its image (check!)
        \[ \sigma_u = (f' \cos v, f'\sin v, g') \]
        \[ \sigma_v = (-f\sin v, f\cos v, 0) \]
        and $\|\sigma_u \times \sigma_v\|^2 =
        f^2(f'^2 + g'^2) \neq 0$.
\end{enumerate}

\subsection{Orientability}

$V, V' \subset \RR^2$ open, $f : V \to V'$ a
diffeomorphism. Then at any $x \in V$, $Df|_x \in
\GL_2(\RR)$. Let $\GL^+_2(\RR) \subset
\GL_2(\RR)$ be the subgroup of matrices of
\emph{positive} determinant.

\begin{definition*}
    We say that $f$ is \emph{orientation
    preserving} if $Df|_x \in \GL^+_2(\RR)$ for
    all $x \in V$.
\end{definition*}

\begin{flashcard}[orientable-surface]
\begin{definition*}
    An abstract smooth surface $\Sigma$ is
    \emph{orientable} if \cloze{it admits an atlas such
    that the transition maps are \fcemph{orientation
    preserving diffeomorphisms} of open sets of
    $\RR^2$. \myskip
    A choice of such an atlas is an
    \emph{orientation} of $\Sigma$ and we say that
    $\Sigma$ is oriented.}
\end{definition*}
\end{flashcard}

\begin{lemma}
    If $\Sigma_1$ and $\Sigma_2$ are abstract
    smooth surfaces and they are diffeomorphic
    then $\Sigma_1$ is orientable if and only if
    $\Sigma_2$ is orientable.
\end{lemma}

\begin{proof}
    Suppose $f : \Sigma_1 \to \Sigma_2$ is a
    diffeomorphism and $\Sigma_2$ is orientable
    and equipped with an oriented atlas
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/b1e0d364a3b611ed.png}
    \end{center}
    Consider the atlas on $\Sigma_1$ given by
    $(f^{-1}U, \psi \circ f|_{f^{-1} U})$, where
    $(U, \psi)$ is a chart of $\Sigma_2$. The
    transition function between two such charts is
    exactly the transition function in the
    $\Sigma_2$ atlas.
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item There is no sensible classification of
        all smooth or topological surfaces, for
        example $\RR^2 \setminus Z$ for $Z$ closed
        realises all kinds of different
        homeomorphic types. \myskip
        By contrast, \emph{compact} smooth
        surfaces up to diffeomorphism are
        classified by Euler characteristic.
    \item There's a definition of orientation
        preserving homeomorphism that needs some
        algebraic topology. The M\"obius band is
        the surface:
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/b98ff8f0a3b711ed.png}
        \end{center}
        It turns out that abstract smooth surface
        is orientable $\iff$ it contains no
        sub-surface homeomorphic to a M\"obius band.
    \item We get other structure by demanding
        transition maps to be such that
        \[ D(\varphi_1 \varphi_2^{-1}) |_x \in G
        \subset \GL_2(\RR) \]
        $G = \GL_1(\CC) \subset \GL_2(\RR) \to$
        Riemann surfaces.
\end{enumerate}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $S^2$ with the atlas of 2 stereographic
        projections, you computed the transition
        map $(u, v) \mapsto \left( \frac{u}{u^2 +
        v^2}, \frac{v}{u^2 + v^2} \right)$ on
        $\RR^2 \setminus \{0\}$. Check: this is
        orientation preserving.
    \item $T^2$, transition maps are translations
        of $\RR^2$, so $T^2$ is orientated.
\end{enumerate}
For surfaces in $\RR^3$ we'd like to have
orientability dictated by some ``ambient
feature''.

\begin{definition*}
    $\Sigma \subset \RR^3$ smooth surface, $p \in
    \Sigma$. Fix an allowable parametrization
    $\sigma : V \to U \subset \Sigma$, $\sigma(0)
    = p$. Then the \emph{tangent plane} $T_p
    \Sigma$ of $\Sigma$ at $p$ is
    $\Im(D\sigma|_0) \subset \RR^3$ a 2D vector
    subspace of $\RR^2$. The \emph{affine}
    tangent plane of $\Sigma$ at $p$, is $p + T_p
    \Sigma \subset \RR^3$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/294ff846a3ba11ed.png}
    \end{center}
\end{definition*}

\begin{hiddenflashcard}[tangent-plane]
    What is the definition of $T_p\Sigma$? \\
    \cloze{$T_p \Sigma = \Im(D\sigma|_0) \subset
    \RR^3$. In particular this passes through $0
    \in \RR^3$ rather than $p$. If this is desired
    instead then, consider the
    \emph{affine tangent plane} which is defined as $p +
    T_p\Sigma$.}
\end{hiddenflashcard}

\begin{lemma}
    $T_p \Sigma$ is well-defined, i.e. it is
    independent of the choice of allowable
    parametrisation near $p$.
\end{lemma}