text:
\begin{document}

\begin{abstract}
We show that every separable \ca{} of real rank zero that tensorially absorbs the Jiang-Su algebra contains a dense set of generators.

It follows that in every classifiable, simple, nuclear \ca{}, a generic element is a generator.
\end{abstract}

\maketitle

\section{Introduction}

The generator problem for \ca{s} asks to determine the minimal number of generators for a given \ca.
One difficulty when studying this problem is that the minimal number of generators is a rather ill-behaved invariant;
in particular, it may increase when passing to ideals or inductive limits.
We showed in \cite{Thi12arX:GenRnk} that a much better behaved invariant is obtained by considering the minimal number $n$ such that generating $n$-tuples are dense.

The most interesting aspect of the generator problem is to determine which \ca{s} are generated by a single operator, and it is a mayor open question if every separable, simple \ca{} is singly generated.
A natural variant of this problem is to describe the class of \ca{s} that contain a dense set of generators.
In previous work, we showed that this class includes all separable AF-algebras (\cite[Theorem~7.3]{Thi12arX:GenRnk}) and all separable, approximately subhomogeneous (ASH) algebras that are $\mathcal{Z}$-stable, that is, that tensorially absorb the Jiang-Su algebra $\mathcal{Z}$ (\cite[Theorem~5.10]{Thi20arX:grSubhom}).

In this paper, we show that every separable, $\mathcal{Z}$-stable \ca{} of real rank zero contains a dense set of generators;
see \autoref{prp:mainThm}.
This includes all separable, nuclear, purely infinite \ca{s} of real rank zero (\autoref{prp:nuclPI}) and in particular every Kirchberg algebra (\autoref{prp:KirchbergAlg}).
Together with the previous result about  $\mathcal{Z}$-stable ASH-algebras, we deduce that every classifiable, simple, nuclear \ca{} contains a dense set of generators;
see \autoref{prp:classifiable-gr1}.
\\

The main tool to prove these results is the generator rank, which was introduced in \cite{Thi12arX:GenRnk}.
A unital, separable \ca{} $A$ has generator rank at most $n$, denoted $\gr(A)\leq n$, if the set of self-adjoint $(n+1)$-tuples that generate $A$ as a \ca{} is dense;
see \autoref{dfn:gr} for the general definition in the nonunital and nonseparable case.
By \cite[Proposition~5.6]{Thi12arX:GenRnk}, $A$ has generator rank zero if and only if $A$ is commutative with totally disconnected spectrum.
Since two self-adjoint elements $a$ and $b$ generate the same sub-\ca{} as the element $a+ib$, it follows that $A$ has generator rank at most one if and only if $A$ contains a dense set of generators.

Our main result (\autoref{prp:mainThm}) shows that separable, $\mathcal{Z}$-stable \ca{s} of real rank zero have generator rank one.
To prove this, we heavily rely on the permanence properties of the generator rank that were established in \cite{Thi12arX:GenRnk}.
In particular, the generator rank does not increase when passing to ideals, quotients, or inductive limits, and we can estimate the generator rank of extensions;
see \autoref{prp:permanence}.

The general strategy is as follows:
Given a unital, separable, $\mathcal{Z}$-stable \ca{} $A$ of real rank zero,
we use that $\mathcal{Z}$ is the inductive limit of generalized dimension-drop algebras $Z_{2^\infty,3^\infty}$ (see the proof of \autoref{prp:gr8} for the definition) to reduce the problem to computing the generator rank of $A\otimes Z_{2^\infty,3^\infty}$.

We obtain the estimate $\gr(A\otimes Z_{2^\infty,3^\infty})\leq 1$ by combining two results:
First, we show that $A\otimes Z_{2^\infty,3^\infty}$ has finite generator rank (we verify that it is at most $8$);
see \autoref{prp:gr8}.
Then, using a delicate construction of generators in $A\otimes Z_{2^\infty,3^\infty}$, we show that $\gr(A\otimes Z_{2^\infty,3^\infty})\leq n+1$ implies that $\gr(A\otimes Z_{2^\infty,3^\infty})\leq n$, for every $n\geq 1$;
see \autoref{prp:construction}.

To verify that $A\otimes Z_{2^\infty,3^\infty}$ has finite generator rank, we use the short exact sequence
\[
0 \to A\otimes C_0((0,1),M_{6^\infty}) \to A\otimes Z_{2^\infty,3^\infty} \to A\otimes(M_{2^\infty}\oplus M_{3^\infty}) \to 0,
\]
and it suffices to show that the ideal and the quotient have finite generator rank.
To estimate the generator rank of $A\otimes C_0((0,1),M_{6^\infty})$, we use that it is an ideal in $A\otimes C_0([0,1],M_{6^\infty})$.
In \autoref{sec:tensI}, we prove an upper bound for the generator rank of such algebras.

The algebra $A\otimes Z_{2^\infty,3^\infty}$ has a natural structure as a $C([0,1])$-algebra, with each fiber over $(0,1)$ isomorphic to $A\otimes M_{6^\infty}$, and with the fibers at $0$ and $1$ isomorphic to $A\otimes M_{3^\infty}$ and $A\otimes M_{2^\infty}$, respectively.
We use a Stone-Weierstra{\ss}-type result that characterizes when a set of self-adjoint elements generates $A\otimes Z_{2^\infty,3^\infty}$:
the elements have to generate each fiber, and they have to suitably separate the points in $[0,1]$.
The assumption of real rank zero is crucial for both:
it is used in the proof of \autoref{prp:UHFabs} to show that the tensor product of $A$ with a UHF-algebra has generator rank one;
and it is used in the proofs of \autoref{prp:approxSepInterval} and \autoref{prp:construction} to construct self-adjoint elements that separate the points in $[0,1]$.

\subsection*{Acknowledgments}

The author thanks Eusebio Gardella for valuable comments and feedback.
I also want to thank the anonymous referee for his useful suggestions.

\subsection*{Notation}

Given a \ca{} $A$, we use $A_\sa$ to denote the set of self-adjoint elements in $A$.
We write $M_d$ for the \ca{} of $d$-by-$d$ complex matrices.
Given a subset $F\subseteq A$ and $a\in A$, we write $a\in_\varepsilon F$ if there exists $b\in F$ such that $\|a-b\|<\varepsilon$.
We set $\NN:=\{0,1,2,\ldots\}$, the natural numbers including $0$.
The spectrum of an operator $a$ is denoted by $\sigma(a)$.

