text:
\chapter{Lattice networks}
\label{sec:lattice-networks}

\headerquote{Observe also (what modern writers almost forgot, but some older writers, such as Euler and Laplace, clearly perceived) that the role of inductive evidence in mathematical investigation is similar to its role in physical research.}{---~G.~P\'{o}lya}

The integer lattices $\bZd \ci \bRd$ are some of the most widely-studied infinite graphs and have an extensive literature; see \cite{DoSn84,Telcs06a}, for example. We begin with some results for the simple lattices; in \S\ref{sec:nonsimple-integer-lattices} we consider the case when \cond is nonconstant. Because the case when $\cond = \one$ is amenable to Fourier analysis, we are able to compute many explicit formulas for many expressions, including $v_x$ and $R(x,y)$. For $d \geq 3$, we even compute $R(x,\iy) = \lim_{y \to \iy} R(x,y)$ in Theorem~\ref{thm:finite-resistance-to-infinity-in-Zd} and give a formula for the monopole $w$. There is a small amount of overlap here with the results of \cite[\S{V.2}]{Soardi94}, where the focus is more on solving Poisson's equation $\Lap u = f$. In \S\ref{sec:magnetism} we employ our formulas in the refinement of an application to the isotropic Heisenberg model of ferromagnetism.

In the present context, we may choose canonical representatives when working pointwise: given $u \in \HE$, we use the representative which tends to 0 at infinity. We take this as a standing assumption for this section, as it allows us to use the Fourier transform without ambiguity or unnecessary technical details. To see that this is justified, note that $\ell^2(\cond)$ is dense\footnote{Technically, the embedded image of $J:\ell^2(\cond) \to \Fin$ is dense in \Fin; see Definition~\ref{def:J-quotient-inclusion-map}.} in \Fin by Theorem~\ref{thm:l2c-in-HE}, and hence dense in \HE for these examples, as it is well-known that there are no nonconstant harmonic functions of finite energy on the integer lattices (we provide a proof in Theorem~\ref{thm:harmonics-on-Zd-are-linear} for completeness). Clearly, $\cond=\one$ implies that $\ell^2(\cond) = \ell^2(\one)$ and that all elements of $\ell^2(\cond)$ vanish at \iy.

\begin{remark}\label{rem:lattices-and-numerical-analysis}
  As mentioned in Remark~\ref{rem:numerical-analysis}, one of the applications of the present investigation is to numerical analysis. Discretization of the real line amounts to considering a graph which is a scaled copy of the integers $\Graph_\epsilon = (\epsilon\bZ, \frac1\epsilon \one)$ where $\epsilon \bZ = \{\epsilon n \suth n\in \bZ\}$. After finding the solution to a given problem, as a function of the parameter $\epsilon$, one lets $\epsilon \to 0$. Let $x_n$ denote the vertex at $\epsilon n$.
  \linenopax
    \begin{align*}
    \xymatrix{
      \dots \ar@{-}[r]_{1/\epsilon}
      & x_{-2} \ar@{-}[r]_{1/\epsilon}
      & x_{-1} \ar@{-}[r]_{1/\epsilon}
      & x_{0} \ar@{-}[r]_{1/\epsilon}
      & x_{1} \ar@{-}[r]_{1/\epsilon}
      & x_{2} \ar@{-}[r]_{1/\epsilon}
      & x_{3} \ar@{-}[r]_{1/\epsilon} & \dots
    }
  \end{align*}
  The \emph{difference operator} $D$ acts on a function on this network by $Df(x_n) := f(x_n)-f(x_{n+1})$. The adjoint of $D$ with respect to $\ell^2$ is $D^\ad f(x_n) = f(x_n) - f(x_{n-1})$. Then $D^\ad D = \Lap$.
  \begin{comment}
   and
  \linenopax
  \begin{align*}
    D^\ad D f(x_n)
    = D f(x_n) - D f(x_{n-1})
    = (f(x_n) - f(x_{n-1})) - (f(x_{n-1}) - f(x_{n-2}))
    = \Lap f(x_{n-1}).
  \end{align*}
  \end{comment}
\end{remark}

\section{Simple lattice networks}
\label{sec:simple-lattice-networks}

\begin{exm}[Simple integer lattices]\label{exm:infinite-lattices}
  The lattice network $(\bZd, \cond)$, with an edge between any two vertices which are one unit apart is called \emph{simple} or \emph{translation-invariant} when $\cond=\one$. The term ``simple'' originates in the literature on random walks.
\end{exm}

  One may compute the energy kernel directly using \eqref{eqn:def:R(x,y)-Lap}, that is, by finding a solution $v_x$ to $\Lap v = \gd_x - \gd_0$ as depicted in Figure~\ref{fig:vx-in-Z1}. Then $R(o,x) = v_x(x)-v_x(o) = x-0 = x$, which is unbounded as $x \to \iy$. This also provides an example of a function $v_x \in \HE$ for which $v_x \notin \ell^2(\cond)$, as discussed in \S\ref{sec:relating-HE-and-L2} and elsewhere.
  \linenopax
  \begin{figure}
    \centering
    \scalebox{1.0}{\includegraphics{vx-in-Z1.pdf}}
    \caption{\captionsize The function $v_x$, a solution to $\Lap v = \gd_x - \gd_0$ in $(\bZ,\one)$.}
    \label{fig:vx-in-Z1}
  \end{figure}
  In Lemma~\ref{thm:vx-on-Zd} we obtain a general formula for $v_x$ on $(\bZd, \one)$. Figure~\ref{fig:fejer-in-Z1} of Example~\ref{exm:Z-not-bounded} shows how this compares to Figure~\ref{fig:vx-in-Z1}.

  To see how the function $v=v_{x_1} = \charfn{[1,\iy)}$ may be approximated by elements of \Fin, define
  \linenopax
  \begin{equation}
    u_n(x_k) =
    \begin{cases}
      1-\frac kn, &1 \leq k \leq n, \\
      0, &\text{else}.
    \end{cases}
  \end{equation}
  The reader can verify that $u_n$ minimizes $\energy(v-u)$ over the set of $u$ for which $\spt(u) \ci [1,n-1]$ and that
  \linenopax
  \begin{align*}
    \energy(v-u_n)
    &= (1-(1-\tfrac1n))^2 + \sum_{k=1}^{n-1} \left((1-\tfrac kn)-(1-\tfrac{k-1}n)\right)^2 + (1-(1-0))^2 \\
    &= \frac1{n^2} + \frac{n-1}{n^2} \limas{n} 0.
  \end{align*}
  The fact that $v \in \clspn{\Fin}$ but $\lim_{k \to \iy} v(x_{k}) \neq \lim_{k \to \iy} v(x_{-k})$ reflects that $(\bZ,\one)$ has two \emph{graph ends}, unlike the other integer lattices; cf. \cite{PicWoess90}.
  Therefore, $(\bZ,\one)$ also provides an example of a network with more than one end which does not support nontrivial harmonic functions.

An explicit formula is given for the potential configuration functions $\{v_x\}$ on the simple $d$-dimensional lattice in Lemma~\ref{thm:vx-on-Zd}. By combining this formula with the dipole formulation of resistance distance
\linenopax
\begin{equation*}
  R(x,y) = v(x)-v(y), \qq \text{for } v=v_x-v_y,
  \text{from \eqref{eqn:def:R(x,y)-Lap},}
\end{equation*}
we are able to compute an explicit formula for resistance distance on the translation-invariant lattice network \bZd in Theorem~\ref{thm:R(x,y)-on-Zd}. Results in this section exploit the Fourier duality $\bZd \simeq \bTd$; \cite{Rud62} is a good reference. We are using Pontryagin duality of locally compact abelian groups; as an additive group of rank $d$, the discrete lattice \bZd is the dual of the $d$-torus \bTd. Conversely, \bTd is the compact group of unitary characters on \bZd (the operation in \bTd is complex multiplication). This duality is the basis for our Fourier analysis in this context. For convenience, we view \bTd as a $d$-cube, i.e., the Cartesian product of $d$ period intervals of length $2\gp$. In this form, the group operation in \bTd is written additively and the Haar measure on \bTd is normalized with the familiar factor of $(2\gp)^{-d}$.

In \cite{Polya21}, P\'{o}lya proved that the random walk on the simple integer lattice is transient if and only if $d \geq 3$; see \cite{DoSn84} for a nice introduction and a proof using \ERNs. In the present context, this can be reformulated as the statement that there exist monopoles on \bZd if and only if $d \geq 3$. We offer a new characterization of this dichotomy, which we recover in Theorem~\ref{thm:monopoles-on-Zd} via a new (and completely constructive) proof. In Remark~\ref{rem:HE(Zd)-lies-in-L2(Zd)} we show that in the infinite integer lattices, functions in \HE may be approximated by functions of finite support.

Sometimes P\'{o}lya's result is restated: the resistance to infinity is finite if and only if $d \geq 3$. There is an ambiguity in this statement which is specific to the nature of resistance metric. One interpretation is that one can construct a \emph{unit flow to infinity}; this is the terminology of \cite{DoSn84} for a current with $\act(\curr) = \gd_x$ and it is clear that this is the induced current of a monopole. Probabilistically, this definition may be rephrased: for a random walk beginning at $x \in \verts$, the expected hitting time of the sphere of (shortest-path) radius $n$ remains bounded as $n \to \iy$. This approach interprets ``infinity'' as the ``set of all points at infinity''.

By contrast, we prove a much stronger result for the simple lattice networks \bZd in  Theorem~\ref{thm:finite-resistance-to-infinity-in-Zd}, where we show $\lim_{y \to \iy} R(x,y)$ is bounded as $y \to \iy$, for any $x \in \verts$. To see the strength of this result, note that the simple ($\cond = \one$) homogeneous trees of degree $d \geq 3$ have finite resistance to infinity, even though $\lim_{y \to \iy} R(x,y) = \iy$ for any $x \in \verts$, and any choice of $y \to \iy$. This is discussed further in Example~\ref{exm:binary-tree:nontrivial-harmonic} of the previous section. The heuristic explanation is that the resistance distance between two places is much smaller when there is high connectivity between them; there is much more connectivity between $x$ and the ``set of all points at infinity'' than between $x$ and a single ``point at infinity''.

