TITLE: Help in solving this differential equation.
QUESTION [0 upvotes]: I'm trying to solve this differential equation but having some trouble as there is a constant in there which would change the solution depending on its value:
$$\ddot{x}+2k\dot{x}+10k^2x=0~~~~~~~(k>0),~(x(0)=0),~ (\dot{x}(0)=u).$$
I need to solve for $x$ in terms of $t$
I got down to trying to solve my characteristic equation which has roots:
$$\lambda_{1,2}=-k\pm\sqrt{k^2-10k}$$ but unsure where to go from here. Surely depending on $k$ this will have either real or complex roots in which case the solution will have a different form so I'm unsure how to proceed.
Any help?

REPLY [0 votes]: Hint:
The caracteristc equation is:
$$
\lambda^2+2k\lambda +10 k^2=0
$$
That has solutions:
$$
\lambda_1=(-1-3i)k \qquad \lambda_1=(-1+3i)k
$$


text:
\begin{document}

\maketitle

In a metric space, such as the real numbers with their standard metric, a set $A$ is open if and only if no sequence with terms outside of $A$ has a limit inside $A$. Moreover, a metric space is compact if and only if every sequence has a converging subsequence. However, in a general topological space these equivalences may fail. Unfortunately this fact is sometimes overlooked in introductory courses on general topology, leaving many students with misconceptions, e.g. that compactness would always be equal to sequence compactness. The aim of this article is to show how sequences might fail to characterize topological properties such as openness, continuity and compactness correctly. Moreover, I will define nets and show how they succeed where sequences fail.

This article grew out of a discussion I had at the University of Leeds with fellow PhD students Phil Ellison and Naz Miheisi. It also incorporates some work I did while enrolled in a topology module taught by Paul Igodt at the Katholieke Universiteit Leuven in 2010.

\section{Prerequisites and terminology}

I will assume that you are familiar with the basics of topological and metric spaces. Introductory reading can be found in many books, such as \cite{kelley} and \cite{munkres}.

I will frequently refer to a topological space $(X, \tau)$ by just the unlying set $X$, when it is irrelevant or clear from the context which topology on $X$ is considered. Remember that any metric space $(X,d)$ has a topology whose basic opens are the open balls \[ B(x,\delta) = \{ y \st d(x,y) < \delta \} \] for all $x \in X$ and $\delta > 0$.

A \textbf{neighbourhood} of a point $x$ in a topological space is an open set $U$ with $x \in U$. Note that some people call $U$ a neighbourhood of $x$ if $U$ just contains an open set containing $x$ \cite[p.~97]{munkres}, but in this article neighbourhoods are always open themselves.

A \textbf{sequence} $(x_n)$ \textbf{converges} to a point $y$ if every neighbourhood of $y$ contains $x_n$ for $n$ large enough. We write $x_n \to y$ and say that $y$ is a \textbf{limit} of the sequence $(x_n)$. If $(x_n)$ converges to $y$, then so does every subsequence of $(x_n)$. If $f: X \to Y$ is a continuous function and $x_n \to y$ in $X$, then also $f(x_n) \to f(y)$ in $Y$. (We say that continuous functions \textbf{preserve} convergence of sequences.) Convergence in a product space is pointwise, i.e. a sequence $(x_n)$ in $\prod_{i \in I} X_i$ converges to $y$ if and only if $x_n(i) \to y(i)$ in $X_i$ for all $i \in I$.

A topological space is \textbf{Hausdorff} if for every two distinct points $x$ and $y$, we can find a neighbourhood of $x$ and a neighbourhood of $y$ that are disjoint. Sequences in general can have more than one limit, but in a Hausdorff space limits (if they exist at all) are unique. Indeed, a sequence cannot be eventually in two disjoint neighbourhoods at once.

A set $X$ is \textbf{countable} when there is a surjection from $\N = \{0, 1, 2, 3, \ldots \}$ onto $X$. So $X$ is countable if and only if $X$ is finite of in bijection with the natural numbers. A countable union of countable sets is still countable. Cantor's famous diagonal argument proves that the unit interval $[0,1]$ and $\R$ are  \textbf{uncountable}.\cite{cantor}

A few of my examples will make use of  \textbf{ordinal numbers}. If you are unfamiliar with ordinal numbers, you can either find background reading in \cite{hrbacek} or you can skip over these examples. I write the first infinite ordinal (i.e. the order type of the natural numbers) as $\omega_0$ and the first uncountable ordinal as $\omega_1$. Because a countable union of countable ordinals is still countable, no countable sequence of countable ordinals can have $\omega_1$ as limit. In other words: the cofinality of $\omega_1$ is $\omega_1$.

\section{Open versus sequentially open}

In a topological space $X$, a set $A$ is \textbf{open} if and only if every $a \in A$ has a neighbourhood contained in $A$. $A$ is \textbf{sequentially open} if and only if no sequence in $X \setminus A$ has a limit in $A$, i.e. sequences cannot \emph{converge out of} a sequentially closed set.

If $X$ is a metric space, then the two notions of open and sequentially open are equivalent. Indeed if $A$ is open, $(x_n)$ is a sequence in $X \setminus A$ and $y \in A$, then there is a neighbourhood $U$ of $y$ contained in $A$. Hence $U$ cannot contain any term of $(x_n)$, so $y$ is not a limit of the sequence and $A$ is sequentially open. Conversely, if $A$ is not open, then there is an $y \in A$ such that every neighbourhood of $y$ intersects $X \setminus A$. In particular we can pick an element \[ x_n \in (X \setminus A) \cap B(y, \frac{1}{n+1} ) \] for all $n \in \N$. The sequence $(x_n)$ in $X \setminus A$ then converges to $y \in A$, so $A$ is not sequentially open.

The implication from open to sequentially open is true in any topological space.

\begin{proposition}
\label{openseqopen}
In any topological space $X$, if $A$ is open, then $A$ is sequentially open.
\end{proposition}

We can just copy the proof for metric spaces, it remains valid in any topological space.

\begin{proof}
Suppose that $A$ is open, let $(x_n)$ be a sequence in $X \setminus A$ and take any $y \in A$. There is a neighbourhood of $y$ contained in $A$, so this neighbourhood doesn't contain any terms of $(x_n)$. Hence the sequence doesn't converge to $y$, as required.
\end{proof}

It is tempting to think that the converse might also hold in any topological space. When this is indeed the case, we call the space sequential.

\begin{definition}
A topological space is \textbf{sequential} when any set $A$ is open if and only if $A$ is sequentially open.
\end{definition}

However, importantly, not every space is sequential.

\begin{proposition}
\label{notseq}
There is a topological space that is not sequential.
\end{proposition}

\begin{proof}
Any of the three examples below constitutes a proof.

\begin{description}
\item{Example 1:} Let $X$ be an uncountable set, such as the set of real numbers. Consider $(X, \tau_{cc})$, the countable complement topology on $X$. Thus $A \subseteq X$ is closed if and only if $A=X$ of $A$ is countable. Suppose that a sequence $(x_n)$ has a limit $y$. Then the neighbourhood \[ \left(\R \setminus \{x_n \st n \in \N \} \right) \cup \{y\}\] of $y$ must contain $x_n$ for $n$ large enough. This is only possible if $x_n = y$ for $n$ large enough. Consequently a sequence in any set $A$ can only converge to an element of $A$, so every subset of $X$ is sequentially open. But as $X$ is uncountable, not every subset is open. So $(X, \tau_{cc})$ is not sequential.
\item{Example 2:} Consider the order topology on the ordinal $\omega_1 +1 = [0, \omega_1]$. Because $\omega_1$ has cofinality $\omega_1$, every sequence of countable ordinals has a countable supremum. Hence no sequence of countable ordinals converges to $\omega_1$, so $\{ \omega_1 \}$ is sequentially open. However, $\{ \omega_1 \}$ is not open as $\omega_1$ is a limit ordinal. So the order topology on $[0, \omega_1]$ is not sequential.
\item{Example 3:} Let $X$ be an uncountable set and let $\{0,1\}$ have the discrete topology. Consider $\Pw(X) = \{0,1\}^X$ with the product topology. Let $\A \subseteq \Pw(X)$ be the collection of all uncountable subsets of $X$. $\A$ is not open; indeed every basic open contains finite sets. However, we claim that $\A$ is sequentially open. Let $(X_n)$ be a sequence of countable subsets of $X$ and suppose that $X_n \to Y$. Then for every $x \in X$ we must have $x \in Y$ if and only if $x \in X_n$ for $n$ large enough. In particular \[ Y \subseteq \bigcup_{n \in \N} X_n.\] But $ \displaystyle \cup_{n \in \N} X_n$ is a countable union of countable sets. Hence a sequence of countable sets can only converge to countable sets, so $\A$ is sequentially open. \qedhere
\end{description}
\end{proof}

Still, a large class of topological spaces is sequential.

\begin{definition}
A \textbf{countable basis at a point $x$} is a countable set \linebreak\mbox{$\{U_n \st n \in \N \}$} of neighbourhoods of $x$, such that for any neighbourhood $V$ of $x$ there is an $n \in \N$ such that $U_n \subseteq V$.

A topological space is \textbf{first countable} if every point has a countable basis.
\end{definition}

Every metric space is first countable, as
\[\left\{ B\left(x, \frac{1}{n+1}\right) \st n \in \N  \right\} \]
is a countable basis at any point $x$. We can prove that every first countable space is sequential by generalizing the proof that every metric space is sequential.

\begin{proposition}
\label{firstcountableseq}
Every first countable space $X$ (and hence every metric space) is sequential.
\end{proposition}

\begin{proof}
Because of Proposition \ref{openseqopen}, we only need to prove that every sequentially open set $A$ is also open. So suppose that $A$ is not open. Then there is an $y \in A$ such that every neighbourhood of $y$ intersects $X \setminus A$. Let $\{U_n \st n \in \N\}$ be a countable basis at $y$. For every $n\in \N$, we can choose
\[ x_n \in (X \setminus A) \cap \left( \bigcap_{i=0}^n U_i \right). \]
Then for every neighbourhood $V$ of $y$, there is an $n \in \N$ such that $U_n \subseteq V$, and hence $x_m \in V$ for every $m \geq n$. So $(x_n)$ is a sequence in $X \setminus A$ that converges to $y \in A$. Therefore $A$ is not sequentially open, as required.
\end{proof}

Sequential spaces are also exactly those spaces $X$ where sequences can correctly define continuity of functions from $X$ into another topological space.

\begin{lemma}
\label{lemmaseq}
Let $X$ be a topological space. Then $A \subseteq X$ is sequentially open if and only if every sequence with a limit in $A$ has all but finitely many terms in $A$.
\end{lemma}
\begin{proof}
We prove that $A \subseteq X$ is \emph{not} sequentially open if and only if there is a sequence with infinitely many terms in $X \setminus A$ and with a limit in $A$.

If $A$ is not sequentially open, then by definition there is a sequence with terms in $X \setminus A$ but with limit in $A$.

Conversely, suppose $(x_n)$ is a sequence with infinitely many terms in $X \setminus A$ that converges to $y \in A$. Then $(x_n)$ has a subsequence in $X \setminus A$ that must still converge to $y \in A$, so $A$ is not sequentially open.
\end{proof}

\begin{proposition}
The following are equivalent for any topological space $X$:
	\begin{enumerate}
	\item $X$ is sequential;
	\item for any topological space $Y$ and function $f: X \to Y$, $f$ is continuous if and only if $f$ preserves convergence (i.e. whenever $x_n \to y$ in $X$, also $f(x_n) \to f(y)$ in $Y$).
	\end{enumerate}
\end{proposition}

\begin{proof}
\begin{description}
\item[$1 \Rightarrow 2$: ] Suppose $X$ is sequential. Any continuous function preserves convergence of sequences, so we only need to prove that if $f: X \to Y$ preserves convergence, then $f$ is continuous. Suppose for contradiction that $f$ is not continuous. Then there is an open $U \subseteq Y$ such that $f^{-1}(U)$ is not open in $X$. As $X$ is sequential, $f^{-1}(U)$ is also not sequentially open, so there is a sequence $(x_n)$ in $X \setminus f^{-1}(U)$ that converges to an $y \in f^{-1}(U)$. However $\left(f(x_n)\right)$ is then a sequence in the closed set $Y \setminus U$, so it cannot have $f(y)$ as a limit. So $f$ does not preserve convergence, as required.
\item[$2 \Rightarrow 1$: ] Suppose that the topological space $(X, \tau)$ is not sequential. Let $(X, \tau_{seq})$ be the topological space where $A \subseteq X$ is open if and only if $A$ is sequentially open in $(X, \tau)$. This is indeed a topology: it is trivial that $\emptyset$ and $X$ are sequentially open, and that any union of sequentially open sets is also sequentially open. It remains to prove that the intersection of two sequentially open sets $A$ and $B$ is sequentially open. Suppose that $(x_n)$ is a sequence with limit $y \in A \cap B$. By Lemma \ref{lemmaseq}, $(x_n)$ must have all but finitely many terms in $A$ and all but finitely many terms in $B$. So $(x_n)$ has all but finitely many terms in $A \cap B$. By Lemmma \ref{lemmaseq} again,  $A \cap B$ is sequentially open.

As $(X, \tau)$ is not already sequential, the topology $\tau_{seq}$ is strictly finer than $\tau$. Hence the identity map
\[id : (X, \tau) \to (X, \tau_{seq}) \]
is not continuous. We claim that $f$ nonetheless preserves convergence. Indeed, suppose $x_n \to y$ in $(X, \tau)$. Every open neighbourhood of $y$ in $(X, \tau_{seq})$ is sequentially open in $(X, \tau)$, so by Lemma \ref{lemmaseq} contains all but finitely many terms of $(x_n)$. Hence also $x_n \to y$ in $(X, \tau_{seq})$, as required.\qedhere
\end{description}
\end{proof}

\section{Sequential spaces as quotients of metric spaces}

First, we recall the definition of a quotient space. Let $X$ be a topological space and let $\sim$ be an equivalence relation on $X$. Consider the set of equivalence classes $X / \sim$ and the projection map $\pi: X \to X/\sim$. We topologize $X / \sim$ by defining $A \subseteq X / \sim$ to be open if and only if $\pi^{-1}(A)$ is open in $X$.

Note that given a surjective function $f: X \to Y$ such that $A \subseteq Y$ is open if and only if $f^{-1}(A)$ is open in $X$, we can consider $Y$ to be a quotient of $X$. Indeed, define an equivalence relation $\sim$ on $X$ such that $x \sim y$ if and only if $f(x) = f(y)$, i.e. the equivalence classes are the fibers of $f$. Then $X / \sim$ is isomorphic to $Y$ by mapping the equivalence class of $x$ to $f(x)$.

