text:
\begin{document}
\title[Simple curves in covering spaces]{Finding simple curves in surface covers is undecidable}
\author{Ingrid Irmer}
\address{Mathematics Department\\
Technion, Israel Institute of Technology\\
Haifa, 32000\\
Israel
}
\email{ingridi@technion.ac.il}
\today

\begin{abstract}
It is shown that various questions about the existence of simple closed curves in normal subgroups of surface groups are undecidable.
\end{abstract}

\maketitle

{\footnotesize
\tableofcontents
}

\section{Introduction}
\label{intro}
Let $S$ be a (not necessarily closed) pointed surface with genus at least 2, and $\tilde{S}$ a regular covering space of $S$. Given a word in $\pi_{1}(S)$ there are numerous algorithms that will determine whether this word represents a simple curve on $S$, for example \cite{BirmanSeries}, \cite{Chillingworth}, \cite{R} and \cite{Z}. However, if we want to decide whether (a) a given word representing a simple curve is contained in $\pi_{1}(\tilde{S})$, or (b) whether there exists a word in $\pi_{1}(\tilde{S})$ representing a simple curve in $S$, it turns out that these two questions are undecidable. The former is a consequence of undecidability of the word problem for finitely presented groups, and follows easily from Corollary \ref{corollary}, and undecidability of the latter is proven in this paper.

\begin{thm}
	\label{maintheorem}
There does not exist an algorithm that determines whether a normal subgroup $\pi_{1}(\tilde{S})\subset \pi_{1}(S)$ contains a word representing a homotopically nontrivial, simple curve on $S$.
\end{thm}

When $\pi_{1}(\tilde{S})$ does not contain a word representing a simple curve in $\pi_{1}(S)$, we will say that the cover $\tilde{S}$ does not have any simple curves. Note that ``simpleness'' here refers to $S$, not to $\tilde{S}$.

\begin{rem}
	Most questions about groups only depend on the groups up to isomorphism. However, Theorem \ref{maintheorem} is not just about the group $\pi_{1}(\tilde{S})$, but depends on how $\pi_{1}(\tilde{S})$ is embedded in the larger group $\pi_{1}(S)$.
\end{rem}

Theorem \ref{maintheorem} arose out of an attempt to devise an algorithm for testing the simple loop conjecture. Let $f: S \rightarrow M$ be a 2-sided immersion of a closed surface $S$ in a 3-manifold $M$, and denote by $f_{\sharp}:\pi_{1}(S) \rightarrow \pi_{1}(M)$ the induced map on the fundamental group. The simple loop conjecture states that if the normal subgroup $ker(f_{\sharp})$ is nontrivial, then there exists an essential simple closed curve in $ker(f_{\sharp})$. The simple loop conjecture has been proven in many special cases, for example Seifert fibered 3-manifolds, \cite{Hass}, graph 3-manifolds, \cite{RubinsteinWang} and 3-manifolds with geometric structures modelled on Sol, \cite{Zemke}.\\

There is a sense in which covering spaces not containing simple curves are generic. One reason for this is that any cover factoring through another cover that does not contain simple curves, will also not contain simple curves; some examples will be given in Section \ref{section2}. The simple loop conjecture is therefore making a very strong claim about maps of surfaces to 3-manifolds, and this paper raises doubts about whether the conjecture is verifiable algorithmically.\\

\textbf{Outline of paper.} Section \ref{section1} will revise some known properties of covering spaces, and Section \ref{section2} will construct examples of covering spaces that do not have simple curves. The techniques used in these examples will be used in Section \ref{section3} to give a proof of Theorem \ref{maintheorem}, by modifying Rabin's construction.

\section*{Acknowlegments}
This paper benefitted from helpful discussions with Andrew Putman (who suggested this problem in another context), Hyam Rubinstein and the group theorists and their visitors at the Technion, in particular Giles Gardam, Tobias Hartnick, Anton Khoroshkin and Michah Sageev. I would like to thank the Technion for its hospitality while this paper was written, and my host Yoav Moriah.

\section{Notation and assumptions}
Since the 2-sphere has no nontrivial covers, and for the torus, homological methods give a simple algorithm for finding all simple curves on a covering space, it will be assumed that the surface $S$ has genus $g$ at least 2.\\

When $S$ is orientable and has empty boundary, the usual presentation $\langle a_{1}, \ldots, a_{2}, b_{1}, \ldots, b_{g}\mid \Pi_{i=1}^{i=g} [a_{i}, b_{i}]\rangle$ is assumed, where the curves $\{a_{i}, b_{i}\}$ are shown in Figure \ref{basis} in the case $g=2$.\\

\begin{figure}
	\centering
	\includegraphics{basis.pdf}
	\caption{A set of simple curves representing a basis for $\pi_{1}(S)$.}
	\label{basis}
\end{figure}

A \emph{curve} on a surface is the image of a map of $S^1$ into the surface. In this convention, a curve is necessarily closed and connected. A curve $\gamma$ is said to be \textit{simple} if $\gamma$ is freely homotopic to an embedding of $S^1$.

\section{Elementary properties of covers}
\label{section1}
This section recalls some properties of covering spaces. For this section and the next, it will be assumed that $S$ is orientable.\\

Given a regular cover $\tilde{S}$ of $S$, there is the following exact sequence, where $D$ denotes the deck transformation group of the cover.

\begin{equation*}
1\rightarrow \pi_{1}(\tilde{S})\rightarrow \pi_{1}(S) \xrightarrow{\phi} D \rightarrow 1
\end{equation*}

The image under $\phi$ of the generating set $\{a_{i}, b_{i}\}$ is a generating set for $D$.

\begin{lem}
	\label{wellknownfact}
	Consider the presentation of $D$ given by
	\begin{equation}
		\langle \phi(a_{1}), \ldots, \phi(a_{g}), \phi(b_{1}), \ldots, \phi(b_{g})\mid r_{1}, \ldots, r_{s}\rangle
	\end{equation}
The subgroup $\pi_{1}(\tilde{S})$ of $\pi_{1}(S)$ is generated by the words $\{r_{i}'\}$ and their conjugates by elements of $\pi_{1}(S)$ that map to nontrivial elements of $D$. Here $r_{i}'$ is the word in the generating set of $\pi_{1}(S)$ obtained from $r_{i}$ by replacing each instance of $\phi(a_{j})$ by $a_{j}$ and $\phi(b_{j})$ by $b_{j}$.
\end{lem}
\begin{proof}
	This claim seems to be well-known. A proof is given in Lemma 4 of \cite{lifts}.
\end{proof}
\begin{cor}[Corollary of Lemma \ref{wellknownfact}]
\label{corollary}
Any finitely presented group $G$ can be embedded in the deck transformation group of a cover of a closed surface of genus $g$.
\end{cor}
\begin{proof}
Suppose $G:=\langle x_{1}, \ldots, x_{s}\mid r_{1}(x_{1}, \ldots, x_{s}), \ldots, r_{q}(x_{1}, \ldots, x_{s})\rangle$, where $s$ is less than $2g$. Let $\{g_{1}, \ldots, g_{2g}\}$ be a generating set of $\pi_{1}(S)$. Then by Lemma \ref{wellknownfact}, $G$ is the deck transformation group of the covering space corresponding to the subgroup of $\pi_{1}(S)$ normally generated by the words $\{r_{i}', g_{s+1}, \ldots, g_{2g}\}$. Here $r_{i}'$ is the word obtained from $r_{i}$ by replacing every instance of $x_{1}$ by the first generator, $g_1$, of $\pi_{1}(S)$, every instance of $x_{2}$ by the second generator of $\pi_{1}(S)$, etc. up to $x_{s}$. By the Freiheitssatz, $\{g_{1}, \ldots, g_{s}\}$ generates a free group, so all we are doing here is writing $G$ as a quotient of a free group in the usual way.\\

Now suppose $G$ has $n$ generators, where $n$ is larger than or equal to $2g$. It follows from the theory of HNN extensions that every countably generated group can be embedded in a group with two generators; a reference is Chapter 4 of \cite{LS}. A detail of this construction that will be needed later is that when embedding $G$ in a 2-generator group,  it is possible to choose a generator of $G$, and then construct the embedding in such a way that this generator is mapped to a generator of the 2-generator group. Embed $G$ in a two generator group $T$. Then $T$ is isomorphic to the deck transformation group of a cover of closed surface of genus $g$, as explained in the previous paragraph.
\end{proof}

\begin{cor}
	\label{corollary2}
	Let $c$ be simple curve represented by the conjugacy class of a word in $\pi_{1}(S)$, and $\tilde{S}\rightarrow S$ a regular cover of $S$. There are pairs of simple curves and covers for which there does not exist an algorithm to determine whether $c$ represents a word in the cover.
\end{cor}

\begin{proof}
To claim that this corollary follows from Corollary \ref{corollary} and the unsolvability of the word problem for finitely presented groups, it is necessary to show that the embedding of $G$ in the deck transformation group does not force us to identify all words that an algorithm is unable to distinguish from the identity with words that could not be simple in $\pi_{1}(S)$. \\

Let $G$ be a finitely presented group, with unsolvable word problem, for which there is no algorithm to decide whether or not $G$ is trivial. Fix a generator $g$ of $G$. In the proof of The proof of Corollary \ref{corollary} we saw that the embedding of $G$ in the deck transformation group $D$ can be done in such a way that $g$ is mapped to the image in $D$ of a simple curve $a_{1}$. If $G$ is the trivial group, then the image of $a_{1}$ is trivial in $D$, and hence by Lemma \ref{wellknownfact}, the simple curve $a_{1}$ represents a word in $\pi_{1}(\tilde{S})$. However, if $G$ is nontrivial, $g$ is a nontrivial element of $G$, and by Lemma \ref{wellknownfact}, $a_{1}$ does not represent a curve in the cover. This problem is therefore undecidable.
\end{proof}

\section{Covering spaces without simple curves}
\label{section2}
This section uses homological arguments and properties of normal subgroups of $[\pi_{1}(S), \pi_{1}(S)]$ to construct covering spaces without simple curves.\\

\textbf{Necessary conditions for curves to be simple. }A primitive homology class is an element of $H_{1}(S;\mathbb{Z})$ that is either trivial or can not be written as a multiple $k\in (\mathbb{Z}\setminus \{\pm 1\})$ of another element of $H_{1}(S;\mathbb{Z})$. On an orientable surface, a necessary condition for a curve to be simple is that it is a representative of a primitive homology class. For curves represented by words in $[\pi_{1}(S), \pi_{1}(S)]$, as proven in \cite{Notsimple}, a necessary condition for being simple is that the word is not contained in $[\pi_{1}(S),[\pi_{1}(S), \pi_{1}(S)]]$.\\

\textbf{Covers without simple curves. }An example of a covering space without simple curves is a covering with deck transformation group that can not be generated by any proper subset of $\{\phi(a_{1}),\ldots,\phi(a_{g}), \phi(b_{1}), \ldots, \phi(b_{g})\}$, and for which the words $\{r_{i}'\}$ in $\pi_{1}(S)$ from Lemma \ref{wellknownfact} are all in $[\pi_{1}(S),[\pi_{1}(S), \pi_{1}(S)]]$.\\

Another source of covers without simple curves  comes from Section 3.5 of \cite{lifts}. In \cite{lifts}, examples of covering spaces are given, for which connected components of pre-images of simple curves do not span the integer homology of the covering space. These covering spaces are constructed by iterating mod $m$ homology covers, as will now be explained briefly.\\

The mod $m$ homology cover is defined by
\begin{equation}
\label{modm}
1\rightarrow \pi_{1}(\tilde{S})\rightarrow \pi_{1}(S) \xrightarrow{\phi} H_{1}(S;\mathbb{Z}_{m}) \rightarrow 1.
\end{equation}
Due to the fact that $\phi$ factors through the Hurewicz homomorphism, it follows from Lemma \ref{wellknownfact} that $[\pi_{1}(S),\pi_{1}(S)]\lhd \pi_{1}(\tilde{S})$. The commutator subgroup of $\pi_{1}(S)$ contains simple curves, so mod $m$ homology covers do have simple curves. However, the only simple curves in mod $m$ homology covers are contained in $[\pi_{1}(S),\pi_{1}(S)]$, because by Lemma \ref{wellknownfact}, none of the other elements of $\pi_{1}(\tilde{S})$ are contained in primitive homology classes in $H_{1}(S; \mathbb{Z})$. As shown in Lemma 3 of \cite{lifts}, none of the simple curves in $[\pi_{1}(S),\pi_{1}(S)]$ are contained in $[\pi_{1}(\tilde{S}),\pi_{1}(\tilde{S})]$, so any simple curves in a mod $m$ homology cover $\tilde{S}$ are nonseparating in $\tilde{S}$. Taking a mod $m$ homology cover $\tilde{\tilde{S}}$ of $\tilde{S}$ therefore kills off any simple curves that were in $\tilde{S}$. It follows that $\tilde{\tilde{S}}$ does not contain simple curves.\\

A characteristic covering space of $S$ is a cover invariant under the induced action of the mapping class group on $\pi_{1}(S)$.
The homomorphism $\phi$ from Equation \eqref{modm} is composed of two homomorphisms; the Hurewicz homomorphism, and the map from $H_{1}(S;\mathbb{Z})\rightarrow H_{1}(S;\mathbb{Z}_{2})$. Since both of these maps are invariant under the induced action of the mapping class group of $S$, it follows that mod $m$ homology covers are characteristic. Composing two mod $m$ homology covers to obtain $\tilde{\tilde{S}}$ therefore gives a regular cover of $S$.\\

\section{Proof of undecidability}
\label{section3}
This section starts with a discussion of properties of mod 2 homology covers. These will then be combined with the simpleness criteria from the previous section, and ideas coming from Rabin's construction \cite{Rabin}, to prove Theorem \ref{maintheorem}.