In the next result, we obtain the Fourier transform of the Laplacian; we recently noticed that this corresponds almost identically to the inverse Fourier transform $H$ of the ``potential kernel'' of \cite[\S{V.2}]{Soardi94}.

\begin{lemma}\label{thm:Fourier-transform-of-Lap-on-Zd}
  On the \ERN $(\bZd, \one)$, the spectral (Fourier) transform of \Lap is multiplication by $S(t) = S(t_1,\dots,t_d) = 4 \sum_{k=1}^d \sin^2\left(\frac{t_k}2\right)$.
  \begin{proof}
    Each point $x$ in the lattice \bZd has $2d$ neighbours, so we need to find the $L^2(\bT^d)$ Fourier representation of
    \linenopax
    \begin{equation}\label{eqn:Fourier-transform-of-Lap-on-Zd}
      \Lap v(x) = (2d\id - T) v(x) = 2d v(x) - \sum_{k=1}^d v(x_1,\dots,x_k \pm 1,\dots x_d).
    \end{equation}
    Here, $t = (t_1, \dots, t_d) \in \bT^d$ and $x = (x_1, \dots, x_d) \in \bZ^d$. The \kth entry of $t$ can be written $t_k = t \cdot \ge_k$ where $\ge_k = [0,\dots,0,1,0,\dots,0]$ has the 1 in the \kth slot. Then moving one step in the lattice by $x \mapsto x+\ge_k$ corresponds to $e^{\ii x \cdot t} \mapsto e^{\ii t_k} e^{\ii x \cdot t}$ under the Fourier transform, and
    \linenopax
    \begin{align*}

      \widehat{\Lap v}(t) &= \left(2d - \sum_{k=1}^d (e^{\ii t_k} + e^{-\ii t_k} )\right) \hat v(t) \\
      &= 2\left(\sum_{k=1}^d (1-\cos(t_k))\right) \hat v(t) \\
      &= 4 \sum_{k=1}^d \sin^2\left(\frac{t_k}2\right) \hat v(t).
      \qedhere
    \end{align*}
  \end{proof}
\end{lemma}

\begin{lemma}\label{thm:vx-on-Zd}
  Let $\{v_x\}_{x \in \bZ^d}$ be the potential configuration on the integer lattice $\bZd$ with $\cond=\one$. Then for $y \in \bZd$,
  \linenopax
  \begin{equation}\label{eqn:vx-on-Zd}
    v_x(y) = \frac1{(2\gp)^d} \int_{\bT^d} \frac{\cos((x-y)\cdot t) - \cos(y\cdot t)}{S(t)} \,dt.
  \end{equation}
  \begin{proof}
    Under the Fourier transform, Lemma~\ref{thm:Fourier-transform-of-Lap-on-Zd} indicates that the equation $\Lap v_x = \gd_x - \gd_o$ becomes $S(t) \hat v_x = e^{\ii x\cdot t} - 1$, whence
    \linenopax
    \begin{align}\label{eqn:vx-on-Zd-as-exponential}
      v_x(y) = \frac1{(2\gp)^d} \int_{\bT^d} e^{-\ii y \cdot t} \frac{e^{\ii x\cdot t} - 1}{S(t)} \,dt.
    \end{align}
    Since we may assume $v_x$ is \bR-valued, the result follows.
  \end{proof}
\end{lemma}

The following result is well-known in the literature (cf.~\cite{DoSn84,Nash-Will59}, e.g.), but usually stated in terms of the current flow induced by the monopole.

\begin{theorem}\label{thm:monopoles-on-Zd}
  The network $(\bZd,\one)$ has a monopole
  \linenopax
  \begin{equation}\label{eqn:monopoles-on-Zd}
    w(x) = -\frac1{(2\gp)^d} \int_{\bTd} \frac{\cos(x \cdot t)}{S(t)}\,dt
  \end{equation}
  if and only if $d \geq 3$, in which case the monopole is unique.
  \begin{proof}
    As in the proof of Lemma~\ref{thm:vx-on-Zd}, we use the Fourier transform to solve $\Lap w = -\gd_o$ by converting it into $S(t) \hat w(t) = -1$.
    This gives \eqref{eqn:monopoles-on-Zd}, and since $\frac{\cos t}{S(t)} \approx \frac1{S(t)} \in L^1(\bTd)$ for $t \approx 0$, the integral is finite iff $d \geq 3$ by the same argument as in the proof of Theorem~\ref{thm:finite-resistance-to-infinity-in-Zd}; see \eqref{eqn:exm:infinite-lattices:spherical-integral}.

    It remains to check that $w \in \HE$. Note that it follows from Theorem~\ref{thm:harmonics-on-Zd-are-linear} that the boundary term of \eqref{eqn:E(u,v)=<u_0,Lapv>+sum(normals)} vanishes, and hence we may compute the energy for $d \geq 3$ via
    \linenopax
    \begin{align}\label{eqn:energy-of-monopole-on-Zd}
      \|w\|_\energy &= \| \Lap^{1/2} w\|_2
      = \int_{\bTd} S(t) \hat w(t)^2\,dt
      = \int_{\bTd} \frac1{S(t)} \,dt
      < \iy.
    \end{align}

    Uniqueness is an immediate corollary of the previous theorem; if $w'$ were another, then $\Lap(w-w') = \gd_o - \gd_o = 0$ and $w-w'$ is constant by  Theorem~\ref{thm:harmonics-on-Zd-are-linear}.
  \end{proof}
\end{theorem}

\begin{remark}
  \version{}{\marginpar{FINISH THIS}}
  Upon comparing \eqref{eqn:monopoles-on-Zd} to \eqref{eqn:vx-on-Zd}, it is easy to see why all networks support finite-energy dipoles: the numerator in the integral for the monopole is of the order of 1 for $t \approx 0$, while the corresponding numerator for the dipole is $o(t)$ for $t \approx 0$.
\end{remark}

\begin{theorem}\label{thm:R(x,y)-on-Zd}
  Resistance distance on the integer lattice $(\bZd,\one)$ is given by
  \linenopax
  \begin{equation}\label{eqn:R(x,y)-on-Zd}
    R(x,y) = \frac1{(2\gp)^d} \int_{\bT^d} \frac{\sin^2((x-y)\cdot \frac t2)} {\sum_{k=1}^d \sin^2\left(\frac{t_k}2\right)} \,dt.
  \end{equation}
  \begin{proof}
    Let $\{v_x\}_{x \in \bZd}$ be the potential configuration on \bZd. Then $v_x-v_y \in \Pot(x,y)$, so by \eqref{eqn:def:R(x,y)-energy} we may use \eqref{eqn:def:R^F(x,y)-Lap}to compute the resistance distance via $R(x,y) = v_x(x) + v_y(y) - v_x(y) - v_y(x)$, since $R^F=R^W$ on \bZd. Using $e_x = e^{\ii x \cdot t}$, substitute in the terms from \eqref{eqn:vx-on-Zd-as-exponential} of Lemma~\ref{thm:vx-on-Zd}:
    \linenopax
    \begin{align}
      R(x,y)
      &= \frac1{(2\gp)^d} \int_{\bTd} \frac{\cj{e_x}(e_x-1) + \cj{e_y}(e_y-1) - \cj{e_x}(e_y-1) - \cj{e_y}(e_x-1)}{S(t)} \,dt \notag \\
      &= \frac1{(2\gp)^d} \int_{\bTd} \frac{1-\cancel{\cj{e_x}} + 1 - \cancel{\cj{e_y}} - e_{y-x} + \cancel{\cj{e_x}} - \cj{e_{y-x}} + \cancel{\cj{e_y}}}{S(t)} \,dt \notag \\
      &= \frac1{(2\gp)^d} \int_{\bTd} \frac{2-2\cos((x-y) \cdot t)}{S(t)} \,dt, \label{eqn:vx-on-Zd-cos}
    \end{align}
    and the formula follows by the half-angle identity.
  \end{proof}
\end{theorem}

\begin{cor}\label{thm:neighbours-in-Zd-are-1/d-apart}
  If $y \nbr x$, then $R(x,y) = \frac1d$ on $(\bZd,\one)$.
  \begin{proof}
    The symmetry of $(\bZd,\one)$ indicates that the distance from $x$ to its neighbour will not depend on which of the $2d$ neighbours is chosen. For $k=1,2,\dots,d$, let $y_k$ be a neighbour of $x$ in the \kth direction. Then  \eqref{eqn:vx-on-Zd} gives
    \begin{equation}\label{eqn:nbr-distance-in-Zd}
      R(x,y_k)
      = \frac1{(2\gp)^d} \int_{\bT^d} \frac{\sin^2(\frac{t_k}2)} {\sum_{k=1}^d \sin^2\left(\frac{t_k}2\right)} \,dt.
    \end{equation}
    Thus, $\sum_{k=1}^d R(x,y_k) = 1$ and $R(x,y_k) = R(x,y_j)$ gives the result.
  \end{proof}
\end{cor}

