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\author{Professor Alejandro Uribe-Ahumada\\ \small\i{Transcribed by Thomas Cohn}}
\title{Math 635 Lecture 30}
\date{3/31/21} % Can also use \today

\begin{document}
\maketitle
\setlength\RaggedRightParindent{\parindent}
\RaggedRight

\par\noindent
Last time, we proved Cartan-Hadamard. To review:\n

\thm{
	(Cartan-Hadamard) If $M$ is a complete, connected Riemannian manifold with $K\le{}0$, then $\forall{}p\in{}M$,\n $\exp_{p}:T_{p}M\to{}M$ is a smooth covering map.\nn
	Proof (sketch): From homework, we know $\exp_{p}$ is a local diffeomorphism. Then we show the following proposition: If $F:\tilde{M}\to{}M$ is a local isometry, with $\tilde{M}$ and $M$ complete, then $F$ is a covering map. Finally, we take $\tilde{M}=T_{p}M$, with the ``pull-back'' metric such that $\exp_{p}$ is a local isometry. Now, we apply the proposition. We just need the metric to be complete, and it is, because rays $t\mapsto{}tv$ (for $v\in{}T_{p}M\cut\set{0}$) are geodesics with respect to the pull-back metric, and because $\exp_{p}$ is a local isometry, they map to geodesics on $M$. So $T_{p}M$ with this metric has the property that all geodesics through $0\in{}T_{p}M$ can be continued, $\forall{}t\in\R$, so $T_{p}M$ is complete.\proven
}

\par\noindent
Why is this such a big deal?\n

\cor{
	If $M$ is as in Cartan-Hadamard, and simply connected, then $M$ is diffeomorphic to $\R^{n}$.\n
}

\par\noindent
In general, for $M$ complete with $K\le{}0$, $M$ is said to be aspherical: $\forall{}n>1$, $\pi_{n}(M)=0$.\n

\par\noindent
Observe: If $M$ is complete, then $M$ cannot be isometrically embedded as a proper open set of some Riemannian manifold $W$. This is true because covering maps are surjective.\n

\prop{
	If $M\subsetneq{}W$, and $M$ is open, then $M$ is not complete.\nn
	Proof: Let $p\in\d{}M$, and $U$ a normal neighborhood of $p$ in $W$. Let $q\in{}M\cap{}U$. Let $\gamma$ be a radial geodesic from $p$ to $q$. Reverse $t$, so we have $\gamma^{-}:q\rightsquigarrow{}p$. For small $t$, $\gamma^{-}$ is a geodesic in $M$. But for larger $t$, $\gamma^{-}(t)\not\in{}M$, so $M$ is not complete.\proven
}

\section*{The Second Fundamental Form}

\par\noindent
(Refer to Chapter 6 of Do Carmo.)\n

\par\noindent
Note: The first fundamental form is just the metric itself.\n

\par\noindent
The setting and notation we'll consider is $\bar{M}$ a Riemannian manifold (known as the ``ambient space''), $M\subseteq\bar{M}$ a submanifold with the induced metric, $\bar\nabla$ the Levi-Civita connection on $\bar{M}$, and $\nabla$ the Levi-Civita connection on $M$. Further, if $p\in{}M$, $v\in{}T_{p}\bar{M}$, we'll write $v=v^{T}+v^{\perp}=v^{T}+v^{N}$, where $v^{T}$ is the orthogonal projection of $v$ onto $T_{p}M$ (known as the ``tangential component'') and $v^{\perp}$ (or $v^{N}$) is its orthogonal complement in $(T_{p}M)^{\perp}$ (known as the ``normal component'').\n

\par\noindent
Recall (from homework): $\forall{}X,Y\in\mf{X}(M)$, $\nabla_{X}Y=(\bar\nabla_{\bar{X}}\bar{Y})^{T}$, where $\bar{X},\bar{Y}\in\mf{X}(\bar{M})$ are (arbitrary) extensions of $X$ and $Y$.\n