We are now ready to prove that the sequential spaces are exactly the quotients of metric spaces. This is a corollary of the following two propositions.

\begin{proposition}
\label{quotientseq}
Any quotient $X / \sim$ of a sequential space $X$ is sequential.
\end{proposition}

\begin{proof}
Suppose that $A \subseteq X / \sim$ is not open. We need to prove that $A$ is not sequentially open either. By definition of quotient space, $\pi^{-1}(A)$ is not open in $X$. As $X$ is sequential, there is a sequence $(x_n)$ in $X \setminus \pi^{-1}(A)$ that converges to some $y \in \pi^{-1}(A)$. But $\pi$ is continuous, so it preserves convergence of sequences. Hence $(\pi(x_n))$ is a sequence in $(X/\sim) \setminus A$ with limit $\pi(y) \in A$. Thus $A$ is not sequentially open, as required.
\end{proof}

\begin{proposition}[Franklin \cite{franklin}]
\label{franklin}
Every sequential space $X$ is a quotient of some metric space.
\end{proposition}

\begin{proof}
Let $\C$ be the set of all sequences $(x_n)$ in $X$ that converge to their first term, i.e. $x_n \to x_0$.

Consider the subspace $Y = \{0\} \cup \{\frac{1}{n+1} \st n \in \N \}$ of $\R$ with the standard metric. Thus, $A \subseteq Y$ is open if and only if $0 \not\in A$ or $A$ contains all but finitely many elements of $Y$. Note that $Y$ is metric as a subspace of a metric space.

Now consider the disjoint sum (i.e. the coproduct in category theory jargon)
\[ Z = \bigoplus_{(x_n) \in \C} \{(x_n)\} \times Y. \]
The underlying set of $Z$ is
\[ \bigcup_{(x_n) \in \C} \{(x_n)\} \times Y \]
 and $A \subseteq Z$ is open if and only if for every $(x_n) \in \C$ the set
\[ \{ y \in Y \st ((x_n), y) \in A\} \]
is open in $Y$. Note that $Z$ is metrizable a disjoint sum of metric spaces.

Next consider the map
\begin{align*}
f: \  Z &\to X \\
((x_n),0) &\mapsto x_0 \\
\left((x_n),\frac{1}{i+1}\right) &\mapsto x_i \quad \quad \quad \mbox{for all $i \in \N$}
\end{align*}
I claim that this map exhibits $X$ as a quotient of $Z$. Indeed $f$ is clearly surjective: for all $x \in X$ the constant sequence at $x$ converges to $x$, so $x=f((x),0)$. Hence it remains to show that $A \subseteq X$ is open if and only if $f^{-1}(A)$ is open in $Z$.

Suppose that $A \subseteq X$ is open. As $X$ is sequential, every sequence $(x_n)$ in $X$ converging to some $a \in A$ must have all but finitely many terms in $A$ by Lemma \ref{lemmaseq}. So if $((x_n), 0) \in f^{-1}(A)$ (which means that $(x_n)$ converges to $x_0 \in A$), then $f^{-1}(A)$ will contain all but finitely many elements of $\{(x_n)\} \times Y$. So for each $(x_n) \in \C$, the set
\[ \{ y \in Y \st ((x_n), y) \in f^{-1}(A) \} \]
is open in $Y$. Hence $f^{-1}(A)$ is open in $Z$.

Conversely, if $A$ is not open in $X$, then there is some sequence $(x_n)$ in $X \setminus A$ that converges to some $a \in A$. But then
\[ \{ y \in Y \st ((x_n), y) \in f^{-1}(A) \} = \{0\} \]
is not open in $Y$, so $f^{-1}(A)$ is not open in $Z$.
\end{proof}

\begin{corollary}
\label{seqquotientmetric}
A topological space is sequential if and only if it is a quotient of a metric space.
\end{corollary}

\begin{proof}
One direction is the above proposition. For the other direction, note that by Proposition \ref{firstcountableseq} every metric space is sequential, so by Proposition \ref{quotientseq} any quotient of a metric space is also sequential.
\end{proof}

We can now also easily prove that sequential is a strictly weaker notion than first countable.

\begin{proposition}
\label{seqnotfirstcountable}
There exists a sequential space which is not first countable.
\end{proposition}

\begin{proof}
Consider $\R$ with the standard topology, and the equivalence relation $\sim$ on $\R$ that identifies all the natural numbers, i.e. the equivalence classes are $\N$ and $\{x\}$ for every $x \in \R \setminus \N$.

The quotient space $\R / \sim$ is sequential as a quotient of a metric space.

I claim that $\R / \sim$ is not first countable, in particular that there is no countable basis at $\N$. Suppose that $\{U_n \st n \in \N \}$ is any countable collection of neighbourhoods of $\N$. For all $n \in \N$, $\pi^{-1}(U_n)$ is a neighbourhood of $n$ in $\R$ with the standard topology, so there is a $\delta_n > 0$ such that $B(n, \delta_n) \subseteq \pi^{-1}(U_n)$. Then consider
\[ \pi\left(\bigcup_{n \in \N} B\left(n, \frac{\delta_n}{2} \right) \right). \]
This is a neighbourhood of $\N$ in $X / \sim$, but it doesn't contain $U_n$ for any $n \in \N$. So $\{U_n \st n \in \N \}$ is not a countable basis at $\N$, as required.
\end{proof}

\section{Nets save the day}

Looking back at the section on open and sequentially open sets, we see that convergence of sequences doesn't give us full information on the topology. For example the discrete topology and the countable complement topology on an uncountable set $X$ have the same converging sequences (namely $x_n \to y$ if and only if $x_n = y$ for $n$ large enough), but the discrete topology is strictly finer than the countable complement topology. The discrete topology is sequential, but the countable complement topology contains sequentially open sets which are not open.

Convergence of sequences works fine when the space is first countable, because a countable basis at a point allows us to \emph{approach} that point nicely with a sequence. However, if a point $x$ does not have a countable basis, then sequences might not succeed in getting \emph{close to} $x$, i.e. eventually in every neighbourhood of $x$. Sequences fall short in two respects: they are \emph{too short} and \emph{too thin}.

Remember Example 2 from the proof of Proposition \ref{notseq}; $[0, \omega_1]$ with the order topology. Even though $\{\omega_1\}$ is not open, a sequence of countable ordinals can only have a countable limit. Because a sequence only has countable many terms, it never can advance \emph{deep enough} in the ordinal numbers to get close to $\omega_1$. A possible solution is to allow sequences indexed by any linearly ordered set, instead of just the natural numbers. Indeed the $\omega_1$-sequence
\begin{align*}
x: \omega_1 &\to [0, \omega_1] \\
	\alpha &\mapsto \alpha
\end{align*}
converges to $\omega_1$, even though every term is countable.

This overcomes the \emph{shortness} of sequences, but is still not enough to solve all difficulties. Indeed, reconsider Example 3 from the proof of Proposition \ref{notseq}; the product space $\Pw(X) = \{0,1\}^X$ where $X$ is uncountable. We take the subspace $\Q$ of $\Pw(X)$ which consists of those subsets of $X$ that are either finite or uncountable. Define $\A \subseteq \Pw(X)$ to be the collection of all uncountable subsets of $X$ like before. $\A$ is still not open in $\Q$ as every basic open contains finite sets. But no sequence of finite sets can converge to an uncountable set, not even if we allow sequences indexed by any linearly ordered set. Indeed, let $(X_i)_{i\in I}$ be any $I$-sequence of finite subsets of $X$, where $I$ is any linearly ordered set. For all $i \in I$, define
\[ Z_i = \{x \in X \st \mbox{$x \in X_j$ for all  $j \geq i$} \}. \]
Every $Z_i$ is finite as $Z_i \subseteq X_i$. Moreover $Z_i \subseteq Z_j$ for $i \leq j$, so there can only be countably many distinct sets $Z_i$. (There can be at most one $Z_i$ with cardinality $1$, at most one with cardinality $2$, etc.) Hence $\cup_{i \in I} Z_i$ is countable as a countable union of countable sets. But if $X_i \to Y$ then we must have $Y = \cup_{i \in I} Z_i$, so $Y$ cannot be uncountable. Intuitively, the problem with sequences here is that they are linearly ordered, so they can only approach a point from \emph{one angle at a time}, whereas to capture the topology we need to consider all \emph{angles of approach} at the same time.

Nets are defined to overcome the shortcomings of sequences. Nets generalize sequences, but they can go both \emph{deeper} and \emph{wider} than sequences. Sequences associate a point to every natural number. Nets are more general, as they can associate a point to every element of a directed set.

\begin{definition}
A \textbf{directed set} is a set $D$ with a preorder relation (i.e. a reflexive and transitive binary relation) such that every two elements have an upper bound.
\end{definition}

Note that we \emph{don't} require that a pair of elements has a \emph{least} upper bound, we just require that some upper bound exists.

{
\setlength{\leftmargini}{5em}
\begin{enumerate}
\item[Example 4]: Every linearly ordered set (such as the $\N$ with the usual order) is a directed set.
\item[Example 5]: Any collection of sets that is closed under binary intersections is a directed set when ordered by reverse inclusion, i.e. $X \le Y$ if and only if $Y \subseteq X$.

In particular, given any point $x$ of a topological space, the collection of all neighbourhoods of $x$ ordered by reverse inclusion is a directed set, which we write as $\Nbh(x)$.
\item[Example 6]: If $D$ and $E$ are directed sets, then so is their product $D \times E$ ordered by $(d_1,e_1) \le (d_2,e_2)$ if and only if $d_1 \le d_2$ in $D$ and $e_1 \le e_2$ in $E$.
\end{enumerate}
}

\begin{definition}
A \textbf{net} in a topological space $X$ is a function $f$ from a directed set $D$ to $X$. We usually write $f(d) = x_d$ for all $d \in D$, an refer to the net by $(x_d)_{d\in D}$.

A net $(x_d)_{d\in D}$ \textbf{converges} to a point $y \in X$ if for every neighbourhood $U$ of $y$, there is a $d \in D$ such that $x_e \in U$ for all $e \ge d$.
\end{definition}

Our motivation for defining nets was the hope that convergence of nets (in contrast to convergence of sequences) would completely determine the topology of the space. We now prove that this is indeed the case.

\begin{proposition}
\label{netsopen}
In any topological space $X$, a set $A \subseteq X$ is open if and only if no net in $X \setminus A$ has a limit in $A$.
\end{proposition}

\begin{proof}
Let $A$ be open, let $(x_d)_{d\in D}$ be a net in $X \setminus A$ and take any $y \in A$. As $A$ is open, there is a neighbourhood of $y$ that is contained in $A$. Hence this neighbourhood does not contain any terms of the net, so $y$ is not a limit of $(x_d)$.

Conversely suppose that $A$ is not open. Then there is an $y \in A$ such that every neighbourhood of $y$ intersects $X \setminus A$. So there is a net $(x_U)_{U \in \Nbh(y)}$ such that \[ x_U \in (X \setminus A) \cap U \] for all neighbourhoods $U$ of $y$. For every neighbourhood $U$ of $y$, $U$ is an element of $\Nbh(y)$ such that $x_V \in U$ for every $V \ge U$ (i.e. $V \subseteq U$). Hence $(x_U)$ is a net in $X \setminus A$ that converges to $y \in A$, as required.
\end{proof}

Like for sequences, convergence of nets is preserved by continuous functions. But again, for nets the converse if also true.

\begin{proposition}
\label{netscontinuous}
Let $f: X \to Y$ be a function between two topological spaces. Then $f$ is continuous if and only if for every net $(x_d)_{d\in D}$ that converges to $y$ in $X$, we have $f(x_d)_{d\in D} \to f(y)$ in $Y$
\end{proposition}

\begin{proof}
Suppose that $f$ is continuous and $x_d \to y$ in X. Take a neighbourhood $U$ of $f(y)$ in $Y$. Then $f^{-1}(U)$ is a neighbourhood of $y$ in $X$. By definition of convergence of nets, there is a $d \in D$ such that $x_e \in f^{-1}(U)$ for all $e \ge d$. So also $f(x_e) \in U$ for all $e \ge d$. This means that $f(x_d) \to f(y)$, as required.

Conversely, suppose that $f$ is not continuous, say $U \subseteq Y$ is open but $f^{-1}(U)$ is not open. By Proposition \ref{netsopen} there is a net $(x_d)_{d\in D}$ in $X \setminus f^{-1}(U)$ that converges to some $y \in f^{-1}(U)$. But then $(f(x_d))_{d\in D}$ is a net in the closed set $Y \subseteq U$ which (again by Proposition \ref{netsopen}) cannot converge to $f(y) \in U$. So $f$ doesn't preserve convergence of nets.
\end{proof}

Like a sequence, a net can have more than one limit, although in Hausdorff spaces every converging sequence has a unique limit. Sequences can also have unique limits in spaces that are not Hausdorff. Consider for example an uncountable set $X$ with the countable complement topology. In Example 1 we saw that limits are unique, but the space is obviously not Hausdorff. On the contrary, nets do succeed in exactly characterizing Hausdorff spaces.

\begin{proposition}
A space $X$ is Hausdorff is and only if no net has two distinct limits.
\end{proposition}

\begin{proof}
Suppose that $X$ is Hausdorff and consider a net $(x_d)_{d\in D}$. Suppose for contradiction that $x$ and $y$ are distinct limits of $(x_d)$. Take disjoint neighbourhoods $U$ of $x$ and $V$ of $y$. By definition of convergence, there is a $d_x$ such that $x_e \in U$ for all $e \ge d_x$ and a $d_y$ such that $x_e \in V$ for all $e \ge d_y$. In particular we have $x_e \in U \cap V$ for an upper bound $e$ of $d_x$ and $d_y$ in the directed set $D$, contradicting the disjointness of $U$ and $V$. Thus $(x_d)$ cannot have two distinct limits.

