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\lhead{Math 4441 Homework 6}
\rhead{<your name>}


\begin{document}
\title{Math 4441 Homework 6}
\author{<your name>}
\date{December 12, 2022}
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\begin{enumerate}
	\item$\!\!\!\!{^{**}}$ For any simple surface $\vec{x}$, prove that $(L_{ij})=(g_{ij})(L^i_j)$.\\
	\emph{Hint: See the last few slides from November 14.}

	\item Let $\vec{x}$ be a simple surface in $\mathbb{R}^3$, and suppose that the Weingarten map $\mathcal{L}_p$ is 0, for all points $p$ in $\mathrm{im}(\vec{x})$.  Show that $\mathrm{im}(\vec{x})$ is contained in a plane.\\
	\emph{Hint: Start by arguing that $\nu\colon\mathrm{im}(\vec{x})\to\mathbb{R}^3$ is a constant map.  Then fix a point $p$ in $\mathrm{im}(\vec{x})$ and argue that $\mathrm{im}(\vec{x})$ lies in the plane through $p$ and perpendicular to the fixed output $\vec{n}_0$ of $\nu$.}
	
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	\item Show that the Weingarten map $\mathcal{L}$ of a surface satisfies the equation
	\[
	\mathcal{L}^2-2H\,\mathcal{L}+K\,\mathrm{id}=0,
	\]
	where $H$ and $K$ are the mean and Gaussian curvatures of the surface, respectively, and $\mathrm{id}$ is the identity map on the tangent space.\\
	\emph{Hint: Recall the Cayley-Hamilton theorem from linear algebra.}
	
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	\item$\!\!\!\!{^{**}}$ Show that there is no simple surface $\vec{x}$ whose first and second fundamental forms are given by
	\[
	(g_{ij}) = \left(\begin{matrix}
	1 & 0\\ 0 & \cos^2 u^1
	\end{matrix}\right)
	\quad\text{and}\quad
	(L_{ij}) = \left(\begin{matrix}
	\cos^2 u^1 & 0\\ 0 & 1
	\end{matrix}\right).
	\]
	\emph{Hint: Show that the second Codazzi-Mainardi equation is violated.  Use the intrinsic formula to compute the Christoffel symbols.}
	
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	\item$\!\!\!\!{^{**}}$ Consider the simple surfaces $\vec{x},\vec{\tilde{x}}\colon(0,\infty)\times(-\infty,\infty)\to\mathbb{R}^3$ defined by
	\[
	\vec{x}(u^1,u^2) = (u^1\,\cos u^2, u^1\,\sin u^2,\ln u^1)
	\quad\text{and}\quad
	\vec{\tilde{x}}(u^1,u^2) = (u^1\,\cos u^2,u^1\,\sin u^2,u^2)
	\]
	Verify that $K(u^1,u^2)=\tilde{K}(u^1,u^2)$ for all $(u^1,u^2)$, but that $(g_{ij})\neq (\tilde{g}_{ij})$.  This suggests that, while Gaussian curvature is preserved by isometry, it cannot detect isometries.
	
	\item$\!\!\!\!{^{**}}$ Suppose a simple surface $\vec{x}$ has $g_{11}\equiv 1$ and $g_{12}\equiv 0$.  Imitate our proof of the \emph{Theorema Egregium} to prove that
	\[
	\dfrac{\partial^2}{\partial (u^1)^2}(\sqrt{g_{22}}) + K\,\sqrt{g_{22}} = 0.
	\]
	\emph{Hint: As in our proof, you'll need to consider the equality of some order-3 mixed partials.}
	
	\item Show that if a simple surface $\vec{x}$ has Gaussian curvature everywhere non-positive, then there are no simple, closed geodesics on $\mathrm{\vec{x}}$.  Why does this not contradict the fact that the parallels of a right circular cylinder are geodesics?\\
	\emph{Hint: Use the Gauss-Bonnet theorem for surfaces with boundary.}
	
	\item Show that if $\mathcal{S}$ is a compact surface without boundary whose Gaussian curvature is everywhere positive, then $\mathcal{S}$ is diffeomorphic to a sphere.  Is the converse of this statement true?
	
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