\defn{
	$\forall{}X,Y\in\mf{X}(M)$, $B(X,Y)=(\bar\nabla_{\bar{X}}\bar{Y})^{\perp}$. So $\forall{}p\in{}M$, $B(X,Y)(p)=(\bar\nabla_{\bar{X}}\bar{Y})(p)-(\nabla_{X}Y)(p)$.\n
}

\par\noindent
Note: $B(X,Y)(p)\in(T_{p}M)^{\perp}$.\n

\newpage
\prop{
	$B(X,Y)(p)$ depends only on $X_{p}$ and $Y_{p}$, and is therefore equivalent to tensors (maps) $\forall{}p\in{}M$:
	\[
		B_{p}:T_{p}M\times{}T_{p}M\to(T_{p}M)^{\perp}
	\]
	Moreover, $\forall{}p\in{}M$, $B_{p}$ is bilinear and symmetric: $\forall{}x,y\in{}T_{p}M$, $B_{p}(x,y)=B_{p}(y,x)$.\nn
	Proof: We'll first show symmetry. Let $X,Y\in\mf{X}(M)$. Then
	\[
		B(X,Y)-B(Y,X)=\restr{\paren{\bar\nabla_{\bar{X}}\bar{Y}-\bar\nabla_{\bar{Y}}\bar{X}}}_{M}-\paren{\nabla_{X}Y-\nabla_{Y}X}=\restr{\comm{\bar{X},\bar{Y}}}_{M}-\comm{X,Y}
	\]
	But we proved in homework (in Math 591) that on $\restr{\comm{\bar{X},\bar{Y}}}_{M}=\comm{X,Y}$, so we have $B(X,Y)-B(Y,X)=0$, and thus, $B$ is symmetric.\nn
	Now, at $p$, we know that any connection depends pointwise in the ``lower'' entry. So $B(X,Y)(p)$ depends (with respect to $X$) only on $X_{p}$. From symmetry, the same is true for $Y$, so $B$ is bilinear.\proven
}

\par\noindent
To get information from $B$, one chooses $\nu_{p}\in(T_{p}M)^{\perp}$ with $\norm{\nu_{p}}=1$.\n

\defn{
	\begin{itemize}[leftmargin=3\parindent, topsep=0pt, itemsep=0pt]
		\item $H_{\nu}:T_{p}M\times{}T_{p}M\to\R$ by taking $H_{\nu}(x,y)=\iprod{B_{p}(x,y),\nu}$.
		\item $\II_{p}(x)=H_{\nu}(x,x)$.
		\item $S_{\nu}:T_{p}M\to{}T_{p}M$, the shape operator, also known as the Weingarten map, is defined by $\forall{}x,y\in{}T_{p}M$,
		\[
			\iprod{S_{\nu}(x),y}=H_{\nu}(x,y)=\iprod{B(x,y),\nu}
		\]
		$S_{\nu}$ exists and is well-defined because $\giprod$ is non-degenerate.
	\end{itemize}\up\n
}

\par\noindent
Observe: Because $B$ is symmetric, $H_{\nu}$ is symmetric, so $S_{\nu}$ is self-adjoint. Thus, $S_{\nu}$ can be diagonalized, i.e., three's a basis $(e_{1},\ldots,e_{n})$ of $T_{p}M$ such that $S_{\nu}(e_{i})=\kappa_{i}e_{i}$, for $\kappa_{i}\in\R$.\n

\defn{
	The $\kappa_{i}$ are the \u{principal curvatures} of $M$ at $p$, and the $e_{i}$ are the \u{principal directions}.\n
}

\par\noindent
Looking ahead, we will prove if $\bar{M}$ is flat (e.g. $\bar{M}=\R^{n}$), then the intrinsic sectional curvature of $M$ satisfies\n $K(e_{i},e_{j})=\kappa_{i}\kappa_{j}$, $\forall{}i\ne{}j$. When $\dim{}M=2$, this is Gauss' ``Theorem Egregium''.

\end{document}