Conversely, suppose that $X$ is not Hausdorff, so there are two distinct points $x$ and $y$ such that any neighbourhood of $x$ intersects any neighbourhood of $y$. So there is a net $(x_{(U,V)})_{(U,V) \in \Nbh(x) \times \Nbh(y)}$ such that \[ x_{(U,V)} \in U \cap V \] for any neighbourhoods $U$ of $x$ and $V$ of $y$. Take any neighbourhood $U_0$ of $x$ and any $(U,V) \in \Nbh(x) \times \Nbh(y)$ with $(U,V) \ge (U_0, X)$. By definition we have $U \subseteq U_0$ and thus $x_{(U,V)} \in U \cap V \subseteq U_0$. This proves that $x_{(U,V)} \to x$ and we can similarly show that $x_{(U,V)} \to y$. So the net $(x_{(U,V)})_{(U,V) \in \Nbh(x) \times \Nbh(y)}$ has two distinct limits, as required.\end{proof}

\section{Compactness and sequential compactness}

Let's now look at compactness of topological spaces. Remember that a space $X$ is \textbf{compact} if and only if every open covering of $X$ (i.e. every collection of open sets whose union is $X$) has a finite subcovering. It is sufficient to consider coverings of basic opens.

Equivalently, $X$ is compact if and only if every collection of closed sets with the finite intersection property (i.e. all finite intersections are nonempty) has a nonempty intersection.

An important theorem by Tychonoff \cite{tychonoff} says that any product of compact spaces is itself compact.

If every sequence in a topological space has a convergent subsequence, then we call the space \textbf{sequentially compact}. A metric space is compact if and only if it is sequentially compact (\cite{munkres}, theorem 28.2). However, bearing in mind the the difference between open and sequentially open, we should be very suspicious of this equivalence holding in general. And indeed, neither direction of the equivalence holds in every topological space.

\begin{proposition}
\label{compactnotseqcompact}
There is a topological space that is compact but not sequentially compact.
\end{proposition}

\begin{proof}
Let $\{0,1\}$ have the discrete topology and consider $\{0,1\}^{[0,1)}$ with the product topology.

By Tychonoff's theorem, $\{0,1\}^{[0,1)}$ is compact as product of compact spaces.

Consider the sequence $(x_n)$ where $x_n(r)$ equals the $n$'th digit in the binary expansion of $r$ for all $r \in [0,1)$, where we never pick an expansion that ends in all $1$s. We claim that $(x_n)$ does not have a convergent subsequence. Indeed if the subsequence $x_{k_n}$ has a limit $y$, then for any $r \in [0,1)$, we must have $x_{k_n}(r) = y(r)$ for $n$ large enough. But there is a real number $r \in [0,1]$ whose unique binary expansion has a $0$ in the $k_n$th position if $n$ is even and a $1$ in the $k_n$th position is $n$ is odd, contradicting that $x_{k_n}$ has any limit. Hence $\{0,1\}^{[0,1)}$ is not sequentially compact.
\end{proof}

\begin{proposition}
\label{seqcompactnotcompact}
There is a topological space that is sequentially compact but not compact.
\end{proposition}

\begin{proof}
Consider the order topology on $\omega_1 = [0, \omega_1)$, the set of all countable ordinals.

The open covering $\A = \{[0, \alpha) \st \alpha < \omega_1\}$ does not have a finite subcovering. Indeed the supremum of the sets in any finite subset $\mathcal{B}$ of $\A$ is a finite union of countable ordinals, and hence itself a countable ordinal which has a successor in $[0, \alpha)$ that is not covered by $\mathcal{B}$.

Now let $(x_n)$ be any sequence in $[0, \omega_1)$. Let $A \subseteq \N$ be the set of all indices $n$ such that $x_n$ is minimal in $\{x_m \st m \geq n \}$. Note that $A$ is infinite as by definition it cannot have a largest element. Let $k_n$ be the $n$'th element of $A$ for all $n \in \N$. The sequence $(x_{k_n})$ is then a subsequence of $(x_n)$, which is nondecreasing and therefore converges to the supremum of its elements. This supremum is countable as a countable union of countable ordinals, so it is indeed an element of $[0, \omega_1)$. Thus every sequence in $[0, \omega_1)$ has a convergent subsequence, as required.
\end{proof}

We can however characterize compact spaces using nets. To do this, we need the notion of a subnet.

\begin{definition}
Let $(x_d)_{d \in D}$ be a net. A \textbf{subnet} of $(x_d)_{d \in D}$ is a net $(x_{f(e)})_{e \in E}$ where $E$ is a directed set and $f: E \to D$ is a function such that:
\begin{enumerate}
\item if $e_1 \le e_2$, then $f(e_1) \le f(e_2)$ (i.e. $f$ is order preserving),
\item for all $d \in D$, there is an $e \in E$ such that $f(e) \ge d$ (i.e. $f(E)$ is cofinal in $D$).
\end{enumerate}
\end{definition}

\begin{proposition}
A topological space $X$ is compact if and only every net has a convergent subnet.
\end{proposition}

\begin{proof}
Suppose $X$ is compact and let $(x_d)_{d \in D}$ be a net. The sets \[ X_d = \{x_e \st e \ge d\} \] have the finite intersection property, so by compactness their closures have a nonempty intersection. Thus we can take an $y \in \cup_{d \in D} \overline{X_d}$. Every neighbourhood $U$ of $y$ intersects $X_d$ for all $d \in D$. In other words, for all $d \in D$ there is an $e \ge d$ such that $x_e \in U$. (Such an $y$ is called a \textbf{cluster point} of the net $(x_d)$.)

Consider the set
\[ E = \{ (d,U) \in D \times \Nbh(y) \st x_d \in U \}, \]
preordered by $(d_1, U_1) \le (d_2, U_2)$ if and only if $d_1 \le d_2$ in $D$ and $U_2 \subseteq U_1$. We claim that this is a directed set. Indeed let $(d_1, U_1)$ and $(d_2, U_2)$ be any pair of elements of $E$. There is an upper bound $e$ of $d_1$ and $d_2$ in $D$, and as $y$ is a cluster point of $(x_d)$, there is an $e' \ge e$ such that $x_{e'} \in U_1 \cap U_2$. Then $(e', U_1 \cap U_2)$ is an upper bound of $(d_1, U_1)$ and $(d_2, U_2)$ in $E$. Therefore $E$ is indeed a directed set.

Define the projection
\begin{align*}
f:\quad E &\to D \\
   (d, U) &\mapsto d.
\end{align*}
This is clearly an order preserving function and it is a surjection as $d = d(d, X)$ for all $d \in D$. So $(x_{f(d,U)})_{(d,U)\in E}$ is a subnet of $(x_d)_{d \in D}$. Moreover, if $U$ is any neighbourhood of $y$, then there is by choice of $y$ a $d \in D$ such that $x_d \in U$. By definition of $E$ we have $x_{f(e,V)} \in U$ for all $(e, V) \ge (d, U)$. Therefore $x_{f(d,U)} \to y$, as required.

\underline{Conversely} suppose that $X$ is not compact. Then there is a collection $\A$ of closed sets with the finite intersection property, but with empty intersection. Let $\D$ be the set of finite subcollections of $A$, ordered by inclusion. This is clearly a directed set. We can choose a net $(x_\B)_{\B \in \D}$ where \[ x_\B \in \cap_{A \in \B} A \] for every finite $\B \subseteq \A$. Suppose for contradiction that $(x_{f(x)})_{e \in E}$ is a subnet of $(x_\B)_{\B \in \D}$ that converges to some $y \in X$. By assumption, there is an $A \in \A$ such that $y \not\in A$. As $A$ is closed, there is neighbourhood $U$ of $y$ such that $U \cap A = \emptyset$, and hence $x_\B \not\in U$ for all $\B \ge \{A\}$. As $f(E)$ is cofinal in $D$, there is an $e_1 \in E$ such that $\{A\} \le f(e_1)$. But there must also be an $e_2 \in E$ such that $x_{f(e)} \in U$ for all $e \ge e_2$. Let $e$ be an upper bound of $e_1$ and $e_2$. Then we must have $x_{f(e)} \in U$, but on the other hand $\{A\} \le f(e_1) \le f(e)$ so we must have $x_{f(e)} \not\in U$, a contradiction. So the net $(x_\B)_{\B \in \D}$ does not have a convergent subnet.
\end{proof}

It is now tempting to accept the following argument: \emph{In a compact space, every nets has a convergent subnet. So every sequence, considered as a net, has a convergent subnet, which is a convergent subsequence. So every compact space is sequentially compact.} But we know from Proposition \ref{compactnotseqcompact} that this is not true. The mistake is the fact that not every subnet of a sequence is a subsequence. In particular note that the function $f$ in the definition of a subnet need not be injective.

\section{Further reading}

\subsection{Nets and filters}

As an alternative to nets in $X$, one can consider filters on $X$. A \textbf{filter} on $X$ is a nonempty collection of subsets of $X$ that is closed under binary intersections and supersets, and does not contain $\emptyset$. A filter $\F$ \textbf{converges} to a point $x$ if $U \in \F$ for every neighbourhood $U$ of $x$. Given a net $(x_d)_{d \in D}$ in $X$ one can consider the filter
\[ \F = \{ A \subseteq X \st \exists d \in D: \mbox{$x_e \in A$ for all $e \geq d$} \} \]
of sets which \emph{eventually} contain all point of the net. This filter has the same limits as $(x_d)_{d \in D}$. Conversely given a filter $\F$ on $X$ one can consider the directed set $\F$ ordered by reversed inclusion. Then $\F$ converges to a point if and only if any net
\[ (x_A)_{A \in \F} \]
with
\[ x_A \in A \]
for all $A \in F$ converges to that point as well. Hence nets and filters are in many ways interchangable. Most of the propositions that I've proved using nets, can equally well be proven using filters. I've only considered nets, because they arize as a generalization of sequences, which was the starting point of this article. However it is instructive to also think about filters, as they give a diferent point of view. Only by combining both points of view, one can get the best insight into the mathematics.

Filters have another advantage. An easy application of Zorn's lemma gives that every filter on $X$ can be refined to an \textbf{ultrafilter}. An ultrafilter on $X$ is a filter which contains either $A$ or $X \setminus A$ for all $A \subseteq X$. The corresponding proposition for nets is that every net has a \textbf{universal} subnet. A net in $X$ is universal (also called an \textbf{ultranet}) if for every $A \subseteq X$, the net is either eventually in $A$ or eventually in $X \setminus A$. There is however no nice direct proof for the fact that every net has a universal subnet, filters are the more natural tool here.

From the fact that every filter has an ultrafilter refinement, an easy proof of Tychonoff's theorem is possible. Indeed, compact spaces can be characterized as those spaces where every ultrafilter has a limit. But then in a product of compact spaces, we can find a limit of any ultrafilter by considering the projections on every component and taking a limit in each of these compact spaces.

A proof of Tychonoff's theorem using just nets is also possible, but is not as elegant \cite{chernoff}.

Nets were introduced in 1922 by E.~H.~Moore and H.~L.~Smith in \cite{mooresmith}. Hence nets were at first called \emph{Moore-Smith sequences}. The theory of nets was further developed by Birkhoff \cite{birkhoff} (most of the propositions of this article first appear in his paper) and by Kelley \cite{kelley1950} (who introduces the terms \emph{net} and \emph{ultranet}, and who proves Tychonoff's theorem using ultrafilters). McShane \cite{mcshane} gives an extensive motivation for the definitions of nets and subnets.

Filters were introduced in 1937 by Cartan \cite{cartan1,cartan2}. Even though filters are now also used in very different contexts, Cartan's motivation for defining them was to generalize the notion of convergence for sequences. Indeed he starts of his article \emph{Th\'eorie des filtres} \cite{cartan1} by writing:
\begin{quote}
Malgr\'e les services rendus en topologie par la consideration des \emph{suites d\'enombrables}, leur emploi n'est pas adapt\'e \`a l'\'etude des espaces g\'en\'eraux. Nous voulons indiquer ici quel est l'instrument qui semble devoir les remplacer.\footnote{Translation from French: In spite of the accomplishments of considering countable sequences in topology, their use is not suitable in the study of general spaces. We want to indicate here which tool apparently should replace them.}
\end{quote}
Bourbaki's book on General Topology \cite{bourbaki} was the first to fully adopt the use of filters. The connections between nets and filters were investigated by Bartle \cite{bartle} and by Bruns and Schmidt \cite{bruns}.

\subsection{Sequential spaces and Fr\'echet-Urysohn spaces}

We have considered sequential spaces. Similar to sequential spaces are the Fr\'echet-Urysohn spaces. A topological space $X$ is \textbf{Fr\'echet-Urysohn} if the closure of any $A \subseteq X$ contains exactly the limits of sequences in $A$. They are also sometimes simply called Fr\'echet spaces, but this might cause confusion as there are other uses for the term \emph{Fr\'echet space}.

Any first countable space is Fr\'echet-Urysohn. This can be proven just like Proposition \ref{firstcountableseq}. The example from the proof of Proposition \ref{seqnotfirstcountable} shows that there is a Fr\'echet-Urysohn which is not first countable. Any Fr\'echet-Urysohn space is sequential, as by definition every sequentially closed set is its own closure. Fr\'echet-Urysohn are exactly those spaces of which every subspace is sequential. However not every subspace of a sequential space is sequential. Indeed, consider
\[Y = \{(x,0) \st x \in \R \setminus \{0\} \}  \cup \{(0,1\} \cup \left\{ \left(\frac{1}{n+1},1\right) \st n \in \N \right\} \]
as a subset of the real plane. Let $(\R, \tau_q)$ be the quotient of $Y$ obtained by projecting onto the first coordinate. Then $(\R, \tau_q)$ is sequential as quotient of a metric space (Corollary \ref{seqquotientmetric}). But the subspace $\R \setminus \{ \frac{1}{n+1} \st n \in \N \}$ has a sequentially open set $\{0\}$ which is not open. So $(\R, \tau_q)$ is a sequential space that is not Fr\'echet-Urysohn.

Sequential spaces and Fr\'echet-Urysohn spaces where most intensively studied in the 1960s. Most of the results mentioned here where obtained by S.~P.~Franklin in \cite{franklin} and \cite{franklin2}. Sequential spaces and Fr\'echet-Urysohn spaces are also covered in Engelking's book \cite{engelking}.

\bibliographystyle{plain}
\bibliography{nets}

\end{document}

config: arxiv


TITLE: Splitting field for $x^4-x^2-2$
QUESTION [1 upvotes]: Am i right to say that the splitting field for $x^4-x^2-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2},i)$ which is of degree 4? i.e. $\{a+b\sqrt{2} + ci+di\sqrt{2} : a,b,c,d\in\mathbb{Q}\}$?