\subsection{Mod 2 homology covers}
Some specific properties of mod 2 homology covers will now be discussed.

\begin{lem}
	\label{m2}
	Let $\tilde{S}\rightarrow S$ be the mod 2 homology cover of $S$. Then $\pi_{1}(\tilde{S})$ is generated by $(2g)^{2}-1$ elements $\{g_{1}=c_{1}^{2}, \ldots, g_{n}=c_{n}^{2}\}$ and their conjugates by elements of $\pi_{1}(S)$ mapping to nontrivial elements of $D$. Here $c_{i}$ is a word representing a simple, nonseparating curve on $S$.
\end{lem}
\begin{proof}
We first claim that a complete set of relations for the deck transformation group $D$ is given by $\{d_{i}^{2}, i=1, \ldots, (2g)^{2}-1\}$, where $d_{i}$ is one of the $(2g)^{2}-1$ nontrivial elements of $D$. Note that this is special to the case $m=2$, as it is only for $m=2$ that every element of $H_{1}(S; \mathbb{Z}_{m})$ is self inverse, which implies the commutation relations:
\begin{equation*}
ab=(ab)^{-1}=b^{-1}a^{-1}=ba
\end{equation*}
$H_{1}(S; \mathbb{Z}_{2})$ is by definition the abelian group with $2g$ generators, for which every group element has order two. This proves the claim.\\

Using Lemma \ref{wellknownfact}, it remains to prove that each of the nontrivial elements of $H_{1}(S;\mathbb{Z}_{2})$ has a simple curve representative. Fix an element $h$ of $H_{1}(S;\mathbb{Z}_{2})$. Then $h$ has as representative a subset of the set of curves $\{a_{i}, b_{j}\}$. If this subset contains both $a_{i}$ and $b_{i}$ for some $i$, replace the curves $a_{i}$ and $b_{i}$ by the simple curve obtained by Dehn twisting $a_{i}$ around $b_{i}$. Therefore, $h$ is represented by a set of at most $g$ simple, pairwise disjoint curves. It is therefore possible to take a connected sum to obtain a simple curve representing $h$. A set of simple curves representing the elements of $H_{1}(S;\mathbb{Z}_{2})$ is denoted by $\{c_{1}, \ldots, c_{(n)}\}$.
\end{proof}

\subsection{Tower of covers}
We now have all the ingredients to prove the main theorem of this paper.
\begin{thm*}
There does not exist an algorithm that determines whether a normal subgroup $\pi_{1}(\tilde{S})\lhd \pi_{1}(S)$ contains a word representing a simple curve on $S$.
\end{thm*}

From now on, for simplicity of notation, the same symbols will be used to denote elements of $\pi_{1}(S)$, and their images in the deck transformation group of the cover.
\begin{proof}
Start off by assuming $S$ is connected, orientable and with empty boundary. Denote by $S_{2}$ the mod 2 homology cover of $S$. In what follows, a tower of covers
\begin{equation*}
\tilde{S}_{2}\rightarrow S_{2}'\rightarrow S_{2}\rightarrow S
\end{equation*}
will be constructed. We first construct the deck transformation group of a cover $S_{2}'$ of $S_{2}$. To describe $S_{2}'$, let $H=\langle h_{1}, h_{2}, \ldots h_{m}\mid R_{1},\ldots, R_{s}\rangle$ be a finitely presented group with unsolvable word problem. The group $H\ast \langle x\rangle$ will be embedded in a two generator group $U$, as will now be explained.\\

Recall the definition of $\{g_{i}\}$ from Lemma \ref{m2}. This subset of a generating set of $\pi_{1}(S_{2})$ will be mapped to generators (denoted by the same symbols) of the deck transformation group of the cover $S_{2}'$ of $S_{2}$. By the Freiheitssatz, the $(2g)^{2}-1$ elements $\{g_{i}\}$ generate a free subgroup of $\pi_{1}(S_{2})$. A cover of $S_{2}$ will now be constructed, with deck transformation group generated by the images of the group elements $\{g_{i}\}$.\\

Let $x:=g_{1}$, $a:=[g_{1},[g_{1},g_{2}]]$ and $b:=[g_{2},[g_{1},g_{2}]]$. The commutator subgroup of $\pi_{1}(S_{2})$ is a free group, so the subgroup $[\pi_{1}(S_{2}),[\pi_{1}(S_{2}), \pi_{1}(S_{2})]]$ of the commutator subgroup is also free, and $a$ and $b$ freely generate a free subgroup of this group. Similarly, the set
\begin{equation*}
\{a, b^{-1}ab, b^{-2}ab^{2}, \ldots, b^{-(m-1)}ab^{m-1}\}
\end{equation*}
freely generates a free subgroup of $\langle a,b \rangle$.\\

Let $\psi:H\ast \langle x\rangle\rightarrow \langle a,b\rangle$ be defined by $\psi(h_{i})=b^{-(i-1)}ab^{i-1}$, and let $U$ be the group with presentation
\begin{equation*}
U=\langle g_{1}, g_{2} \mid R_{1}' , \ldots, R_{s}'\rangle
\end{equation*}
where $R_{i}'$ is the word in the generators $g_{1}$ and $g_{2}$ obtained by taking $R_{i}$ and replacing every instance of $h_j$ with the word in $\langle g_{1}, g_{2}\rangle$ given by $\psi(h_{j})$.\\

\begin{rem}
It is necessary here that $\langle g_{i} \rangle$ generates a free group. As a result, the above construction gives an embedding of $H$ in a larger group, in the usual way. The same is true for the embeddings of groups that follow.
\end{rem}

Now let
\begin{equation*}
J:=\langle U, g_{3}, g_{4}\mid g_{3}g_{1}g_{3}^{-1}=g_{1}^{2},\ \  g_{4}^{-1}g_{2}g_{4}=g_{2}^{2}\rangle
\end{equation*}
\begin{equation*}
K_{0}:=\langle J, g_{5}\mid g_{5}^{-1}g_{3}g_{5}=g_{3}^{2},\ \  g_{5}^{-1}g_{4}g_{5}=g_{4}^{2}\rangle
\end{equation*}
\begin{equation*}
K_{1}:=\langle K_{0}, g_{6}\mid g_{6}^{-1}g_{5}g_{6}=g_{5}^{2}\rangle
\end{equation*}
\begin{equation*}\vdots\end{equation*}
\begin{equation*}
K_{(2g)^{2}-8}:=\langle K_{(2g)^{2}-7},\ \  g_{(2g)^{2}-4}\mid g_{(2g)^{2}-4}^{-1}g_{(2g)^{2}-5}g_{(2g)^{2}-4}=g_{(2g)^{2}-5}^{2}\rangle
	\end{equation*}
	\begin{equation*}
	Q:=\langle g_{(2g)^{2}-3}, \ \ g_{(2g)^{2}-2}, \ \ g_{(2g)^{2}-1}\mid g_{(2g)^{2}-2}^{-1}g_{(2g)^{2}-3}g_{(2g)^{2}-2}=g_{(2g)^{2}-3}^{2},\ \  g_{(2g)^{2}-1}^{-1}g_{(2g)^{2}-2}g_{(2g)^{2}-1}=g_{(2g)^{2}-2}^{2}\rangle
	\end{equation*}
	\begin{equation*}
	D_{w}:=\langle K_{(2g)^{2}-7}\ast Q\mid g_{(2g)^{2}-3}=g_{(2g)^{2}-4},\ \  g_{(2g)^{2}-1}=[w,x]\rangle
	\end{equation*}
where $w$ is a word in the generators $a$ and $b$ of $H$. Note that, given a presentation for $H$, it is possible to write down a presentation $D_{w}=\langle g_{1}, \ldots, g_{(2g)^{2}-1}\mid r_{1}, \ldots, r_{M}\rangle$ for $D_{w}$.\\

By Lemma \ref{wellknownfact}, $D_{w}$ is isomorphic to the deck transformation group of a cover $S_{2}'$ of $S_{2}$, as follows:
\begin{equation*}
\langle g_{1}, \ldots, g_{(2g)^{2}-1}, d_{i}(g_{1}), \ldots, d_{i}(g_{(2g)^{2}-1}) \mid r_{k}(g_{1}, \ldots, g_{(2g)^{2}-1}),  d_{j}(g_{1})=d_{j}(g_{2})=\ldots=d_{j}(g_{(2g)^{2}-1})=1\rangle
\end{equation*}
where $i=1, \ldots, (2g)^{2}-1$,  $j=2, \ldots,  (2g)^{2}-1$, $k=1, \ldots, M$, and $d_{i}$ is an element of the deck transformation group of the cover $S_{2}\rightarrow S$.\\

Claim: The following are equivalent:
\begin{enumerate}
\item{$w=1\in H$}
\item{$S_{2}' \rightarrow S_{2}$ is the trivial cover.}
\end{enumerate}
To show that 1 implies 2, recall that by construction, $[w,x]$ is only trivial in the deck transformation group when $w$ is the trivial word in $H$. When $w$ is the trivial word in $H$, this implies $g_{(2g)^{2}-1}$ is trivial, from which it then follows that $g_{(2g)^{2}-2}$ is trivial, and hence $g_{(2g)^{2}-3}, \ldots, g_{1}$ are all trivial. In this case $S_{2}^{1}$ is the trivial cover of $S_{2}$. \\

To show that 2 implies 1, when $w$ is not the trivial word in $H$, $D_{w}=\langle g_{1}, \ldots, g_{(2g)^{2}-1}\mid r_{1}, \ldots, r_{M}\rangle$ is a nontrivial group. This follows from normal form theorems for HNN extensions and free products with amalgamation arising from the work of \cite{HNN}. A reference is Section 2 of Chapter IV of \cite {LS}. These normal form theorems also imply that none of the generators $g_{1}, \ldots, g_{(2g)^{2}-1}$ are trivial in $D_{w}$. This proves the claim.\\

Now please note that $S_{2}'$ is a regular cover of $S_{2}$, but $\pi_{1}(S_{2}')$ is not a normal subgroup of $\pi_{1}(S)$.
We now want to find a regular cover $\tilde{S}_{2}$ of $S$ factoring through $S_{2}'$. A necessary condition is that the deck transformation group of the cover $\tilde{S}_{2}\rightarrow S_{2}$ is invariant under the induced action of the deck transformation group of the cover $S_{2}\rightarrow S$. Without this condition, it is not possible to find a deck transformation group, hence the cover could not be regular. We therefore consider the covering spaces $S_{2,i}$ of $S_{2}$, where $S_{2,i}$ has deck transformation group $\tilde{D}_{w, i}$ given by
	\begin{equation*}
	\label{compose}
	\langle d_{j}(g_{1}), \ldots, d_{j}(g_{(2g)^{2}-1}) \mid r_{k}(d_{i}(g_{1}), \ldots, d_{i}(g_{(2g)^{2}-1})),  d_{j}(g_{1})=\ldots=d_{j}(g_{(2g)^{2}-1})=I\rangle
	\end{equation*}
where $i=0,1,\ldots, (2g)^{2}-1$, the index $j=0,1, \ldots, i-1, i+1, \ldots,  (2g)^{2}-1$, the index $k=1, \ldots, M$, and $d_{i}$ is an element of the deck transformation group of the cover $S_{2}\rightarrow S$, where $d_{0}=1$.\\

The group $\pi_{1}(\tilde{S}_{2})$ is defined to be the largest normal subgroup of $\pi_{1}(S)$ in the intersection of all the groups $\pi_{1}(S_{2,i})$. Denote by $N$ the intersection in $\pi_{1}(S)$ of the groups $\pi_{1}(S_{2,i})$. To show that $\pi_{1}(\tilde{S}_{2})$ is nontrivial, we first show that $N$ is nonempty. This is a consequence of Lemma \ref{wellknownfact}, as words of the form $r_{k}(d_{i}(g_{1}), \ldots, d_{i}(g_{(2g)^{2}-1}))$, for $i=0,1,\ldots, (2g)^{2}-1$ and $k=1, \ldots, M$ are in the intersection.\\

To show that $N$ contains a nontrivial normal subgroup, first note that $N$ is invariant under conjugation by elements $\{a_{i}, b_{j}\}$ that are mapped to generators of the deck transformation group. This is because conjugation by such elements merely permutes the groups whose intersection is $N$. Since $N$ is invariant under conjugation by the generators of $\pi_{1}(S)$, it follows that $N$ is normal in $\pi_{1}(S)$, and hence $N=\pi_{1}(\tilde{S}_{2})$.\\

Given a presentation for $D_{w}$, a presentation for $\pi_{1}(\tilde{S}_{2})$ can be obtained. The relations of this presentation consist of the relations of $D_{w}$, the conjugates by $\{a_{i}, b_{j}\}$ of the relations of $D_{w}$, and the image of the relation $\Pi_{i=1,\ldots, g}[a_{i},b_{i}]$ in $\pi_{1}(S)$.\\

By construction, $\pi_{1}(\tilde{S}_{2})$ is a nontrivial cover of $S_{2}$ if and only if $w$ is a nontrivial word in $H$. In the former case, $\tilde{S}_{2}\rightarrow S$ is the cover $S_{2}\rightarrow S$, which we have seen does contain simple curves. In the latter case, none of the words in $\pi_{1}(\tilde{S}_{2})$ represent simple curves on $S$, as these words are all in $[\pi_{1}(S), [\pi_{1}(S), \pi_{1}(S)]]$ or are representatives of nonprimitive homology classes. \\

The cover $\tilde{S}_{2}\rightarrow S$ therefore has simple curves if and only if $w$ is a trivial word in $H$. An algorithm deciding whether covers have simple curves or not could therefore be used to solve the word problem for finitely presented groups. For connected, orientable surfaces with empty boundary, the theorem follows by contradiction.	\\

If $S$ has nonempty boundary, an identical construction proves the theorem, with one small difference being that a generating set for $\pi_{1}(S)$ would then consist of curves $\{a_{i}, b_{j}\}$ as before, as well as some curves freely homotopic to boundary curves. Also, by first factoring through a double cover, the above construction applies when $S$ is not orientable. Similarly, the proof can be generalised to surfaces that are not connected in the obvious way.
\end{proof}

\bibliography{bib}
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\end{document}

config: arxiv


TITLE: prove the L’Hospitals rule without Cauchy's Mean Value Theorem
QUESTION [1 upvotes]: I am trying to prove the L’Hospitals rule in the following using the mean value theorem of the differential calculus instead of Cauchy's Mean Value Theorem(the generalized mean value theorem of differential calculus).
Consider two functions $f(x)$ and $g(x)$, continuous on a closed interval $[a, b]$ of the X-axis, and differentiable on the interior of that interval. We assume that $g'(x)$ is positive and $f(a) = g(a) = 0$. The ordinary mean value theorem of differential calculus applied separately to $f(x)$ and $g(x)$ furnishes the expression:
$$\frac{f(x) - f(a)}{g\left( x \right) - g\left( a \right)}=\frac{f'(c_1)}{g^{'}\left( c_2 \right)}.$$
where $c_1$ and $c_2$ are suitable intermediate values in the open interval $(a, x)$.
After taking limit on both sides and substituting $f(a) = g(a) = 0 $,  one got
$$\lim _{x\rightarrow a+}\frac{f(x) - f(a)}{g\left( x \right) - g\left( a \right)}=\lim _{x\rightarrow a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}$$
However, L’Hospital’s Rule needs $$\lim _{x\rightarrow a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g^{'}\left( x \right)}$$
so I wonder whether $\lim _{x\rightarrow a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g^{'}\left( x \right)}$ hold in this case ?

Some progress I have made in solving the problem :

For every x, there always exist $c_1$ and $c_2$ in (a, x), so one
can denote  $c_1=m(x)$ and $c_2=n(x)$, then $$\lim _{x\rightarrow
a+}\frac{f(x)}{g\left( x \right)}=\lim _{x\rightarrow
a+}\frac{f'(c_1)}{g^{'}\left( c_2 \right)}=\lim _{x\rightarrow
a+}\frac{f'(m(x))}{g^{'}\left( n(x) \right)}$$(As @Bernard pointed out in the comment - if there are several $c_1$s
or  $c_2$s in $(a, x)$,  then I would choose any one of them, then  $c_1$ and $c_2$ are functions of $x$ .), so the problem
becomes : whether $\lim _{x\rightarrow
a+}\frac{f'(m(x))}{g^{'}\left( n(x) \right)}=\lim _{x\rightarrow
a+}\frac{f'(x)}{g^{'}\left( x \right)}$ hold in this case ?
As $x$ approaches $a$, $c_1$ and $c_2$ also approach $a$ because it
always lies between $a$ and $x$.