\section{The generator rank}
\label{sec:grPrelim}

In this section, we recall the definition and basic properties of the generator rank $\gr$ and its precursor $\grPre$ from \cite{Thi12arX:GenRnk}.

\begin{dfn}[{\cite[Definitions~2.1, 3.1]{Thi12arX:GenRnk}}]
\label{dfn:gr}
Let $A$ be a \ca{}.
Then $\grPre(A)$ is the defined as the smallest integer $n\geq 0$ such that for every $a_0,\ldots,a_n\in A_\sa$, $\varepsilon>0$ and $c\in A$, there exist $b_0,\ldots,b_n\in A_\sa$ such that
\[
\|b_j-a_j\|<\varepsilon \ \text{ for } j=0,\ldots,n, \andSep
c\in_\varepsilon C^*(b_0,\ldots,b_n).
\]
If no such $n$ exists, then $\grPre(A)=\infty$.
The \emph{generator rank} of $A$ is defined as $\gr(A):=\grPre(\widetilde{A})$, where $\widetilde{A}$ denotes the minimal unitization of $A$.
\end{dfn}

For the definition and the basic properties of the real rank of \ca{s}, we refer to \cite[Section~V.3.2, p.452ff]{Bla06OpAlgs}.

\begin{rmk}
Let $A$ be a \ca.
If $A$ is unital, then $\gr(A)=\grPre(A)$ by definition.
By \cite[Theorem~5.5]{Thi20arX:grSubhom}, we also have $\gr(A)=\grPre(A)$ whenever $A$ is subhomogeneous.
By \cite[Theorem~3.13]{Thi12arX:GenRnk}, we have $\gr(A)=\max\{\rr(A),\grPre(A)\}$.
In particular, if $A$ has real rank zero, then $\gr(A)=\grPre(A)$.

In general, however, it is unclear if $\gr(A)=\grPre(A)$;
see \cite[Question~3.16]{Thi12arX:GenRnk}.
\end{rmk}

For separable \ca{s}, the generator rank and its precursor can be described by the denseness of generating tuples.

\begin{thm}[{\cite[Theorem~3.4]{Thi12arX:GenRnk}}]
\label{prp:grCharSep}
Let $A$ be a \emph{separable} \ca{} and $n\in\NN$.
Then $\grPre(A)\leq n$ if and only if for every $a_0,\ldots,a_n\in A_\sa$ and $\varepsilon>0$, there exist $b_0,\ldots,b_n\in A_\sa$ such that
\[
\|b_j-a_j\|<\varepsilon \ \text{ for } j=0,\ldots,n, \andSep
A=C^*(b_0,\ldots,b_n).
\]

Analogously, we obtain a characterization of $\gr(A)\leq n$ by denseness of generating tuples in $\widetilde{A}$.
\end{thm}

We will repeatedly use the following permanence properties of the generator rank, which were shown in Theorems~6.2 and~6.3 of \cite{Thi12arX:GenRnk}.

\begin{thm}
\label{prp:permanence}
Let $I\subseteq A$ be a closed, two-sided ideal in a \ca{} $A$.
Then
\[
\max\big\{ \gr(I),\gr(A/I) \big\}
\leq \gr(A)
\leq \gr(I)+\gr(A/I)+1.
\]
Moreover, if $A=\varinjlim_\lambda A_\lambda$ is an inductive limit, then
\[
\gr(A) \leq \liminf_\lambda\gr(A_\lambda).
\]
\end{thm}

It is natural to expect that the generator rank of the direct sum of two \ca{s} is the maximum of the generator ranks of the summands.
We have verified this in the case that both summands have real rank zero (see \autoref{prp:gr_sum_rr0} below), and whenever one of the summands is subhomogeneous (\cite[Proposition~5.8]{Thi20arX:grSubhom}).
However, in general this remains unclear;
see \cite[Question~6.4]{Thi12arX:GenRnk}.

\begin{prp}[{\cite[Lemma~7.1]{Thi12arX:GenRnk}}]
\label{prp:gr_sum_rr0}
Let $A, B$ be \ca{s} of real rank zero.
Then $\gr(A\oplus B)=\max\{\gr(A),\gr(B)\}$.
\end{prp}

\begin{prp}[{\cite[Proposition~5.6]{Thi12arX:GenRnk}}]
\label{prp:gr0}
A \ca{} $A$ has generator rank zero if and only if $A$ is commutative with totally disconnected spectrum.
\end{prp}

\begin{rmk}
\label{rmk:gr1}
Let $A$ be a separable \ca.
Since two self-adjoint elements $a$ and $b$ generate the same sub-\ca{} as the element $a+ib$, and since the set of generators in $A$ is always a $G_\delta$-subset, we have $\grPre(A)\leq 1$ if and only if the set of (non-self-adjoint) generators in $A$ is a dense $G_\delta$-subset;
see \cite[Remark~3.7]{Thi12arX:GenRnk}.

If $A$ has real rank zero, then $\grPre(A)=\gr(A)$.
Thus, a separable \ca{} $A$ of real rank zero has generator rank at most one if and only if a generic element in $A$ is a generator.
\end{rmk}

We end this section with a standard result that will be used in the proof of \autoref{prp:construction}.
We include a proof for completeness.
For the definition and basic results of $C(X)$-algebras, we refer to \S{2} in \cite{Dad09CtsFieldsOverFD}.

Given an element $a$ in a $C(X)$-algebra, and $x\in X$, we let $a(x)$ denote the image of $a$ in the fiber at $x$.
We will repeatedly use that the map $X\to\RR$, $x\mapsto\|a(x)\|$, is upper semicontinuous, and that $\|a\|=\sup_{x\in X}\|a(x)\|$;
see \cite[Lemma~2.1]{Dad09CtsFieldsOverFD}.