REPLY [1 votes]: Since the roots of $\;x^4-x^2-2=(x^2-2)(x^2+1)\;$ are $\;\pm\sqrt2\,,\,\,\pm i\;$ , the answer to your question is yes.


text:
\begin{document}

\title{Totally quasi-umbilic timelike surfaces in $\mathbb{R}^{1,2}$}

\author{Jeanne N. Clelland}
\address{Department of Mathematics, 395 UCB, University of
Colorado,
Boulder, CO 80309-0395}
\email{Jeanne.Clelland@colorado.edu}

\subjclass[2000]{Primary(51B20, 53C42), Secondary(53C10, 53A55)}
\keywords{timelike surfaces, quasi-umbilic, method of moving frames}
\thanks{This research was supported in part by NSF grant DMS-0908456.}

\begin{abstract}
For a regular surface in Euclidean space $\mathbb{R}^3$, umbilic points are precisely the points where the Gauss and mean curvatures $K$ and $H$ satisfy $H^2=K$; moreover, it is well-known that the only totally umbilic surfaces in $\R^3$ are planes and spheres.  But for timelike surfaces in Minkowski space $\mathbb{R}^{1,2}$, it is possible to have $H^2=K$ at a non-umbilic point; we call such points {\em quasi-umbilic}, and we give a complete classification of totally quasi-umbilic timelike surfaces in $\mathbb{R}^{1,2}$.
\end{abstract}

\maketitle

\section{Introduction}\label{intro-sec}

For a regular surface $\Sigma$ in Euclidean space $\bb{R}^3$, it is well-known (see, e.g., \cite{doCarmo76}) that the shape operator at any point $\bx \in \Sigma$ is a self-adjoint linear operator $S_{\bx}:T_{\bx}\Sigma \to T_{\bx}\Sigma$ and is therefore diagonalizable over $\bb{R}$.  But for a timelike surface $\Sigma$ in 3-dimensional Minkowski space $\R^{1,2}$ (cf. Definition \ref{R12-def}), this is no longer necessarily true; the shape operator is still self-adjoint, but because the metric on $T_{\bx}\Sigma$ is now indefinite, it can have any of three algebraic types: diagonalizable over $\bb{R}$, diagonalizable over $\bb{C}$ but not $\bb{R}$, or non-diagonalizable over $\bb{C}$ with a single null eigenvector (see, e.g., \cite{Magid05}, \cite{Magid09}, \cite{Milnor83}).

An {\em umbilic point} of a regular surface $\Sigma$ in either Euclidean or Minkowski space is a point $\bx \in \Sigma$ for which the second fundamental form $\text{II}_{\bx}$ of $\Sigma$ is a scalar multiple of the first fundamental form $\text{I}_{\bx}$.  In the Euclidean case, the Gauss curvature $K$ and mean curvature $H$ of $\Sigma$ satisfy $H^2 - K \geq 0$, and the umbilic points are precisely those points where $H^2 - K = 0$.  But for a timelike surface in Minkowski space, the quantity $H^2 - K$ can take on any real value.  Specifically:
\begin{enumerate}
\item If the shape operator is diagonalizable over $\bb{R}$, then $H^2 - K \geq 0$, with $H^2-K$=0 precisely at umbilic points.
\item If the shape operator is diagonalizable over $\bb{C}$ but not $\bb{R}$, then $H^2 - K < 0$.
\item If the shape operator is non-diagonalizable over $\bb{C}$, then $H^2-K=0$.
\end{enumerate}
Because of this relationship between $H$ and $K$ in Case (3), we make the following definition:

\begin{definition}\label{quasi-umbilic-def}
Let $\Sigma$ be a regular timelike surface in $\bb{R}^{1,2}$.  A point $\bx \in \Sigma$ will be called {\em quasi-umbilic} if the shape operator $S_{\bx}:T_{\bx}\Sigma \to T_{\bx}\Sigma$ is non-diagonalizable over $\bb{C}$.
\end{definition}

\begin{definition}\label{tqu-def}
A regular timelike surface $\Sigma$ in $\bb{R}^{1,2}$ will be called {\em totally quasi-umbilic} if every point $\bx \in \Sigma$ is quasi-umbilic.
\end{definition}

The goal of this paper is to give a classification of totally quasi-umbilic timelike surfaces in $\R^{1,2}$.  Our main result is:

\begin{theoremvoid}[cf. Theorem \ref{main-thm}]
Let $\Sigma$ be a totally quasi-umbilic, regular timelike surface in $\R^{1,2}$. Then $\Sigma$ is a ruled surface whose rulings are all null lines, with the additional property that any null curve $\alpha$ in $\Sigma$ which is transverse to the rulings is nondegenerate (i.e., the vectors $\alpha', \alpha''$ are linearly independent at each point).  Conversely, given any nondegenerate null curve $\alpha(u)$ in $\R^{1,2}$ and any null vector field $\bar{\f}_2(u)$ along $\alpha$ which is linearly independent from $\alpha'(u)$ for all $u$, the immersed ruled surface
\[ \bx(u,v) = \alpha(u) + v \bar{\f}_2(u) \]
is totally quasi-umbilic, possibly containing a curve of umbilic points.
\end{theoremvoid}

The paper is organized as follows: in \S \ref{Minkowski-review-sec} we briefly review basic notions from the geometry of timelike surfaces in $\R^{1,2}$. In \S \ref{fundamental-forms-sec}, we use Cartan's method of moving frames to derive and solve a system of PDEs whose solutions provide a complete description of the Maurer-Cartan forms for totally quasi-umbilic surfaces in local null coordinates.  In \S \ref{main-result-sec}, we solve the resulting Maurer-Cartan equations to find explicit local parametrizations for all such surfaces and thereby prove Theorem \ref{main-thm}.  Finally, in \S \ref{examples-sec}, we give some examples of totally quasi-umbilic surfaces.

\section{Brief review of geometry of timelike surfaces in $\R^{1,2}$}\label{Minkowski-review-sec}

\begin{definition}\label{R12-def}
Three-dimensional {\em Minkowski space} is the manifold $\R^{1,2}$ defined by
\[ \R^{1,2} = \{\bx = (x^0, x^1, x^2): x^0, x^1, x^2 \in \bb{R} \}, \]
with the indefinite inner product $\langle \cdot, \cdot \rangle$ defined on each tangent space $T_{\bx}\R^{1,2}$ by
\[ \langle \bv, \bw \rangle = v^0 w^0 - v^1 w^1 - v^2 w^2. \]
\end{definition}

\begin{definition}
A nonzero vector $\bv \in T_{\bx}\bb{R}^{1,2}$ is called:
\begin{itemize}
\item {\em spacelike} if $\langle \bv, \bv \rangle < 0$;
\item {\em timelike} if $\langle \bv, \bv \rangle > 0$;
\item {\em lightlike} or {\em null} if $\langle \bv, \bv \rangle = 0$.
\end{itemize}
\end{definition}

For a given real number $c$, the ``sphere" consisting of all vectors $\bv$ with $\langle \bv, \bv \rangle = c$ is:
\begin{itemize}
\item a hyperboloid of one sheet if $c<0$;
\item a hyperboloid of two sheets if $c>0$;
\item a cone, called the {\em light cone} or {\em null cone} if $c=0$. \end{itemize}
(See Figure \ref{Minkowski-spheres}; note that the $x^0$-axis is drawn as the vertical axis.)

\begin{figure}[h]
\includegraphics[width=2in]{spacelike-sphere-fig}
\includegraphics[width=2in]{timelike-sphere-fig}
\includegraphics[width=2in]{light-cone-fig}
\caption{``Spheres" in $\R^{1,2}$ with $c < 0, \ c > 0, \ c = 0$}
\label{Minkowski-spheres}
\end{figure}

\begin{definition}
A regular surface $\Sigma \subset \bb{R}^{1,2}$ is called:
\begin{itemize}
\item {\em spacelike} if the restriction of $\langle \cdot, \cdot \rangle$ to each tangent plane $T_{\bx}\Sigma$ is negative definite;
\item {\em timelike} if the restriction of $\langle \cdot, \cdot \rangle$ to each tangent plane $T_{\bx}\Sigma$ is indefinite;
\item {\em lightlike} if the restriction of $\langle \cdot, \cdot \rangle$ to each tangent plane $T_{\bx}\Sigma$ is degenerate.
\end{itemize}
\end{definition}

The Minkowski isometry group consists of all transformations $\varphi: \R^{1,2} \to \R^{1,2}$ of the form
\[ \varphi(\bx) = A\bx + \mathbf{b}, \]
where $A \in O(1,2)$ and $\mathbf{b} \in \R^{1,2}$.  Cartan's method of moving frames can be used to compute local invariants for timelike surfaces under the action of this isometry group.  Given a regular timelike surface $\Sigma \subset \R^{1,2}$, an {\em adapted orthonormal frame} at a point $\bx \in \Sigma$ consists of three mutually orthogonal vectors $(\e_0, \e_1, \e_2)$ in $T_{\bx}\R^{1,2}$ such that:
\begin{itemize}
\item $\langle \e_0, \e_0 \rangle = 1, \ \ \ \langle \e_1, \e_1 \rangle = \langle \e_2, \e_2 \rangle = -1$;
\item $\e_0, \e_1$ span the tangent plane $T_{\bx}\Sigma$.
\end{itemize}
An {\em adapted orthonormal frame field} along $\Sigma$ consists of smooth vector fields $(\e_0, \e_1, \e_2)$ along $\Sigma$ such that for each $\bx \in \Sigma$, the vectors $(\e_0(\bx), \e_1(\bx), \e_2(\bx))$ form an adapted orthonormal frame at $\bx$.
If the surface $\Sigma$ is given an orientation, then the unit normal vector field $\e_2$ along $\Sigma$ is completely determined. In this case, it is convenient to consider only {\em oriented} orthonormal frame fields, defined by the condition that $\e_0 \times \e_1 = -\e_2$, where the Minkowski cross product is given by
\begin{equation}
 \bv \times \bw = (v^1 w^2 - v^2 w^1, \, v^0 w^2 - v^2 w^0, \, v^1 w^0 - v^0 w^1 ) \label{Minkowski-cross-product}
\end{equation}
for $\bv = (v^0, v^1, v^2), \ \bw = (w^0, w^1, w^2)$.

\begin{remark}
These sign conventions are chosen in order to satisfy the orientation condition
\[ \langle \bu, \bv \times \bw \rangle = \det \begin{bmatrix} \bu & \bv & \bw  \end{bmatrix} \]
for any three vectors $\bu, \bv, \bw$, as well as the condition that the standard basis vectors
\[ \underline{\e}_0 = (1,0,0), \qquad \underline{\e}_1 = (0,1,0), \qquad \underline{\e}_2 = (0,0,1) \]
form an oriented orthonormal frame.
\end{remark}

Given an oriented adapted orthonormal frame field $(\e_0, \e_1, \e_2)$ along $\Sigma$, any other oriented adapted orthonormal frame field $(\tilde{\e}_0, \tilde{\e}_1, \tilde{\e}_2)$ along $\Sigma$ is given by
\begin{equation*}
 \begin{bmatrix} \tilde{\e}_0 & \tilde{\e}_1 & \tilde{\e}_2
\end{bmatrix} = \begin{bmatrix} \e_0 & \e_1 & \e_2 \end{bmatrix}
\begin{bmatrix} \cosh \theta & \sinh \theta & 0 \\[0.05in] \sinh \theta &
\cosh \theta & 0 \\[0.05in] 0 & 0 & 1  \end{bmatrix}
\end{equation*}
for some real-valued function $\theta$ on $\Sigma$.

As in the Euclidean case, the normal vector field $\e_2$ of an oriented adapted orthonormal frame field along $\Sigma$ defines the {\em Gauss map} from $\Sigma$ to the Minkowski sphere $\langle \bv, \bv \rangle = 1$, and at any point $\bx \in \Sigma$, the differential $d\e_2$ of the Gauss map is a self-adjoint linear operator
\[ d\e_2: T_{\bx}\Sigma \to T_{\bx}\Sigma. \]
The {\em shape operator} of $\Sigma$ at any point $\bx \in \Sigma$ is defined to be
\[ S_{\bx} = -d\e_2: T_{\bx}\Sigma \to T_{\bx}\Sigma. \]
The first and second fundamental forms $\text{I}_{\bx}, \text{II}_{\bx}:T_{\bx}\Sigma \to \R$ are defined by
\begin{align*}
\text{I}_{\bx}(\bv) & = \langle \bv, \bv \rangle, \\
\text{II}_{\bx}(\bv) & = \langle S_{\bx}(\bv), \bv \rangle,
\end{align*}
and the Gauss curvature $K$ and mean curvature $H$ at $\bx$ are defined to be
\[ K(\bx) = \det \left( S_{\bx} \right), \qquad  H(\bx) =  \tfrac{1}{2} \text{tr} \left( S_{\bx} \right). \]
A point $\bx \in \Sigma$ is called {\em umbilic} if the quadratic form $\text{II}_{\bx}$ is a scalar multiple of $\text{I}_{\bx}$. It is straightforward to show that $\bx$ is umbilic if and only if $S_{\bx}$ is a scalar multiple of the identity; moreover, if $\bx$ is an umbilic point, then $H(\bx)^2 - K(\bx) = 0$.

The shape operator $S_{\bx}$ is a self-adjoint linear operator; i.e., for any vectors $\bv, \bw \in T_{\bx}\Sigma$, we have
\[ \langle S_{\bx}(\bv), \bw \rangle = \langle \bv,  S_{\bx}(\bw) \rangle. \]
This is equivalent to the condition that the matrix representation of $S_{\bx}$ with respect to any orthonormal basis $(\e_0, \e_1)$ for $T_{\bx}\Sigma$ is {\em skew-symmetric}.  So unlike in the Euclidean case, where $S_{\bx}$ is always diagonalizable over $\R$, any of three possibilities may occur: $S_{\bx}$ may be diagonalizable over $\bb{R}$, diagonalizable over $\bb{C}$ but not $\bb{R}$, or non-diagonalizable over $\bb{C}$ with a repeated real eigenvalue and a 1-dimensional eigenspace.

If $S_{\bx}$ is not diagonalizable over $\bb{C}$, then there exists a basis for $T_{\bx}\Sigma$ with respect to which the matrix for $S_{\bx}$ has the form
\[ \begin{bmatrix} \lambda & 1 \\[0.05in] 0 & \lambda \end{bmatrix} \]
for some $\lambda \in \R$.  Thus we have $K(\bx) = \lambda^2, \ H(\bx) = \lambda$, and so $H(\bx)^2 - K(\bx) = 0$.  Because of this similarity with umbilic points, we call such points $\bx$ {\em quasi-umbilic} (cf. Definition \ref{quasi-umbilic-def}).  Moreover, the unique eigenvector for $S_{\bx}$ is necessarily a null vector; to see this, note that with respect to any orthonormal basis $(\e_0, \e_1)$ for $T_{\bx}\Sigma$, the matrix for $S_{\bx}$ is skew-symmetric and so has the form
\[ \begin{bmatrix} a & b \\[0.05in] -b & c \end{bmatrix}. \]
The condition that $S_{\bx}$ has a repeated eigenvalue is
\[ (a-c)^2 = 4 b^2, \]
and $S_{\bx}$ has a 1-dimensional eigenspace (and hence $\bx$ is quasi-umbilic) if and only if $a \neq c$. In this case the repeated eigenvalue is $\lambda = \tfrac{1}{2}(a+c)$, with eigenvector $\e_0 - \e_1$, which is clearly null.