REPLY [0 votes]: Since we have the same restrictions of domain for $f, g$ we can say that without loss of generality that $c_1 = c_2$.
Fully the proof would be as follows:
If $x \in (a,b)$, then by the Mean Value Theorem applied to our given restrictions of $f,g$ to $[a, x]$, there exists some $\xi_x \in (a,x)$ with $$\frac{f'(\xi_x )}{g'(\xi_x )}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f(x)}{g(x)}$$
Since $\xi_x \rightarrow a$ as $x\rightarrow a_+$ we have $$\lim_{x\rightarrow a_+} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a_+} \frac{f'(\xi_x)}{g'(\xi_x)}$$

REPLY [0 votes]: You have to use Cauchy's Mean Value Theorem; that is
$$\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c)}{g'(c)}$$
for some $c$ such that $a<c<x$. Thus when $x\to a^+$, one has $\to a^+$. So
$$\lim _{x\rightarrow a+}\frac{f(x)}{g(x)}=\lim _{x\rightarrow a+}\frac{f'(c)}{g'(c)}=\lim_{c\rightarrow a+}\frac{f'(c)}{g'(c)}=\lim _{x\rightarrow a+}\frac{f'(x)}{g'(x)}.$$


TITLE: Normal distribution, easy question
QUESTION [0 upvotes]: I have a little doubt: if I have the following distribution $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(|x|-\mu)^2}$$
$1)$ can I say that it correspond to the $N(\mu,\sigma)$, also if I have |x| (but integral starting from $0$).
$2)$ for solving the integral how could I procede: I could start in this way
$$2\int_{0}^\infty \frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(x-\mu)^2} $$ but from here I can only deduce that the previous is $\le1$, right?

REPLY [0 votes]: If you are asking whether $$\frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(|x|-\mu)^2} I_{x>0}$$ is the density of a normal random variable, it is not. This density vanishes on $(-\infty, 0)$ so it is not normal.
$\int_0^{\infty} e^{-\frac 1  {2\sigma} (x-\mu)^{2}}dx=\int_{-\mu}^{\infty} e^{-\frac 1  {2\sigma} x^{2}}dx$ which can be expressed in terms of the standard normal distribution function $\Phi (x)$: it is $\sqrt {\frac 2  {\pi}} [{1-\Phi(-\frac {\mu} {\sqrt \sigma})}]$.


TITLE: Any local maximum of F in the interior of a simplex is also a global maximum?
QUESTION [1 upvotes]: Let G be a graph with vertices $1,2,...,n$ and $F(x)=\displaystyle{\sum_{ij\in\rm{E(G)}}x_ix_j}$.
Let S be the subset of $\mathbb{R}^n$ given by $x_i\ge 0$, $\sum x_i=1$. We're interested in $\displaystyle{\max_{x\in S}}$ $F(x)$.
Why is any local maximum of F in the interior of S also a global maximum?

I was reading this paper, http://math.ca/cjm/v17/cjm1965v17.0533-0540.pdf.
 In the middle of the third page, there is a remark saying:
Any local maximum of F in the interior of S is also a global maximum.
I don't see why this is true. Is this obvious? Or well-known?

REPLY [1 votes]: Does the problem actually depend at all on the graph theoretic context, or are we just asking about the sum of any subset of products of coordinate pairs? And is there any way for a quadratic function on a connected open linear manifold to have more than one isolated local extremum?


TITLE: Laplace Transform - Why it works
QUESTION [4 upvotes]: The definition of the Laplace Transform is $$F(s) = \int^{\infty}_0 f(t)e^{-st}  dt$$
It is very useful in terms of solving linear, constant coefficient ordinary differential equations, but why exactly does it work? Why does taking the Laplace transform of each term in in the differential equation, using the Laplace Transform's linearity to get each individual term in the differential equation in its own Laplace Transform work in solving differential equations?
Why does it work?
I am not so sure why I am hesitant to accept this. Having seen generating functions of recursive sequences as a way to describe the sequence of numbers  and aids in obtaining a (closed) formula for the recursive sequence - this is a similar analogy, we are essentially "transforming" sequences to a power series, but I am more accepting of this idea than the Laplace transform.

REPLY [0 votes]: The two key properties that make Laplace transform so useful for solving certain ODEs are linearity and the relationship between the Laplace transform of a function and its derivative https://en.wikipedia.org/wiki/Laplace_transform $$\mathcal{L} \left\{f(t)\right\} = \int_{0^-}^\infty e^{-st} f(t)\, dt \\
                                 \left[\frac{f(t)e^{-st}}{-s} \right]_{0^-}^\infty -
                                       \int_{0^-}^\infty \frac{e^{-st}}{-s} f'(t) \, dt \\
                                   \left[-\frac{f(0^-)}{-s}\right] + \frac 1 s \mathcal{L} \left\{f'(t)\right\}$$ where use is made of integration by parts.
By this property, applied to each term of a linear, constant-coefficient ODE, you can communte for a differential equation to an algebraic one, hopefully easier to solve.
So, this is the "mechanics" of it: apply the transform to both sides of an ODE, use linearity, and convert from derivatives to powers.
As to "why" it works, one is tempted to say that it does so, as a result of the fact each operation above can be shown to be valid (although there might be a "circular argument" flavour hanging around here).
If you convicne yoursleve that linearity and the above relationship between transforms of a function and its derivative apply, by looking at the simple proofs, there should not be additional conceptual obstacles towards appreciating how it works. If I missed some point I would gladly expand upon it.


TITLE: Field of definition of dominant morphisms
QUESTION [8 upvotes]: Let $k$ be an algebraically closed field and $k_0$ a sub-field. Let $X,Y$ be two projective varieties defined over $k_0$. Suppose that that there exists a dominant morphism $f$ between $X_k=X\otimes k$ and $Y_k=Y\otimes k$.
Such a dominant morphism cannot always been defined over a finite extension of $k_0$. Take by exemple an elliptic curve $X=Y$ defined over $k_0=\mathbb{Q}$, $k=\mathbb{C}$ and $f$ a translation by a point not in the algebraic closure of $\mathbb{Q}$.
But it is still natural to ask the following:
With the above hypothesis on the existence of $f$, do there always exists a dominant morphism $g$ between $X$ and $Y$ defined over a finite extension $k_1$ of $k_0$ ?
According to the appendix 1 of Mumford’s book « Abelian Varieties », the answer is yes if $X$ and $Y$ are Abelian varieties. For the proof, using a translation, one reduces to the case of an homomorphism. Then the (dense set of)  torsion elements of $X$ are sent to torsion elements of $Y$, therefore the homomorphism is defined over a finite extension.
The last implication is a bit obscure for me, maybe the following can help:
field of definition of isogenies of abelian varieties
?
Anyway I would like to understand the general case, in which one can also replace the word "dominant" by "isomorphism" or "birational morphism"...

REPLY [9 votes]: The general technique goes as follows.  Let $\{A_i\}$ be a directed system of rings with limit $A$; e.g., $A=k$ and $\{A_i\}$ the set of finitely generated $k_0$-subalgebras of $k$.  Let $X$ and $Y$ be schemes of finite presentation over some $A_{i_0}$, and define $X_i = X \otimes_{A_{i_0}} A_i$ for $i \ge i_0$, $X_A = X \otimes_{A_{i_0}} A$, and similarly for $Y_i$ (with $i \ge i_0$) and $Y_A$.  There is a natural map of sets
$$\varinjlim {\rm{Hom}}_{A_i}(X_i,Y_i) \rightarrow {\rm{Hom}}_A(X_A,Y_A)$$
and it is bijective by EGA IV$_3$, 8.8.2(i).
So given an $A$-morphism $f:X_A \rightarrow Y_A$, there exists some $i_1$ and an $A_{i_1}$-morphism $f_{i_1}:X_{i_1} \rightarrow Y_{i_1}$ which descends $f$.  But we want more: for various properties $\mathbf{P}$ of morphisms of schemes, if $f$ satisfies $\mathbf{P}$ then we want $f_i := f_{i_1} \otimes_{A_{i_1}} A_i$ to also satisfy $\mathbf{P}$ for some $i \ge i_1$.  The property "isomorphism" is immediate by applying the preceding formalism to the inverse $A$-morphism too (using bijectivity in the displayed map of sets above).  See IV$_3$ 8.10.5 for the tip of the iceberg on many possibilities for $\mathbf{P}$. By IV$_3$, 8.6.3, open subschemes of $X_A$ with finitely presented complement also descend to such open subschemes of some $X_i$.
For $i \ge i_1$, the (set-theoretic) image
$Z_i = f_i(X_i) \subset Y_i$ is constructible and if $j \ge i$ then $Z_j$ is the preimage of $Z_i$ under $Y_j \rightarrow Y_i$ and likewise $f(X_A)$ is the preimage of $Z_i$ in $Y_A$. Define $E_i$ to be the set of of $s \in S_i := {\rm{Spec}}(A_i)$ such that $(Z_i)_s$ is dense in $(X_i)_s$,
and likewise for an analogous subset $E \subset S := {\rm{Spec}}(A)$. The density or not of a constructible subset of a scheme of finite type over a field is insensitive to extension of the ground field, so  for $j \ge i$ the preimage of $E_i$ under $S_j \rightarrow S_i$ is $E_j$ (due to the analogue for $Z_i \subset Y_i$ and $Z_j \subset Y_i$ under $Y_i \rightarrow Y_i$ noted above) and likewise the preimage of $E_i$ under $S \rightarrow S_i$ is $E$. But $E_i$ is constructible in $S_i$ by IV$_3$ 9.5.3 and likewise for $E$ inside $S$. Thus, if $E=S$ then $E_i = S_i$ for all large $i$ due to IV$_3$ 8.3.5.
Thus, in the initial setup of interest, if $X_k \rightarrow Y_k$ is dominant then we get a finitely generated $k_0$-subalgebra $R \subset k$ and an $R$-morphism $h:X_R \rightarrow Y_R$ which is dominant between fibers over all points of Spec($R$).
In a similar manner, using the "spreading out" for open subschemes (with finitely presented complement) mentioned above, if there are dense open subschemes of $X_k$ and $Y_k$ between which $f$ restricts to an isomorphism, then by increasing $R$ further we can arrange that there exist fiberwise dense open subschemes $U \subset X_R$ and $V \subset Y_R$ between which $h$ restricts to an isomorphism.
Finally, thanks to the Nullstellensatz over $k_0$, using the pullback of  $h$ between fibers over any closed point of such a Spec($R$) allows us to conclude for the properties of dominance, isomorphism, or birational morphism.
This overall method is sometimes called the "Principle of finite extensions" (i.e., whatever happens after some extension of the ground field already happens over a finite extension), and complete proofs of virtually every such property you could ever imagine wanting is rigorously documented in remarkable and useful generality in EGA IV$_3$, $\S8$-$\S11$ (skip $\S10$) and IV$_4$, $\S17$-$\S18$.


text:
\begin{document}

\title{Compact differences of composition operators}

\author{Katherine Heller}
\address{Katherine Heller, Department of Mathematics, P. O. Box 400137,
University of Virginia, Charlottesville, VA 22904}
\email{kch3c@virginia.edu}

\author{Barbara D. MacCluer}
\address{Barbara D. MacCluer, Department of Mathematics, P.O. Box 400137,
University of Virginia, Charlottesville, VA 22904}
\email{bdm3f@virginia.edu}

\author{Rachel J. Weir}
\address{Rachel J. Weir, Department of Mathematics,
Allegheny College, Meadville, PA 16335}
\email{rweir@allegheny.edu}

\thanks{The third named author would like to thank the Allegheny College Academic Support Committee
for funding provided during the development of this paper.}

\date{June 8, 2010}

\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}

\newtheorem{remark}{\bf Remark}[section]
\newtheorem{example}[remark]{\bf Example}
\newtheorem{definition}[remark]{\bf Definition}

\newcommand{\p}{\varphi}
\newcommand{\cp}{C_{\varphi}}
\newcommand{\D}{{\mathbb D}}
\newcommand{\B}{{\mathcal B}}
\newcommand{\com}{{[C_{\varphi}^*,C_{\varphi}]}}

 \newcommand{\ebox}{\hspace*{1em}\linebreak[0]\hspace*{1em}\hfill\rule[-1ex]{
.1in}{.1in}\\}
 \newcommand{\pf}{\noindent \bf Proof.\rm\ \ }

\begin{abstract}
When $\varphi$ and $\psi$ are linear-fractional self-maps of the unit
ball $B_N$ in ${\mathbb C}^N$, $N\geq 1$, we show that the difference
$C_{\varphi}-C_{\psi}$ cannot be non-trivially compact on either the
Hardy space $H^2(B_N)$ or any weighted Bergman space $A^2_{\alpha}(B_N)$.
Our arguments emphasize geometrical properties of the inducing maps
$\varphi$ and $\psi$.
\end{abstract}

\maketitle

\section{Introduction}
For a domain $\Omega$ in ${\mathbb C}^N$, where $N\geq 1$, and an analytic
map $\varphi:\Omega\rightarrow \Omega$, we define the composition
operator $C_{\varphi}$ by $C_{\varphi}(f)=f\circ\varphi$, where $f$ is analytic
in $\Omega$.  In the case that $\Omega=\D$, the unit disk in ${\mathbb C}$,
every composition operator acts boundedly on the Hardy space
$$H^2(\D)=\left\{f\mbox{ analytic in }\D:\|f\|_{H^2}^2\equiv\lim_{r\rightarrow 1^-}\int_0^{2\pi}|f(re^{i\theta})|^2\frac{d\theta}{2\pi}<\infty\right\}$$
and weighted Bergman spaces
$$A^2_{\alpha}(\D)=\left\{f\mbox{ analytic in }\D:\|f\|_{\alpha}^2\equiv(\alpha+1)\int_{\D}|f(z)|^2(1-|z|^2)^{\alpha}dA(z)<\infty\right\},$$
where $\alpha>-1$ and $dA$ is normalized area measure.
If $\D$ is replaced by the unit ball $B_N$ in ${\mathbb C}^N$, $N>1$, it is no longer the
case that every composition operator is bounded on the Hardy or weighted Bergman space
of the ball (these spaces are defined in Section~3). However for a large class of maps $\varphi$,
including the rich class of linear-fractional maps, boundedness does continue to hold.

Many properties of composition operators have been studied over the past four decades; the
monographs \cite{cmbook} and \cite{Sbook} give an overview of the work before the mid-1990's.
Recently there has been considerable interest in studying algebras of composition operators,
often modulo the ideal of compact operators (see, for example, \cite{jury1}, \cite{jury2}, \cite{kmm1},
\cite{kmm}, \cite{kmm2}).  In this direction, the question of when a difference
$C_{\varphi}-C_{\psi}$ is compact naturally arises.