\begin{lma}
\label{prp:genCenter}
Let $X$ be a compact, Hausdorff space, let $A$ be a unital $C(X)$-algebra, and let $B\subseteq A$ be a sub-\ca.
Then the following are equivalent:
\begin{enumerate}
\item[(a)]
We have $C(X)\subseteq B$;
\item[(b)]
$B$ separates the points in $X$ in the sense that for every distinct $x_0,x_1\in X$ there exists $b\in B$ with $b(x_0)=0$ and $b(x_1)=1$.
\end{enumerate}
\end{lma}
\begin{proof}
Assuming~(b), we need to verify~(a).
We first show that $1\in B$.
For every $y\in X$, there exists $b_y\in B$ with $b_y(y)=1$.
Replacing $b_y$ by $b_yb_y^*$, we may assume that $b_y$ is positive.
Set $U_y:=\{z\in X : \|1-b_y(z)\|<\tfrac{1}{2}\}$, which is open since $\|1-b_y(\cdot)\|$ is upper semicontinuous.
Hence, the compact set $X$ is covered by $(U_y)_{y\in X}$, which allows us to choose $y_1,\ldots,y_N\in X$ such that $X$ is covered by $\bigcup_{j=1}^N U_{y_j}$.
Set $b:=\sum_{j=1}^N b_{y_k}\in B$.
Let $f\colon\RR\to[0,1]$ be a continuous function with $f(0)=0$ and such that $f$ takes value $1$ on $[\tfrac{1}{2},\infty)$.
Set $c:=f(b)\in B$.
For each $x$, we have $b(x)\geq\tfrac{1}{2}$ and therefore $c(x)=1$.
It follows that $c=1$, since $\|c-1\|=\sup_{x\in X}\|c(x)-1\|=0$.

\emph{Claim~1: Let $F\subseteq X$ be closed and $x\in X\setminus F$.
Then there exists a positive $b\in B$ with $b|_F=0$ and $b(x)=1$.}
To prove the claim, let $y\in F$.
By assumption, there is $b_y\in B$ such that $b_y(y)=0$ and $b_y(x)=1$.
As above, we may assume that $b_y$ is positive, and we may then find $y_1,\ldots,y_N\in F$ such that $F$ is covered by the sets $U_j:=\{z\in X : \|b_{y_j}(z)\|<\tfrac{1}{2}\}$ for $j=1,\ldots,N$.
Let $g\colon\RR\to[0,1]$ be a continuous function with $g(1)=1$ and such that $g$ takes the value $0$ on $(-\infty,\tfrac{1}{2}]$.
For each $j$, set $c_j:=g(b_{y_j})\in B$.
Then $c_j$ vanishes on $U_{y_j}$, and $c_j(x)=1$.
Thus, $b:=(c_1c_2\cdots c_N)(c_1c_2\cdots c_N)^*$ has the desired properties, which proves the claim.

\emph{Claim~2: Let $F\subseteq G\subseteq X$ with $F$ closed and $G$ open.
Then there exists $b\in B$ with $0\leq b\leq 1$, $b|_{X\setminus G}=0$ and $b|_F=1$.}
To prove the claim, set $C:=X\setminus G$.
Applying Claim~1, and using that $1\in B$, for each $y\in C$ we obtain a positive $b_y\in B$ such that $b_y|_F=1$ and $b_y(y)=0$.
As in Claim~1, we find $y_1,\ldots,y_M$ such that $C$ is covered by the sets $\{z\in X : \|b_{y_j}(z)\|<\tfrac{1}{2}\}$ for $j=1,\ldots,N$.
With $g$ as above, the element $d:=g(b_{y_1})\cdots g(b_{y_M})$ satisfies $d|_C=0$ and $d|_F=1$.
Set $b:=h(dd^*)$, where $h\colon\RR\to\RR$ is the function $h(t)=\min\{t,1\}$.
Then $b$ has the claimed properties.

\emph{Claim~3: Let $f\colon X\to[0,\infty)$ be continuous, and let $\varepsilon>0$.
Then there exists $b\in B$ with $\|b-f\|\leq 2\varepsilon$.}
Set $b_0:=1$.
For each $k\geq 1$, set
\[
F_k := \big\{ x\in X : f(x)\geq k\varepsilon \big\}, \andSep
G_k := \big\{ x\in X : f(x)>(k-1)\varepsilon \big\}.
\]
Then $G_k$ is open, $F_k$ is closed, and $F_k\subseteq G_k$.
Using Claim~2, choose $b_k\in B$ such that $0\leq b_k\leq 1$, $b_k|_{X\setminus G_k}=0$ and $b_k|_{F_k}=1$.

Note that $b_k=0$ for large enough $k$, which allows us to set $b:=\varepsilon\sum_{k=0}^\infty b_k$.
For each $x\in X$, we have $f(x)\leq b(x)\leq f(x)+2\varepsilon$ by construction, and thus $\|b(x)-f(x)\|\leq 2\varepsilon$.
Hence, $\|b-f\|=\sup_{x\in X}\|b(x)-f(x)\|\leq 2\varepsilon$.

It follows from Claim~3 that $B$ contains every positive function in $C([0,1])$, which implies the result.
\end{proof}

\section{Generator rank of C([0,1],A)}
\label{sec:tensI}

Throughout this section, we let $A$ denote a unital, separable \ca{} of real rank zero and generator rank at most one, and we set $B:=A\otimes C([0,1])$.
We consider $B$ with its natural $C([0,1])$-algebra structure, with each fiber isomorphic to $A$.
The goal is to verify $\gr(B)\leq 6$, which we accomplish in a series of Lemmas.

\begin{lma}
\label{prp:approxGenFibers}
Let $x_1,\ldots,x_4\in B_\sa$ and $\varepsilon>0$.
Then there exist $y_1,\ldots,y_4\in B_\sa$ such that
\[
\| y_k - x_k \| < \varepsilon\ \text{ for  } k=1,\ldots,4, \andSep A=C^*(y_1(t),\ldots,y_4(t))\ \text{ for all } t\in[0,1].
\]
\end{lma}
\begin{proof}
Using that each $x_j\colon[0,1]\to A$ is uniformly continuous, choose $N\geq 1$ such that $\|x_j(t)-x_j(t')\|<\tfrac{\varepsilon}{3}$ for all $t,t'\in[0,1]$ with $|t-t'|\leq\tfrac{1}{N}$.