\begin{theorem}\label{thm:finite-resistance-to-infinity-in-Zd}
  The metric space $((\bZd,\one),R)$ is bounded if and only if $d \geq 3$, in which case
  \linenopax
  \begin{align}\label{eqn:limR(x,y)-on-Zd}
    \lim_{y \to \iy} R(x,y)
    = \frac2{(2\gp)^d} \int_{\bT^d} \frac{1} {S(t)}\,dt
    \qq\text{for } d \geq 3.
  \end{align}
  \begin{proof}
    This result hinges upon the convergence properties of the integrand for $R(x,y)$ as computed in Lemma~\ref{thm:R(x,y)-on-Zd}. In particular, to see that $1/S(t) \in L^1(\bT^d)$ one only needs to check for $t \approx 0$, where
    \linenopax
    \begin{align*}
      \frac{1}{S(t)} = O\left(\frac{1}{\sum t_k^2}\right), \qq \text{as } t \to 0.
    \end{align*}
    Switching to spherical coordinates, $1/S(\gr) = O\left(\gr^{-2}\right)$, as $\gr \to 0$, and one requires
    \linenopax
    \begin{equation}\label{eqn:exm:infinite-lattices:spherical-integral}
      \int_0^1 |\gr^{-2}| \gr^{d-1} \, dS_{d-1} < \iy,
    \end{equation}
    where $dS_{d-1}$ is the usual $(d-1)$-\dimnl spherical measure. Of course, \eqref{eqn:exm:infinite-lattices:spherical-integral} holds precisely when $-2+d-1>-1$, i.e., when $d > 2$. Similarly, the function $\cos((x-y) \cdot t)/S(t) \in L^1(\bT^d)$ iff $d \geq 3$. Therefore, \eqref{eqn:vx-on-Zd-cos} gives
    \linenopax
    \begin{align*}
      R(x,y)
      = \frac2{(2\gp)^d} \int_{\bT^d} \frac{1} {S(t)}\,dt
       -\frac2{(2\gp)^d} \int_{\bT^d} \frac{\cos((x-y) \cdot t)}{S(t)} \,dt,
      \qq\text{for } d \geq 3.
    \end{align*}
    Now replace $y$ with a sequence of vertices tending to infinity as in Definition~\ref{def:path-to-infinity}. By the Riemann-Lebesgue lemma, the second integral vanishes and for any such $y \to \iy$, we have \eqref{eqn:limR(x,y)-on-Zd}. Note that this is independent of $x \in \verts$, as one would expect from the translational invariance of the network, since $\cond = \one$.
  \end{proof}
\end{theorem}

\begin{defn}\label{def:distance-to-infty}
  Denote $R_\iy := \lim_{y \to \iy} R(o,y)$, as it is clear from the previous result that the limit does not depend on the choice of $y$.
\end{defn}

\begin{cor}\label{thm:proximity-of-infinity}
  For $d \geq 3$, there exists $x \in \bZd$ for which $R(o,x) > R_\iy$.
  \begin{proof}
     From \eqref{eqn:R(x,y)-on-Zd}, it is clear that $R(o,x) \leq R(o,\iy)$ if and only if
    \linenopax
    \begin{align*}
      \frac1{(2\gp)^d} \int_{\bTd} \frac{1-\cos(x \cdot t)}{S(t)}\,dt
      \leq \frac1{(2\gp)^d} \int_{\bTd} \frac{1}{S(t)}\,dt,
    \end{align*}
    which is equivalent to
    \linenopax
    \begin{align}\label{eqn:black-hole}
      \frac1{(2\gp)^d} \int_{\bTd} \frac{\cos(x \cdot t)}{S(t)} \geq 0.
    \end{align}
    However, \eqref{eqn:black-hole} cannot hold for all $x \in \bZd$, as such an inequality would mean that all Fourier coefficients of $w$ are nonnegative, in violation of Heisenberg's uncertainty principle.
  \end{proof}
\end{cor}

\begin{remark}\label{rem:black-hole}
  Corollary~\ref{thm:proximity-of-infinity} leads to the paradoxical conclusion that given $x \in \verts$, there may be a $y$ which is ``further from $x$ than infinity''. This is the case for $d=3$; numerical computation of \eqref{eqn:limR(x,y)-on-Zd} gives
  \begin{equation}\label{eqn:R_iy-in-Z3}
    \lim_{y \to \iy} R(x,y) \approx 0.5054620038965394,
    \qq \text{in } \bZ^3,
  \end{equation}
  and for $y=(1,1,1)$,
  \linenopax
  \begin{equation}\label{eqn:R(x,y)>R(x,infty)}
    R\left(o,y\right) \approx 0.5334159062457338.
  \end{equation}
  In fact, numerical computations indicate the following extremely bizarre situation:
  \linenopax
  \begin{align*}
    R(x,y_{2k}) < R(x,\iy) < R(x,y_{2k+1}), \qq \text{for } y_n := (n,n,0).
  \end{align*}
\end{remark}

\begin{figure}
  \centering
  \scalebox{0.70}{\includegraphics{Z3-black-hole.pdf}}
  \caption{\captionsize In $\bZ^3$, it may happen that $R(x,y) > R(x,\iy)$, where $R(x,\iy) = \lim_{z \to \iy} R(x,z)$. This phenomenon is represented here schematically as a ``black hole''.}
  \label{fig:Z3-black-hole}
\end{figure}

\begin{remark}\label{rem:Bochner-on-Zd}
  An application of Bochner's Theorem (see Theorem~\ref{thm:Bochner's-theorem}) yields a unique Radon probability measure \prob on \bTd such that
  \linenopax
  \begin{align*}
    \int_{\bTd} e^{\ii t \cdot x} \,d\prob(t) = e^{-\frac12 R(o,x)},
    \q \forall x \in \bZd.
  \end{align*}
\end{remark}

\begin{cor}\label{thm:vx-in-L2(Zd)-for-big-d}
  For $(\bZd,\one)$, $v_x \in \ell^2(\bZd)$ if and only if $d \geq 3$.
  \begin{proof}
    By computations similar to those in the proof of Theorem~\ref{thm:finite-resistance-to-infinity-in-Zd}, one can see that in absolute values, the integrand $\left|(e^{\ii x\cdot t} - 1)/S(t)\right|$ of \eqref{eqn:vx-on-Zd-as-exponential} is in $L^2(\bTd)$ if and only if $d \geq 3$, in which case Parseval's theorem applies.
  \end{proof}
\end{cor}

\begin{cor}\label{thm:w-in-L2(Zd)-for-big-d}
  For $(\bZd,\one)$, the monopole $w \in \ell^2(\bZd)$ if and only if $d \geq 5$.
  \begin{proof}
    The proof is almost identical to that of Corollary~\ref{thm:vx-in-L2(Zd)-for-big-d}, except that the integrand is $1/S(t)$, which is in $L^2(\bTd)$ if and only if $d \geq 5$.
  \end{proof}
\end{cor}

\begin{exm}\label{exm:Z-not-bounded}
  To see why $R$ is not bounded on $(\bZ,\one)$, one can evaluate \eqref{eqn:vx-on-Zd} explicitly via the Fej\'{e}r kernel:
  \linenopax
  \begin{align*}
    R(x,y)
    &= \frac1{(2\gp)} \int_{\bT} \frac{\sin^2((x-y)\frac t2)} {\sin^2\left(\frac{t}2\right)} \,dt \\
    &= \frac1{(2\gp)} \int_{\bT} |x-y|\frac{\sin^2((x-y)\frac t2)} {|x-y| \sin^2\left(\frac{t}2\right)} \,dt \\
    &= |x-y| \frac1{(2\gp)} \int_{\bT} F_N(t) \,dt \\
    &= |x-y|.
  \end{align*}
  where $F_N(t)$ is the Fej\'{e}r kernel with $N=|x-y|$; see Figure~\ref{fig:fejer-in-Z1}. Of course, this was to be expected because $R$ coincides with shortest path metric on trees.
  \begin{figure}
    \centering
    \scalebox{0.90}{\includegraphics{fejer.pdf}}
    \caption{\captionsize The function $v_x$, for $x=1,2,3$ in \bZ, as obtained from the Fej\'{e}r kernel. See Example~\ref{exm:Z-not-bounded}.}
    \label{fig:fejer-in-Z1}
  \end{figure}
\end{exm}

The following result is well known; we include it for completeness and the novelty of the proof.
\begin{theorem}\label{thm:harmonics-on-Zd-are-linear}
  $h$ is a harmonic function on $(\bZd,\one)$ if and only if $h$ is linear (or affine). Consequently, $\HE = \clspn{\Fin}$ for \bZd.
  \begin{proof}
    \version{}{\marginpar{Use proof from Rem.~\ref{rem:HE(Zd)-lies-in-L2(Zd)}.}}
    From $\Lap h = 0$, the Fourier transform gives $S(t) h(t)=0$. By the formula of Lemma~\ref{thm:Fourier-transform-of-Lap-on-Zd} for $S(t)$, this means $\hat h$ can only be supported at $t=0$ and hence that $\hat h$ is a distribution which is a linear combination of derivatives of the Dirac mass at $t=0$; see \cite{Rud91} for this structure theorem from the theory of distributions.

    Denoting this by $\hat h(t) = P(\gd_o)$, where $\gd_o$ is the Dirac mass at $t=0$, and $P$ is some polynomial. The inverse Fourier transform gives $h(x) = P(x)$. If the degree of $P$ is 2 or higher, then $\Lap h$ will have a constant term of $-2d$ (cf.~\eqref{eqn:Fourier-transform-of-Lap-on-Zd}) and hence cannot vanish identically.

    It is clear that a linear function on \bZd has infinite energy; consequently \Harm is empty on this network and the second conclusion follows.
  \end{proof}
\end{theorem}

\begin{exm}[Nontrivial harmonic functions on $\bZd \cup \bZd$]
  \label{exm:Nontrivial-harmonic-functions-on-2Zd}
  Consider the disjoint union of two copies of \bZd, with $\cond = \one$ and $d \geq 3$. Now connect the origins $o_1,o_2$ of the two lattices with a single edge of conductance $\cond_{o_1o_2} = 1$. Let $w_1 \in \HE$ be a monopole on the first copy of \bZd, as ensured by Theorem~\ref{thm:monopoles-on-Zd}. We may assume $w_1$ is normalized so that $w(o_1)=1$, and then extend $w_1$ to the rest of the network by letting $\tilde w_1(x) = 0$ for all $x$ in the second copy of \bZd. Similarly, let $w_2$ be a function which is a monopole on the second \bZd, satisfies $w(o_2)=1$, and extend it to $\tilde w_2$ by defining $\tilde w_2(x) = 0$ for $x$ in the first copy of \bZd. Now one can check that $\Lap \tilde w_1 = \Lap \tilde w_2 = -\gd_{o_2}$. Note that the unit drop in $\tilde w_1$ across the edge $\cond_{o_1o_2}$ moves the Dirac mass of $\Lap w_1$ to the second copy of \bZd. Now define
  \linenopax
  \begin{align}\label{eqn:harmonic-functions-on-2Zd}
    h := w_1 - w_2.
  \end{align}
  It is easy to check that $h \in \HE$ and that $h \in \Harm$.
\end{exm}