The main result of this paper shows that if $\varphi$ and $\psi$ are linear-fractional
self-maps of $B_N$, then $C_{\varphi}-C_{\psi}$ cannot be non-trivially compact; i.e.
if the difference is compact,
either $C_{\varphi}$ and $C_{\psi}$ are individually compact (this happens precisely
when $\|\varphi\|_{\infty}<1$ and $\|\psi\|_{\infty}<1$), or $\varphi=\psi$.
While our focus is on the several variable case,
we begin with a simplified proof of this result in one variable.
The fact that a difference of linear-fractional
composition operators cannot be non-trivially compact on
$H^2(\D)$ or $A^2_{\alpha}(\D)$ was first obtained by
P. Bourdon \cite{bour} and J. Moorhouse \cite{moor} as a consequence of
results on the compactness of a difference of more general composition
operators in one variable.  Our approach here is self-contained, and
takes a geometric perspective,
which will allow us to generalize our arguments to several variables.
The analogy between the one and several variable arguments is not perfect,
owing to a number of phenomena that are present when $N>1$
but do not occur when $N=1$.  Nevertheless, our geometric approach when
$N=1$ leads us to a tractable way to proceed when $N>1$, and highlights the
new phenomena which must be addressed.
Since our arguments
are essentially the same for either the Hardy or weighted Bergman spaces,
in what follows we will let ${\mathcal H}$ denote any of these spaces.
Our starting point, in either the disk or the ball, will be the following
necessary condition for compactness of $C_{\varphi}-C_{\psi}$.

\begin{thm}\label{phdist}[\,\cite{mac-weir},\cite{moor}\,]
Suppose $\varphi, \psi$ are holomorphic self-maps of $\D$ (respectively, $B_N$) and suppose
that there exists a sequence of points $z_n$ tending to the boundary
of  $\D$ ($B_N$) along which
\begin{equation}\label{phexpress}
\rho(\varphi(z_n), \psi(z_n))\left(\frac{1-|z_n|^2}{1-|\varphi(z_n)|^2}+\frac{1-|z_n|^2}{1-|\psi(z_n)|^2}\right)
\end{equation}
does not converge to zero, where $\rho(\varphi(z_n), \psi(z_n))$ is defined by
\begin{equation}\label{defofrho}
1-(\rho(\varphi(z_n), \psi(z_n)))^2=\frac{(1-|\varphi(z_n)|^2)(1-|\psi(z_n)|^2)}{|1-\langle\varphi(z_n),\psi(z_n)\rangle|^2}.
\end{equation}
Then $C_{\varphi}-C_{\psi}$ is not compact on ${\mathcal H}$.
\end{thm}

For $z$ and $w$ in $\D$ or $B_N$, the quantity $\rho(z,w)$ will be referred to as the pseudohyperbolic distance
between $z$ and $w$,
so that the first factor in~(\ref{phexpress}) is the pseudohyperbolic
distance between $\varphi(z_n)$ and $\psi(z_n)$.  In the disk, the pseudohyperbolic
distance has the simpler expression
\begin{equation}\label{phdisksimp}
\rho_{\D}(z, w)=\left|\frac{z-w}{1-\overline{z}w}\right|.
\end{equation}

 \section{Results in one variable}

Throughout this section ${\mathcal H}$ denotes either the Hardy space $H^2(\D)$ or
a weighted Bergman space $A^2_{\alpha}(\D)$, as defined in the previous section.

\begin{thm}\label{mainforD}
Suppose that $\varphi$ and $\psi$ are linear-fractional self-maps of $\D$.
If the difference $C_{\varphi}-C_{\psi}$ is compact on ${\mathcal H}$, then
either $\varphi=\psi$, or both $\|\varphi\|_{\infty}$ and $\|\psi\|_{\infty}$
are strictly less than $1$, so that $C_{\varphi}$ and $C_{\psi}$ are individually compact.
\end{thm}

The key step in our proof of Theorem~\ref{mainforD} is contained in the following
result.

\begin{thm}\label{Dsamefirst}
Suppose $\varphi$ and $\psi$ are non-automorphism linear-fractional self-maps of $\D$ with
$\varphi(\zeta)=\psi(\zeta)\in\partial\D$ and $\varphi'(\zeta)=\psi'(\zeta)$
for some $\zeta\in\partial \D$.
If  $\varphi\neq \psi$ then $C_{\varphi}-C_{\psi}$ is not compact on ${\mathcal H}$.
\end{thm}
\begin{proof}
By pre- and post- composing with rotations, we may assume
without loss of generality that $\zeta =1$ and $\varphi(\zeta)=\psi(\zeta)=1$.
Since $\varphi$ and $\psi$ are linear-fractional, we may also assume without loss
of generality that $\varphi(\D)\subseteq\psi(\D)$, so that $\tau_1\equiv\psi^{-1}\circ\varphi$
is a well-defined linear-fractional self-map of $\D$.  Note that $\tau_1(1)=1$ and $\tau_1'(1)=1$.
Since $\varphi\neq \psi$, $\tau_1$ is not the identity.  Thus $\tau_1$ is conjugate
via the Cayley transform
$$C(z)=i\frac{1+z}{1-z}$$
to a translation $w\rightarrow w+b$ of the upper half-plane ${\mathbb H}=\{w:\mbox{Im }w>0\}$ for some
$b\neq 0$ with $\mbox{Im }b\geq 0$.
Moreover, it is easy to see that
$$\psi\circ\tau_1=\tau_2\circ\psi$$ for some
linear-fractional $\tau_2$ which is also conjugate to a translation in the upper half-plane.
Specifically, if $\tau_1$ is conjugate to translation by $b$, then $\tau_2$
is conjugate to translation by $c=b/|\psi'(1)|$; see Lemma 5 in \cite{kmm}.
Since $b\neq 0$, so also $c\neq 0$.

For any positive number $k$, the line
$\{\mbox{Im }w=k\}$ corresponds under the Cayley transform
to $E_k\equiv\{z:|1-z|^2=\frac{1}{k}(1-|z|^2)\}$, which is an
internally tangent circle
in $\D$ passing through $1$.  The radius of this circle is equal to $(k+1)^{-1}$.
By choosing $k$ sufficiently large, this circle will be contained in $\psi(\D)\cup \{1\}$.
Fix such a $k$ and choose points $w_n$ on $\{\mbox{Im }w=k\}$ with $w_n\rightarrow\infty$.
The corresponding points $v_n=C^{-1}(w_n)$ in the disk tend to $1$ along the circle
$E_k$, and each $v_n$ is the image under
$\psi$ of some $z_n$ belonging to
the internally tangent circle $\psi^{-1}(E_k)=E_{k'}$.
Notice that $z_n\rightarrow 1$ as $n\rightarrow\infty$.

Next we compute the pseudohyperbolic distance between $\varphi(z_n)$ and $\psi(z_n)$.
To simplify the computations, we define the pseudohyperbolic distance
in the upper half-plane ${\mathbb H}$ by
$$\rho_{\mathbb H}(u,v)=\rho_{\D}(C^{-1}u,C^{-1}v)$$
for $u$ and $v$ in ${\mathbb H}$.  Using this definition and ~(\ref{phdisksimp}) it is straightforward
to see that
$$\rho_{\mathbb H}(u,v)=\left|\frac{u-v}{u-\overline{v}}\right|.$$
Since $\varphi=\psi\circ\tau_1=\tau_2\circ\psi$,
\begin{eqnarray*}
\rho_{\D}(\varphi(z_n),\psi(z_n))&=&\rho_{\D}(\tau_2(\psi(z_n)),\psi(z_n))=\rho_{\D}(\tau_2(v_n),v_n)\\
&=&\rho_{{\mathbb H}}(C(\tau_2(v_n)),C(v_n))=\rho_{{\mathbb H}}(C(\tau_2(C^{-1}(w_n)),CC^{-1}(w_n))\\&=&
\rho_{{\mathbb H}}(w_n+c,w_n)\\&=&\left|\frac{c}{2i\mbox{Im }w_n+c}\right|=\left|\frac{c}{2ik+c}\right|.
\end{eqnarray*}
Thus for all $n$, the pseudohyperbolic distance between $\varphi(z_n)$ and $\psi(z_n)$ is
a positive constant.

Turning to the second factor in Equation~(\ref{phexpress}), we have
$$\frac{1-|z_n|^2}{1-|\psi(z_n)|^2}=\frac{k'(|1-z_n|^2)}{k(|1-\psi(z_n)|^2)}$$
by the geometry of the sequence $\{z_n\}$ already noted.
Thus since $\psi$ is differentiable at $1$ with $\psi'(1)\neq 0$ and $\psi(1)=1$,
$$\lim_{n\rightarrow\infty}\frac{1-|z_n|^2}{1-|\psi(z_n)|^2}=\frac{k'}{k|\psi'(1)|^2}\neq 0.$$
Thus we have shown that
$$\rho(\varphi(z_n),\psi(z_n))\left(\frac{1-|z_n|^2}{1-|\psi(z_n)|^2}\right)$$
has a positive limit as $n\rightarrow 1$ (where $z_n\rightarrow 1$).
Theorem~\ref{phdist} guarantees that $C_{\varphi}-C_{\psi}$ is not compact on ${\mathcal H}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{mainforD}.]

If either $\|\varphi\|_{\infty}<1$ or $\|\psi\|_{\infty}<1$, then the compactness
of the difference $C_{\varphi}-C_{\psi}$ implies the compactness of each operator individually.
Thus we may assume $\|\varphi\|_{\infty}=\|\psi\|_{\infty}=1$.  Suppose $\varphi(\zeta)=\eta$
where $\zeta,\eta$ are in $\partial \D$.  Since $\varphi$ and $\psi$ are linear fractional,
both $\varphi'(\zeta)$ and $\psi'(\zeta)$ exist and are non-zero.  If either $\varphi(\zeta)\neq \psi(\zeta)$
or $\varphi'(\zeta)\neq \psi'(\zeta)$, then by Theorem 9.16 of \cite{cmbook} the essential norm
of $C_{\varphi}-C_{\psi}$ satisfies
$$\|C_{\varphi}-C_{\psi}\|_e^2\geq |\varphi'(\zeta)|^{\beta}$$
for some positive number $\beta$ depending on the particular choice of ${\mathcal H}$ in question;
when ${\mathcal H}=H^2(\D)$ we may take $\beta=1$, and when ${\mathcal H}=A^2_{\alpha}(\D)$, $\beta=\alpha+2$.
This gives a positive lower bound on the essential norm of the difference,
so that if the difference is compact we must have $\varphi(\zeta)=\psi(\zeta)$ and
$\varphi'(\zeta)=\psi'(\zeta)$.  Note that this argument also shows that if
$C_{\varphi}-C_{\psi}$ is compact but non-zero, neither $\varphi$ nor
$\psi$ can be an automorphism. An appeal
to Theorem~\ref{Dsamefirst} finishes the proof.

\end{proof}

\section{Results in several variables}

In this section
${\mathcal H}$ will denote either the Hardy space $H^2(B_N)$ or a weighted Bergman space $A^2_{\alpha}(B_N)$,
where $B_N$ is
the ball
$$\{(z_1,z_2,\cdots,z_N)\in {\mathbb C^N}:\sum_{j=1}^N|z_j|^2<1\}$$
in ${\mathbb C}^N$, $N>1$.
These Hilbert spaces are defined by
$$H^2(B_N)=
\{f\mbox{ analytic on }B_N:\|f\|^2\equiv
\sup_{0<r<1}\int_{\partial B_N}|f(r\zeta)|^2d\sigma(\zeta)<\infty\},
$$
where $\sigma$ is normalized Lebesgue surface area measure on $\partial B_N$,
and for $\alpha>-1$,
$$A^2_{\alpha}(B_N)=\{f\mbox{ analytic on }B_N:\|f\|_{\alpha}^2\equiv
\int_{B_N}|f(z)|^2w_{\alpha}(z)d\nu(z)<\infty\},$$
where
$$w_{\alpha}(z)=\frac{\Gamma(N+\alpha+1)}{\Gamma(N+1)\Gamma(\alpha+1)}(1-|z|^2)^{\alpha},$$
and $\nu$
is normalized Lebesgue volume measure on $B_N$.
In $H^2(B_N)$ the reproducing kernel for the bounded linear
functional of evaluation at $w\in B_N$ is
\begin{equation}\label{H2ker}
K_w(z)=\frac{1}{(1-\langle z,w\rangle)^N}
\end{equation}
which has norm $(1-|w|^2)^{N/2}$.
The reproducing kernel for evaluation at $w$
in $A^2_{\alpha}(B_N)$ is
\begin{equation}\label{A2ker}
K_w(z)=\frac{1}{(1-\langle z, w\rangle)^{N+1+\alpha}}.
\end{equation}

By a linear-fractional map of $B_N$ we mean a map that is
analytic in $B_N$ and of the form
\begin{equation}\label{defoflinfrac}
\varphi(z)=\frac{Az+B}{\langle z,C\rangle +d},
\end{equation}
where $A$ is an $N\times N$ matrix, $B$ and $C$ are $N\times 1$ column
vectors, $d$ is a complex scalar, and $\langle \cdot,\cdot \rangle$ is
the usual inner product on ${\mathbb C}^N$.  If $\varphi$ maps
$B_N$ into itself, then necessarily
$|d|>|C|$, so that in particular $d\neq 0$, and $\varphi$ is
analytic in a neighborhood of $\overline{B_N}$.  When
$\varphi$ is a linear-fractional self-map of $B_N$, $C_{\varphi}$ is
bounded on $H^2(B_N)$ and $A^2_{\alpha}(B_N)$ for
$\alpha>-1$ (\cite{cowenmac}).  Moreover, every automorphism of
$B_N$ is linear-fractional (\cite{Ru}, Theorem 2.2.5).
An important geometric property of linear-fractional maps is that
they take affine sets into affine sets (\cite{cowenmac}, Theorem 7).  By an affine set in ${\mathbb C}^N$ we
mean the translate of a complex subspace; the dimension of the affine set is
the dimension of the subspace.  An ``affine subset of $B_N$'' is the intersection
of $B_N$ with an affine set in ${\mathbb C}^N$.