It is well-known that totally umbilic surfaces in Euclidean space $\R^3$ are necessarily contained in either planes or spheres.  The analogous result for timelike surfaces in $\R^{1,2}$ is:

\begin{theorem}
Suppose that a regular timelike surface $\Sigma \subset \R^{1,2}$ is totally umbilic. Then $\Sigma$ is contained in either a plane or a hyperboloid of one sheet (i.e., a Minkowski ``sphere").
\end{theorem}

For the remainder of this paper, we will consider the classification problem for totally quasi-umbilic timelike surfaces (cf. Definition \ref{tqu-def}).

\section{Null frames and Maurer-Cartan forms}\label{fundamental-forms-sec}

We will approach the classification problem for totally quasi-umbilic timelike surfaces via the method of moving frames.  It turns out to be more convenient to use null frames rather than orthonormal frames, and so we make the following definition:

\begin{definition}
An {\em adapted null frame field} along a timelike surface $\Sigma \subset \R^{1,2}$ consists of linearly independent, smooth vector fields $(\f_1, \f_2, \f_3)$ along $\Sigma$ such that for each $\bx \in \Sigma$,
\begin{itemize}
\item $\f_1(\bx), \f_2(\bx)$ are null vectors which span the tangent space $T_{\bx}\Sigma$ at each point $\bx \in \Sigma$;
\item $\langle \f_1(\bx), \f_2(\bx) \rangle = 1$;
\item $\f_3(\bx)$ is orthogonal to $T_{\bx}\Sigma$ at each point $\bx \in \Sigma$, with $\langle \f_3(\bx), \f_3(\bx) \rangle = -1$.
\end{itemize}
\end{definition}
For instance, if $(\e_0, \e_1, \e_2)$ is an adapted orthonormal frame field along $\Sigma$, then the vector fields
\[
\f_1  = \tfrac{1}{\sqrt{2}}(\e_0 + \e_1), \qquad
\f_2  = \tfrac{1}{\sqrt{2}}(\e_0 - \e_1), \qquad
\f_3  = \e_2
\]
form an adapted null frame field along $\Sigma$.

Let $\eta^i, \eta^i_j, \ 1 \leq i,j \leq 3$ be the Maurer-Cartan forms on $\Sigma$ associated to an adapted null frame field along $\Sigma$.  These forms are defined by the equations (note: all indices range from 1 to 3, and we use the Einstein summation convention)
\begin{align}
d\bx & = \f_i \eta^i, \label{MC-forms-def}   \\
d\f_i & =  \f_j \eta^j_i, \notag
\end{align}
and they satisfy the Maurer-Cartan structure equations
\begin{align}
d\eta^i & = - \eta^i_j \& \eta^j,  \label{structure-eqns} \\
d\eta^i_j & = - \eta^i_k \& \eta^k_j. \notag
\end{align}
(See \cite{IL03} for a discussion of Maurer-Cartan forms and their structure equations.)
Differentiating the inner product relations
\begin{alignat*}{3}
\langle \f_1, \f_1 \rangle & = 0, & \qquad \langle \f_2, \f_2 \rangle & = 0, & \qquad \langle \f_3, \f_3 \rangle & = -1,   \\[0.1in]
\langle \f_1, \f_2 \rangle & = 1, & \qquad \langle \f_1, \f_3 \rangle & = 0, & \qquad \langle \f_2, \f_3 \rangle & = 0
\end{alignat*}
yields the following relations among the Maurer-Cartan forms:
\begin{gather}
\eta^1_2 = \eta^2_1 = \eta^3_3 = 0, \label{MC-forms-relations}  \\[0.1in]
\eta^2_2 = -\eta^1_1, \qquad \eta^1_3 = \eta^3_2, \qquad \eta^2_3 = \eta^3_1. \notag
\end{gather}

From the equation
\[ d\bx = \f_1 \eta^1 + \f_2 \eta^2 + \f_3 \eta^3 \]
and the fact that $d\bx$ takes values in $T_{\bx}\Sigma$, it follows that $\eta^3 = 0$.  Moreover, $\eta^1$ and $\eta^2$ are linearly independent 1-forms which form a basis for the cotangent space $T^*_{\bx}\Sigma$ at each point $\bx \in \Sigma$.

Differentiating the equation $\eta^3 = 0$ yields
\[ 0 = d\eta^3 = -\eta^3_1 \& \eta^1 - \eta^3_2 \& \eta^2. \]
By Cartan's Lemma (see \cite{IL03}), there exist functions $k_{ij} = k_{ji}$ on $\Sigma$  such that
\begin{equation}
 \begin{bmatrix} \eta^3_1 \\[0.1in] \eta^3_2 \end{bmatrix} = -\begin{bmatrix}
k_{11} & k_{12} \\[0.1in] k_{12} & k_{22} \end{bmatrix} \begin{bmatrix}
\eta^1 \\[0.1in] \eta^2 \end{bmatrix}. \label{kij-def}
\end{equation}
(The minus sign is included for convenience in what follows.)
Compare the matrix in \eqref{kij-def} to that of the shape operator at any point $\bx \in \Sigma$; using \eqref{MC-forms-def}, \eqref{MC-forms-relations}, and \eqref{kij-def}, we have:
\begin{align*}
 S_{\bx} & = -d\f_3 \\
 & = -\f_1 \eta^1_3 - \f_2 \eta^2_3 \\
 & = -\f_1 \eta^3_2 - \f_2 \eta^3_1 \\
 & = \f_1 (k_{12} \eta^1 + k_{22} \eta^2) + \f_2 (k_{11} \eta^1 + k_{12} \eta^2) \\
 & = (k_{12} \f_1 + k_{11} \f_2) \eta^1 + (k_{22} \f_1 + k_{12} \f_2) \eta^2.
\end{align*}
This means that
\[ S_{\bx}(\f_1) = k_{12} \f_1 + k_{11} \f_2, \qquad  S_{\bx}(\f_2) = k_{22} \f_1 + k_{12} \f_2, \]
and so the matrix of $S_{\bx}$ with respect to the basis $(\f_1, \f_2)$ for $T_{\bx}\Sigma$ is
\[ \begin{bmatrix} k_{12} & k_{22} \\[0.1in] k_{11} & k_{12} \end{bmatrix}. \]
Thus we have
\begin{gather*}
 H = k_{12}, \qquad K = k_{12}^2 - k_{11} k_{12}, \\
 H^2 - K = k_{11}k_{22}.
\end{gather*}
It follows that $\Sigma$ is totally quasi-umbilic if and only if $k_{11} k_{22} \equiv 0$ on $\Sigma$ and $k_{11}, k_{22}$ are not both zero at any point $\bx \in \Sigma$.  Without loss of generality, we may assume that $k_{22}=0, \, k_{11} \neq 0$.

Now we will examine how the functions $k_{ij}$ transform if we make a change of adapted null frame field.  So suppose that $(\tilde{\f}_1, \tilde{\f}_2, \tilde{\f}_3)$ is any other adapted null frame field along $\Sigma$.  Then we must have
\begin{equation}
 \begin{bmatrix} \tilde{\f}_1 & \tilde{\f}_2 & \tilde{\f}_3
\end{bmatrix} = \begin{bmatrix} \f_1 & \f_2 & \f_3 \end{bmatrix}
\begin{bmatrix} \eps_1 e^{\theta} & 0 & 0 \\[0.1in] 0 &
\eps_1 e^{-\theta} & 0 \\[0.1in] 0 & 0 & \eps_2  \end{bmatrix} \label{frame-change-eqn}
\end{equation}
for some function $\theta$ on $\Sigma$, where $\eps_1, \eps_2 = \pm 1$.  (Note that we cannot exchange $\f_1$ and $\f_2$ due to our assumption that $k_{11} \neq 0, \ k_{22} = 0$.)  Under such a transformation, the Maurer-Cartan forms $\tilde{\eta}^i, \tilde{\eta}^i_j$ associated to the frame field $(\tilde{\f}_1, \tilde{\f}_2, \tilde{\f}_3)$ satisfy the conditions:
\begin{gather*}
\begin{bmatrix} \tilde{\eta}^1 \\[0.1in] \tilde{\eta}^2 \end{bmatrix} =
\begin{bmatrix} \eps_1 e^{-\theta} & 0 \\[0.1in] 0 & \eps_1 e^{\theta} \end{bmatrix}
\begin{bmatrix} \eta^1 \\[0.1in]  \eta^2 \end{bmatrix} = \begin{bmatrix} \eps_1 e^{-\theta} \eta^1 \\[0.1in]  \eps_1 e^{\theta} \eta^2 \end{bmatrix}, \\[0.2in]
\begin{bmatrix} \tilde{\eta}^1_3 \\[0.1in] \tilde{\eta}^2_3 \end{bmatrix} =
\eps_2 \begin{bmatrix} \eps_1 e^{-\theta} & 0 \\[0.1in] 0 & \eps_1 e^{\theta} \end{bmatrix}
\begin{bmatrix} \eta^1_3 \\[0.1in]  \eta^2_3 \end{bmatrix} = \begin{bmatrix} \eps_1 \eps_2 e^{-\theta} \eta^1_3 \\[0.1in]  \eps_1 \eps_2 e^{\theta} \eta^2_3 \end{bmatrix} .
\end{gather*}
Therefore,
\[ \begin{bmatrix} \tilde{\eta}^3_1 \\[0.1in]  \tilde{\eta}^3_2 \end{bmatrix} = \begin{bmatrix} \tilde{\eta}^2_3 \\[0.1in]  \tilde{\eta}^1_3 \end{bmatrix} = \begin{bmatrix} \eps_1 \eps_2 e^{\theta} \eta^2_3 \\[0.1in]  \eps_1 \eps_2 e^{-\theta} \eta^1_3  \end{bmatrix} = \begin{bmatrix} \eps_1 \eps_2 e^{\theta} \eta^3_1 \\[0.1in]  \eps_1 \eps_2 e^{-\theta} \eta^3_2  \end{bmatrix}, \]
and the functions $\tilde{k}_{ij}$ associated to the frame field $(\tilde{\f}_1, \tilde{\f}_2, \tilde{\f}_3)$ are given by
\begin{equation*}
\begin{bmatrix} \tilde{k}_{11} & \tilde{k}_{12} \\[0.1in] \tilde{k}_{12} &
\tilde{k}_{22} \end{bmatrix} = \eps_2
\begin{bmatrix} \eps_1 e^{\theta} & 0 \\[0.1in] 0 & \eps_1 e^{-\theta} \end{bmatrix} \begin{bmatrix} k_{11} & k_{12} \\[0.1in] k_{12} & k_{22} \end{bmatrix} \begin{bmatrix} \eps_1 e^{\theta} & 0 \\[0.1in] 0 & \eps_1 e^{-\theta} \end{bmatrix} =
\eps_2 \begin{bmatrix} e^{2\theta} k_{11} & k_{12} \\[0.1in] k_{12} & e^{-2\theta} k_{22} \end{bmatrix} .
\end{equation*}
Since $k_{11} \neq 0$, it follows that there exists an adapted null frame field $(\f_1, \f_2, \f_3)$ along $\Sigma$ with $k_{11}\equiv 1$, and this frame field is unique up to the discrete transformation
$(\tilde{\f}_1, \tilde{\f}_2, \tilde{\f}_3) = (\eps_1 \f_1, \eps_1 \f_2, \f_3)$.
For such an adapted null frame field, the first and second fundamental forms of $\Sigma$ are:
\begin{gather}
\text{I} = \langle d\bx, d\bx \rangle =  2 \eta^1 \eta^2, \label{fundamental-forms} \\
\text{II} = - \langle d\f_3, d\bx \rangle =  -(\eta^3_1 \eta^1 + \eta^3_2 \eta^2) =
(\eta^1)^2 + 2H \eta^1 \eta^2.  \notag
\end{gather}
(Recall that $H = k_{12}$.)

Now let $u,v$ be local null coordinates on $\Sigma$; note that these are well-defined up to transformations of the form
\[ u \to \tilde{u}(u), \qquad v \to \tilde{v}(v), \]
with $\tilde{u}'(u), \tilde{v}'(v) \neq 0$.  Then there exist functions $f, g$ on $\Sigma$ such that
\begin{equation}
 \eta^1 = e^f\, du, \qquad \eta^2 = e^g\, dv, \label{MC-forms-1}
\end{equation}
and therefore
\begin{gather}
\text{I} = 2 e^{f+g}\, du\, dv, \label{fundamental-forms-coords} \\
\text{II} = e^{2f}\, du^2 + 2He^{f+g}\, du\, dv. \notag
\end{gather}
From \eqref{kij-def}, we also have
\begin{equation}
\eta^3_1 = \eta^2_3 = -(e^f\, du + H e^g\, dv), \qquad
\eta^3_2 = \eta^1_3 = -H e^f\, du. \label{MC-forms-2}
\end{equation}
The structure equations \eqref{structure-eqns} for $d\eta^1$ and $d\eta^2$ can be used to show that
\begin{equation}
 \eta^1_1 = -\eta^2_2 = g_u\, du - f_v\, dv, \label{MC-forms-3}
\end{equation}
and according to \eqref{MC-forms-relations}, the remaining Maurer-Cartan forms are zero.

Finally, the remaining structure equations in \eqref{structure-eqns}---which may be interpreted as the Gauss and Codazzi equations for timelike surfaces in $\R^{1,2}$---imply that the functions $f, g, H$ satisfy the following PDE system:
\begin{gather}
H_v = 0, \label{PDE1} \\
H_u = 2f_v e^{f-g}, \label{PDE2} \\
(f+g)_{uv} = H^2 e^{f+g}. \label{PDE3}
\end{gather}
Conversely, any solution to this system gives rise (at least locally) to the Maurer-Cartan forms \eqref{MC-forms-1}, \eqref{MC-forms-2}, \eqref{MC-forms-3} of a totally quasi-umbilic timelike surface.