Let $j\in\{0,\ldots,N\}$.
Then $x_1(\tfrac{2j}{2N})$ and $x_2(\tfrac{2j}{2N})$ belong to $A_\sa$.
Using that $A$ is unital and separable with $\gr(A)\leq 1$, we can apply \autoref{prp:grCharSep} to obtain $y_{1}^{(2j)},y_{2}^{(2j)}\in A_\sa$ such that
\[
\left\| y_{1}^{(2j)} - x_1(\tfrac{2j}{2N}) \right\| < \frac{\varepsilon}{3}, \quad
\left\| y_{2}^{(2j)} - x_2(\tfrac{2j}{2N}) \right\| < \frac{\varepsilon}{4}, \andSep
A = C^*(y_{1}^{(2j)},y_{2}^{(2j)}).
\]
Analogously, we obtain $y_3^{(2j)},y_4^{(2j)}\in A_\sa$ such that
\[
\left\| y_3^{(2j)} - x_3(\tfrac{2j}{2N}) \right\| < \frac{\varepsilon}{3}, \quad
\left\| y_4^{(2j)} - x_4(\tfrac{2j}{2N}) \right\| < \frac{\varepsilon}{3}, \andSep
A = C^*(y_3^{(2j)},y_4^{(2j)}).
\]

For $j\in\{0,\ldots,N\}$ let $f_j,g_j\colon[0,1]\to[0,1]$ be continuous functions such that:
\begin{enumerate}
\item
$f_j$ takes the value $1$ on $[\tfrac{2j}{2N},\tfrac{2j+1}{2N}]$ for $j=0,\ldots,N-1$;
\item
the collection $(f_j)_{j=0,\ldots,N}$ is a partition of unity subordinate to the family $\{ [0,\tfrac{2}{2N}),(\tfrac{1}{2N},\tfrac{4}{2N}), \ldots,(\tfrac{2N-3}{2N},\tfrac{2N}{2N}), (\tfrac{2N-1}{2N},\tfrac{2N}{2N}] \}$;
\item
$g_j$ takes the value $1$ on $[\tfrac{2j-1}{2N},\tfrac{2j}{2N}]$ for $j=1,\ldots,N$;
\item
the collection $(g_j)_{j=0,\ldots,N}$ is a partition of unity subordinate to the family $\{ [0,\tfrac{1}{2N}),(\tfrac{0}{2N},\tfrac{3}{2N}), \ldots,(\tfrac{2N-4}{2N},\tfrac{2N-1}{2N}), (\tfrac{2N-2}{2N},\tfrac{2N}{2N}] \}$.
\end{enumerate}
The functions are depicted in the following picture:
\[
\makebox{
\begin{tikzpicture}[xscale=0.5]
\draw [help lines, ->] (-0.1,0) -- (14.2,0);
\draw [help lines, ->] (0,-0.1) -- (0,1.5);
\draw [thick] (0,1) -- (1,1);
\draw [thick] (1,1) -- (2,0);
\draw [thick] (1,0) -- (2,1);
\draw [thick] (2,1) -- (3,1);
\draw [thick] (3,1) -- (4,0);
\draw [thick] (-0.1,1) -- (0,1);
\node [left] at (0,1) {$1$};
\node [right] at (0,1.3) {$f_0$};
\node [right] at (2,1.3) {$f_1$};

\draw [thick] (0,0) -- (0,-0.1);
\draw [thick] (1,0) -- (1,-0.1);
\draw [thick] (2,0) -- (2,-0.1);
\node [below] at (0,-0.05) {$0$};
\node [below] at (1,-0.05) {$\tfrac{1}{2N}$};
\node [below] at (2,-0.05) {$\tfrac{2}{2N}$};
\fill (5,0.5) ellipse (0.1 and 0.05);
\fill (5.5,0.5) ellipse (0.1 and 0.05);
\fill (6,0.5) ellipse (0.1 and 0.05);
\draw [thick] (7,0) -- (8,1);
\draw [thick] (8,1) -- (9,1);
\draw [thick] (9,1) -- (10,0);
\draw [thick] (9,0) -- (10,1);
\draw [thick] (10,1) -- (11,1);
\draw [thick] (11,1) -- (12,0);
\draw [thick] (11,0) -- (12,1);
\node [right] at (7.5,1.3) {$f_{N-2}$};
\node [right] at (9.5,1.3) {$f_{N-1}$};
\node [right] at (12,1.0) {$f_{N}$};
\draw [thick] (11,0) -- (11,-0.1);
\draw [thick] (12,0) -- (12,-0.1);
\node [below] at (11,-0.05) {$\tfrac{2N-1}{2N}$};
\node [below] at (12,-0.05) {$1$};

\draw [help lines, ->] (-0.1,-2.2) -- (14.2,-2.2);
\draw [help lines, ->] (0,-2.3) -- (0,-0.7);
\draw [thick] (0,-1.2) -- (1,-2.2);
\draw [thick] (0,-2.2) -- (1,-1.2);
\draw [thick] (1,-1.2) -- (2,-1.2);
\draw [thick] (2,-1.2) -- (3,-2.2);
\draw [thick] (2,-2.2) -- (3,-1.2);
\draw [thick] (3,-1.2) -- (4,-1.2);
\draw [thick] (4,-1.2) -- (5,-2.2);
\draw [thick] (-0.1,-1.2) -- (0,-1.2);
\node [left] at (0,-1.2) {$1$};
\node [right] at (-0.1,-1.1) {$g_0$};
\node [above] at (1.5,-1.2) {$g_1$};
\node [above] at (3.5,-1.2) {$g_2$};
\draw [thick] (0,-2.2) -- (0,-2.3);
\draw [thick] (1,-2.2) -- (1,-2.3);
\draw [thick] (2,-2.2) -- (2,-2.3);
\node [below] at (0,-2.25) {$0$};
\node [below] at (1,-2.25) {$\tfrac{1}{2N}$};
\node [below] at (2,-2.25) {$\tfrac{2}{2N}$};
\fill (6,-1.7) ellipse (0.1 and 0.05);
\fill (6.5,-1.7) ellipse (0.1 and 0.05);
\fill (7,-1.7) ellipse (0.1 and 0.05);
\draw [thick] (8,-2.2) -- (9,-1.2);
\draw [thick] (9,-1.2) -- (10,-1.2);
\draw [thick] (10,-1.2) -- (11,-2.2);
\draw [thick] (10,-2.2) -- (11,-1.2);
\draw [thick] (11,-1.2) -- (12,-1.2);
\node [above] at (9.5,-1.2) {$g_{N-1}$};
\node [above] at (11.5,-1.2) {$g_{N}$};
\draw [thick] (11,-2.2) -- (11,-2.3);
\draw [thick] (12,-2.2) -- (12,-2.3);
\node [below] at (11,-2.25) {$\tfrac{2N-1}{2N}$};
\node [below] at (12,-2.25) {$1$};
\end{tikzpicture}
}
\]

Set
\[
y_1 := \sum_{j=0}^{N} y_1^{(2j)} f_{j}, \quad
y_2 := \sum_{j=0}^{N} y_2^{(2j)} f_{j}, \quad
y_3 := \sum_{j=0}^{N} y_3^{(2j)} g_{j}, \quad
y_4 := \sum_{j=0}^{N} y_4^{(2j)} g_{j}.
\]