\begin{cor}\label{thm:monopole-on-Zd-vanishes-at-infty}
\label{thm:monopole-on-Zd-as-resistance-distance}
  If $w$ is the monopole on $(\bZd,\one)$, $d \geq 3$, then
  \begin{equation}\label{eqn:monopole-on-Zd-as-resistance-distance}
    w(x) = \tfrac12 (R(o,x) - R(o,\iy)),
  \end{equation}
  and consequently $\lim_{x \to \iy} w(x) =0$.
  \begin{proof}
    Subtract \eqref{eqn:limR(x,y)-on-Zd} from \eqref{eqn:vx-on-Zd-cos} and compare to \eqref{eqn:monopoles-on-Zd}. For the latter statement, one can take the limit of \eqref{eqn:monopole-on-Zd-as-resistance-distance} as $x \to \iy$ directly or apply the Riemann-Lebesgue lemma to \eqref{eqn:monopoles-on-Zd}.
  \end{proof}
\end{cor}

\begin{cor}\label{thm:energy-of-monopole-on-Zd}
  If $w$ is the monopole on $(\bZd,\one)$, then
  \begin{equation}\label{eqn:energy-of-monopole-on-Zd}
    \energy(w) = \frac12 \lim_{y \to \iy} R(x,y).
  \end{equation}
  \begin{proof}
    Compare \eqref{eqn:energy-of-monopole-on-Zd} to \eqref{eqn:limR(x,y)-on-Zd} and note that $\Harm=\{0\}$, so $w$ is unique.
  \end{proof}
\end{cor}

\begin{remark}\label{rem:HE(Zd)-lies-in-L2(Zd)}
  For $(\bZd,\one)$, it is instructive to work out directly why \Fin is dense in \HE. That is, let us suppose that the boundary term vanishes for every $v \in \HE$, and use this to prove that every function which is orthogonal to \Fin must be constant (and hence 0 in \HE). This shows that \Fin is dense in \HE in the energy norm.

  \begin{proof}[``Proof'']
    If $v \in \HE$, then $\|v\|_\energy = \la v, \Lap v\ra_\cond < \iy$, the Fourier transform sends $v \mapsto \hat v(t) = \sum_\bZ v_n e^{\ii n \cdot t}$ and
    \linenopax
    \begin{align}\label{eqn:thm:HE(Zd)-lies-in-L2(Zd):orthog-to-Dirac}
      \la v, \Lap v\ra_\cond
      \mapsto (2\gp)^{-d} \int_{\bTd} \hat v(t) S(t) \hat v(t) \,dt
      <\iy,

    \end{align}
    where $S(t) = 4\sum_{k=1}^d \sin^2\left(\tfrac{t_k}2\right)$, as in Lemma~\ref{thm:Fourier-transform-of-Lap-on-Zd}.
    Then note that the Schwarz inequality gives
    \linenopax
    \begin{align*}
      \left(\int_{\bTd} S(t) \hat v(t) \,dt\right)^2
      \leq \int_{\bTd} S(t) \,dt \int_{\bTd} S(t) \hat v(t)^2 \,dt,
    \end{align*}
    so that $S(t) \hat v(t) \in L^1(\bTd)$.
    From the other hypothesis, $v \perp \Fin$ means that $\la \gd_x, v \ra_\energy = 0$ for each $x \in \verts$, whence Parseval's equation gives
    \linenopax
    \begin{align*}
      0 &= \la \gd_{x_m}, v \ra_\energy
      = \la \gd_{x_m}, \Lap v \ra_\cond
      \mapsto (2\gp)^{-d} \int_{\bTd} e^{\ii m \cdot t} S(t) \hat v(t) \,dt
      = 0, \forall m.
    \end{align*}
    This implies that $S(t) \hat v(t) =0$ in $L^1(\bTd)$, and hence $\hat v$ can only be supported at $t=0$. From Schwartz's theory of distributions, this means
    \linenopax
    \begin{align*}
      \hat v(t) = f_0(t) + c_0 \gd_0(t) + c_1 D \gd_0(t) + c_2 D^{(2)} \gd_0(t) + \dots,
    \end{align*}
    where $f_0$ is an $L^1$ function and all the other terms are derivatives of the Dirac mass at $t=0$ ($D^{(2)}$ is a differential operator of rank 2, etc.).

    If $\hat v$ is just a function, then it is 0 a.e. and we are done. If the distribution $\gd_0(t)$ is a component of $\hat v$, then $\sF^{-1}(\gd_0) = \one$, which is zero in \HE. In one dimension, the distribution $\gd_0'(t)$ cannot be a component of $\hat v$ because  $\sF^{-1}(\gd_0')(x_m) = m$, and this function does not have finite energy (the computation of the energy picks up a term of 1 on every edge of the lattice \bZd). The computation is similar for higher derivatives of $\gd_0$, but they diverge even faster. For higher dimensions, note that $D_1 \gd_{(0,0)} = D \gd_0 \otimes \gd_0$ and $\energy(D \gd_0 \otimes \gd_0) = \energy(D \gd_0) \energy(\gd_0)$ (this is a basic fact about quadratic forms on a Hilbert space), and so this devolves into same argument as for the 1-\dimnl case.
  \end{proof}
\end{remark}

Remark~\ref{rem:HE(Zd)-lies-in-L2(Zd)} does not hold for general graphs; see Example~\ref{exm:binary-tree:nontrivial-harmonic}. Also, the end of the proof shows why $\tilde{\Lap}^{1/2} (\widehat{v+k}) = \tilde{\Lap}^{1/2} \hat v$, as mentioned in Remark~\ref{rem:LapTilde-doesn't-see-constants}; the addition of a constant corresponds on the Fourier side to the addition of a Dirac mass outside the support of \gc.

The case of $(\bZd, \one)$, for $d=1$ is a tree and hence very simple with $R(x,y) = |x-y|$, and for $d \geq 3$ may be fairly well understood by the formulas given above. However, the case $d=2$ seems to remain a bit mysterious. It appears that $R(x,y) \approx \log(1+|x-y|)$; we now give two results in this direction.

\begin{remark}\label{rem:asymptotic-neighobur-difference-R(x,y)}
  From Theorem~\ref{thm:finite-resistance-to-infinity-in-Zd} it is clear that for $d \geq 3$, if $y_n \nbr z_n$ and both tend to \iy, one has $\lim_{n \to \iy} (R(x,y_n)-R(x,z_n)) =0$. In fact, this remains true in $\bZ^2$ but not \bZ. For \bZ,
  \linenopax
  \begin{align*}
    y_n \nbr z_n \q\implies\q |R(x,y_n)-R(x,z_n)|=1 \limas{n} 1 \neq 0.
  \end{align*}
  A little more work is required for $\bZ^2$, where we work with $x=o$ for simplicity:
  \linenopax
  \begin{align*}
    R(o,z) - R(o,y)
    &= \frac1{(2\gp)^2} \int_{\bT^2} \frac{\cos(y \cdot t) - \cos(y \cdot t + t_k)} {S(t)} \,dt \\
    &= \frac1{(2\gp)^2} \int_{\bT^2} \frac{\cos(y \cdot t)(1-\cos t_k) + \sin(y \cdot t) \sin t_k} {S(t)} \,dt.
  \end{align*}
  One can check that $\frac{1-\cos t_k}{S(t)},\frac{\sin t_k}{S(t)} \in L^1(\bT^2)$ by converting to spherical coordinates and making the estimate
  \linenopax
  \begin{align*}
    \int_{\bT^2} \frac{\gr}{\gr^2} \gr \, d\gr \, d\gq < \iy.
  \end{align*}
  Now the Riemann-Lebesgue Lemma shows that $R(o,y) - R(o,z)$ tends to 0 as $y$ (and hence also $z$) tends to \iy.
\end{remark}

\begin{theorem}\label{thm:gradR(x,y)-vanishes-at-infty}
  On $(\bZ^2, \one)$, the gradient of $R$ vanishes at infinity, i.e.,
  \begin{equation}\label{eqn:gradR(x,y)-vanishes-at-infty}
    \lim_{y \to \iy} \nabla R(x,y) = 0.
  \end{equation}
\end{theorem}

\begin{theorem}\label{thm:R(x,y)-in-Z2-as-contour-integral}
  On $(\bZ^2, \one)$, resistance distance is given by
  \begin{equation}\label{eqn:R(x,y)-in-Z2-as-contour-integral}
    R(x,y) = \int_0^x A(o,t) \,dt + \int_0^\iy B(x,t) \,dt,
  \end{equation}
  \begin{equation}\label{eqn:A1-and-A2}
    \text{where}\q
    A(s,t) := \frac{\del}{\del t_1}
    B(s,t) := \frac{\del}{\del t_2}
  \end{equation}
\end{theorem}

\section{Noncompactness of the transfer operator}

\begin{exm}[\Trans may not be the uniform limit of finite-rank operators]
  \label{exm:nonuniformly-converging-transfer-operator}
  Let \Graph be the integers \bZ with edges only between vertices of distance one apart (as in Example~\ref{exm:infinite-lattices} with $d=1$), with $\cond \equiv 1$. Then the transfer operator $T := \gs^+ + \gs^-$ consists of the sum of two unilateral shifts, for which the finite truncations (as described just above) are the banded matrices
  \begin{equation}\label{eqn:exm:nonuniformly-converging-transfer-operator}
    \Trans_N =
    \left[\begin{array}{rrrrrrrrr}
      0 &1 &0 &0 &\dots &0 &0 &0 \\
      1 &0 &1 &0 &\dots &0 &0 &0 \\
      0 &1 &0 &1 &\dots &0 &0 &0 \\
      0 &0 &1 &0 &\dots &0 &0 &0 \\
      \vdots &\vdots &\vdots &\vdots & \ddots &\vdots &\vdots &\vdots \\
      0 &0 &0 &0 &\dots &0 &1 &0 \\
      0 &0 &0 &0 &\dots &1 &0 &1 \\
      0 &0 &0 &0 &\dots &0 &1 &0
    \end{array}\right].
  \end{equation}
  Then consider the vectors
  \begin{equation}\label{eqn:exm:nonuniformly-converging-transfer-operator:vectors}
    \gx_n := (\underbrace{0,\dots,0}_{n\text{ zeros}}, 1, \tfrac12, \tfrac13, \tfrac14, \dots), \qq
    \|\gx_n\|^2 = 2\sum_{k=1}^\iy \frac1{k^2} = \frac{\gp^2}3.
  \end{equation}
  Then $\Trans_N$ does not converge to \Trans uniformly, because for $n=N$,
  \linenopax
  \begin{align*}
    \la \gx_n, (\Trans - \Trans_n)\gx_n\ra_\cond
    = \sum_{|k|>n} \gx_k \gx_{k+1}
    &= \sum_{|k|>n} \frac1{(k-n)(k-n+1)}
    = \sum_{|k| \geq 1} \frac1{k(k+1)} \\
    &\geq \sum_{|k| \geq 1} \left(\frac1{k+1}\right)^2
    \approx \frac{\gp^2}6,
  \end{align*}
  which is bounded away from 0 as $n \to \iy$.
\end{exm}