Equation \eqref{PDE1} implies that $H$ is a function of $u$ alone.  In order to solve the remaining equations, we will divide into cases based on whether $H(u)$ is zero or nonzero.

\subsection{Case 1: $H(u) \equiv 0$}

Then equations \eqref{PDE2} and \eqref{PDE3} reduce to:
\begin{gather*}
f_v = 0 ,  \\
(f+g)_{uv} = 0.
\end{gather*}
Thus $f = f(u)$ is a function of $u$ alone, and
\[ f + g = \phi(u) + \psi(v) \]
for some functions $\phi(u), \psi(v)$.

The first and second fundamental forms of $\Sigma$ now take the form
\begin{gather*}
\text{I} = 2 e^{\phi(u)} e^{\psi(v)}\, du\, dv, \\
\text{II} = e^{2 f(u)}\, du^2.
\end{gather*}
By making a change of coordinates $(u,v) \to (\tilde{u}, \tilde{v})$ with
\[ d\tilde{u} = e^{\phi(u)}\, du, \qquad d\tilde{v} = e^{\psi(v)}\, dv \]
we can assume that $\phi(u) = \psi(v) = 0$, and so
\begin{gather}
\text{I} = 2\, du\, dv, \label{case1-fundamental-forms} \\
\text{II} = e^{2 f(u)}\, du^2. \notag
\end{gather}
From \eqref{fundamental-forms}, it follows that
\begin{equation}
 \eta^1 = e^{f(u)}\, du, \qquad \eta^2 = e^{-f(u)}\, dv, \label{case1-MC-forms-1}
\end{equation}
and equations \eqref{MC-forms-2}, \eqref{MC-forms-3} become:
\begin{gather}
\eta^3_1 = \eta^2_3 = -e^f\, du , \qquad
\eta^3_2 = \eta^1_3 = 0, \label{case1-MC-forms-2} \\
 \eta^1_1 = -\eta^2_2 = -f'(u)\, du . \label{case1-MC-forms-3}
\end{gather}
Thus solutions of the PDE system \eqref{PDE1}-\eqref{PDE3} with $H(u) \equiv 0$ depend on one arbitrary function of one variable---i.e., the function $f(u)$.

\subsection{Case 2: $H(u) \neq 0$}

Let $h(u) = \ln |H(u)|$.  Then we can write equation \eqref{PDE3} as
\[ (f+g)_{uv} = e^{f + g + 2 h(u)}, \]
or equivalently,
\[ \left( f+g + 2 h(u) \right)_{uv} = e^{f + g + 2 h(u)}. \]
Thus the function
\[ z(u,v) = f(u,v) + g(u,v) + 2h(u) \]
is a solution to Liouville's equation
\begin{equation}
  z_{uv} = e^z. \label{Liouville's-eqn}
\end{equation}
The general solution to \eqref{Liouville's-eqn} is
\begin{equation}
z(u,v) = \ln \left( \frac{2 \phi'(u) \psi'(v)}{(\phi(u) + \psi(v))^2} \right), \label{Liouville-gen-soln}
\end{equation}
where $\phi(u), \psi(v)$ are arbitrary functions with $\phi'(u), \psi'(v) \neq 0$; thus we have
\begin{equation}
f(u,v) + g(u,v) + 2h(u) = \ln \left( \frac{2 \phi'(u) \psi'(v)}{(\phi(u) + \psi(v))^2} \right) \label{apply-Liouville-gen-soln}
\end{equation}
for some functions $\phi(u), \psi(v)$ with $\phi'(u), \psi'(v) \neq 0$.

\begin{remark}
The expression \eqref{Liouville-gen-soln} for the general solution to \eqref{Liouville's-eqn} can be derived from the B\"acklund transformation
\begin{equation}\label{Liouvillewave}
\begin{aligned}
z_u - w_u &=  2e^{\frac{(z+w)}{2}}, \\
z_v + w_v &= e^{\frac{(z-w)}{2}},
\end{aligned}
\end{equation}
which relates solutions $z(u,v)$ of \eqref{Liouville's-eqn} to solutions $w(u,v)$ of the wave equation $w_{uv}=0$ (see, e.g., \cite{CI09}).  Substituting the general solution
\[ w(u,v) = \rho(u) + \sigma(v) \]
of the wave equation into \eqref{Liouvillewave} and solving the resulting overdetermined PDE system for $z(u,v)$ yields \eqref{Liouville-gen-soln}, where
\[ \phi(u) = \int e^{\rho(u)}\, du, \qquad \psi(v) = \int \tfrac{1}{2} e^{-\sigma(v)}\, dv. \]
\end{remark}

By making the change of variables $\tilde{u} = \phi(u)$, $\tilde{v} = \psi(v)$, we can write \eqref{apply-Liouville-gen-soln} as
\[ f(u,v) + g(u,v) + 2h(u) = \ln \left( \frac{2}{(u+v)^2} \right). \]
Thus we have
\begin{align}
f + g & = \ln \left( \frac{2}{(u+v)^2} \right) - 2 h(u) \notag \\
& = \ln \left( \frac{2}{(H(u))^2(u+v)^2} \right). \label{fg-expr}
\end{align}
Now we can write equation \eqref{PDE2} as
\begin{align*}
H'(u) & = 2 f_v e^{2f} e^{-(f+g)} \\
& = \frac{(e^{2f})_v (H(u))^2}{(f+g)_{uv}} \text{\ (from equation \eqref{PDE3})};
\end{align*}
therefore,
\[ \frac{H'(u)}{(H(u))^2} (f+g)_{uv} = (e^{2f})_v. \]
Integrating with respect to $v$ yields
\[ \frac{H'(u)}{(H(u))^2} (f+g)_{u} = (e^{2f}) + k(u) \]
for some function $k(u)$.  But from equation \eqref{fg-expr}, we have
\[ (f+g)_u = -2\left( \frac{H'(u)}{H(u)} + \frac{1}{u+v} \right). \]
Therefore,
\begin{align*}
e^{2f} & = \frac{H'(u)}{(H(u))^2} (f+g)_{u} - k(u) \\
& =  \frac{-2H'(u)}{(H(u))^2} \left( \frac{H'(u)}{H(u)} + \frac{1}{u+v} \right) - k(u) \\
& = \frac{-2H'(u)}{(H(u))^2 (u+v)} - \tilde{k}(u),
\end{align*}
where
\[ \tilde{k}(u) = k(u) + 2\frac{(H'(u))^2}{(H(u))^3}. \]

Thus the Maurer-Cartan forms are given by \eqref{MC-forms-1}, \eqref{MC-forms-2}, \eqref{MC-forms-3}, where
\begin{gather}
H = H(u) \neq 0, \notag \\
f = \tfrac{1}{2} \ln \left(\frac{-2H'(u)}{(H(u))^2 (u+v)} - k(u) \right), \label{fgH-case2} \\
g = \ln \left( \frac{2}{(H(u))^2(u+v)^2} \right) - \tfrac{1}{2} \ln \left(\frac{-2H'(u)}{(H(u))^2 (u+v)} - k(u) \right), \notag
\end{gather}
and $k(u)$ is an arbitrary function of $u$.  Thus solutions of the PDE system \eqref{PDE1}-\eqref{PDE3} with $H(u) \neq 0$ depend on two arbitrary functions of one variable---i.e., the functions $h(u), k(u)$.

\section{Explicit parametrizations}\label{main-result-sec}

In this section we use the Maurer-Cartan forms obtained in \S \ref{fundamental-forms-sec} to derive explicit local parametrizations for all totally quasi-umbilic timelike surfaces.  The Maurer-Cartan equations \eqref{MC-forms-def} can be written in matrix form as
\begin{equation}
 \begin{bmatrix} d\bx & d\f_1 & d\f_2 & d\f_3 \end{bmatrix} = \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.1in] \eta^1 & \eta^1_1 & \eta^1_2 & \eta^1_3 \\[0.1in] \eta^2 & \eta^2_1 & \eta^2_2 & \eta^2_3 \\[0.1in] \eta^3 & \eta^3_1 & \eta^3_2 & \eta^3_3 \end{bmatrix}. \label{MC-matrix-eqn}
\end{equation}

\subsection{Case 1: $H(u) \equiv 0$}  In this case, equation \eqref{MC-matrix-eqn} takes the form
\begin{equation*}
 \begin{bmatrix} d\bx & d\f_1 & d\f_2 & d\f_3 \end{bmatrix} = \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.1in] e^{f(u)}\, du & -f'(u)\, du & 0 & 0 \\[0.1in] e^{-f(u)}\, dv & 0 & f'(u)\, du & -e^{f(u)}\, du \\[0.1in] 0 & -e^{f(u)}\, du & 0 & 0 \end{bmatrix}.
\end{equation*}
This is equivalent to the compatible, overdetermined PDE system
\begin{gather}
 \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix}_u  = \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.1in] e^{f(u)} & -f'(u) & 0 & 0 \\[0.1in] 0 & 0 & f'(u) & -e^{f(u)} \\[0.1in] 0 & -e^{f(u)} & 0 & 0 \end{bmatrix}, \label{case1-u-pde} \\[0.2in]
 \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix}_v  = \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.1in] 0 & 0 & 0 & 0 \\[0.1in] e^{-f(u)} & 0 & 0 & 0 \\[0.1in] 0 & 0 & 0 & 0 \end{bmatrix}  \label{case1-v-pde}
\end{gather}
for the vector-valued functions $\bx, \f_1, \f_2, \f_3$ on $\Sigma$.
These equations can be integrated explicitly, and the general solution for $\bx(u,v)$ is
\[
\bx(u,v) = \bx^0 +  u e^{f(0)} \f^0_1 + \left[ v + \tfrac{1}{2} \int_0^u \left( \int_0^{\sigma} e^{2f(\tau)} d\tau \right)^2 d\sigma \right] e^{-f(0)}\f^0_2
- \left[ \int_0^u \left( \int_0^{\sigma} e^{2f(\tau)} d\tau \right) d\sigma \right] \f^0_3,
\]
where $\bx^0, \f_1^0, \f_2^0, \f_3^0$ represent the initial conditions
\[ \begin{bmatrix} \bx(0,0) & \f_1(0,0) & \f_2(0,0) & \f_3(0,0) \end{bmatrix} = \begin{bmatrix} \bx^0 & \f^0_1 & \f^0_2 & \f^0_3 \end{bmatrix}. \]
Thus $\Sigma$ is a cylindrical surface of the form
\begin{equation}
 \bx(u,v) = \alpha(u) + v \f^0_2, \label{cylinder-surface}
\end{equation}
where the $v$-coordinate curves are null lines parallel to the null vector $\f^0_2$, and the generating curve $\alpha(u)$ is the null curve
\[
 \alpha(u) = \bx^0 +  u e^{f(0)} \f^0_1 + \left[ \tfrac{1}{2} \int_0^u \left( \int_0^{\sigma} e^{2f(\tau)} d\tau \right)^2 d\sigma \right]  e^{-f(0)}\f^0_2 - \left[ \int_0^u \left( \int_0^{\sigma} e^{2f(\tau)} d\tau \right) d\sigma \right] \f^0_3 .
\]

We can give a slightly simpler parametrization for $\Sigma$ (not by null coordinates) as \eqref{cylinder-surface} with
\[ \alpha(u) = \bx^0 + u e^{f(0)} \f^0_1 - \left[ \int_0^u \left( \int_0^{\sigma} e^{2f(\tau)} d\tau \right) d\sigma \right] \f^0_3. \]
Then $\alpha$ may be viewed as the graph in the $\f_1^0 \f_3^0$-plane of an arbitrary function $F(u)$ with $F''(u) < 0$.  We summarize this result, noting that
\begin{itemize}
\item the $\f_1^0 \f_3^0$-plane is a lightlike plane,
\item $\f_3^0$ spans the unique direction in this plane which is orthogonal to $\f_2^0$, and
\item $\alpha'(u)$ is always linearly independent from $\f_3^0$,
\end{itemize}
 as:

\begin{proposition}\label{H=0-prop}
Let $\Sigma$ be a totally quasi-umbilic, regular timelike surface in $\R^{1,2}$ with mean curvature $H \equiv 0$.  Then $\Sigma$ is a cylinder over a convex curve $\alpha$ which is contained in a lightlike plane, with null rulings transverse to the plane of $\alpha$; moreover, the tangent vector to $\alpha$ is nowhere orthogonal to the rulings.  Conversely, any such cylinder is totally quasi-umbilic with mean curvature $H \equiv 0$.
\end{proposition}

\begin{remark}
Since the surfaces of Proposition \ref{H=0-prop} have $K \equiv 0$ as well, this proposition shows that there exists a family of non-planar surfaces in $\R^{1,2}$, parametrized by one arbitrary function of one variable, all of which have Gauss and mean curvature identically equal to zero!
\end{remark}

\subsection{Case 2: $H(u) \neq 0$} In this case, the PDE system represented by equation \eqref{MC-matrix-eqn} is considerably more complicated and generally cannot be completely solved analytically, but there is still enough information in \eqref{MC-matrix-eqn} to give a geometric description of the solution surfaces.  The $v$-component of the corresponding PDE system is
\begin{equation}
\begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix}_v  = \begin{bmatrix} \bx & \f_1 & \f_2 & \f_3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.1in] 0 & -f_v & 0 & 0 \\[0.1in] e^g & 0 & f_v & -He^g \\[0.1in] 0 & -He^g & 0 & 0 \end{bmatrix},  \label{case2-v-pde}
\end{equation}
where $f, g, H$ are given by \eqref{fgH-case2}.  In particular, we have
\[ (\f_2)_v = f_v \f_2, \]
and so
\[ \f_2(u,v) = e^{f} \bar{\f}_2(u) \]
for some null vector field $\bar{\f}_2(u)$ along $\Sigma$ which is a function of $u$ alone.  This means that along any $v$-coordinate curve $\bx(u_0, v)$ in $\Sigma$, the vectors $\f_2(u_0, v)$ are all parallel.  Moreover, we have
\[ \bx_v(u,v) = e^g \f_2(u,v) = e^{f+g} \bar{\f}_2(u); \]
it follows that the $v$-coordinate curve $\bx(u_0, v)$ is a null line parallel to $\bar{\f}_2(u_0)$.  So if we let $\alpha$ be the null curve
\[ \alpha(u) = \bx(u, 0) \]
and define $\bar{\f}_2(u)$ to be the null vector field
\[ \bar{\f}_2(u) = \f_2(u, 0) \]
along $\alpha$, then we can reparametrize $\Sigma$ (not necessarily by null coordinates) as
\[ \bx(u,v) = \alpha(u) + v\bar{\f}_2(u). \]
Therefore, any such surface is ruled, and the rulings are all null lines.  Moreover, the $u$-component of the PDE system corresponding to \eqref{MC-matrix-eqn}

implies that:
\begin{itemize}
\item The vectors $\alpha'(u)$ and $\alpha''(u)$ are linearly independent for all $u$; we call such a null curve {\em nondegenerate}.
\item The vectors $\bar{\f}_2(u)$ and $\bar{\f}'_2(u)$ are linearly independent if and only if $H(u) \neq 0$.
\end{itemize}