The goal of this section
is to obtain the following result.

\begin{thm}\label{scvmain}
If $\varphi$ and $\psi$ are linear-fractional self maps of $B_N$ with
$C_{\varphi}-C_{\psi}$ compact on ${\mathcal H}$, then either
$\varphi=\psi$, or $\|\varphi\|_{\infty}$ and $\|\psi\|_{\infty}$ are both
strictly less than $1$, so that $C_{\varphi}$ and $C_{\psi}$ are compact
on ${\mathcal H}$.
\end{thm}

The key geometric arguments of the last section can be
extended to several variables.  The Cayley upper half space
${\mathbb H}_N$ is defined by
$${\mathbb H}_N=\{(w_1,w'):\mbox{Im }w_1>|w'|^2\}.$$
where $w'=(w_2,..., w_N)$ and $|w'|^2=|w_2|^2+\cdots+|w_N|^2$.
Its boundary is of course
$$\{(w_1,w'):\mbox{Im }w_1=|w'|^2\}.$$
Let $e_1=(1,0,\cdots,0)=(1,0')$.
The (linear-fractional) Cayley transform
$$C(z)=i\frac{e_1+z}{1-z_1}$$
is a biholomorphic map of the ball $B_N$ onto ${\mathbb H}_N$; its inverse
is
$$C^{-1}(w)=\left(\frac{w_1-i}{w_1+i},\frac{2w'}{w_1+i}\right).$$
If $b=(b_1,b')\in\mathbb{C}^N$, an ${\mathbb H}$-translation is a map of the form
\begin{equation}\label{h-translations}
h_b(w_1,w')=(w_1+2i\langle w',b'\rangle+b_1,w'+b').
\end{equation}
If $\mbox{Im }b_1\geq |b'|^2$ it maps  ${\mathbb H}_N$ into itself.
It is an automorphism of ${\mathbb H}_N$ if $\mbox{Im }b_1= |b'|^2$.
The following two facts, which generalize results we used in the
previous section, are easily checked:
\begin{itemize}
\item An ${\mathbb H}$-translation $h_b$  maps
the set $\Gamma_k\equiv\{(w_1,w'):\mbox{Im }w_1-|w'|^2=k\}$ into the corresponding
set $\{(w_1,w'):\mbox{Im }w_1-|w'|^2=\tilde{k}\}$
where
$\tilde{k}=k+\mbox{Im }b_1-|b'|^2$.
\item For any $k> 0$,
$$C^{-1}(\Gamma_k)= E(k,e_1)\equiv\{z\in B_N:|1-z_1|^2= \frac{1}{k}(1-|z|^2)\}.$$
\end{itemize}
The set $E(k,e_1)$ is an internally tangent ellipsoid at $e_1=(1,0')$;
a computation shows that $E(k,e_1)$ consists of the points $(z_1,z')$ satisfying
$$\left|z_1-\frac{k}{1+k}\right|^2+\frac{1}{1+k}|z'|^2=\left(\frac{1}{1+k}\right)^2.$$
In particular, for $t$ real, points of the form $(t+i(k+|w'|^2), w')$ in $\Gamma_k$ pull back under
$C^{-1}$ to points on the ellipsoid $E(k,e_1)$.  For fixed $w'$, these pull-back
points tend to $e_1$ as $t\rightarrow\infty$, and for fixed $t$, they tend to $e_1$
as $|w'|\rightarrow\infty$.

Recall that the pseudohyperbolic metric $\rho_{B_N}(\cdot,\cdot)$ on $B_N$ is defined by
$$1-\rho_{B_N}(z,w)^2=\frac{(1-|z|^2)(1-|w|^2)}{|1-\langle z,w\rangle|^2}.$$
For points $v, u$ in ${\mathbb H}_N$, write $\rho_{{\mathbb H}}(v,u)$ for
$\rho_{B_N}(C^{-1}v,C^{-1}u)$;
we will call this the pseudohyperbolic metric on ${\mathbb H}_N$.
Since the pseudohyperbolic
metric $\rho_{B_N}$ is easily seen to be automorphism invariant, it follows that
for any automorphism $\Lambda$ of ${\mathbb H}_N$,
$$\rho_{{\mathbb H}}(\Lambda v,\Lambda u)=\rho_{{\mathbb H}}(v,u).$$
In the next theorem, we will use this observation
with $\Lambda$
an automorphic ${\mathbb H}$-translation.

By a parabolic linear-fractional
map in $B_N$ fixing $e_1$ we mean a linear-fractional map $\tau$ of $B_N$ into $B_N$ with
$\tau(e_1)=e_1$ and $D_1\tau_1(e_1)=1$, but fixing no other point
in $\overline{B_N}$.
By \cite{braccietal} any parabolic linear-fractional self-map $\varphi$ of $B_N$ that fixes $e_1$
is conjugate to a self-map of ${\mathbb H}_N$ of the
form
$$\Phi(w_1,w')=(w_1+2i\langle w', \delta\rangle +b,Aw'+\gamma)$$
(where $\delta$ and $\gamma$ are in ${\mathbb C}^{N-1}$, $b\in {\mathbb C}$,
and certain conditions hold, including
$|A|\leq 1$).  Note that the ${\mathbb H}$-translations of Equation~(\ref{h-translations})
are a special case of this family of maps.
Conjugating $\Phi$ by the Cayley transform $C$ we see that the first coordinate function
of $C^{-1}\Phi C$ is
\begin{equation}\label{firstcoordpara}
\frac{(2i-b)z_1-2\langle z',\delta\rangle +b}
{-bz_1-2\langle z',\delta\rangle+2i+b}.
\end{equation}
We will need this explicit formula in the proof of the next result.

\begin{thm}\label{parabolics}
Suppose $\varphi$ and $\psi$ are parabolic linear-fractional self-maps of
the ball fixing $e_1$, so that
$$D_1\varphi_1(e_1)=1=D_1\psi_1(e_1).$$  If $\varphi\neq\psi$, then
$C_{\varphi}-C_{\psi}$ is not compact on ${\mathcal H}$.
\end{thm}
\begin{proof}

We will show that for distinct maps $\varphi$ and $\psi$ as in the hypothesis, there exists
a sequence of points $\{z^{(n)}\}$ in $B_N$ such that
\begin{itemize}
\item[(a)] $z^{(n)}\rightarrow e_1$ as $n\rightarrow \infty$.
\item[(b)] For all $n$, $\rho_{B_N}(\varphi(z^{(n)}),\psi(z^{(n)}))$ has a strictly positive constant value.
\item[(c)] $(1-|\varphi(z^{(n)})|^2)/(1-|z^{(n)}|^2)$ has finite positive limit as $n\rightarrow \infty$.
\end{itemize}
An appeal to Theorem~\ref{phdist} will then complete the proof.

We will use the corresponding upper case letters for a self-map of $B_N$ conjugated to ${\mathbb H}_N$,
so that $\Phi=C\varphi C^{-1}$ and $\Psi=C\psi C^{-1}$.  These maps have the forms
$$\Phi(w_1,w')=(w_1+2i\langle w',\delta_1\rangle+b_1,A_1w'+\gamma_1) $$
and
$$\Psi(w_1,w')=(w_1+2i\langle w',\delta_2\rangle+b_2,A_2w'+\gamma_2) $$
for some $\delta_i$ and $\gamma_i$ in ${\mathbb C}^{N-1}$, scalars $b_i$
and $(N-1)\times (N-1)$ matrices $A_i$, $i=1,2$.

Fix a sequence of points $\{w^{(n)}\}= \{(w_1^{(n)},c)\}$ in ${\mathbb H}_N$,
where $c$ is a constant in ${\mathbb C}^{N-1}$ to be determined,
satisfying
\begin{itemize}
\item[(i)] $\mbox{Im }w_1^{(n)}-|c|^2=k$
\item[(ii)] $C^{-1}(w_1^{(n)},c)\rightarrow e_1$
\end{itemize}
where $k>0$ is fixed but arbitrary.
By (i), the points $C^{-1}(w^{(n)})$ lie on the
ellipsoid
$E(k,e_1)$.
We can ensure that condition (ii) holds by requiring that $\mbox{Re}\ w_1^{(n)}\rightarrow
\infty$.

Let $$P_1^{(n)}=\Phi(w^{(n)})=(w_1^{(n)}+2i\langle c,\delta_1\rangle+b_1,A_1c+\gamma_1)
$$
and
$$P_2^{(n)}=\Psi(w^{(n)})=(w_1^{(n)}+2i\langle c,\delta_2\rangle+b_2,A_2c+\gamma_2).
$$
Since the pseudohyperbolic metric is automorphism invariant, we have
$$
\rho_{\mathbb H}(P_1^{(n)},P_2^{(n)})=\rho_{\mathbb H}(h(P_1^{(n)}),
h(P_2^{(n)}))$$
where $h$ is the automorphic ${\mathbb H}$-translation given by
$$h(v_1,v')=(v_1-\mbox{Re } w_1^{(n)},v').$$
Thus for any positive integer $n$,
the distance $\rho_{\mathbb H}(P_1^{(n)},P_2^{(n)})$ is equal to
$$\rho_{\mathbb H}((i(k+|c|^2)+2i\langle c,\delta_1\rangle+b_1,A_1c+\gamma_1),
(i(k+|c|^2)+2i\langle c,\delta_2\rangle+b_2,A_2c+\gamma_2)).
$$
Note that this quantity is independent of the particular point
$w^{(n)}$ in our sequence chosen to satisfy of
(i) and (ii), and that if $\varphi\neq \psi$
(so that not all of $b_1=b_2, \delta_1=\delta_2, A_1=A_2$ and $\gamma_1=\gamma_2$
hold) we may certainly choose $c$ so that this quantity is not $0$.
Thus for such a choice, condition (ii) gives the existence of a sequence of
points $\{z^{(n)}\}$ in $B_N$ tending to $e_1$ along $E(k,e_1)$ for which
$$\rho_{B_N}(\varphi(z^{(n)}),\psi(z^{(n)}))$$ is a positive constant value;
the $z^{(n)}$'s being just the inverse images under the Cayley transform
$C$ of our chosen
points $w^{(n)}$ in ${\mathbb H}_N$.  Hence conditions (a) and (b) hold.

For property (c), first note that the images under $\Phi$ of points of the form
$(w_1,c)$ with $\mbox{Im }w_1-|c|^2=k$  look like
$$(w_1+2i\langle c,\delta_1\rangle+b_1,A_1c+\gamma_1),$$
and for these points we see that
$$\mbox{Im }(w_1+2i\langle c,\delta_1\rangle+b_1)-|A_1c+\gamma_1|^2=k+|c|^2+
\mbox{Im}(2i\langle c,\delta_1\rangle+b_1)-|A_1c+\gamma_1|^2,$$
which is constant, say $k'$.  In other words, the image under $\varphi$ of our points
$z^{(n)}$ lie on some ellipsoid $E(k',e_1)$
and
$$|1-\varphi_1(z^{(n)})|^2= \frac{1}{k'}(1-|\varphi(z^{(n)})|^2).$$
Moreover, since the points $z^{(n)}$ lie on $E(k,e_1)$, we have
$$|1-z^{(n)}_1|^2= \frac{1}{k}(1-|z^{(n)}|^2).$$
Thus
\begin{equation}\label{relatingtoder}
\frac{1-|\varphi(z^{(n)})|^2}{1-|z^{(n)}|^2}=\frac{k'}{k}\frac{|1-\varphi_1(z^{(n)})|^2}{|1-z^{(n)}_1|^2}.
\end{equation}

Since $\varphi_1$ is differentiable at $e_1$,
we have a Taylor series expansion of $\varphi_1$ in a neighborhood of
$e_1$:
\begin{eqnarray*}
\varphi_1(z)&=&\varphi_1(e_1)+D_1\varphi_1(e_1)(z_1-1)+\sum_{j=2}^{N}D_j\varphi_1(e_1)z_j+
\frac{1}{2!}D_{11}\varphi_1(e_1)(z_1-1)^2\\&+&\sum_{j=2}^ND_{1j}\varphi_1(e_1)(z_1-1)z_j+\sum_{k,j=2;k\neq j}^ND_{kj}\varphi_1(e_1)z_kz_j
+\frac{1}{2!}\sum_{j=2}^ND_{jj}\varphi_1(e_1)z_j^2+\cdots.
\end{eqnarray*}
Recall that by
hypothesis $D_1\varphi_1(e_1)=1$.  Direct computation using Equation~(\ref{firstcoordpara}) shows that $D_j\varphi_1(e_1)=0,$
for $j=2,...,N$ (this also follows more generally from the fact that
$e_1$ is a fixed point of $\varphi$; see Lemma 6.6 in \cite{cmbook}) and $D_{kj}\varphi_1(e_1)=0$ for $k,j=2,...,N$.
Thus
$$\varphi_1(z)-1=(z_1-1)+\frac{1}{2!}\left[D_{11}\varphi_1(e_1)(z_1-1)^2
+2\sum_{j=2}^ND_{1j}\varphi_1(e_1)(z_1-1)z_j\right]
+\cdots,$$
where the $+\cdots$ indicates higher order terms
of the form
$$\frac{D^{\nu}\varphi_1(e_1)(z-e_1)^{\nu}}{\nu!},$$
and $\nu$ is a multi-index of order at least $3$.  Since
for $z\in B_N$ we have
$$\frac{|z_2|^2+\cdots+|z_N|^2}{|1-z_1|}\leq \frac{1-|z_1|^2}{|1-z_1|}\leq 2\frac{1-|z_1|}{|1-z_1|}\leq 2$$
we see that
$$\frac{1-\varphi_1(z)}{1-z_1}\rightarrow 1 \mbox{ as }z\rightarrow e_1 \mbox{ in }B_N.$$
By~(\ref{relatingtoder}) this implies that
$$\frac{1-|\varphi(z^{(n)})|^2}{1-|z^{(n)}|^2}$$ has a positive finite limit as $n\rightarrow \infty$,
and property (c) holds as desired.

\end{proof}

To prove Theorem~\ref{scvmain} we will use the preceding result and
the following qualitative generalization of Theorem 9.16 in \cite{cmbook},
specialized to linear-fractional maps.  In the statement we use
the notation $\psi_{\zeta}$ for the coordinate of $\psi$ in the
$\zeta-$direction, that is $\psi_{\zeta}(z)=\langle \psi(z),\zeta\rangle$.
Moreover the derivative of $\psi_{\zeta}$ in the $\eta$ direction,
denoted $D_{\eta}\psi_{\zeta}$,
is defined by $D_{\eta}\psi_{\zeta}(z)\equiv \langle \psi'(z)\eta,\zeta\rangle$.
Note that when $\zeta=\eta=e_1$ this is just $D_1\psi_1(z)$.

For $\eta\in\partial B_N$, write $[\eta]$ for the complex
line containing $\eta$ and $0$; that is, the one-dimensional
subspace of $\overline{B_N}$ consisting of
all points of the form $\{\alpha \eta:\alpha \in \mathbb{C}\}$.
In particular, the complex line $[e_1]$ intersected with $B_N$
consists of all points in the ball
whose last $N-1$ coordinates are $0$.

\begin{thm}\label{samefirst}
Suppose $\varphi$ and $\psi$ are linear-fractional self-maps of $B_N$,
and suppose $\varphi(\zeta)=\zeta$ for some
$\zeta\in\partial B_N$.
If either $\psi(\zeta)\neq \zeta$, or $\psi(\zeta)=\zeta$ and $D_{\zeta}\varphi_{\zeta}(\zeta)\neq
D_{\zeta}\psi_{\zeta}(\zeta)$, then $C_{\varphi}-C_{\psi}$ is not compact.
\end{thm}

\begin{proof}
The argument follows that of Theorem 9.16 in \cite{cmbook}.
Without loss of generality we may assume $\zeta=e_1$.