\subsection{The Paley-Wiener space $H_s$}
The transfer operator is not compact in \HE, as Example~\ref{exm:nonuniformly-converging-transfer-operator} shows. However, by introducing the correct weights we can obtain a compact operator, i.e., the transfer operator is compact when considered as acting on the correct Hilbert space. To this end, we make the identification between $\gx \in \ell^2(\bZ)$ and $f(z) = \sum_{n \in \bZ} \gx_n z^n \in L^2(\bT)$ via Fourier series, so that we may use analytic continuation and introduce the following spaces.

\begin{defn}\label{def:s-norm}
  For an function $f \in L^2(\bT)$ given by $f(z) = \sum_{n \in \bZ} \gx_n z^n$, we define
  \begin{equation}\label{eqn:def:s-norm}
    \|f\|_s := \left(\sum_{n \in \bZ} e^{s|n|} |\gx_n|^2 \right)^2,
  \end{equation}
  and consider the space.
  \begin{equation}\label{eqn:Hs-space}
    H_s := \{f : \bT \to \bT \suth \|f\|_s < \iy\}.
  \end{equation}
  For $s=0$ we recover good old $\ell^2(\cond)$, but for $s>0$, we have the subspace of $\ell^2(\cond)$ which consists of those functions with an analytic continuation to the annulus $\{z \suth 1 - s < |z| < 1 + s\}$ about \bT. In general, we have $H_s \ci L^2(\bT) \ci H_{-s} = H_s^\ast$.
\end{defn}

\begin{theorem}\label{thm:Trans-is-compact-on-Hs}
  The transfer operator is a compact operator on $H_s$.
  \begin{proof}
    Using $\Lap = \cond-\Trans$, we show that there exist solutions to $\Lap v = \gd_\ga - \gd_\gw$ by construction, using spectral theory. The Laplacian may be represented as the infinite symmetric banded matrix
    \linenopax
    \begin{align*}
      \left[\begin{array}{cccccc}
        \dots &\cond(x_1) &-\cond(x_1, y_1) &\dots \\
        \dots &-\cond(x_1, y_1) &\cond(x_2) &-\cond(x_2, y_2) &\dots \\
        \dots & &-\cond(x_2, y_2) &\cond(x_3) &-\cond(x_3, y_3) &\dots \\
        \dots & & &\ddots &\ddots &\ddots
      \end{array}\right]
    \end{align*}
    The symmetry is immediate from the symmetry of $\cond_{xy}$, of course.

    Using the same notation as in Example~\ref{exm:nonuniformly-converging-transfer-operator}, we must check that $\Trans_N \to \Trans$ uniformly in $H_s$, so we examine $D_N := \Trans - \Trans_N$:
\version{}{\marginpar{How to justify \eqref{eqn:thm:Trans-is-compact-on-Hs:deriv1}? This is supposed to be for any old \gx, right?}}
    \linenopax
    \begin{align}
      \la \gx, D_N \gx \ra
      &= \sum_{|n| \geq N} e^{s|n|} \cj{\gx_n} \gx_{n+1} \notag \\
      &\leq \left(\sum_{|n| \geq N} e^{s|n|} |\gx_{n}|^2\right)^{1/2}
        \left(\sum_{|n| \geq N} e^{s|n|} |\gx_{n+1}|^2\right)^{1/2} \notag \\
      &= \left(\sum_{|n| \geq 0} e^{s|n+N|} |\gx_{n+N}|^2\right)^{1/2}
        \left(\sum_{|n| \geq 0} e^{s|n+N|} |\gx_{n+N+1}|^2\right)^{1/2} \notag \\
      &= e^{-s/2} \left(e^{sN} |\gx_N|^2 + \sum_{|n| \geq N+1} e^{sn} |\gx_{n}|^2\right)^{1/2}
        \left(\sum_{|n| \geq N+1} e^{sn} |\gx_{n}|^2\right)^{1/2}
        \label{eqn:thm:Trans-is-compact-on-Hs:deriv1} \\
      \frac{\la \gx, D_N \gx\ra}{\|\gx\|_s}
      &= e^{-s/2} \frac{\left(\sum_{|n| \geq N+1} e^{sn} |\gx_{n}|^2\right)^{1/2}}{\left(e^{sN} |\gx_N|^2 + \sum_{|n| \geq N+1} e^{sn} |\gx_{n}|^2\right)^{-1/2}}
      \limas{N}&\; 0. \notag
    \end{align}
\version{}{\marginpar{The hat here indicates passing to the Calkin Alg, i.e, modding by compacts, right?}}
    Then \Trans is compact in $H_s$, and in fact, \Trans is trace class! Then $\hat{cpt} = 0$, so $\Lap = \id - \Trans$ implies $\hat{\Lap} = \hat{\Cond}$.
  \end{proof}
\end{theorem}

\section{Non-simple integer lattice networks}
\label{sec:nonsimple-integer-lattices}

In this section, we illustrate some of the phenomena that may occur on integer lattices when the conductances are allowed to vary.
\version{}{\marginpar{Add a decay example for $\lim \Graph$.}}
Many of these examples serve to demonstrate certain definitions or general properties discussed in previous sections.

\begin{exm}[Symmetry of the graph vs. symmetry of the network]
  \label{exm:2-dimnl-integer-lattice}
  Consider the 2-\dimnl integer lattice; the case $d=2$ in Example~\ref{exm:infinite-lattices}, and think of these points as living in the complex plane, so each vertex is $m+n\ii$, where $m$ and $n$ are integers and $\ii=\sqrt{-1}$. It is possible to define the conductances in such a way that a function $v(z)$ has finite energy, but $v(\ii z)$ does not (this is just precomposing with a symmetry of the graph: rotation by by $\frac\gp2$). However, $v(z)$ is in $\ell^2(\one)$ if and only if $v(\ii z)$ is in $\ell^2(\one)$. Thus, $\ell^2(\one)$ does not see the graph.

  Define the conductances by
  \linenopax
  \begin{align*}
    \cond_{xy} =
    \begin{cases}
      1, & y=x+1, \\
      2^{|\Im(y)|}, & y=x+\ii,
    \end{cases}
  \end{align*}
  so that the conductances of horizontal edges are all 1 and the conductances of vertical edges grow like $2^k$. Now consider the function
  \linenopax
  \begin{align*}
    v(z) = v(x+\ii y) =
    \begin{cases}
      2^{-|Re(x)|}, & y=0, \\
      0, & y \neq 0,
    \end{cases}
  \end{align*}
  When computing the energy $\energy(v)$, the only contributing terms are the edges along the real axis, and the edges immediately adjacent to the real axis:
  \linenopax
  \begin{align*}
    \energy(v(z))
    &= \mathrm{horizontal} + \mathrm{vertical} \\
    &= 2(1/2 + 1/4 + 1/8 + ...) + 4(1/2 + 1/4 + 1/8 + ...)
    = 6,
  \end{align*}
  which is finite. However,
  \begin{align*}
    \energy(v(\ii z))
    = 2(1 + 1 + 1 + ...) + 4(1/2 + 1/4 + 1/8 + ...) = \iy.
  \end{align*}
\end{exm}

\begin{exm}[An example where $\ell^2 \nsubseteq \HE$]
  \label{exm:1-dimnl-integer-lattice}
  Let $\bZ$ have $\cond_{n-1,n} = n$. Consider $\ell^2(\verts,\gn)$ where \gn is the counting measure. The Dirac functions $\gd_{x_k}$ satisfy $\|\gd_{x_k}\|_\unwt = 1$, so $\{\gd_{x_k}\}$ is a bounded \seq in $\ell^2(\unwtdsp)$. However, the Laplacian is
  \linenopax
  \begin{align*}
    \Lap_\unwt = \left[
    \begin{array}{rrr}
      \ddots \\ -n & 2n+1 & -(n+1) \\ &&\ddots
    \end{array}\right]
  \end{align*}
  and $\energy(\gd_{x_k}) = \la \gd_{x_k}, \Lap \gd_{x_k}\ra = 2k+1 \limas{k} \iy$. So we cannot have the bound $\|v\|_\energy \leq K \|v\|_\unwt$, for any constant $K$.

  \version{}{\marginpar{Why doesn't unitary equivalence provide an $\ell^2_\cond$ counterexample?}}
  This is ``corrected'' by using the measure \cond instead. In this case, $\|\gd_k\|_\cond = 2k+1$ so that $\{\gd_k\}$ is not bounded and we must use $\{\gd_k/\sqrt{\cond(k)}\}$. But then the Laplacian is
  \linenopax
  \begin{align*}
    \Lap_\cond = \left[
    \begin{array}{rrr}
      \ddots \\ -\tfrac{n}{2n+1} & 1 & -\tfrac{n+1}{2n+1} \\ &&\ddots
    \end{array}\right]
  \end{align*}
  and $\energy\left(\frac{\gd_{x_k}}{\sqrt{\cond(x_k)}}\right) = \frac1{\cond(x_k)} \energy(\gd_x) = 1$.
\end{exm}

\begin{exm}\label{exm:Unbounded-functions-of-finite-energy}
  It is quite possible to have unbounded functions of finite energy. Consider the network $(\Graph, \cond) = (\bZ,\one)$ with vertices at each integer and unit conductances to nearest nearest neighbours. Then it is simple to show that $u(n) = \sum_{i=1}^n \frac1n$ and $v(n) = \log |1+n|$ are unbounded and have finite energy --- use the identity $\sum_{n=1}^\iy \frac1{n^2}=\frac{\gp^2}6$. For $v$, note that $\log|1+n| - \log|1+(n-1)| = \log\left|1+\frac1n\right| \leq \frac1n$.
\end{exm}

\begin{exm}[An unbounded function with finite energy]
  \label{exm:unbounded-finite-energy-fn}
  Let $\bZ$ have $\cond_{n-1,n} = \frac1{n^2}$. Then the function $f(n)=n$ is clearly unbounded, but
  \linenopax
  \[\energy(f)
    = \sum \frac1{n^2}(f(n)-f(n-1))^2
    = \sum \frac1{n^2}
    = \frac{\gp^2}3 < \iy.\]
  Conclusion: it is possible to have unbounded functions of finite energy if \cond decays sufficiently fast.