Conversely, suppose that $\Sigma$ is a timelike ruled surface whose rulings are all null lines.  Let $\alpha$ be a null curve contained in $\Sigma$ which is transverse to the rulings, and suppose further that $\alpha$ is nondegenerate.  Define vector fields $\bar{\f}_1(u), \bar{\f}_2(u), \bar{\f}_3(u)$ along $\alpha$ by the conditions that:
\begin{itemize}
\item $\bar{\f}_1(u) = \alpha'(u)$;
\item $\bar{\f}_2(u)$ spans the ruling of $\Sigma$ passing through $\alpha(u)$, and $\langle \bar{\f}_1(u), \bar{\f}_2(u) \rangle = 1$;
\item $\bar{\f}_3(u)$ is a unit normal vector to $\Sigma$ at $\alpha(u)$.
\end{itemize}
Then we can parametrize $\Sigma$ as
\[ \bx(u,v) = \alpha(u) + v \bar{\f}_2(u), \]
and there exists an adapted null frame field $(\f_1, \f_2, \f_3)$ along $\Sigma$ with the property that
\[ \f_1(u,0) = \bar{\f}_1(u), \qquad \f_2(u,0) = \bar{\f}_2(u), \qquad \f_3(u,0) = \bar{\f}_3(u). \]

The $u$-component of the Maurer-Cartan equation \eqref{MC-matrix-eqn} for the adapted null frame field $(\f_1, \f_2, \f_3)$ evaluated along $\alpha$ implies that there exist functions $h_{ij}(u)$ such that
\[ \begin{bmatrix} \bar{\f}'_1(u) & \bar{\f}'_2(u) & \bar{\f}'_3(u) \end{bmatrix} = \begin{bmatrix} \bar{\f}_1(u) & \bar{\f}_2(u) & \bar{\f}_3(u) \end{bmatrix} \begin{bmatrix} h_{11}(u) & 0 & h_{32}(u) \\[0.1in] 0 & -h_{11}(u) & h_{31}(u) \\[0.1in] h_{31}(u) & h_{32}(u) & 0 \end{bmatrix}, \]
and the condition that $\alpha$ is nondegenerate implies that $h_{31}(u) \neq 0$ for all $u$.

In order to show that $\Sigma$ is totally quasi-umbilic, we will compute the shape operator $S_{\bx}$ at any point $\bx \in \Sigma$ with respect to the basis $(\bx_u, \bx_v)$ for $T_{\bx}\Sigma$.  First we compute:
\begin{align*}
\bx_u & = \bar{\f}_1(u) - v h_{11}(u) \bar{\f}_2(u) + v h_{32}(u) \bar{\f}_3(u), \\
\bx_v & = \bar{\f}_2(u).
\end{align*}
It is straightforward to check that the unit normal vector field $\f_3$ to $\Sigma$ is given by
\[ \f_3(u,v) = \bar{\f}_3(u) + v h_{32}(u) \bar{\f}_2(u). \]
So the action of the shape operator on the basis $(\bx_u, \bx_v)$ is given by:
\begin{align*}
S_{\bx}(\bx_u) & = -(\f_3)_u \\
   & = -\bar{\f}_3'(u) - v(h'_{32}(u) \bar{\f}_2(u) + h_{32}(u) \bar{\f}'_2(u)) \\
   & = -h_{32}(u) \bar{\f}_1(u) + (-h_{31}(u) - v h'_{32}(u) + v h_{32}(u) h_{11}(u)) \bar{\f}_2(u) - v (h_{32}(u))^2 \bar{\f}_3(u)       \\
   & =  -h_{32}(u) ( \bar{\f}_1(u) - v h_{11}(u) \bar{\f}_2(u) + v h_{32}(u) \bar{\f}_3(u)) - (h_{31}(u) + v h'_{32}(u)) \bar{\f}_2(u)     \\
   & = -h_{32}(u) \bx_u - (h_{31}(u) + v h'_{32}(u)) \bx_v, \\
S_{\bx}(\bx_v) & = -(\f_3)_v \\
& = -h_{32}(u) \bar{\f}_2(u) \\
& = -h_{32}(u) \bx_v.
\end{align*}
Therefore, the matrix of $S_{\bx}$ with respect to the basis $(\bx_u, \bx_v)$ is
\[ \begin{bmatrix} -h_{32}(u) & 0 \\[0.1in] -h_{31}(u) - v h'_{32}(u) & -h_{32}(u) \end{bmatrix} . \]
Since $h_{31}(u) \neq 0$, it follows that $\Sigma$ is quasi-umbilic almost everywhere, and if $h'_{32}(u) \neq 0$, then $\Sigma$ contains a curve of umbilic points defined by the equation $v = -\frac{h_{31}(u)}{h'_{32}(u)}$.

Together with Proposition \ref{H=0-prop}, this proves the following classification theorem:

\begin{theorem}\label{main-thm}
Let $\Sigma$ be a totally quasi-umbilic, regular timelike surface in $\R^{1,2}$. Then $\Sigma$ is a ruled surface whose rulings are all null lines, with the additional property that any null curve $\alpha$ in $\Sigma$ which is transverse to the rulings is nondegenerate.  Conversely, given any nondegenerate null curve $\alpha(u)$ in $\R^{1,2}$ and any null vector field $\bar{\f}_2(u)$ along $\alpha$ which is linearly independent from $\alpha'(u)$ for all $u$, the immersed ruled surface
\[ \bx(u,v) = \alpha(u) + v \bar{\f}_2(u) \]
is totally quasi-umbilic, possibly containing a curve of umbilic points.
\end{theorem}

\section{examples}\label{examples-sec}

In light of Theorem \ref{main-thm}, any totally quasi-umbilic, regular timelike surface $\Sigma$ has a parametrization of the form
\[ \bx(u,v) = \alpha(u) + v \bar{\f}_2(u), \]
where $\alpha$ is a nondegenerate null curve whose tangent vector $\bar{\f}_1(u) = \alpha'(u)$ is a null vector which is linearly independent from the null vector $\bar{\f}_2(u)$.  The nondegeneracy of $\alpha$ implies that $\bar{\f}_1'(u)$ must be linearly independent from $\bar{\f}_1(u)$.

For convenience, we can reparametrize the curve $\alpha(u)$ and scale the vector field $\bar{\f}_2(u)$ so as to arrange that both $\bar{\f}_1(u)$ and $\bar{\f}_2(u)$  have {\em Euclidean} length 2; i.e., we can assume that
\[ \bar{\f}_1(u) = (1, \cos \theta_1(u), \sin \theta_1(u)), \qquad
\bar{\f}_2(u) = (1, \cos \theta_2(u), \sin \theta_2(u)) \]
for some functions $\theta_1(u), \theta_2(u)$.  (These vectors will no longer necessarily satisfy the condition that $\langle \bar{\f}_1(u), \bar{\f}_2(u) \rangle = 1$, but this condition is not important at this point.)  Then the linear independence conditions of the previous paragraph are simply that $\theta_1'(u) \neq 0$ and $\theta_1(u) \neq \theta_2(u)$ for all $u$.  Moreover, $\Sigma$ has Gauss and mean curvature $H \equiv K \equiv 0$ if and only if $\theta_2'(u) \equiv 0$.

\subsection{Case 1: $H(u) \equiv 0$}  In this case $\theta_2(u)$ must be constant, while $\theta_1(u)$ must be either strictly increasing or strictly decreasing and never equal to $\theta_2(u)$.

\begin{example}\label{example-1}
Let
\[   \theta_1(u) = 2\tan^{-1} u , \qquad  \theta_2(u) = \pi . \]
Then we have $\theta_1'(u) > 0$ for all $u$, and since $-\pi < \theta_1(u) < \pi$, the linear independence requirements are satisfied.  The corresponding null vector fields are given by
\begin{align*}
\bar{\f}_1(u) & = \left(1, \cos \left(2 \tan^{-1} u \right), \sin \left(2 \tan^{-1} u \right) \right) \\
& = \left(1, \frac{1 - u^2}{1 + u^2}, \frac{2u}{1 + u^2} \right)
\end{align*}
and
$\bar{\f}_2(u) = (1, -1, 0). $
Thus we have
\[ \alpha(u) = \int \bar{\f}_1(u)\, du = \left(u,\, 2 \tan^{-1} u - u,\, \ln (1 + u^2) \right), \]
and the corresponding surface $\Sigma$ is given by
\begin{align*}
\bx(u,v) & = \alpha(u) + v \bar{\f}_2(u) \\
& = \left(u + v, \, 2 \tan^{-1} u - u - v, \, \ln (1 + u^2) \right).
\end{align*}
Note that we could rewrite $\bx(u,v)$ as
\begin{align*}
\bx(u,v) & = \left( \tan^{-1} u, \, \tan^{-1} u, \ln(1 + u^2) \right) +  (u - \tan^{-1} u + v, -(u - \tan^{-1} u + v), 0) \\
& = \left( \tan^{-1} u, \, \tan^{-1} u, \ln(1 + u^2) \right) + \tilde{v} (1, -1, 0),
\end{align*}
where $\tilde{v} = u - \tan^{-1} u + v.$  This realizes $\Sigma$ as a cylinder over the convex curve
\[ \alpha(u) = (\tan^{-1} u) (1, 1, 0) + \ln(1 + u^2) (0,0,1) \]
in the lightlike plane spanned by the vectors $(1,1,0)$ and $(0,0,1)$.  Making the substitution $\sigma = \tan^{-1} u$ shows that this curve is the graph of the function $F(\sigma) = \ln(\sec^2 \sigma)$.  (See Figure \ref{example-1-fig}; note that the $x^0$-axis is drawn as the vertical axis.)
\begin{figure}[h]
\includegraphics[width=3in]{example-1-fig}
\caption{The surface of Example \ref{example-1}}
\label{example-1-fig}
\end{figure}

\end{example}

\subsection{Case 2: $H(u) \neq 0$} In this case, $\theta_1(u)$ and $\theta_2(u)$ must both be either strictly increasing or strictly decreasing, and we must have $\theta_1(u) \neq \theta_2(u)$ for all $u$.

\begin{example}\label{example-2}
Let
\[ \theta_1(u) = u, \qquad \theta_2(u) = u + \pi. \]
The corresponding null vector fields are given by
\[ \bar{\f}_1(u) = (1,\, \cos u,\, \sin u), \qquad \bar{\f}_2(u) = (1,\, -\cos u,\, -\sin u). \]
Thus we have
\[ \alpha(u) = \int \bar{\f}_1(u)\, du = \left(u,\, \sin u,\, -\cos u \right), \]
and the corresponding surface $\Sigma$ is given by
\begin{align*}
\bx(u,v) & = \alpha(u) + v \bar{\f}_2(u) \\
& = \left(u + v, \, \sin u - v \cos u, \, -\cos u - v \sin u \right).
\end{align*}
(See Figure \ref{example-2-fig} for embedded and immersed portions of this surface; note that the $x^0$-axis is drawn as the vertical axis.)
\begin{figure}[h]
\includegraphics[width=2.5in]{example-2-fig-a}
\includegraphics[width=2.5in]{example-2-fig-b}
\caption{Embedded and immersed portions of the surface of Example \ref{example-2}}
\label{example-2-fig}
\end{figure}

In order to compute shape operator $S_{\bx}$, first we compute
\begin{align*}
\bx_u & = (1, \, \cos u + v \sin u, \, \sin u - v \cos u) \\
\bx_v & = (1, \, -\cos u,\, -\sin u).
\end{align*}
The unit normal vector to $\Sigma$ is given by
\begin{align*}
 \f_3(u,v) & = \frac{\bx_u \times \bx_v}{\sqrt{-\langle \bx_u \times \bx_v, \bx_u \times \bx_v\rangle}} \\
 &  =  \left(-\frac{v}{2}, \, -\sin u + \frac{v}{2} \cos u, \, \cos u + \frac{v}{2} \sin u \right).
\end{align*}
(Note that the cross product here denotes the Minkowski cross product \eqref{Minkowski-cross-product}.)  Now the action of the shape operator on the basis $(\bx_u, \bx_v)$ for $T_{\bx}\Sigma$ is given by:
\begin{align*}
S_{\bx}(\bx_u) & = -(\f_3)_u \\
   & = \left( 0, \, \cos u + \tfrac{1}{2} \sin u, \, \sin u - \tfrac{1}{2} \cos u \right) \\
   & = \tfrac{1}{2} \bx_u - \tfrac{1}{2} \bx_v, \\
S_{\bx}(\bx_v) & = -(\f_3)_v \\
& = \left( \tfrac{1}{2},\,  -\tfrac{1}{2} \cos u, \, \tfrac{1}{2} \sin u \right) \\
& = \tfrac{1}{2} \bx_v.
\end{align*}
Therefore the matrix of $S_{\bx}$ with respect to the basis $(\bx_u, \bx_v)$ for $T_{\bx}\Sigma$ is
\[ \begin{bmatrix} \tfrac{1}{2} & 0 \\[0.1in] -\tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} . \]
Thus we see that $\Sigma$ has constant Gauss and mean curvatures $K \equiv \tfrac{1}{4}$, $H \equiv \tfrac{1}{2}$.  Moreover, $\Sigma$ contains no umbilic points.
\end{example}

\begin{example}\label{example-3}
Let
\[ \theta_1(u) = u^3 + u, \qquad \theta_2(u) = u^3 + u + \pi. \]
The corresponding null vector fields are given by
\[ \bar{\f}_1(u) = (1,\, \cos (u^3 + u),\, \sin (u^3 + u)), \qquad \bar{\f}_2(u) = (1,\, -\cos (u^3 + u),\, -\sin (u^3 + u)). \]
Define
\[ \alpha(u) = \int_0^u \bar{\f}_1(s)\, ds = \left(u,\, \int_0^u \cos (s^3 + s)\, ds,\, \int_0^u \sin (s^3 + s)\, ds \right), \]
and the corresponding surface $\Sigma$ is given by
\begin{align*}
\bx(u,v) & = \alpha(u) + v \bar{\f}_2(u) \\
& =  \left(u + v,\, \int_0^u \cos (s^3 + s)\, ds - v\cos(u^3 + u),\, \int_0^u \sin (s^3 + s)\, ds - v\sin(u^3 + u) \right).
\end{align*}

In order to compute shape operator $S_{\bx}$, first we compute
\begin{align*}
\bx_u & = (1, \, \cos (u^3 + u) + v (3 u^2 + 1) \sin (u^3 + u), \, \sin (u^3 + u) - v (3 u^2 + 1) \cos (u^3 + u)) \\
\bx_v & = (1, \, -\cos (u^3 + u),\, -\sin (u^3 + u)).
\end{align*}
The unit normal vector to $\Sigma$ is given by
\begin{align*}
 \f_3(u,v) & = \frac{\bx_u \times \bx_v}{\sqrt{-\langle \bx_u \times \bx_v, \bx_u \times \bx_v\rangle}} \\
& =  \left( 0, \, -\sin (u^3 + u), \, \cos (u^3 + u) \right) + \frac{v}{2} (3u^2 + 1) \left( -1, \, \cos (u^3 + u), \, \sin (u^3 + u) \right).

