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\lhead{Math 4441 Activity 11}
\rhead{<your name(s)>}

\begin{document}
\title{Math 4441 Activity 11}
\author{<your name(s)>}
\date{November 16, 2022}
\maketitle
\vspace{5em}

\setcounter{section}{11}

\begin{exercise}\label{activity-11:cylinder-is-locally-flat}
Consider the following surfaces:
\[
\vec{x}(u^1,u^2) := \left(R\,\cos\left(\dfrac{u^1}{R}\right),R\,\sin\left(\dfrac{u^1}{R}\right),u^2\right),
\quad
-\pi<u^1<\pi,-\infty<u^2<\infty,
\]
for some $R>0$, and
\[
\vec{y}(u^1,u^2) := (u^1,u^2,0),
\quad
-\infty<u^1,u^2<\infty.
\]
These give a right circular cylinder of radius $R$ and a plane, respectively.
\begin{enumerate}[label=(\alph*)]
	\item Check that the two surfaces have the same matrix of metric coefficients $(g_{ij})$.  This tells us that the surfaces have the same intrinsic geometry.
	
	\item Show that the Weingarten maps of $\vec{x}$ and $\vec{y}$ are represented by the matrices
	\[
	(L_i^j)_{\vec{x}} = \left(\begin{matrix}
	-1/R & 0\\ 0 & 0
	\end{matrix}\right)
	\quad\text{and}\quad
	(L_i^j)_{\vec{y}} = \left(\begin{matrix}
	0 & 0\\ 0 & 0
	\end{matrix}\right),
	\]
	in the bases $\{\vec{x}_1,\vec{x}_2\}$ and $\{\vec{y}_1,\vec{y}_2\}$, respectively.  This tells us that the Weingarten map is \emph{not} intrinsic --- it depends on the embedding into $\mathbb{R}^3$, rather than just the first fundamental form.
\end{enumerate}
\end{exercise}

\begin{exercise}
Use \verb|weingarten.nb| to visualize the Gauss map of the right circular cylinder given in Exercise~\ref{activity-11:cylinder-is-locally-flat}.  Based on what you see, use \textbf{complete sentences} to give a geometric interpretation of the fact that $(L_i^j)$ has the form
\[
(L_i^j) = \left(\begin{matrix}
* & 0\\ 0 & 0
\end{matrix}\right),
\]
with $*$ nonzero.
\end{exercise}
\begin{solution}
<your solution>
\end{solution}

\begin{exercise}
Consider the following three simple surfaces, all with domain $U=(-\infty,\infty)\times(-\infty,\infty)$:
\begin{align*}
(u^1,u^2) &\mapsto \left(u^1,u^2,\tfrac{1}{2}((u^1)^2+(u^2)^2)\right)\\
(u^1,u^2) &\mapsto \left(u^1,u^2,\tfrac{1}{2}((u^1)^2-(u^2)^2)\right)\\
(u^1,u^2) &\mapsto \left(u^1,u^2,-\tfrac{1}{2}((u^1)^2+(u^2)^2)\right).
\end{align*}
At $(u^1,u^2)=(0,0)$, the matrices $(L_i^j)$ for these surfaces are given by
\[
\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right),
\quad
\left(\begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix}\right),
\quad
\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right),
\]
but \textbf{not necessarily in this order}!  Use \verb|weingarten.nb| to investigate these surfaces (they're pre-written) and match each surface to its Weingarten map at $(u^1,u^2)=(0,0)$.  Justify your answers with \textbf{complete sentences}.\\
\emph{Hint: The notebook should default the values of $u^1$ and $u^2$ to zero.  Holding one of them there, investigate what happens as the other varies.  Don't forget the pesky minus sign in the definition of the Weingarten map!}
\end{exercise}

\begin{exercise}
The simple surface $\vec{x}\colon(-\pi,\pi)\times(-\pi,\pi)\to\mathbb{R}^3$ defined by
\[
\vec{x}(u^1,u^2) = (\cos u^1(5-2\sin u^2),\sin u^1(5-2\sin u^2),2\cos u^2)
\]
has image lying on a torus (i.e., donut) in $\mathbb{R}^3$).  The matrix representation of the Weingarten map for this surface has the form
\[
(L_i^j) = \left(\begin{matrix}
f(u^2) & 0\\ 0 & -1/2
\end{matrix}\right),
\]
for some function $f\colon(-\pi,\pi)\to\mathbb{R}$.  Use \verb|weingarten.nb| to investigate this surface (it's pre-written) and determine the values of $u^2$ for which $f(u^2)<0$, $f(u^2)=0$, or $f(u^2)>0$.  Explain your conclusions in \textbf{complete sentences}.\\
\emph{Hint: In the notebook, you can set $u^2=0$ and then notice what happens as $u^1$ varies.  Then try setting $u^2$ to some positive or negative value and vary $u^1$ again.}
\end{exercise}

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