To verify that $\|y_1-x_1\|<\varepsilon$, we estimate $\|y_1(t)-x_1(t)\|$ for $t\in[0,1]$.
For $t=1$, we have $y_1(1)=y_1^{(2N)}$, and so
\[
\left\| y_1(1) - x_1(1) \right\|
= \left\| y_1(1) - y_1^{(2N)} \right\| < \frac{\varepsilon}{3} \leq \frac{2}{3}\varepsilon.
\]
Given $t\in[0,1)$, let $j\in\{1,\ldots,N\}$ satisfy $t\in[\tfrac{2j-2}{2N},\tfrac{2j}{2N})$.
Since $|t-\tfrac{2j-2}{2N}|,|t-\tfrac{2j}{2N}|<\tfrac{2}{2N}$, we have
\begin{align*}
\left\| y_1^{(2j-2)} - x_1(t) \right\|
&\leq \left\| y_1^{(2j-2)} - x_1(\tfrac{2j-2}{2N}) \right\|
+ \left\| x_1(\tfrac{2j-2}{2N}) - x_1(t) \right\|
< \frac{\varepsilon}{3} + \frac{\varepsilon}{3}
\leq \frac{2}{3}\varepsilon, \\
\left\| y_1^{(2j)} - x_1(t) \right\|
&\leq \left\| y_1^{(2j)} - x_1(\tfrac{2j}{2N}) \right\|
+ \left\|  x_1(\tfrac{2j}{2N}) - x_1(t) \right\|
< \frac{\varepsilon}{3} + \frac{\varepsilon}{3}
\leq \frac{2}{3}\varepsilon.
\end{align*}
Using that $y_1(t)= y_1^{(2j-2)}f_{j-1}(t)+y_1^{(2j)}f_j(t)$, and that $f_{j-1}(t)+f_{j}(t)=1$, we get
\[
\|y_1(t)-x_1(t)\|
\leq \left\| y_1^{(2j-2)} - x_1(t) \right\|f_{j-1}(t)
+ \left\| y_1^{(2j)} - x_1(t) \right\|f_{j}(t)
\leq\frac{2}{3}\varepsilon.
\]

Hence, $\|y_1-x_1\|=\sup_{t\in[0,1]}\|y_1(t)-x_1(t)\|\leq\tfrac{2}{3}\varepsilon<\varepsilon$.
Analogously, one shows that $\|y_k-x_k\|<\varepsilon$ for $k=2,3,4$.

It remains to verify that $\{y_1,\ldots,y_4\}$ generates $A$ in each fiber.
Given $t\in[0,1]$, choose $l\in\{0,\ldots,2N-1\}$ such that $t\in[\tfrac{l}{2N},\tfrac{l+1}{2N}]$.
If $l$ is even, set $j=\tfrac{l}{2}$.
Then
\[
y_1(t) = y_1^{(2j)}, \andSep
y_2(t) = y_2^{(2j)}.
\]
If $l$ is odd, set $j=\tfrac{l+1}{2}$.
Then
\[
y_3(t) = y_3^{(2j)}, \andSep
y_4(t) = y_4^{(2j)}.
\]

Thus, either $\{y_1(t),y_2(t)\}$ or $\{y_3(t),y_4(t)\}$ generates $A$, and it follows that $A=C^*(y_1(t),\ldots,y_4(t))$ in either case.
\end{proof}

\begin{lma}
\label{prp:approxSepInterval}
Let $x_1,x_2,x_3\in B_\sa$ and $\varepsilon>0$.
Then there exist $y_1,y_2,y_3\in B_\sa$ with $\|y_k-x_k\|<\varepsilon$ for $k=1,2,3$, and such that $C^*(y_1,y_2,y_3)$ separates the points in $[0,1]$ in the sense of \autoref{prp:genCenter}(b).
\end{lma}
\begin{proof}
Using that each $x_j\colon[0,1]\to A$ is uniformly continuous, choose $N\geq 1$ such that $\|x_j(t)-x_j(t')\|<\tfrac{\varepsilon}{4}$ for all $t,t'\in[0,1]$ with $|t-t'|\leq\tfrac{1}{N}$.

Let $j\in\{0,\ldots,N\}$.
Using that $A$ has real rank zero, we find invertible elements $y_1^{(3j)},y_2^{(3j)},y_3^{(3j)}\in A_\sa$ with finite spectra such that
\[
\left\| y_1^{(3j)} - x_1(\tfrac{3j}{3N}) \right\| < \frac{\varepsilon}{4}, \quad
\left\| y_1^{(3j)} - x_2(\tfrac{3j}{3N}) \right\| < \frac{\varepsilon}{4}, \andSep
\left\| y_1^{(3j)} - x_3(\tfrac{3j}{3N}) \right\| < \frac{\varepsilon}{4}.
\]

By perturbing the elements, if necessary, we may assume that the spectra of $y_k^{(3j)}$ and $y_{k'}^{(3j')}$ are disjoint whenever $k\neq k'$ or $j\neq j'$.
Choose $\mu>0$ such that any two distinct points in $\{0\}\cup\bigcup_{k,j}\sigma(y_k^{(3j)})$ have distance at least $2\mu$.
We may assume that $\mu<\tfrac{\varepsilon}{4}$.