\end{align*}

A computation shows that the action of the shape operator on the basis $(\bx_u, \bx_v)$ for $T_{\bx}\Sigma$ is given by:
\begin{align*}
S_{\bx}(\bx_u)
   & = \tfrac{1}{2} (3u^2 + 1) \bx_u  + \left( 3uv - \tfrac{1}{2} (3u^2 + 1) \right)  \bx_v, \\[0.1in]
S_{\bx}(\bx_v)
& = \tfrac{1}{2} (3u^2 + 1) \bx_v.
\end{align*}
Therefore the matrix of $S_{\bx}$ with respect to the basis $(\bx_u, \bx_v)$ for $T_{\bx}\Sigma$ is
\[ \begin{bmatrix} \tfrac{1}{2} (3u^2 + 1) & 0 \\[0.1in] 3uv - \tfrac{1}{2} (3u^2 + 1) & \tfrac{1}{2} (3u^2 + 1) \end{bmatrix} . \]
Thus we see that $\Sigma$ has Gauss and mean curvatures
\[ K = \tfrac{1}{4} (3u^2 + 1)^2, \qquad H = \tfrac{1}{2} (3u^2 + 1). \]
Moreover, $\Sigma$ contains a curve of umbilic points defined by the equation
\[ 3uv - \tfrac{1}{2} (3u^2 + 1) = 0. \]
See Figure \ref{example-3-fig} for various views of embedded and immersed portions of this surface, including the curve of umbilic points; note that the $x^0$-axis is drawn as the vertical axis.
\begin{figure}[h]
\includegraphics[width=2.5in]{example-3-fig-a}
\includegraphics[width=2.5in]{example-3-fig-b}
\includegraphics[width=2.5in]{example-3-fig-c}
\includegraphics[width=2.5in]{example-3-fig-d}
\caption{Embedded and immersed portions of the surface of Example \ref{example-3}}
\label{example-3-fig}
\end{figure}

\end{example}

\begin{example}\label{example-4}
The previous two examples were immersed surfaces; we can obtain an embedded example by perturbing the totally umbilic hyperboloid $\langle \bv, \bv \rangle = 1$.  We begin with the standard parametrization
\[ \bx(u,v) = (0,\, \cos u ,\, \sin u) + v (1,\, -\sin u , \, \cos u) \]
for the hyperboloid.  (Unlike our previous examples, the generating curve $\alpha(u) = (0,\, \cos u ,\, \sin u)$ is not a null curve, but this is not crucial.)  We will modify this example by replacing the parameter $u$ in the ruling vector field by the function $u + \tfrac{1}{2} \sin u$.

\begin{remark}
Any small perturbation of the parameter $u$ would suffice, as long as the perturbation is $2\pi$-periodic and has a derivative of magnitude less than 1.  Alternatively, we could perturb the parameter $u$ in the generating curve, or perform some combination of the two perturbations.
\end{remark}

Thus we will consider the surface $\Sigma$ with parametrization
\[ \bx(u,v) = (0,\, \cos u ,\, \sin u) + v \left( 1,\, -\sin (u + \tfrac{1}{2} \sin u) , \, \cos (u + \tfrac{1}{2} \sin u) \right). \]

Computations similar to those of the previous example show that $\Sigma$ is totally quasi-umbilic, with a curve of umbilic points defined by the equation
\begin{multline*}
 [8 \sin(u - \sin u) + 8 \sin u - \sin(2u + \sin u) + \sin (2u - \sin u) - 2\sin(\sin u)] v \\
   +  [ 8\cos(u + \tfrac{1}{2} \sin u) + 8 \cos(u - \tfrac{1}{2} \sin u)] = 0. \end{multline*}
See Figure \ref{example-4-fig} for two views of this surface, including the curve of umbilic points (which has two components); note that the $x^0$-axis is drawn as the vertical axis.
\begin{figure}[h]
\includegraphics[width=2.5in]{example-4-fig-a}
\includegraphics[width=2.5in]{example-4-fig-b}
\caption{The surface of Example \ref{example-4}}
\label{example-4-fig}
\end{figure}

\end{example}

\bibliographystyle{amsplain}
\bibliography{quasi-umbilic-bib}

\end{document}


TITLE: Separability of a normed space E
QUESTION [1 upvotes]: I'm stucked with the following problem:
Let $E$ a normed space. Show that $S=\{x\in E\hspace{0.1cm}|\hspace{0.1cm} ||x||=1\}$ is separable if and only if, $E$ is separable.
Could be used the Hann-Banach theorem? and where? I don't see this clearly.
Thanks !

REPLY [1 votes]: Suppose $S$ is separable, with countable dense subset $S'\subseteq S$. Then, for each $s \in S$ and $\epsilon > 0$, there exists $s' \in S'$ such that $\|s-s'\| < \epsilon$. Let $E'=\{ \rho s' : \rho \in \mathbb{Q}, s' \in S\}=\bigcup_{\rho \in\mathbb{Q},\;\rho > 0}\rho S'$ where $\rho S'=\{ \rho s' : s' \in S\}$. $E'$ is countable because it is the countable union of countable sets. To see that $E'$ is dense in $E$, let $x \in E\setminus\{0\}$ be fixed, and let $\epsilon > 0$ be given. Then there exists $\rho > 0$ in $\mathbb{Q}$ such that $\|x\| < \rho < \|x\|+\epsilon/2$. And there exists $s'\in S$ such that $\|\frac{1}{\|x\|}x -s'\| < \frac{\epsilon}{2\rho}$. So,
$$
\begin{align}
    \|x-\rho s'\| & =\|x\|\| \frac{1}{\|x\|}x-\frac{\rho}{\|x\|}s'\|\\
       & < \|x\|\|\frac{1}{\|x\|}x-s'\|+\|x\|\|s'-\frac{\rho}{\|x\|}s'\|\\
       & < \frac{\epsilon\|x\|}{2\rho}+\|x\|(\frac{\rho}{\|x\|}-1) \\
       & < \frac{\epsilon}{2} +(\rho-\|x\|) \\
       & < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.
\end{align}
$$
Therefore, if $S$ is separable, then $E$ is separable. To show that $S$ is separable if $E$ is separable, let $\{ e_{n} \}$ be a countable dense subset of $E$, discarding $0$ if it is in this set. To show that $\{ \frac{1}{\|e_{n}\|}e_{n}\}$ is a dense subset of $S$, let $s \in S$ and $0 < \epsilon < 1$ be given. Then there exists $k$ such that $\|s-e_{k}\| < \epsilon/2$, which guarantees $e_{k} \ne 0$ with $|1-\|e_{k}\||=|\|s\|-\|e_{k}\||<\|s-e_{k}\|<\epsilon/2$. Hence,
$$
\begin{align}
      \|s - \frac{1}{\|e_k\|}e_k\| & \le \|s-e_k\|+\|e_k-\frac{1}{\|e_k\|}e_k\| \\
        & \le \frac{\epsilon}{2}+\|e_k\||1-1/\|e_k\|| \\
        & = \frac{\epsilon}{2}+|1-\|e_{k}\|| \\
        & < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.
\end{align}
$$


TITLE: Is the limit $x \to - \infty$ of the cumulative distribution function always equal to $0$?
QUESTION [0 upvotes]: I've the probability density function:
$$
f_X(x) =
\begin{cases}
\frac{1}{2} x^2 \, e^{-x} & \text{if } x \geq 0 \\
0 & \text{if } x < 0
\end{cases}
$$
The CDF is:
$$
CDF_X(x) = 1 - \left( \frac{x^2}{2} + x + 1 \right) \, e^{-x}
$$
Now if I want to check the goodness of the CDF I should check the limits for $x \to \pm \infty$; the limit for $x \to + \infty$ is 1 and that's ok. But the limit for $x \to - \infty$ is $- \infty$.
Maybe in this case I should consider the fact that $f_X(x)$ is non zero for $x \geq 0$ so I should consider the limit for $x \to 0$ (in this case the result is $0$ which is correct)?

REPLY [2 votes]: Yes, distribution functions must have these properties:

$\lim_{t\to \infty }F(t)=1$
$\lim_{t\to-\infty}F(t)=0$
$F$ must be right continuous and increasing

In your case, you have that
$$
\begin{align*}
F_X(c)&=\int_{(-\infty ,c]}f_X(t)\,d t\\
&=\int_{(-\infty ,c]}\mathbf{1}_{[0,\infty )}(t)\frac1{2}t^2 e^{-t}\,d t\\
&=\int_{[0,c]}\frac1{2}t^2 e^{-t}\,d t\\
&=\left(1-e^{-c}\left(\frac{c^2}{2}+c+1\right)\right)\mathbf{1}_{[0,\infty )}(c)
\end{align*}
$$


TITLE: What is the volume of an n-dimension cube?
QUESTION [1 upvotes]: What is the volume of an n-dimension cube? Consider the length of each side to be $a$. How to solve this problem?

REPLY [1 votes]: Perhaps it would be easier to imagine the volume of the cube to be the "n-dimensional area". Let's consider what happens in 1D, 2D and 3D first:
n=1:
In the first dimension, the volume would be just the side length or a=$a^1$.
For n=2:
This would be the square whose side length is a and the area is $a^2$
For n=3:
This would be the cube with volume of a^3.
For n>3:
I could do an induction proof on the conclusion, however a more intuitive pattern starts to emerge:
having the dimension be $0<n\in\Bbb Z$, would most certainly mean that Volume(n-dimensional cube with side length a)=V(a,n)=$a^n$.
Keep in mind that the units would be $(units)^n$ for the n-dimensional area of an n-dimensional shape with the above restrictions, The dimension should a counting
The reason I said that an induction proof would not be as good is because I am not at the level of education to do this. I know how to do a very basic induction proof,  but I am unconfident for this one. I guess it would go like this:
A line has n units which I will call u which are line segments so u+u+u+...+u=nu similarly, a square has square units, but also a certain amount of squares in one line segment like row so this would be nu+nu+...nu n times getting us $n^2$ u. We have this dimension being d=2 with n d=1 shapes “within” it. number, ie a positive integer excluding zero.
The reason I said that an induction proof would not be as good is because I am not at the level of education to do this. I know how to do a very basic induction proof,  but I am unconfident for this one. I guess it would go like this: A line has n units which I will call u which are line segments so A=u+u+u+...+u=nu similarly, a square has square units, but also a certain amount of squares in one line segment like row so this would be A=nu+nu+...nu n times getting us $n^2$ u. We have this dimension being d=2 with n d=1 shapes “within” it. This means that our “hypothesis” is a d=N+1 cube has n of the d=N cubes “within” it using the same dimension units. The abbreviation u represents any general unit to the n-th power.
All in all, a d=N+1 cube can be drawn by connecting the corresponding vertices of a d=N cube assuming the connecting edge length is equal to the d=N cube edge length. This means that the hypothesis is “argued out” and so is informally correct with A=($n^{N}+...+n^{N})*u$ n times getting us A=n^{N+1}*units^{N+1}.

If a unit was instead defined by,say a n-tetrahedron, these formulas would not be correct for the space as it would likely be transformed.
I am not completely sure about the zeroth dimension nor fractal/fractional dimensions nor complex dimensions nor other types of dimensions here like negative dimensions.
Here are some links to these other dimensions:
Has the notion of having a complex amount of dimensions ever been described? And what about negative dimensionality?,
https://en.m.wikipedia.org/wiki/Fractal_dimension,
Imaginary and Complex Dimensions
Maybe there are other dimensions?


TITLE: Problem with an integration
QUESTION [1 upvotes]: What do I have to do, or what technique do I have to use, to perform the following integral?
$$ \frac{1}{4\pi^{2}}\iint_{-\infty}^{\infty}\mathrm{d}x\mathrm{d}y (x^{2}+y^{2}) \left| E(x,y) \right|^{2} =
 \dfrac{1}{4\pi^{2}}\iint_{-\infty}^{\infty}\mathrm{d}k_{x}\mathrm{d}k_{y} \left( \left| \frac{\partial A}{\partial x} \right|^{2} + \left| \dfrac{\partial A}{\partial y} \right|^{2} \right)$$
where:
$$ E(x,y) = \iint_{-\infty}^{\infty} A(k_{x},k_{y})  \exp\left[i\left(k_{x}x+k_{y}y+  z \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right)\right] \mathrm{d}k_{x}\mathrm{d}k_{y}
 $$

REPLY [3 votes]: Assuming that what you're asking is how to get the identity in the first line, given the Fourier decomposition in the second line: Note that multiplying by $x$ in real space corresponds to applying $\mathrm i\partial/\partial k_x$ in Fourier space. You can write $x^2|E(x,y)|$ as $|xE(x,y)|^2$, and then use Parseval's theorem to replace this by the integral over the squared magnitude of the Fourier transform, where the Fourier transform is given by
$$\mathrm i\frac{\partial}{\partial k_x}\left(A(k_x,k_y)\exp\left[\mathrm iz \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right]\right)\;.$$
Differentiating $A$ gives you the term you want in the result, and differentiating the other factor gives a factor
$$\frac{\partial}{\partial k_x}\left(k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)=\frac{k_x}{k}-\frac{k_x}{k}+\frac{(k_x^2+k_y^2)k_x}{2k^3}=\frac{k_x}{2k}\frac{k_x^2+k_y^2}{k^2}\;.$$
I guess the last term is dropped because of your approximation $k_{x}^{2}+k_{y}^{2} \ll k^{2}$?