\version{}{\marginpar{tree example}  It is natural to wonder if there exist unbounded harmonic functions of finite energy. We do not know the answer but have so far failed to produce any. The equation
  \linenopax
  \begin{align*}
    \energy(h) = \sum_{\bd \Graph} h \dn h < \iy
  \end{align*}
  leads us to expect it is not possible.}
\end{exm}

  We've seen that there are no harmonic functions of finite energy on $(\bZd,\cond)$, when $\cond = \one$. However, the situation is very different when \cond is not bounded.

\begin{theorem}\label{thm:harmonic-functions-on-summable-resistance-integers}
  $\Harm \neq 0$ for $(\bZ,\cond)$ iff $\sum \cond_{xy}^{-1} < \iy$. In this case, \Harm is spanned by a single bounded function.
  \begin{proof}
    \fwd Fix $u(0)=0$, define $u(1) = \frac1{\cond_{01}}$ and let $u(n)$ be such that
    \linenopax
    \begin{align}\label{eqn:harmonic-recipe-on-Z1}
      u(n) - u(n-1) = \frac1{\cond_{n-1,n}},
      \q\forall n.
    \end{align}
    Now $u$ is harmonic:
    \linenopax
    \begin{align*}
      \Lap u(n)
      &= \cond_{n-1,n}(u(n)-u(n-1)) - \cond_{n,n+1}(u(n+1)-u(n)) \\
      &= \cond_{n-1,n}\frac1{\cond_{n-1,n}} - \cond_{n,n+1}\frac1{\cond_{n,n+1}}
       = 0,
    \end{align*}
    and $u$ is of finite energy
    \linenopax
    \begin{align*}
      \energy(u)
      &= \sum_{n \in \bZ} \cond_{n-1,n}(u(n)-u(n-1))^2
       = \sum_{n \in \bZ} \frac1{\cond_{n-1,n}}
       < \iy.
    \end{align*}
    Note that once the values of $u(0)$ and $u(1)$ are fixed, all the other values of $u(n)$ are determined by \eqref{eqn:harmonic-recipe-on-Z1}. Therefore, \Harm is 1-dimensional.

    \bwd If $\Lap u(n) = \cond_{n-1,n}(u(n)-u(n-1)) - \cond_{n,n+1}(u(n+1)-u(n)) = 0$ for every $n$, then
    \linenopax
    \begin{align*}
      \cond_{n-1,n}(u(n)-u(n-1)) = \cond_{n,n+1}(u(n+1)-u(n)) = a,
    \end{align*}
    for some fixed $a$ (the amperage of a sourceless current). Then
    \linenopax
    \begin{align}\label{eqn:finite-energy-harmonic-implies-bounded-on-Z}
      \energy(u)
      = \sum_{n \in \bZ} \cond_{n-1,n}(u(n)-u(n-1))^2

      = a^2 \sum_{n \in \bZ} \frac{1}{\cond_{n-1,n}}
      < \iy,
    \end{align}
    since $u \in \Harm \ci \HE$. Note that \eqref{eqn:finite-energy-harmonic-implies-bounded-on-Z} implies $u$ is bounded: $\energy(u) = a \sum_{n \in \bZ}(u(n)-u(n-1))$ and $\sum_{n \geq 1} (u(n)-u(n-1)) = \lim_{n \to \iy} u(n) - u(0)$. The function $u$ is monotonic because it is harmonic, so the sum is absolutely convergent.
  \end{proof}
\end{theorem}

We will now explore a specific example of this kind of network, where explicit computations are tractable.

  \begin{figure}
    \centering
    \scalebox{1.0}{\includegraphics{vx-on-Zc.pdf}}
    \caption{\captionsize The functions $v_1$, $v_2$, and $v_3$ on $(\bZ,c^n)$. Also, the monopole $w_o$ and the projection $f_1 = \Pfin v_1$. See Lemma~\ref{thm:repkernels-on-geometric-integers}.}
    \label{fig:vx-on-Zc}
  \end{figure}

\begin{exm}\label{exm:geometric-integers}\label{def:geometric-integers}
  For a fixed constant $c>1$, let $(\bZ,c^n)$ denote the network with integers for vertices, and with geometrically increasing conductances defined by
  \linenopax
    \begin{align*}
    \cond_{n-1,n} = c^{\max\{|n|,|n-1|\}},
  \end{align*}
  so that the network under consideration is
  \linenopax
    \begin{align*}
    \xymatrix{
      \dots \ar@{-}[r]^{c^3}
      & -2 \ar@{-}[r]^{c^2}
      & -1 \ar@{-}[r]^{c}
      & 0 \ar@{-}[r]^{c}
      & 1 \ar@{-}[r]^{c^2}
      & 2 \ar@{-}[r]^{c^3}
      & 3 \ar@{-}[r]^{c^4}
      & \dots
    }
  \end{align*}
  We fix $o=0$.
\end{exm}

\begin{lemma}\label{thm:monopole-on-geometric-integers}\label{thm:repkernels-on-geometric-integers}
  On $(\bZ,c^n)$, the energy kernel is given by
  \linenopax
  \begin{align*}
    v_n(k) =
    \begin{cases}
      0, &k \leq 0, \\
      \frac{1-r^{k+1}}{1-r}, &1 \leq k \leq n, \\
      \frac{1-r^{n+1}}{1-r}, &k \geq n,
    \end{cases}
    n > 0,
  \end{align*}
  and similarly for $n < 0$.
  Furthermore, the function $w_o(n) = ar^{|n|}$, $a:= \frac{r}{2(1-r)}$, defines a monopole, and $h(n) = \operatorname{sgn}(n) (1-w_o(n))$ defines an element of \Harm.
  \begin{proof}
    It is easy to check that $\Lap w_o(0) = 2c(a-ar) = 1$, and that $\Lap w_o(n) = c^n(ar^n-ar^{n-1}) + c^{n+1}(ar^n-ar^{n+1}) = 0$ for $n \neq 0$. The reader may check that $\energy(w_o) = \frac r{2(1-r)}$ so that $w_o \in \HE$. The computations for $v_x$ and $h$ are essentially the same. See Figure~\ref{fig:vx-on-Zc}.

  \end{proof}
\end{lemma}

  In Figure~\ref{fig:vx-on-Zc}, one can also see that $f_1 = \Pfin v_1$ induces a current flow of 1 amp from 1 to 0, with  $\frac{1+r}2$ amps flowing down the 1-edge path from 1 to 0, and the remaining current of $\frac{1-r}2$ amps flowing down the ``pseudo-path'' from 1 to $+\iy$ and then from $-\iy$ to 0.

  \begin{figure}
    \centering
    \scalebox{1.0}{\includegraphics{Dirac-projected-to-Fin2.pdf}}
    \caption{\captionsize The projection of the Dirac mass $-\gd_o$ onto $\Fin_2$; see Example~\ref{exm:Fin_2-not-dense-in-Fin} and also Lemma~\ref{thm:monopole-on-geometric-integers} and Lemma~\ref{thm:repkernels-on-geometric-integers}.}
    \label{fig:Dirac-projected-to-Fin2}
  \end{figure}

\begin{exm}[Geometric half-integer model]\label{def:geometric-half-integers}
  It is also interesting to consider $(\bZ_+,c^n)$, as this network supports a monopole, but has $\Harm = 0$.
  \linenopax
  \begin{align*}
    \xymatrix{
      0 \ar@{-}[r]^{c}
      & 1 \ar@{-}[r]^{c^2}
      & 2 \ar@{-}[r]^{c^3}
      & 3 \ar@{-}[r]^{c^4}
      & \dots
    }
  \end{align*}
  As in Lemma~\ref{thm:monopole-on-geometric-integers}, it is straightforward to check that $w_o(n) = ar^{|n|}$, $a:= \frac{r}{(1-r)}$, defines a monopole on the geometric half-integer model $(\bZ_+,c^n)$. However, it is also easy to check by induction that $\Harm = 0$ for this model.

  For $k=2,3,\dots$, the network $(\bZ_+,k^n)$ can be thought of as the ``projection'' of the homogeneous tree of degree $k$ $(\sT_k, \tfrac1k \one)$ under a map which sends $x$ to $n \in \bZ$ iff there are $n$ edges between $x$ and $o$.
\end{exm}

\begin{exm}[Decomposition in \Gdd]\label{exm:monopolar-decomposition}
  In Remark~\ref{rem:comparison-to-Royden-decomp}, we discussed the Hilbert space \Gdd and its inner product $\la u, v \ra_\gdd := u(o)v(o) + \la u, v \ra_\energy$. Since $(\bZ_+,c^n)$ and $(\bZ,c^n)$ are both transient for $c>1$ (but only the latter contains harmonic functions), it is interesting to consider $P_{\Gddo}\one$ for these models (see Remark~\ref{rem:transient-iff-1-splits}). The projections $v = P_{\Gddo}\one$ and $u = \one-v = P_{\Gddo}^\perp\one$ on $(\bZ,c^n)$ are given by
  \linenopax
  \begin{align}\label{eqn:v=P_Do1-on-Z}
    v(x) = 2 - 2a + ar^{|x|}
    \qq\text{and}\qq
    u(x) = 2a - 1 - a r^{|x|},
  \end{align}
  where with $a = \frac1{3-2c}$ and $r=c^{-1}$, and one can check that $v \in \MP_o^+$ and $u \in \MP_o^-$; see Definition~\ref{def:poles-and-antipoles} and Lemma~\ref{thm:positive-monopoles-in-Do}. In particular, $\Lap v = (1-v_o)\gd_0$ and $\Lap u = -u_o\gd_0$ (as usual, $o=0$). Now consider the representative of $w \in \MP_o$ given by
  \linenopax
  \begin{align}\label{eqn:monopolar-charfn}
    w(x) = (2-a) \charfn{[-\iy,0]} + \left(1 + 2a(r^{|x|} - c)\right)\charfn{[1,\iy)} \,.
  \end{align}
  A straightforward computation shows that $w = v + h$ with $h \in \GHD_o$.