For $j\in\{0,\ldots,N\}$ let $f_j,g_j,h_j\colon[0,1]\to[0,1]$ be continuous functions such that:
\begin{enumerate}
\item
$f_j$ takes the value $1$ on $[\tfrac{3j}{3N},\tfrac{3j+2}{3N}]$ for $j=0,\ldots,N-1$;
\item
the collection $(f_j)_{j=0,\ldots,N}$ is a partition of unity subordinate to the family $\{ [0,\tfrac{3}{3N}),(\tfrac{2}{3N},\tfrac{6}{2N}), \ldots,(\tfrac{3N-4}{3N},\tfrac{3N}{3N}), (\tfrac{3N-1}{3N},\tfrac{3N}{3N}] \}$;
\item
$g_j$ takes the value $1$ on $[\tfrac{3j-2}{3N},\tfrac{3j}{2N}]$ for $j=1,\ldots,N$;
\item
the collection $(g_j)_{j=0,\ldots,N}$ is a partition of unity subordinate to the family $\{ [0,\tfrac{1}{3N}),(\tfrac{0}{3N},\tfrac{4}{3N}), \ldots,(\tfrac{3N-6}{3N},\tfrac{3N-2}{3N}), (\tfrac{3N-3}{3N},\tfrac{3N}{3N}] \}$.
\item
$h_0$ takes the value $1$ on $[\tfrac{0}{3N},\tfrac{1}{3N}]$;
$h_N$ takes the value $1$ on $[\tfrac{3N-1}{3N},\tfrac{3N}{3N}]$;
and $h_j$ takes the value $1$ on $[\tfrac{3j-1}{3N},\tfrac{3j+1}{3N}]$ for $j=1,\ldots,N-1$;
\item
the collection $(h_j)_{j=0,\ldots,N}$ is a partition of unity subordinate to the family $\{ [0,\tfrac{2}{3N}),(\tfrac{1}{3N},\tfrac{5}{2N}), \ldots,(\tfrac{3N-5}{3N},\tfrac{3N-1}{3N}), (\tfrac{3N-2}{3N},\tfrac{3N}{3N}] \}$.
\end{enumerate}
The functions are depicted in the following picture:
\[
\makebox{
\begin{tikzpicture}[xscale=0.5]
\draw [help lines, ->] (-0.1,0) -- (14.2,0);
\draw [help lines, ->] (0,-0.1) -- (0,1.5);

\draw [thick] (0,1) -- (2,1);
\draw [thick] (2,1) -- (3,0);
\draw [thick] (2,0) -- (3,1);
\draw [thick] (3,1) -- (5,1);
\draw [thick] (5,1) -- (6,0);
\fill (7,0.5) ellipse (0.1 and 0.05);
\fill (7.5,0.5) ellipse (0.1 and 0.05);
\fill (8,0.5) ellipse (0.1 and 0.05);
\draw [thick] (9,0) -- (10,1);
\draw [thick] (10,1) -- (12,1);
\draw [thick] (12,1) -- (13,0);
\draw [thick] (12,0) -- (13,1);
\draw [dotted] (13,0) -- (13,1);

\draw [thick] (-0.1,1) -- (0,1);
\node [left] at (0,1) {$1$};
\node [above] at (1,1) {$f_0$};
\node [above] at (4,1) {$f_1$};
\node [above] at (11,1) {$f_{N-1}$};
\node [right] at (13,1.0) {$f_{N}$};

\draw [thick] (0,0) -- (0,-0.1);
\draw [thick] (1,0) -- (1,-0.1);
\draw [thick] (2,0) -- (2,-0.1);
\draw [thick] (3,0) -- (3,-0.1);
\draw [thick] (4,0) -- (4,-0.1);
\draw [thick] (5,0) -- (5,-0.1);
\draw [thick] (6,0) -- (6,-0.1);
\draw [thick] (12,0) -- (12,-0.1);
\draw [thick] (13,0) -- (13,-0.1);
\node [below] at (0,-0.05) {$0$};
\node [below] at (1,-0.05) {$\tfrac{1}{3N}$};
\node [below] at (2,-0.05) {$\tfrac{2}{3N}$};
\node [below] at (3,-0.05) {$\tfrac{3}{3N}$};
\node [below] at (4,-0.05) {$\tfrac{4}{3N}$};
\node [below] at (5,-0.05) {$\tfrac{5}{3N}$};
\node [below] at (6,-0.05) {$\tfrac{6}{3N}$};
\node [below] at (12,-0.05) {$\tfrac{3N-1}{3N}$};
\node [below] at (13,-0.05) {$1$};

\draw [help lines, ->] (-0.1,-2.2) -- (14.2,-2.2);
\draw [help lines, ->] (0,-2.3) -- (0,-0.7);

\draw [thick] (0,-1.2) -- (1,-2.2);
\draw [thick] (0,-2.2) -- (1,-1.2);
\draw [thick] (1,-1.2) -- (3,-1.2);
\draw [thick] (3,-1.2) -- (4,-2.2);
\fill (5,-1.7) ellipse (0.1 and 0.05);
\fill (5.5,-1.7) ellipse (0.1 and 0.05);
\fill (6,-1.7) ellipse (0.1 and 0.05);
\draw [thick] (7,-2.2) -- (8,-1.2);
\draw [thick] (8,-1.2) -- (10,-1.2);
\draw [thick] (10,-1.2) -- (11,-2.2);
\draw [thick] (10,-2.2) -- (11,-1.2);
\draw [thick] (11,-1.2) -- (13,-1.2);
\draw [dotted] (13,-2.2) -- (13,-1.2);

\draw [thick] (-0.1,-1.2) -- (0,-1.2);
\node [left] at (0,-1.2) {$1$};
\node [above] at (0.5,-1.3) {$g_0$};
\node [above] at (2,-1.2) {$g_1$};
\node [above] at (9,-1.2) {$g_{N-1}$};
\node [above] at (12,-1.2) {$g_{N}$};

\draw [thick] (0,-2.2) -- (0,-2.3);
\draw [thick] (1,-2.2) -- (1,-2.3);
\draw [thick] (2,-2.2) -- (2,-2.3);
\draw [thick] (3,-2.2) -- (3,-2.3);
\draw [thick] (4,-2.2) -- (4,-2.3);
\draw [thick] (10,-2.2) -- (10,-2.3);
\draw [thick] (11,-2.2) -- (11,-2.3);
\draw [thick] (12,-2.2) -- (12,-2.3);
\draw [thick] (13,-2.2) -- (13,-2.3);
\node [below] at (0,-2.25) {$0$};
\node [below] at (1,-2.25) {$\tfrac{1}{3N}$};
\node [below] at (2,-2.25) {$\tfrac{2}{3N}$};
\node [below] at (3,-2.25) {$\tfrac{3}{3N}$};
\node [below] at (4,-2.25) {$\tfrac{4}{3N}$};
\node [below] at (10,-2.25) {$\tfrac{3N-3}{3N}$};
\node [below] at (13,-2.25) {$1$};

\draw [help lines, ->] (-0.1,-4.4) -- (14.2,-4.4);
\draw [help lines, ->] (0,-4.5) -- (0,-2.9);