First suppose $\psi(e_1)\neq e_1$.  If we can find
a sequence of
points $z^{(n)}$ tending to $\partial B_N$, so that
$$\left\|(C_{\varphi}-C_{\psi})^*\left(\frac{K_{z^{(n)}}}{\|K_{z^{(n)}}\|}\right)\right\|$$
is bounded away from $0$,
where $K_{z^{(n)}}$ denotes the kernel function in ${\mathcal H}$ at $z^{(n)}$ (see Equations~(\ref{H2ker})
and~(\ref{A2ker})),
then $C_{\varphi}-C_{\psi}$ is not compact,
since the normalized kernel functions $K_{z^{(n)}}/\|K_{z^{(n)}}\|$ tend weakly to $0$ as $z^{(n)}\rightarrow
\partial B_N$.
Using the fact that for any bounded composition operator $C_{\tau}$ we have
$C_{\tau}^*(K_z)=K_{\tau(z)}$, we see that
$$\|(C_{\varphi}-C_{\psi})^*(K_z)\|^2=\|K_{\varphi(z)}\|^2+\|K_{\psi(z)}\|^2-2\mbox{Re }K_{\varphi(z)}(\psi(z)),$$
and thus
\begin{equation}\label{adjkercomp}
\|(C_{\varphi}-C_{\psi})^*(K_z/\|K_z\|)\|^2\geq \left(\frac{1-|z|^2}{1-|\varphi(z)|^2}\right)^{\beta}
-2\mbox{Re }\frac{K_{\varphi(z)}(\psi(z))}{\|K_z\|^2}
\end{equation}
where $\beta=N$ in $H^2(B_N)$ and $\beta =N+1+\alpha $ in $A^2_{\alpha}(B_N)$.
With our assumption that $\psi(e_1)\neq e_1$, it is easy to see that as $z\rightarrow e_1$ radially, the second
term on the right hand side of Equation~(\ref{adjkercomp}) tends to $0$.  By Julia-Caratheodory
theory (see for example, \cite{Ru}, Section 8.5), the first term
tends to the positive value $(D_1\varphi_1(e_1))^{-\beta}$.
This shows that $C_{\varphi}-C_{\psi}$ is not compact if $ \psi(e_1)\neq \varphi(e_1)$.

Now suppose $ \psi(e_1)= \varphi(e_1)=e_1$ but $D_1\psi_1(e_1)\neq D_1\varphi_1(e_1)$.
As before, if we can find a sequence of points $z^{(n)}$ in $B_N$ tending to $e_1$
along which
$$\|(C_{\varphi}-C_{\psi})^*(K_{z^{(n)}}/\|K_{z^{(n)}}\|)\|$$
is bounded away from $0$, then we can conclude that $C_{\varphi}-C_{\psi}$ is not
compact.
The sequence $\{z^{(n)}\}$ will chosen so that $z^{(n)}=(z_1^{(n)},0')$ where
$$\frac{|1-z_1^{(n)}|}{1-|z^{(n)}|^2}=\frac{|1-z_1^{(n)}|}{1-|z_1^{(n)}|^2}=M$$
for a fixed and suitably large value of $M$; that is, the points $z^{(n)}$
will approach $e_1$ along the boundary of a non-tangential approach region, of large aperture,
in the complex line $[e_1]$.  To analyze the second term on the right hand side
of Equation~(\ref{adjkercomp}), we first consider
\begin{equation}\label{threeterm}
\frac{1-\langle \varphi(z),\psi(z)\rangle}{1-|z|^2}=
\frac{1-|\varphi(z)|^2}{1-|z|^2}+\frac{\langle \varphi(z)-e_1,\varphi(z)-\psi(z)\rangle}{1-|z|^2}
+\frac{\langle e_1,\varphi(z)-\psi(z)\rangle}{1-|z|^2}.
\end{equation}
The third term on the right hand side of Equation~(\ref{threeterm}) has modulus
$$\frac{|\varphi_1(z)-\psi_1(z)|}{1-|z|^2}=
\left|\frac{1-\psi_1(z)}{1-z_1}-\frac{1-\varphi_1(z)}{1-z_1}\right|
\frac{|1-z_1|}{1-|z|^2},$$
and if $z$ is chosen to approach $e_1$ in $[e_1]$ along
the curve $|1-z_1|/(1-|z_1|^2)=M$ this will tend to
$|D_1\psi_1(e_1)-D_1\varphi_1(e_1)|M$. Since $D_1\varphi_1(e_1)\neq D_1\psi_1(e_1)$
by assumption, this can be made
as large as desired by choosing $M$ large.

The first term on
the right hand side of Equation~(\ref{threeterm}) tends
to $|D_1\varphi_1(e_1)|$ along any sequence of points approaching
$e_1$ non-tangentially in $[e_1]$.  We claim that the second
term in~(\ref{threeterm}) tends to $0$ along any such sequence.  To see this, it's
enough to show that
$$\frac{|\varphi(z)-e_1||\varphi(z)-\psi(z)|}{1-|z|^2}$$
tends to $0$ as $z$ approaches $e_1$ non-tangentially in $[e_1]$.
Since
$$|\varphi(z)-\psi(z)|\leq |\varphi(z)-e_1|+|e_1-\psi(z)|$$
it suffices to show
that
$$\frac{|\varphi(z)-e_1|^2}{1-|z|^2}$$
and
$$\frac{|\varphi(z)-e_1||\psi(z)-e_1|}{1-|z|^2}$$
both tend to $0$.  We have
$$\frac{|\varphi(z)-e_1|^2}{1-|z|^2}=\frac{|\varphi_1(z)-1|^2+|\varphi'(z)|^2}{|1-z_1|}\ \frac{|1-z_1|}{1-|z|^2}$$
where $\varphi'(z)$ denotes the $(N-1)$-tuple $(\varphi_2(z),\ldots,\varphi_N(z))$.
We're considering points $z^{(n)}=(z_1^{(n)},0')$ tending to $e_1$ for which
$|1-z_1^{(n)}|/(1-|z_1^{(n)}|^2)$ is some constant value $M$.  Along such a sequence,
$|1-\varphi_1(z)|/|1-z_1|$ tends to $|D_1\varphi_1(e_1)|$, so
that
$$\frac{|\varphi_1(z)-1|^2}{|1-z_1|}\frac{|1-z_1|}{1-|z|^2}\rightarrow 0.$$
By the Julia-Caratheodory theorem (\cite{Ru}, Theorem 8.5.6),
we also have for $2\leq k\leq N$,
$$\frac{|\varphi_k(z)|^2}{|1-z_1|}\rightarrow 0$$
along any non-tangential sequence approaching $e_1$ in $[e_1]$.
Since a similar analysis applies to show that
$$\frac{|\psi(z)-e_1|^2}{1-|z|^2}\rightarrow 0$$
along the sequences under consideration,
it follows that
$$\frac{|\varphi(z)-e_1||\psi(z)-e_1|}{1-|z|^2}\rightarrow 0$$
as desired.

Thus we have shown the following:  If $\varphi(e_1)=\psi(e_1)=e_1$ but
$D_1\varphi_1(e_1)\neq D_1\psi_1(e_1)$,
then given $\epsilon>0$ there exists $M<\infty$ so that if
$z^{(n)}=(z_1^{(n)},0')$ approaches $e_1$ as $n\rightarrow\infty$, where
$$\frac{|1-z_1^{(n)}|}{1-|z_1^{(n)}|^2}=M,$$
then
$$\limsup_{n\rightarrow\infty}\frac{|K_{\varphi(z^{(n)})}(\psi(z^{(n)}))|}{\|K_{z^{(n)}}\|^2}<\epsilon.$$
By Equation~(\ref{adjkercomp}) this
says $C_{\varphi}-C_{\psi}$ is not compact.

\end{proof}

Remark:  It's clear that a version of Theorem~\ref{samefirst} holds, with essentially
the same proof, when $\varphi$ and $\psi$ are more general analytic self-maps
of $B_N$, if in the statement of the theorem the values of $\varphi,\psi, D_{\zeta}\varphi_{\zeta}$
and $D_{\zeta}\psi_{\zeta}$ at $\zeta$ are replaced by their radial limits there.
Also a version of the result, with the hypothesis $\varphi(\zeta)=\zeta$ replaced
by $\varphi(\zeta)=\eta$ for $\zeta,\eta\in\partial B_N$ can be formulated.  Since
we do not need these more general versions, we leave the precise statements
to the interested reader.

\begin{prop}\label{niceform2}
Suppose that $\tau$ is a linear-fractional self-map
of $B_N$ such that the restriction of $\tau$ to the
the complex
line $[e_1]$ in $\overline{B_N}$ is the identity on $[e_1]\cap \partial B_N$.
Then $\tau(z_1,z_2,\cdots, z_N)=(z_1,A'z')$, where
$z'$ denotes $(z_2, \cdots, z_N)$ and $A'$ is an $(N-1)\times(N-1)$
matrix.
\end{prop}
\begin{proof}
Since $\varphi$ is linear-fractional we have
$$\varphi(z)=\frac{Az+B}{c_1z_1+c_2z_2+\cdots c_Nz_N+1}$$
for some $N\times N$ matrix $A=(a_{jk})$, $N\times 1$ matrix $B$,
and constants $c_k$.  By hypothesis we must have
$$a_{k1}\lambda+b_k=0$$
for $2\leq k\leq N$ and all $\lambda \in {\mathbb C}$ with $|\lambda|=1$.  Thus
$a_{k1}=b_k=0$ for $2\leq k \leq N$.  From $\varphi(e_1)=e_1$ and
$\varphi(-e_1)=-e_1$ we see that
$b_1=c_1$ and $a_{11}=1$.  Using this, and the requirement
that $a_{11}\lambda +b_1=\lambda(c_1\lambda+1)$ for
$|\lambda|=1$, we must have $b_1=0$ and thus also $c_1=0$.

Moreover, since $\varphi$ fixes $e_1$, we have by Lemma 6.6 of \cite{cmbook} that
\begin{equation}\label{der1}
D_k\varphi_1(e_1)=0\mbox{ for }k=2,3,\cdots,N.
\end{equation}
Since $\varphi(-e_1)=-e_1$, we may apply the same lemma to $-\varphi(-z)$ to
see that
\begin{equation}\label{der2}
D_k\varphi_1(-e_1)=0 \mbox{ for }k=2,3,\cdots,N.
\end{equation}
Equations~(\ref{der1}) and~(\ref{der2}) together say
$$a_{1k}=c_k=0\mbox{ for }k=2,\cdots,N.$$
This completes the proof.
\end{proof}

To move from Theorem~\ref{parabolics}, which deals with parabolic maps,
to the full result of Theorem~\ref{scvmain}, we will need the notion
of the Krein adjoint of the linear-fractional map $\varphi$.  If $\varphi$
is as
given in Equation~(\ref{defoflinfrac}), its Krein adjoint is defined to be the linear
fractional map
\begin{equation}\label{kreinadjdef}
\sigma_{\varphi}(z)=\frac{A^*z-C}{\langle z,-B\rangle +\overline{d}}.
\end{equation}
This will be a self-map of $B_N$ whenever $\varphi$ is, and
when $\varphi$ is an automorphism, its Krein adjoint is equal
to $\varphi^{-1}$.  Moreover, $\varphi$ and $\sigma_{\varphi}$ have
the same fixed  points on $\partial B_N$; for these and other basic facts, see \cite{cowenmac}
and \cite{mac-weir}.
Properties of the map $\varphi\circ\sigma_{\varphi}$ appear in the next
result.

\begin{thm}\label{univalentcase}
Suppose $\varphi$ and $\psi$ are linear fractional maps with
$\|\varphi\|_{\infty}=\|\psi\|_{\infty}=1$.  Assume further that
at least one of the maps $\varphi, \psi$ is univalent.
If
$C_{\varphi}-C_{\psi}$ is compact on ${\mathcal H}$,
then $\varphi=\psi$.
\end{thm}
\begin{proof}
By the symmetric roles of $\varphi$ and $\psi$ we may assume
that $\varphi$ is univalent.
The hypothesis $\|\varphi\|_{\infty}=1$ implies that there exists $\zeta$
in $\partial B_N$ with $|\varphi(\zeta)|=1$.
Composing on the left and right
by unitaries, there is no loss of generality in assuming $\zeta=e_1$ and $\varphi(e_1)=e_1$.
By Theorem~\ref{samefirst}, we must have
$\psi(e_1)=e_1$ as well.

Let $\sigma_{\varphi}$ be the
Krein adjoint of $\varphi$ as defined in Equation~(\ref{kreinadjdef}).  Since $C_{\varphi}-C_{\psi}$ is compact
and $C_{\sigma_{\varphi}}$ is bounded,
$$C_{\sigma_{\varphi}}(C_{\varphi}-C_{\psi})=C_{\varphi\circ\sigma_{\varphi}}-C_{\psi\circ\sigma_{\varphi}}$$
is also compact.  Set
$\tau=\varphi\circ\sigma_{\varphi}$ and $\xi=\psi\circ\sigma_{\varphi}$.
We have
$$\tau(e_1)=\varphi\circ\sigma_{\varphi}(e_1)=e_1$$
and thus by Theorem~\ref{samefirst}, $\xi(e_1)=e_1$ and $D_1\tau_1(e_1)=D_1\xi_1(e_1)$.
A computation shows that $D_1\tau_1(e_1)=1$, (details of this computation can
be found in the proof of Theorem 2 in \cite{mac-weir}) so that
$$D_1\xi_1(e_1)=1.$$

We claim that $\tau=\xi$.  To see this, note that it is immediate by Theorem~\ref{parabolics}
if neither $\tau$ nor $\xi$ have any fixed point in the open ball $B_N$.
Suppose next that $\tau$
has a fixed point in the open ball
and lying in the complex line $[e_1]$.
Restricting $\tau$ to the intersection of $[e_1]$ and the ball,
we see that $\tau$ must be the identity on
$[e_1]\cap \overline{B_N}$, since $D_1\tau_1(e_1)=1$ (see, for example,
Problem 2.38 in \cite{cmbook}, p. 60).
By Proposition~\ref{niceform2} we see that $\tau$ has
the form $\tau(z_1,z')=(z_1,Az')$ for some $(N-1)\times(N-1)$ matrix
$A$. We label the entries of $A$ as $a_{jk}$ for $j,k=2,\cdots, N$.
Since $C_{\tau}-C_{\xi}$ is compact, we appeal to Theorem~\ref{samefirst}
to see that, since $\tau$ is the identity at each point of $[e_1]\cap\partial B_N$,
so is $\xi$.  Applying Proposition~\ref{niceform2} again, we see that $\xi(z_1,z')=(z_1,Mz')$
for an $(N-1)\times(N-1)$ matrix $M=(m_{jk})$, $j,k=2,\cdots, n$.
Our goal is to show that $A=M$.

Fix $j$ with $2\leq j\leq N$, and consider the pseudohyperbolic
distance $\rho(\tau(\omega_t),\xi(\omega_t))$ at points of the form
\begin{equation}\label{path}
\omega_t=(t,0,\cdots,\sqrt{1-t},0\cdots,0)=(t,0',\sqrt{1-t},0''),
\end{equation}
for $0<t<1$, where the $\sqrt{1-t}$ appears in the $j^{th}$ component.
These points lie in the ball $B_N$.
A computation shows that
$$1-\rho^2(\tau(\omega_t),\xi(\omega_t))$$ is equal to
$$\frac{[1-t^2-(1-t)\sum_{k=2}^N|a_{kj}|^2][1-t^2-(1-t)\sum_{k=2}^N|m_{kj}|^2]}
{|1-t^2-(1-t)\sum_{k=2}^Na_{kj}\overline{m_{kj}}|^2},
$$ and further computation shows that as $t\uparrow 1$ this has limit
equal to
\begin{equation}\label{lhopital}
\frac{(2-\lambda^2)(2-\gamma^2)}{(2-\sum_ka_{kj}\overline{m_{kj}})(2-\sum_k\overline{a_{kj}}m_{kj})}
\end{equation}
where
$$\lambda=\left(\sum_{k=2}^N|a_{kj}|^2\right)^{1/2}$$ and
$$\gamma=\left(\sum_{k=2}^N|m_{kj}|^2\right)^{1/2}.$$
Observe that $\lambda$ and $\gamma$ are at most $1$, since
$\tau$ and $\xi$ map the ball into itself.
Write $Z=(a_{2j},a_{3j}\cdots,a_{Nj})$ and $W=(m_{2j},m_{3j},\cdots,m_{Nj}),$
so that $\lambda=\|Z\|$ and $\gamma=\|W\|$.
Moreover, the denominator in~(\ref{lhopital}) is
$|2-\langle Z,W\rangle|^2$,
and by the Cauchy-Schwarz inequality,
$|\langle Z,W\rangle|\leq \lambda\gamma$ with equality only if either $Z=c W$
for some $c\in {\mathbb C}$ or one of $Z,W$ is $0$.