  The function $v = P_{\Gddo}\one$ was computed for $(\bZ,c^n)$ in \eqref{eqn:v=P_Do1-on-Z} by using the formula $\energy(u) = u_o - u_o^2$, from Lemma~\ref{thm:parabolic-u(o)-parameter}, where $u := P_{\Gddo}^\perp \one = \one - v$ and $u_o = u(o)$. For a general network $(\Graph,\cond)$, this formula implies that $(u_o, \energy(u))$ lies on a parabola with $u_o \in [0,1)$ and maximum at $(\frac12,\frac14)$. From \eqref{eqn:v=P_Do1-on-Z}, it is clear that the network $(\bZ,c^n)$ provides an example of how $u_o = 1 - \frac1{2c-1}$ can take any value in $[0,1)$. Note that $c=1$ corresponds to $\energy(u)=0$, which is the recurrent case.
\end{exm}

\begin{exm}[Star networks]\label{exm:star}
  Let $(\sS_m,c^n)$ be a network constructed by conjoining $m$ copies of $(\bZ_+,c^n)$ by identifying the origins of each; let $o$ be the common origin.
\end{exm}

\begin{exm}[$\Fin_2$ not dense in \Fin]
  \label{exm:Fin_2-not-dense-in-Fin}
  On $(\bZ,2^n)$, $\Fin_2 = \cl {\spn\{\gd_x-\gd_o\}}$ is not dense in $\Fin$ (see Definition~\ref{def:Fin2-and-Fin1}). To illustrate this, we compute the projection of $-\gd_o$ to $\Fin_2$. This may be accomplished by computing
  \linenopax
  \begin{align*}
    u_n := \left[\text{projection of } -\gd_o \text{ to } \spn\{\gd_x-\gd_o \suth |x| \leq n\}\right],
  \end{align*}
  and then taking the limit as $n \to \iy$. The result is depicted in Figure~\ref{fig:Dirac-projected-to-Fin2}.
  We leave the computation of the case of general geometric conductance $(\bZ,c^n)$ as an exercise.
\end{exm}

\section{Defect spaces}
\label{sec:defect-on-geometric-integers}

We will construct a defect vector $u \in \HE$ satisfying $\Lap u = -u$ on $(\bZ,c^n)$, $c>1$, the 1-dimensional integer lattice with geometrically growing conductances. We do this in two stages: (i) construct a defect vector on $(\bZ_+,c^n)$, and (ii) combine two copies of this defect vector to obtain an example on $(\bZ,c^n)$.

\begin{exm}[Defect on the positive integers]
  \label{exm:defective-positive-integers}\label{exm:integer-lattice-with-defect}
  We consider $(\bZ_+,\cond)$ where
  \linenopax
  \begin{align*}
    \cond_{n-1,n} = c^n, \q &n \geq 1,
  \end{align*}
  for some fixed $c>1$. Thus, the network under consideration is
  \linenopax
    \begin{align*}
    \xymatrix{
      0 \ar@{-}[r]^{c}
      & 1 \ar@{-}[r]^{c^2}
      & 2 \ar@{-}[r]^{c^3}
      & 3 \ar@{-}[r]^{c^4} & \dots
    }
  \end{align*}
  Now recursively define a system of polynomials in $r=1/c$ by
  \linenopax
  \begin{align*}
    \left[\begin{array}{cc}
      p_n \\ q_n
    \end{array}\right]
    =
    \left[\begin{array}{cc}
      1 & 1 \\ r^n & 1+r^n
    \end{array}\right]

    \cdots
    \left[\begin{array}{cc}
      1 & 1 \\ r^2 & 1+r^2
    \end{array}\right]
    \left[\begin{array}{cc}
      1 & 1 \\ r & 1+r
    \end{array}\right]
    \left[\begin{array}{cc}
      0 \\ 1
    \end{array}\right]
  \end{align*}
  We will show that $u(n) := q_n$ satisfies $\Lap u = -u$ and has finite energy. It will be helpful to note that
  \linenopax
  \begin{align}\label{eqn:p_n-is-diff}
    p_n = c^n (u(n) - u(n-1)),
  \end{align}
  and hence
  \linenopax
  \begin{align*}
    p_{n+1} = p_n + q_n,
    \q\text{and}\q
    q_{n+1} = q_n + r^{n+1} p_{n+1}.
  \end{align*}
  Now, $\Lap u = -u$ because
  \linenopax
  \begin{align*}
     \Lap u(n)
     = p_n - p_{n+1}
     = -q_n = -u(n).
  \end{align*}
  We will need the following lemma to show that $u \in \HE$.

\begin{lemma}\label{thm:geometric-bounds-on-p,q}
  There is an $m$ such that $p_n \leq n^m$ and $q_n \leq (n+1)^m - n^m$ for $n \in \bZ_+$.
  \begin{proof}
    We prove both bounds simultaneously by induction, so assume both bounds hold for $n$ and prove
    \linenopax
    \begin{align*}
      p_{n+1} &\leq (n+1)^m, \q \text{and} \\
      q_{n+1} &\leq (n+2)^m - (n+1)^m.
    \end{align*}
    The estimate for $p_{n+1} = p_n + q_n$ is immediate from the inductive hypotheses.
    For the $q_{n+1}$ estimate, choose an integer $m$ so that
    \linenopax
    \begin{align*}
      m(m-1)
      \geq \max\{t^2 r^t \suth t \geq 0\}
      = \left(\frac{2}{e \log c}\right)^2.
    \end{align*}
    Then $(n+1)^2 r^{n+1} \leq m(m-1)$ for all $n$, so
    \linenopax
    \begin{align*}
      2 + r^{n+1}
      \leq 2 + \frac{m(m-1)}{(n+1)^2}
      \leq \left(\frac{n}{n+1}\right)^m + \left(\frac{n+2}{n+1}\right)^m,
    \end{align*}
    by using the binomial theorem to expand $\left(\frac{n}{n+1}\right)^m = \left(1-\frac{1}{n+1}\right)^m$ and $\left(\frac{n+2}{n+1}\right)^m = \left(1+\frac{1}{n+1}\right)^m$. Multiplying by $(n+1)^m$ gives
    \linenopax
    \begin{align*}

      \left((n+1)^m - n^m\right) + r^{n+1}(n+1)^m &\leq (n+2)^m - (n+1)^m,
    \end{align*}
    which is sufficient because the left side is an upper bound for $q_{n+1} = q_n + r^{n+1} p_{n+1}$.
  \end{proof}
\end{lemma}

\begin{lemma}\label{thm:defect-vector-has-finite-energy}
  The defect vector $u(n) := q_n$ has finite energy and is bounded.
  \begin{proof}
    Applying Lemma~\ref{thm:geometric-bounds-on-p,q} to the formula for \energy yields
    \linenopax
    \begin{align*}
      \energy(u)
      &= \sum_{n=1}^\iy c^n(u(n)-u(n-1))^2
      = \sum_{n=1}^\iy r^n p_n^2
      \leq \sum_{n=1}^\iy r^n n^{2m}
      = \operatorname{Li}_{-2m}(r) < \iy,
    \end{align*}
    since a polylogarithm indexed by a negative integer is continuous on $\bR$, except for a single pole at 1 (but recall that $r \in (0,1)$).
  \end{proof}
\end{lemma}

  Lemma~\ref{thm:defect-vector-has-finite-energy} ensures that the defect vector is bounded; in the example in Figure~\ref{fig:defect-vector}, the defect vector has a limiting value of $\approx 4.04468281$, although the function value does not exceed 4 until $x=10$. The first few values of the function are
  \linenopax
  \begin{align*}
    u &= \left[ \tfrac{3}{2}, \tfrac{17}{8}, \tfrac{173}{64}, \tfrac{3237}{1024}, \tfrac{114325}{32768}, \tfrac{7774837}{2097152}, \tfrac{1032268341}{268435456}, \tfrac{270040381877}{68719476736}, \tfrac{140010315667637}{35184372088832}, \dots\right] \\
      &\approx \left[{1.5}, {2.125}, {2.7031}, {3.1611}, {3.4889}, {3.7073}, {3.8455}, \
{3.9296}, {3.9793}, {4.0080}, \dots \right]
  \end{align*}

  \begin{figure}
    \centering
    \scalebox{1.0}{\includegraphics{defect.pdf}}
    \caption{\captionsize A Mathematica plot of the defect vector $u$ on $(\bZ_+,2^n)$; see Example~\ref{exm:defective-positive-integers} and Lemma~\ref{thm:defect-vector-has-finite-energy}. The left plot shows $u(x)$ for $x=0,1,\dots,10$, and the plot on the right shows data points for $u(x)$, $x=10,11,12,\dots$.}
    \label{fig:defect-vector}
  \end{figure}