\draw [thick] (0,-3.4) -- (1,-3.4);
\draw [thick] (1,-3.4) -- (2,-4.4);
\draw [thick] (1,-4.4) -- (2,-3.4);
\draw [thick] (2,-3.4) -- (4,-3.4);
\draw [thick] (4,-3.4) -- (5,-4.4);
\fill (6,-3.9) ellipse (0.1 and 0.05);
\fill (6.5,-3.9) ellipse (0.1 and 0.05);
\fill (7,-3.9) ellipse (0.1 and 0.05);
\draw [thick] (8,-4.4) -- (9,-3.4);
\draw [thick] (9,-3.4) -- (11,-3.4);
\draw [thick] (11,-3.4) -- (12,-4.4);
\draw [thick] (11,-4.4) -- (12,-3.4);
\draw [thick] (12,-3.4) -- (13,-3.4);
\draw [dotted] (13,-4.4) -- (13,-3.4);

\draw [thick] (-0.1,-3.4) -- (0,-3.4);
\node [left] at (0,-3.4) {$1$};
\node [above] at (0.5,-3.4) {$h_0$};
\node [above] at (3,-3.4) {$h_1$};
\node [above] at (10,-3.4) {$h_{N-1}$};
\node [above] at (12.5,-3.4) {$h_{N}$};

\draw [thick] (0,-4.4) -- (0,-4.5);
\draw [thick] (1,-4.4) -- (1,-4.5);
\draw [thick] (2,-4.4) -- (2,-4.5);
\draw [thick] (3,-4.4) -- (3,-4.5);
\draw [thick] (4,-4.4) -- (4,-4.5);
\draw [thick] (5,-4.4) -- (5,-4.5);
\draw [thick] (11,-4.4) -- (11,-4.5);
\draw [thick] (12,-4.4) -- (12,-4.5);
\draw [thick] (13,-4.4) -- (13,-4.5);
\node [below] at (0,-4.45) {$0$};
\node [below] at (1,-4.45) {$\tfrac{1}{3N}$};
\node [below] at (2,-4.45) {$\tfrac{2}{3N}$};
\node [below] at (3,-4.45) {$\tfrac{3}{3N}$};
\node [below] at (4,-4.45) {$\tfrac{4}{3N}$};
\node [below] at (5,-4.45) {$\tfrac{5}{3N}$};
\node [below] at (11,-4.45) {$\tfrac{3N-2}{3N}$};
\node [below] at (13,-4.45) {$1$};
\end{tikzpicture}
}
\]

Define $y_1,y_2,y_3\colon[0,1]\to A$ by
\begin{align*}
y_1(t) &:= \sum_{j=0}^{N} \left( y_1^{(3j)} + t\mu \right) f_{j}(t), \quad
y_2(t) := \sum_{j=0}^{N} \left( y_2^{(3j)} + t\mu \right) g_{j}(t), \\
y_3(t) &:= \sum_{j=0}^{N} \left( y_3^{(3j)} + t\mu \right) h_{j}(t),
\end{align*}
for $t\in[0,1]$.

To verify that $\|y_1-x_1\|<\varepsilon$, we estimate $\|y_1(t)-x_1(t)\|$ for $t\in[0,1]$, similarly as in the proof of \autoref{prp:approxGenFibers}.
For $t=1$, we obtain
\[
\left\| y_1(1) - x_1(1) \right\|
= \left\|  y_1^{(3j)} + \mu - x_1(1) \right\|
< \frac{\varepsilon}{4} + \frac{\varepsilon}{4}
\leq \frac{3}{4}\varepsilon.
\]
Given $t\in[0,1)$, let $j\in\{1,\ldots,N\}$ satisfy $t\in[\tfrac{3j-3}{3N},\tfrac{3j}{3N})$.
Using at the second step that $|t-\tfrac{3j}{3N}|<\tfrac{3}{3N}$, we have
\begin{align*}
\left\| y_1^{(3j)} + t\mu - x_1(t) \right\|
&\leq \mu + \left\| y_1^{(3j)} - x_1(\tfrac{3j}{3N}) \right\|
+ \left\| x_1(\tfrac{3j}{3N}) - x_1(t) \right\|

\leq \frac{3}{4}\varepsilon.
\end{align*}
Analogously, we obtain $\| y_1^{(3j-3)} + t\mu - x_1(t) \|\leq \frac{3}{4}\varepsilon$.
We deduce that
\[
\left\| y_1(t) - x_1(t) \right\|
= \left\| \left( \left(y_1^{(3j-3)} + t\mu\right)f_{j-1}(t) + \left( y_1^{(3j)} + t\mu\right) f_j(t) \right) - x_1(t) \right\|
\leq \frac{3}{4}\varepsilon.
\]
Hence, $\|y_1-x_1\|=\sup_{t\in[0,1]}\|y_1(t)-x_1(t)\|\leq\tfrac{3}{4}\varepsilon<\varepsilon$.
Analogously, one shows that $\|y_k-x_k\|<\varepsilon$ for $k=2,3$.

It remains to verify that $C^*(y_1,y_2,y_3)$ separates the points in $[0,1]$ in the sense of \autoref{prp:genCenter}(b).
Let $s,t\in[0,1]$ be distinct.
Choose $k\in\{1,2,3\}$ such that both $s$ and $t$ are contained in the set
\[
\bigcup_{j\in\ZZ} \left[ \frac{3j+k-1}{3N},\frac{3j+k+1}{3N} \right]
\]
Let us consider the case $k=1$.
Choose $j,j'\in\{0,\ldots,N\}$ such that $s\in[\tfrac{3j}{3N},\frac{3j+2}{3N}]$ and $t\in[\tfrac{3j'}{3N},\frac{3j'+2}{3N}]$.
Then
\[
y_1(s) = y_1^{(3j)} + s\mu , \andSep
y_1(t) = y_1^{(3j')} + t\mu.
\]
Since $s\neq t$, and by choice of $\mu$, it follows that $y_1(s)$ and $y_1(t)$ have finite spectra that are disjoint and do not contain $0$.
Hence, there exists a continuous function $f\colon\RR\to[0,1]$ such that $f(y_1(s))=0$ and $f(y_1(t))=1$.
Then the element $b:=f(y_1)\in B$ satisfies $b(s)=0$ and $b(t)=1$.

The cases $k=2,3$ are analogous, using $y_2$ and $y_3$.
\end{proof}

\begin{prp}
\label{prp:tensI}
Let $A$ be a unital, separable \ca{} of real rank zero and generator rank at most one.
Set $B:=A\otimes C([0,1])$.
The we have $\gr(B)\leq 6$.
\end{prp}
\begin{proof}
We show that every $7$-tuple in $B_\sa$ can be approximated by a tuple that generates $B$.
Since $B$ is separable and unital, this verifies $\gr(B)\leq 6$;
see \autoref{prp:grCharSep}.