We investigate the condition under
which the expression in~(\ref{lhopital}) is equal to $1$.
We have
$$\frac{(2-\lambda^2)(2-\gamma^2)}{|2-\langle Z,W\rangle|^2}
\leq \frac{(2-\lambda^2)(2-\gamma^2)}{(2-|\langle Z,W\rangle|)^2}
\leq \frac{(2-\lambda^2)(2-\gamma^2)}{(2-\lambda\gamma)^2}\leq 1,$$
with the last inequality following from its equivalence to
$(\lambda-\gamma)^2\geq 0$.
Thus if the expression in~(\ref{lhopital}) is equal to $1$,
we must have $\lambda=\gamma$ and
$$|2-\langle Z,W\rangle|=2-|\langle Z,W\rangle|=2-\|Z\|\|W\|.$$
Together these force $Z=W$, which says that the
 $(j-1)^{st}$ column of $A$ is the
same as the $(j-1)^{st}$ column of $M$.
Thus if $A\neq M$, $\rho(\tau,\xi)$ has a strictly positive limit along some path
as in Equation~(\ref{path}).
The ratio
$$\frac{1-|z|^2}{1-|\tau(z)|^2}$$
has the positive limit $(2-\lambda^2)^{-1}$ along the same path.
Applying Theorem~\ref{phdist} we have
a contradiction to the hypothesis that $C_{\tau}-C_{\xi}$ is compact.
Thus $A=M$,
verifying the claim under the assumption that $\tau$ has a fixed point
in $[e_1]\cap B_N$.

Finally, suppose $\tau$ has a fixed point in the intersection of the open ball
and the complex line through $\eta$ and $e_1$ for some $\eta\in\partial B_N$, but
not in $[e_1]$.  Since the automorphisms act doubly transitively on
$\partial B_N$, we may find an automorphism $\Phi$ of the ball,
fixing $e_1$ so that $\widetilde{\tau}\equiv\Phi^{-1}\tau\,\Phi$
fixes $e_1$ and a point of $[e_1]\cap B_N$.
A computation shows that $D_1\widetilde{\tau}_1(e_1)=1$; this
computation is aided by the
fact that
$$D_1\widetilde{\tau}_1(e_1)=\langle \widetilde{\tau}\,'(e_1)e_1,e_1\rangle$$
and the observation that since
$\tau, \Phi$ and $\Phi^{-1}$ all fix $e_1$, we have
$$D_k\tau_1(e_1)=0, D_k\Phi_1(e_1)=0, D_k\Phi_1^{-1}(e_1)=0\mbox{ for all }k=2,3,\ldots,N$$
(\cite{cmbook}, Lemma 6.6).
Conjugating $\xi$ by $\Phi$ as well to get
$\widetilde{\xi}\equiv\Phi^{-1}\xi\Phi$,
we apply the previous argument to see that $\widetilde{\tau}=\widetilde{\xi}$,
and hence $\tau=\xi$, in this case as well.

Thus compactness of $C_{\varphi}-C_{\psi}$ implies that
$\tau=\xi,$
or equivalently
\begin{equation}\label{tauxi}
\varphi\circ\sigma_{\varphi}=\psi\circ\sigma_{\varphi}
\end{equation}
on $B_N$, where
$\sigma_{\varphi}$ is the Krein adjoint of $\varphi$.
From this we see that $\varphi$ and $\psi$ agree on the range of $\sigma$.
Since we are assuming that $\varphi$ is univalent, so is $\sigma_{\varphi}$ (\cite{cowenmac})
and it follows that $\varphi=\psi$, since the
range of $\sigma_{\varphi}$ is an open set in $B_N$.
\end{proof}

The final step is to remove the hypothesis of univalence in the last result
to obtain the full proof of Theorem~\ref{scvmain}, which we turn to next.
It will be helpful to recast our Hilbert space ${\mathcal H}$
as weighted Hardy spaces, defined below, and consider
restriction and extension
operators on these weighted Hardy spaces.

If $f$ is analytic in
$B_N$, then $f$ has a homogeneous
expansion
\[ f= \sum_s f_s, \]
where, for each $z \in B_N$, we have
\begin{equation}\label{homogeneousfnc}
 f_s(z) = \sum_{|\alpha|=s} c_{\alpha} z^{\alpha}.
 \end{equation}
Here, $\alpha = (\alpha_1, \dots, \alpha_N)$ and $|\alpha| = \alpha_1 + \cdots +
\alpha_N$. The Hardy space $H^2(B_N)$ is equivalently
defined as
\begin{equation}\label{h2altdef}
\{f\mbox{ analytic in }B_N:\sum_{s=0}^{\infty}\|f_s\|_2^2<\infty\},
\end{equation}
where $\| \cdot \|_{2}$ is the norm in $L^2(\sigma)$.
The sum in~(\ref{h2altdef}) is $\|f\|_{H^2(B_N)}^2$.
More generally, given a suitable sequence of
positive numbers $\{ \beta(s) \}$, the weighted Hardy space
$H^2(\beta, B_N)$ is the set of
functions $f$ which are analytic in $B_N$ and which satisfy
\[ \|f\|_{H^2(\beta, B_N)}^2\equiv \sum_{s=0}^{\infty} \|f_s\|_{2}^2\  \beta(s)^2 <
\infty. \]
 Note that,
since the monomials
$z^{\alpha}$ are orthogonal on $L^2(\sigma)$ (\cite{Ru}, Section 1.4),
\[ \|f_s\|_{2}^2 =  \sum_{|\alpha|=s} |c_{\alpha}|^2 \| z^{\alpha} \|^2_{2}
= \sum_{|\alpha|=s}
|c_{\alpha}|^2 \frac{(N-1)! \alpha!}{(N-1+s)!} . \]

The next result realizes the weighted Bergman spaces $A^2_{\gamma}(B_N)$ as weighted
Hardy spaces.

\begin{lemma} \label{lem:equiv1}
\begin{itemize}
\item[(a)]
Let $\gamma>-1$ and set $\beta(s)^2 = (s+1)^{-(\gamma+1)}$. We have
$H^2(\beta,B_N)=A^2_{\gamma}(B_N)$, with equivalent norms.
\item[(b)] Let $K $ be an integer with $1 \leq K < N$, and let
\[ \beta(s)^2=\frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!}(s+1)^{-(\gamma+1)}  \]
where $\gamma\geq -1$.
Then $H^2(\beta,B_K)=A^2_{N-K+\gamma}(B_K)$, with equivalent norms.
\end{itemize}
\end{lemma}

\begin{proof}
We prove (a) first.  Let $f$ be analytic in $B_K$ with homogeneous expansion $f= \sum f_s$, where
$f_s$ is as in Equation~(\ref{homogeneousfnc}).
If $\beta(s)^2=(s+1)^{-(\gamma+1)}$, we have
\[ \|f\|_{H^2(\beta, B_N)}^2 = \sum_{s=0}^{\infty} \sum_{|\alpha|=s}
|c_{\alpha}|^2 \frac{\alpha!
(N-1)!}{\Gamma(N+s)} \cdot \frac{1}{(s+1)^{\gamma+1}}  \]
and
\[ \|f\|_{A^2_{\gamma}(B_N)}^2 = \sum_{s=0}^{\infty} \sum_{|\alpha|=s} |c_{\alpha}|^2
\frac{ \alpha!
\Gamma(N+\gamma+1)}{\Gamma(N+s+\gamma+1)} . \]
This last formula follows from the fact that
\[ \|z^{\alpha} \|_{A^2_{\gamma}(B_K)}^2 = \frac{\alpha!\Gamma(N+\gamma+1)}{\Gamma(N+|\alpha|+\gamma+1)} \]
(see Lemma~1.11 in \cite{Zhubook}.)

The result in (a) will follow if we can show that
$$ \left( \frac{\alpha! (N-1)!}{\Gamma(N+s)} \cdot \frac{1}{(s+1)^{\gamma+1}} \right)
\cdot \left( \frac{\Gamma(N+s+\gamma+1)}{
\alpha! \Gamma(N+\gamma+1)} \right) $$
is bounded above and below by positive constants, depending only
on $N$ and $\gamma$, for all $s\geq 0$.
This follows easily from the fact that, by Stirling's formula,
$$\lim_{s\rightarrow \infty}\frac{\Gamma(N+s+\gamma+1)}{(s+1)^{\gamma+1}\Gamma(N+s)}=1.$$

From (a) we know that $A^2_{N-K+\gamma}(B_K) = H^2(\widetilde
\beta, B_K)$ where $\widetilde
\beta(s)^2 = (s+1)^{-(N-K+\gamma+1)}$.
Thus it suffices to show that
$$\left[\frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!}(s+1)^{-(\gamma+1)}\right]\cdot\left[(s+1)^{(N-K+\gamma+1)}\right]$$
is bounded
above and below by positive constants (depending on $N$ and $K$)
for all $s\geq 0$.  Straightforward estimates show that
$$\frac{(N-1)!}{(K-1)!}\geq \frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!}\cdot (s+1)^{N-K}\geq
\frac{(N-1)!}{(K-1)!} \left(\frac{1}{N+1}\right)^{N-K}$$
for all $s\geq 0$, and the desired result follows.

\end{proof}

Since the proof of Theorem~\ref{scvmain} ultimately relies on an inductive argument,
we will work with certain restriction and extension operators on weighted Hardy spaces.
These are defined next.

Let $K, N \in \mathbb{N}$ with $1 \leq K < N$. Given a sequence $\{ \beta(s) \}$
of positive numbers, we define the associated sequence $\{ \widetilde \beta(s) \}$ by
\[ \widetilde \beta(s)^2 = \frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!} \beta(s)^2. \]
We can then define the extension operator $E: H^2(\widetilde \beta, B_K) \rightarrow H^2(\beta, B_N)$ by
\[ (Ef)(z_1, \dots, z_N) = f(z_1, \dots, z_K), \]
and the restriction operator $R: H^2(\beta, B_N)\rightarrow H^2(\widetilde \beta, B_K)$ by
\[ (Rf)(z_1, \dots, z_K) = f(z_1, \dots, z_K, 0'). \]
The next result establishes properties of these operators;
it is an extension of Proposition 2.21 in \cite{cowenmac} which applies to the case $K=1$.
\begin{lemma}\label{restrictextend}
\begin{itemize}
\item[(a)] The extension operator $E$ is an isometry of $H^2(\widetilde \beta, B_K)$ into $H^2(\beta, B_N)$.

\item[(b)] The restriction operator $R$ is a norm-decreasing map of $H^2(\beta, B_N)$ onto $H^2(\widetilde \beta, B_K)$.

\end{itemize}
\end{lemma}

\begin{proof}

For (a), let $f \in H^2(\widetilde \beta, B_K)$ with homogeneous expansion $f= \sum f_s$
with $f_s$ as in Equation~(\ref{homogeneousfnc}) for $z\in {\mathbb C}^K$.
Then $Ef = \sum f_s$ also, and writing $\|f_s\|_{2,K}$ for the norm
of $f_s$ in $L^2(\partial B_K,\sigma)$ we have
\begin{align*}
\| f \|^2_{H^2(\widetilde \beta, B_K)} & = \sum_s \|f_s\|^2_{2,K} \widetilde \beta(s)^2 \\
 & = \sum_s \sum_{|\alpha|=s} |c_{\alpha}|^2 \frac{(K-1)! \alpha!}{(K-1+s)!} \widetilde \beta(s)^2 \\
 & = \sum_s \sum_{|\alpha|=s} |c_{\alpha}|^2 \frac{(K-1)! \alpha!}{(K-1+s)!} \cdot \frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!} \beta(s)^2 \\
 & = \sum_s \sum_{|\alpha|=s} |c_{\alpha}|^2 \frac{(N-1)! \alpha!}{(N-1+s)!} \beta(s)^2 \\
 & = \sum_s \|f_s\|^2_{2,N} \beta(s)^2 \\
 & = \| Ef \|^2_{H^2(\beta, B_N)}.
\end{align*}
Therefore, $E$ is an isometry.

For (b) let $f \in H^2(\beta, B_N)$ have homogeneous expansion
$f=\sum f_s$ with $f_s$ as in Equation~(\ref{homogeneousfnc}) for $z\in B_N$.
For each nonnegative integer $s$, let $A_s = \{ \alpha: \alpha = (\alpha', 0') \}$, where $\alpha'$ is a multi-index with $K$ entries and $0'$ denotes the zero vector in $\mathbb{C}^{N-K}$, and let $B_s$ consist of all other multi-indices $\alpha$ with $N$ entries satisfying $|\alpha|=s$. Writing
\[ f_s(z) = \sum_{A_s} c_{\alpha} z^{\alpha} + \sum_{B_s} c_{\alpha} z^{\alpha}, \]
it follows that
\[ (Rf)(z_1, \dots, z_K) = \sum_s \sum_{A_s} c_{\alpha} z_1^{\alpha_1} \cdots z_K^{\alpha_K}, \]
and
\begin{align*}
\|Rf\|^2_{H^2(\widetilde \beta, B_K)} & = \sum_s \sum_{A_s} |c_{\alpha}|^2 \frac{(K-1)! \alpha!}{(K-1+s)!} \widetilde \beta(s)^2 \\
 & = \sum_s \sum_{A_s} |c_{\alpha}|^2 \frac{(N-1)! \alpha!}{(N-1+s)!} \beta(s)^2 \\
 & \leq  \sum_s \sum_{|\alpha|=s} |c_{\alpha}|^2 \frac{(N-1)! \alpha!}{(N-1+s)!} \beta(s)^2 \\
 & = \| f \|^2_{H^2(\beta, B_N)}.
\end{align*}
Therefore, $R$ is norm-decreasing. To see that $R$ is surjective, let $f \in H^2(\widetilde \beta, B_K)$ with
\[ f(z_1, \dots, z_K) = \sum_s \sum_{|\alpha'|=s} c_{\alpha'} z_1^{\alpha_1'} \cdots z_K^{\alpha_K'}, \]
where $\alpha'$ is a multi-index with $K$ entries.
Then $f=RF$, where
\[ F(z) = \sum_s \sum_{A_s} c_{\alpha} z^{\alpha} . \]
and $c_{\alpha} \equiv c_{\alpha'}$ for $\alpha = (\alpha', 0')$ as before. Also,
\begin{align*}
\|F\|^2_{H^2(\beta, B_N)} & = \sum_s \sum_{A_s} |c_{\alpha}|^2 \frac{(N-1)! \alpha!}{(N-1+s)!} \beta(s)^2 \\
 & = \sum_s  \sum_{A_s} |c_{\alpha}|^2 \frac{(K-1)! \alpha!}{(K-1+s)!} \widetilde \beta(s)^2 \\
 & = \sum_s  \sum_{|\alpha'|=s} |c_{\alpha'}|^2 \|z_1^{\alpha_1'} \cdots z_K^{\alpha_K'}\|^2_{2,K} \\
 & = \|f\|^2_{H^2(\widetilde \beta, B_K)} < \infty,
\end{align*}
so $F \in H^2(\beta, B_N)$.