  While we are unable to provide a nice closed-form formula for the defect vector, we can provide generating functions for it, using the $p_n = p_n(r)$ and $q_n = q_n(r)$ obtained just above. Define
  \linenopax
  \begin{align*}
    P(x) = \sum_{n=0}^\iy p_n(r) x^n
    \q\text{and}\q
    Q(x) = \sum_{n=0}^\iy q_n(r) x^n.
  \end{align*}
  Multiplying both sides of $p_{n+1} = p_n + q_n$ by $x^{n+1}$ and summing from $n=0$ to \iy,
  \linenopax
  \begin{align}\label{eqn:genfun-for-P-1}
    P(x) = xP(x) + xQ(x),
  \end{align}
  where we have used the fact that $p_0=0$.
  Meanwhile, multiplying both sides of $q_{n+1} = q_n + r^{n+1}p_{n+1}$ by $x^{n+1}$ and summing from $n=0$ to \iy,
  \linenopax
  \begin{align}\label{eqn:genfun-for-Q-1}
    Q(x) -1 = xQ(x) + P(rx).
  \end{align}
  Write \eqref{eqn:genfun-for-P-1} in the form $(1-x)P(x) = xQ(x)$ and substituting in $(1-x)Q(x) = 1 + P(rx)$ from \eqref{eqn:genfun-for-Q-1}, to get $1 + P(rx) = (1-x)Q(x) = \tfrac{(1-x)^2}{x} P(x)$ or
  \linenopax
  \begin{align*}
    P(x)
    = \tfrac{x}{(1-x)^2} + \tfrac{x}{(1-x)^2}P(rx)
    &= \tfrac{x}{(1-x)^2} + \tfrac{x(rx)}{(1-x)^2(1-rx)^2} + \tfrac{x(rx)}{(1-x)^2(1-rx)^2}P(r^2x) \\
    \cdots
    &= \sum_{n=0}^\iy \prod_{k=0}^n \frac{r^k x}{(1-r^kx)^2}
    = \sum_{n=0}^\iy\frac{r^{n(n+1)/2} x^n}{\prod_{k=0}^n (1-r^kx)^2}.
  \end{align*}
  Note that $r \in (0,1)$, so $P(r^kx) \limas{k} P(0) = 0$, again since $p_0=0$. Now \eqref{eqn:genfun-for-P-1} gives $Q(x) = \frac{1-x}x P(x)$, whence
  \linenopax
  \begin{align*}
    Q(x) = \sum_{n=0}^\iy\frac{r^{n(n+1)/2} x^{n-1}}{\prod_{k=1}^n (1-r^kx)^2}.
  \end{align*}
\end{exm}

\begin{exm}[Defect on the integers]
  \label{exm:defective-integers}
  We consider $(\bZ,\cond)$ as in Definition~\ref{def:geometric-integers}:
  \linenopax
  \begin{align*}
    \xymatrix{
      \dots \ar@{-}[r]^{c^3}
      & {-2} \ar@{-}[r]^{c^2}
      & {-1} \ar@{-}[r]^{c}
      & {0} \ar@{-}[r]^{c}
      & {1} \ar@{-}[r]^{c^2}
      & {2} \ar@{-}[r]^{c^3}
      & {3} \ar@{-}[r]^{c^4} & \dots
    }
  \end{align*}

  Proceeding as in Example~\ref{exm:defective-positive-integers}, one uses $\Lap u(0) = -u(0)$ to compute
  \linenopax
  \begin{align*}
    2c(u(0) - u(1)) = -u(0)
    \q\implies\q
    u(1) = \left(1 + \tfrac 1{2c}\right) u(0),
  \end{align*}
  and obtain the initial values $p_1 = \frac12$ and $q_1 = 1+\frac r2$.
  Therefore, for \bZ we instead use the polynomials defined by
  \linenopax
  \begin{align*}
    \left[\begin{array}{cc}
      p_n \\ q_n
    \end{array}\right]
    =
    \left[\begin{array}{cc}
      1 & 1 \\ r^n & 1+r^n
    \end{array}\right]

    \cdots
    \left[\begin{array}{cc}
      1 & 1 \\ r^2 & 1+r^2
    \end{array}\right]
    \left[\begin{array}{cc}
      1 & 1 \\ r & 2+r
    \end{array}\right]
    \left[\begin{array}{cc}
      0 \\ \tfrac12
    \end{array}\right]
  \end{align*}
  The other computations are essentially identical to those for $(\bZ_+,c^n)$.
\end{exm}

\version{}{\marginpar{Add $w_x \in \ran \LapV^\ad$}}

\section{Remarks and references}
\label{sec:Remarks-and-References-lattice-examples}

The infinite lattices offer a second attractive family of examples; and they are especially relevant for lattice-spin models in physics, as discussed in Chapter~\ref{sec:Magnetism-and-long-range-order}. The book \cite{Soardi94} by Soardi is a nice introduction to the subject, and a classical introductory reference is \cite{Spitzer}. Of the results in the literature of relevance to the present chapter, the references \cite{GovN51, GoSt06, ScZi09, Martinelli99, ChLi07, Lig99, Lig95, Lig93} are especially relevant.

The geometric integers of Example~\ref{exm:defective-positive-integers} came about from our desire to apply von NeumannÕs theory of unbounded operators and their deficiency indices \cite{vN32a, vN32b, vN32c, DuSc88} to the metric geometry of infinite weighted graphs $(\Graph, \cond)$. Starting with $(\Graph, \cond)$ there are two natural Hilbert spaces $\ell^2(\verts)$ (where \verts is the vertex-set) and the energy Hilbert space \HE. An intriguing aspect of \S\ref{sec:defect-on-geometric-integers} is that the boundary features of $(\Graph, \cond)$ deriving from deficiency indices cannot be accounted for with the use of the more naive of the two Hilbert spaces $\ell^2(\verts)$; \HE is forced upon us.

The geometric integers discussed in Example~\ref{exm:defective-positive-integers} is called a \emph{weighted linear graph} in \cite{AlFi09} and is studied in conjunction with birth and death processes; see the references cited therein.


TITLE: show a function is riemann integrable
QUESTION [1 upvotes]: Define $\phi(x)=x\sin\frac{1}{x}$ for $0<x\leq 1 $ and $\phi(0)=0$. If $f:[-1,1] \rightarrow \mathbb{R}$ is Riemann integrable, show that $f \circ \phi$ is also Riemann integrable.
Denote $D(f)$ to be the set of discontinuities of $f$.
Since $D(f \circ \phi) \subset D(f) \cup D(\phi)=D(f) \cup \{0\}$ and by Riemann Lebesgue theorem, we have $D(f)$ is a zero set. Since subset of zero set is again zero set, by Riemann Lebesgue theorem, $f \circ \phi$ is Riemann integrable.
Question: Is $D(f \circ \phi) \subset D(f) \cup D(\phi)$ correct in this question?
EDIT: Is $\phi(D(f \circ \phi)) = D(f)$ true here?

REPLY [1 votes]: No, $D(f\circ\phi)\subset D(f)\cup D(\phi)$ does not hold. See, for example, $\phi(x)=x+1$ on $[0,1]$ and $f$ a function which is discontinuous at $1.5$., for example,
$$f(x) = \begin{cases}0 & \text{if }x\in[1,1.5)\\1&\text{if }x\in[1.5,2)\end{cases}$$
In that case, you have $$(f\circ \phi)(x) = \begin{cases}0 & \text{if }x\in[0,0.5)\\1&\text{if }x\in[0.5,1)\end{cases}$$
so $D(f\circ\phi) = \{0.5\}\not\subset\{1.5\} = D(f)\cup D(\phi).$


TITLE: Abelian varieties over $p$-adic fields
QUESTION [3 upvotes]: Theorem : Let $A$ be an abelian varieties of dimension $d$ over a field $k$, non-archimedian valued complete, i.e. $\mathbb{Q}_p$, then $A(k)$ contains a subgroup of finite index analytically isomorphic and homeomorphic to $I \oplus I \oplus \ldots \oplus I$ ($d$ summands) where $I$ is the ring of integers of $k$.
This is a theorem written in a paper by Arthur Mattock in 1955. I was wondering if someone knows where one can find any other proof of his theorem.
P.S. the name of the paper is Abelian varieties over $p$-adic ground fields.

REPLY [6 votes]: Answered in the comments by user74230:

You omitted the necessary hypothesis that $k$ has characteristic $0$. Over any such field, Serre's book "Lie groups and Lie algebras" proves something much more general: every analytic group manifold over such $k$ has a canonically associated Lie algebra (vanishing bracket in the commutative case) and has an open subgroup isomorphic to to the "Campbell-Baker-Hausdorff" group law. For trivial Lie bracket this is what you seek (up to scaling the coordinates), but only finite index for $A(k)$ when $k$ is finite degree over $\mathbb{Q}_{p}$.


text:
\section{Annihilating Polynomial of Minimal Degree is Irreducible}
Tags: Minimal Polynomials

\begin{theorem}
Let $L / K$ be a [[Definition:Field Extension|field extension]].
Let $\alpha \in L$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
Let $f \in K[x]$ be a [[Definition:Nonzero Polynomial|nonzero polynomial]] of [[Definition:Smallest Natural Number|minimal]] [[Definition:Degree of Polynomial|degree]] such that $\map f \alpha = 0$.
Then $f$ is [[Definition:Irreducible Polynomial|irreducible]] in $K \sqbrk x$.
\end{theorem}

\begin{proof}
{{AimForCont}} that $f$ is not [[Definition:Irreducible Polynomial|irreducible]].
Then there exist non-[[Definition:Constant Polynomial|constant polynomials]] $g, h \in K \sqbrk x$ such that $f = g h$.
We have:
:$0 = \map f \alpha = \map g \alpha \map h \alpha$
Since $L$ is a [[Definition:Field (Abstract Algebra)|field]], it is an [[Definition:Integral Domain|integral domain]].
Therefore, as $\map g \alpha, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$.
This contradicts the minimality of the [[Definition:Degree of Polynomial|degree]] of $f$.
Hence the result, by [[Proof by Contradiction]].
{{qed}}
\end{proof}

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TITLE: obtain $y=\cos2\alpha$ knowing $x=\tan\alpha$
QUESTION [0 upvotes]: obtain $y=\cos2\alpha$ knowing $x=\tan\alpha$
So obviously I tried the double angle identity to see where it gets me:
$$\cos2\alpha = \cos^2\alpha-\sin^2\alpha=1-2\sin^2\alpha$$
$$1-2\sin^2\alpha = 1-2\tan^2\alpha\cos^2\alpha=1-2x^2\cos^2\alpha$$
I'm not sure how to continue in order show $y$ with $x$ without the cos or sin functions..
any tips or hints?

I'm not a student yet, this is one of the Tel-Aviv university perparation exercises for the entry test so this shouln't too hard.

REPLY [1 votes]: $$y=\cos2\alpha=2\cos^2\alpha-1=\frac{2\cos^2\alpha}{\cos^2\alpha+\sin^2\alpha}-1=\frac2{1+\tan^2\alpha}-1=\frac2{1+x^2}-1=\frac{1-x^2}{1+x^2}.$$