\end{proof}

Recall that by an \textit{affine subset of dimension $k$} in $B_N$, we mean the intersection
of $B_N$ with a translate of a $k$-dimensional subspace of ${\mathbb C}^N$.

\begin{proof}[Proof of Theorem~\ref{scvmain}]
Suppose that $\varphi$ and $\psi$ are linear-fractional maps with
$\|\varphi\|_{\infty}=\|\psi\|_{\infty}=1$. We will show that if
$C_{\varphi}-C_{\psi}$ is compact on ${\mathcal H}$, then $\varphi=\psi$.
The hypothesis $\|\varphi\|_{\infty}=1$ implies that there exists a point
$\zeta$
in $\partial B_N$ with $|\varphi(\zeta)|=1$. Composing on the left and right
by unitaries, there is no loss of generality in assuming $\zeta=e_1$ and
$\varphi(e_1)=e_1$. By
Theorem~\ref{samefirst}, we must have
$\psi(e_1)=e_1$ as well.

The argument proceeds exactly as in the proof of Theorem~\ref{univalentcase}
up to the point where the relationship
\begin{equation}\label{tauxi2}
\varphi\circ\sigma_{\varphi}=\psi\circ\sigma_{\varphi},
\end{equation}
is obtained.  Since Theorem~\ref{univalentcase} covers the case that
at least one of $\varphi$ and $\psi$ is one-to-one, we now
only consider the case that neither is univalent.
This implies that there is a smallest $k_1$ with $1\leq k_1<N$ so that $\varphi(B_N)$
is contained in an affine set of dimension $k_1$, and
there is a smallest $k_2$ with $1\leq k_2<N$ so that $\psi(B_N)$ is contained
in an affine set of dimension $k_2$.
Since the roles of $\varphi$ and $\psi$ can be reversed,
there is no loss of generality in assuming $k_1\geq k_2$.

Our first goal is to show that equality
$k_1=k_2$
holds as a consequence of Equation~(\ref{tauxi2}).
By Proposition 13 in \cite{cowenmac}, $\sigma_{\varphi}(B_N)$ is
also contained in an $k_1$-dimensional affine set.
Set $\sigma_{\varphi}(0)=p$, and let $\phi_p$ be
an automorphism of $B_N$ sending $p$ to $0$ and
satisfying $\phi_p=\phi_p^{-1}$.
Since $\phi_p\circ\sigma_{\varphi}$ fixes the origin,
and maps the ball into a $k_1$-dimensional affine set,
we may write, as in \cite{cowenmac},
$$\phi_p\circ\sigma_{\varphi}=L\circ\tau$$
where $L$ is linear of rank $k_1$ and $\tau$ is a one-to-one linear fractional
map.  Specifically, if
$$\phi_p\circ\sigma_{\varphi}=\frac{Az}{\langle z,C\rangle +1}$$
we can choose
$$L(z)=Az\mbox{   and   }\tau(z)=\frac{z}{\langle z,C\rangle +1}.$$
(Note that $L$ and $\tau$ need not be self-maps of $B_N$, though
their composition is, and
both are defined and analytic on a neighborhood of the closed
ball).  From this it follows that
$\sigma_{\varphi}=\phi_p\circ L\circ\tau$.  Taking Krein adjoints
on both sides we have
$\varphi=\sigma_{{\tau}}\circ L^*\circ\sigma_{\phi_p}$,
where $\sigma_{\tau}(z)=z-C$.
We have $\sigma_{\phi_p}=\phi_p^{-1}=\phi_p$, so that
\begin{equation}\label{dimcomp}
\varphi\circ\sigma_{\varphi}=\sigma_{\tau}\circ L^*\circ\sigma_{\phi_p}\circ\phi_p\circ L\circ\tau=
\sigma_{\tau}\circ L^*L\circ\tau,
\end{equation} where $L^*L$ is linear
with rank $k_1$, $\tau$ is univalent,
and $\sigma_{\tau}$ is a translation.  Thus the image of the ball $B_N$ under
$\varphi\circ\sigma_{\varphi}$ cannot be contained in a $k$ dimensional
affine set for any $k<k_1$, and the relation $\varphi(\sigma_{\varphi}(B_N))=\psi(\sigma_{\varphi}(B_N))
\subseteq\psi(B_N)$ says that strict inequality $k_1>k_2$ is impossible, and therefore $k_1=k_2$ as desired.
We denote the common value of $k_1$ and $k_2$ by $K$.

Thus $\varphi(B_N)$ is contained in a $K$-dimensional
affine set $A_1$ and is not contained in any
affine set of smaller dimension, and $\psi(B_N)$ is contained in a $K$-dimensional
affine set $A_2$, and is not contained in any affine set of
smaller dimension.  We have
$$A_2\supseteq \psi(B_N)\supseteq \psi(\sigma_{\varphi}(B_N))=\varphi(\sigma_{\varphi}(B_N))$$
where $A_1\supseteq\varphi(\sigma_{\varphi}(B_N))=\sigma_{\tau}L^*L\tau(B_N)$
for linear $L^*L$ of rank $K$ and univalent linear fractional $\tau$ and $\sigma_{\tau}$.
This forces $A_1=A_2$; that is, the range of $\varphi$ and the range
of $\psi$ are contained in the same $K$-dimensional affine set, which we will simply denote $A$. Note
that $e_1 \in \overline{A}$.

Our goal is to show that $\varphi=\psi$. Let $\zeta\in\partial B_N$
with $\zeta\neq e_1$.  Let $\Lambda_{\zeta}$ be a $K$-dimensional affine subset of
$B_N$, containing $e_1$ and
$\zeta$ in its boundary, whose intersection with $B_N$ is a $K$-dimensional ball.
We will write $\widetilde{B_K}$ for
$\{(z_1,z_2,\ldots,z_K,0'')\equiv(z',0'')\in B_N\}$, where
$0''$ denotes the $0$ in ${\mathbb C}^{N-K}$.  Let $\rho_1$ be an automorphism
of $B_N$ fixing $e_1$ with $\rho_1(\widetilde{B_K})=\Lambda_{\zeta}$ and let $\rho_2$ be an automorphism
of $B_N$ fixing $e_1$ with $\rho_2(A)=\widetilde{B_K}$; such automorphisms exist because of the two-fold
transitivity of the automorphisms on $\partial B_N$.
Note that $\rho_2\circ\varphi\circ\rho_1$ and $\rho_2\circ\psi\circ\rho_1$ are linear-fractional
self-maps of $B_N$ with $\rho_2\circ\varphi\circ\rho_1(\widetilde{B_K})\subseteq \widetilde{B_K}$
and $\rho_2\circ\psi\circ\rho_1(\widetilde{B_K})\subseteq \widetilde{B_K}$.
Let $\pi$ be the projection of ${\mathbb C}^N$ onto ${\mathbb C}^K$ defined
by $\pi(z',z'')=z'$ and define maps $\mu$ and $\nu$ on $B_K$
by $$\mu(z')=\pi\circ\rho_2\circ\varphi\circ\rho_1(z',0'')$$
and
$$\nu(z')=\pi\circ\rho_2\circ\psi\circ\rho_1(z',0'').$$
These are linear-fractional self-maps of
$B_K$ fixing $(1,0\ldots,0)\in\partial B_K$.

Write ${\mathcal H}$ as $H^2(\beta, B_N)$ with $\beta(s)^2=(s+1)^{-(\gamma+1)}$,
where $\gamma=-1$ if ${\mathcal H}=H^2(B_N)$
and $\gamma=\alpha$ if
${\mathcal H}=A^2_{\alpha}(B_N)$, up to an
equivalent norm.
We claim that $C_{\mu}-C_{\nu}$ is compact on the weighted Hardy space
$H^2(\widetilde{\beta},B_K)$ where
$$\widetilde{\beta}(s)^2=\frac{(N-1)!(K-1+s)!}{(K-1)!(N-1+s)!}\beta(s)^2.  $$
Since, by
Lemma~\ref{lem:equiv1}, $H^2(\widetilde{\beta},B_K)=A^2_{N-K+\gamma}(B_K)$ with equivalent norms, it will follow
that $C_{\mu}-C_{\nu}$ is compact on $A^2_{N-K+\gamma}(B_K)$.

To prove the claim it suffices to show that if $\{f_n\}$ is a bounded sequence
on $H^2(\widetilde{\beta},B_K)$ and
$f_n\rightarrow 0$ almost uniformly on $B_K$, then
$$\|(C_{\mu}-C_{\nu})f_n\|_{H^2(\widetilde{\beta},B_K)}\rightarrow 0.$$
Let $f_n$ be such a sequence in $H^2(\widetilde{\beta},B_K)$.  Define functions
$F_n$ on $B_N$ by
$$F_n=Ef_n,$$
where $E: H^2(\widetilde{\beta}, B_K) \rightarrow H^2(\beta,B_N)$ is the extension operator
defined by
\[ (Ef)(z_1, \dots, z_N) = f(z_1, \dots, z_K). \]
By Lemma~\ref{restrictextend}, $E$ is an isometry, so the functions $F_n$ form a
bounded sequence in ${\mathcal H}$
and $F_n\rightarrow 0$ uniformly on compact subsets of $B_N$.  Since
$C_{\rho_2\circ\varphi\circ\rho_1}-C_{\rho_2\circ\psi\circ\rho_1}=C_{\rho_1}(C_{\varphi}-C_{\psi})C_{\rho_2}$ is
compact on ${\mathcal H}$, we have
$$\|F_n\circ\rho_2\circ\varphi\circ\rho_1-F_n\circ\rho_2\circ\psi\circ\rho_1\|_{H^2(\beta,B_N)} \rightarrow 0$$
Define the restriction operator $R: H^2(\beta,B_N)\rightarrow H^2(\widetilde{\beta}, B_K)$ by
\[ (Rf)(z_1, \dots, z_K) = f(z_1, \dots, z_K, 0''). \]
By Lemma~\ref{restrictextend}, $R$ is a norm-decreasing map of $H^2(\beta,B_N)$ onto
$H^2(\widetilde{\beta}, B_K)$ and
so
$$\|R(F_n\circ\rho_2\circ\varphi\circ\rho_1)-R(F_n\circ\rho_2\circ\psi\circ\rho_1)\|_{H^2(\widetilde{\beta}, B_K)}
\rightarrow 0$$
But $R(F_n\circ\rho_2\circ\varphi\circ\rho_1)=f_n\circ\mu$ on $B_K$ and
$R(F_n\circ\rho_2\circ\psi\circ\rho_1)=f_n\circ\nu$ on $B_K$,
so the claim is verified, and $C_{\mu}-C_{\nu}$ is compact on $A^2_{ N-K+\gamma}(B_K)$.
Since $K<N$ and $\mu$ and $\nu$ are linear-fractional self-maps of $B_K$
with $\|\mu\|_{\infty}=\|\nu\|_{\infty}=1$, by induction this forces
$\mu = \nu$, which in turn says that $\varphi=\psi$ on the affine set
$\Lambda_{\zeta}$ containing $\zeta$ and $e_1$.
Since $\zeta $ is an arbitrary
point in $\partial B_N$, this says $\varphi=\psi$ in $B_N$.

\end{proof}


TITLE: If $1 \leq \|T\| \leq 2$ and $T^{-1}$ exists, is it true that $1/2 \leq \|T^{-1}\|\leq 1$?
QUESTION [0 upvotes]: Let $X$ and $Y$ be Banach spaces. Suppose that $T:X \rightarrow Y$ is a bounded linear bijective operator with operator norm
$$1 \leq \|T\| \leq 2.$$
Since $T$ is bijective, $T^{-1}$ exists and bounded.

Question: Is it true that $1/2 \leq \|T^{-1}\|\leq 1$?

I have a feeling that this is true but I have no idea how to prove it.
Any hint would be appreciated.

REPLY [7 votes]: This is not true, and counterexamples can be found in $\mathbb R^2$.
Consider, for example, the matrix
$$
 A=\begin{bmatrix}
  1 & 0 \\
  0 & \frac13
 \end{bmatrix}
$$
with the euklidian norm.


TITLE: Information on this Improper Integral
QUESTION [1 upvotes]: I was working with integrals of the form
$$\int_{2}^{\infty} \frac{\ln ^{k}(x) (1-2x)}{x^2 (x-1)^2}dx$$
And started toying around with it in Wolfram alpha, and noticed that ever $k$ I plugged in, Wolfram had an answer. But when leaving it as $k$ , it says there is no answer in any standard functions. I find that curious because it must be get the previous answers from somewhere? Is there in fact a closed form for this when $k$ is a natural number?

REPLY [1 votes]: I tried with other CAS and faced the same situation as you did; apparently, there is no general formula.
However, if we expand as Taylor series,we can notice that
$$\frac{1-2 x}{x^2(x-1)^2 }=-\sum_{n=2}^\infty \frac n {x^{n+1}}$$ and then
$$\int \frac {n\log^k(x)} {x^{n+1}}\,dx=-n^{-k} \Gamma (k+1,n \log (x))$$
$$\int_2^\infty \frac {n\log^k(x)} {x^{n+1}}\,dx=n^{-k} \,\Gamma (k+1,n \log (2))$$ where appears the incomplete gamma function.
$$\int_{2}^{\infty} \frac{\ln ^{k}(x) (1-2x)}{x^2 (x-1)^2}\,dx=-\sum_{n=2}^\infty n^{-k} \,\Gamma (k+1,n \log (2))$$
Edit
From a numerical point of view, the summation seems to converge quite fast. Let us consider the partial sum
$$S_p=-\sum_{n=2}^p n^{-k} \,\Gamma (k+1,n \log (2))$$ For $k=10$, we should get
$$\left(
\begin{array}{cc}
 p & S_p \\
 2 & -3543.749089 \\
 3 & -3605.202406 \\
 4 & -3608.662579 \\
 5 & -3609.033817 \\
 6 & -3609.093604 \\
 7 & -3609.106306 \\
 8 & -3609.109596 \\
 9 & -3609.110581 \\
 10 & -3609.110910 \\
 11 & -3609.111029 \\
 12 & -3609.111075 \\
 13 & -3609.111094 \\
 14 & -3609.111101 \\
 15 & -3609.111105 \\
 16 & -3609.111106 \\
 17 & -3609.111107 \\
 18 & -3609.111107 \\
 19 & -3609.111107 \\
 20 & -3609.111108
\end{array}
\right)$$