Let $x_1,\ldots,x_7\in B_\sa$ and $\varepsilon>0$.
Applying \autoref{prp:approxGenFibers} for $x_1,\ldots,x_4$, we obtain $y_1,\ldots,y_4\in B_\sa$ such that
\[
\| y_j - x_j \|<\varepsilon \ \text{ for } j=1,\ldots,4, \andSep
A=C^*(y_1(t),\ldots,y_4(t)) \ \text{ for all } t\in[0,1].
\]
Applying \autoref{prp:approxSepInterval} for $x_5,x_6,x_7$, we obtain $y_5,y_6,y_7\in B_\sa$ such that $\|y_k-x_k\|<\varepsilon$ for $k=5,6,7$, and such that $C^*(y_5,y_6,y_7)$ separates the points in $[0,1]$ in the sense of \autoref{prp:genCenter}(b).

Set $D:=C^*(y_1,\ldots,y_7)\subseteq B$.
Then $D$ exhausts each fiber of $B$, and moreover separates the points in $[0,1]$.

Hence, $D=B$ by \cite[Lemma~3.2]{ThiWin14GenZStableCa}, which shows that $\{y_1,\ldots,y_7\}$ generates $B$, as desired.
\end{proof}

\begin{rmk}
The proof of \autoref{prp:tensI} can be generalized to show the following:
If $A$ is a unital, separable \ca{} of real rank zero, then $B:=A\otimes C([0,1])$ satisfies $\gr(B)\leq 2\gr(A)+4$.
\end{rmk}

\section{Establishing finite generator rank}
\label{sec:finiteGr}

In this section we prove that unital, separable, $\mathcal{Z}$-stable \ca{s} of real rank zero have finite generator rank;
see \autoref{prp:gr8}.
In the next section, we will successively reduce the upper bound for the generator rank of such algebras down to one.

We start with a lemma that simplifies the computation of the generator rank for \ca{s} that absorb a strongly self-absorbing \ca.
For the definition and basic results of strongly self-absorbing \ca{s} we refer to \cite{TomWin07ssa}.
Given a strongly self-absorbing \ca{} $D$, a \ca{} $A$ is said to be \emph{$D$-stable} if $A\cong A\otimes D$.
Since every strongly self-absorbing \ca{} is nuclear, we do not need to specify the tensor product.
Typical examples of strongly self-absorbing \ca{s} are UHF-algebras of infinite type, the Jiang-Su algebra $\mathcal{Z}$, and the Cuntz algebras $\mathcal{O}_\infty$ and $\mathcal{O}_2$.

\begin{lma}
\label{prp:genDabsorbing}
Let $D$ be a strongly self-absorbing \ca{}, let $A$ be a separable, $D$-stable \ca, and $n\in\NN$.
Then the following are equivalent:
\begin{enumerate}
\item
We have $\grPre(A)\leq n$;
\item
For every $x_0,\ldots,x_n\in A_\sa$ and $\varepsilon>0$, there exist $y_0,\ldots,y_n\in (A\otimes D)_\sa$ such that
\[
\quad \quad
\| y_j - (x_j\otimes 1) \|<\varepsilon \ \text{ for } j=0,\ldots,n, \andSep
A\otimes D = C^*(y_0,\ldots,y_n);
\]
\item
For every $x_0,\ldots,x_n\in A_\sa$, $\varepsilon>0$, and $z\in A$, there exist $y_0,\ldots,y_n\in (A\otimes D)_\sa$ such that
\[
\quad \quad
\| y_j - (x_j\otimes 1) \|<\varepsilon\ \text{ for } j=0,\ldots,n, \andSep
z\otimes 1\in_\varepsilon C^*(y_0,\ldots,y_k).
\]
\end{enumerate}
\end{lma}
\begin{proof}
Since $A\cong A\otimes D$, we have $\grPre(A)=\grPre(A\otimes D)$.
Using that $A$ is separable, it follows from \autoref{prp:grCharSep} that~(1) implies~(2).
It is clear that~(2) implies~(3).
Assuming~(3), let us verify~(1).
To show $\grPre(A\otimes D)\leq n$, let $a_0,\ldots,a_n\in (A\otimes D)_\sa$, $\varepsilon>0$ and $c\in A\otimes D$.
We need to find $b_0,\ldots,b_n\in (A\otimes D)_\sa$ such that
\[
\| b_j - a_j \| < \varepsilon \text{ for } j=0,\ldots,n, \andSep
c\in_\varepsilon C^*(b_0,\ldots,b_n).
\]

Since $D$ is strongly self-absorbing and $A$ is separable and $D$-stable, there exists a ${}^*$-isomorphism $\Phi\colon A\to A\otimes D$ that is approximately unitarily equivalent to the inclusion $\iota\colon A\to A\otimes D$ given by $\iota(a)=a\otimes 1$, that is, there exists a sequence $(u_m)_m$ of unitaries in $A\otimes D$ such that $\lim_{m\to\infty} u_m\Phi(a)u_m^* = \iota(a)$ for all $a\in A$;
see \cite[Theorem~2.2]{TomWin07ssa}.
Set
\[
x_j := \Phi^{-1}(a_j)\ \text{ for } j=0,\ldots,n, \andSep
z:=\Phi^{-1}(c).
\]
Using that $u_m^*(x_j\otimes 1)u_m\to a_j$ for $j=0,\ldots,n$, and $u_m^*(z\otimes 1)u_m\to c$, we can choose $m$ such that
\[
\left\| u_m^*(x_j\otimes 1)u_m-a_j \right\| < \frac{\varepsilon}{2}\ \text{ for } j=0,\ldots,n, \andSep
\left\| u_m^*(z\otimes 1)u_m-c \right\| < \frac{\varepsilon}{2}.
\]
By assumption~(3), we obtain $y_0,\ldots,y_n\in (A\otimes D)_\sa$ such that
\[
\| y_j - (x_j\otimes 1) \| < \frac{\varepsilon}{2}\ \text{ for } j=0,\ldots,n, \andSep
z\otimes 1 \in_{\varepsilon/2} C^*(y_0,\ldots,y_{n}).
\]
For $j=0,\ldots,n$, we have
\[
\|u_m^* y_j u_m - a_j\|
\leq \|u_m^* y_j u_m - u_m^* (x_j\otimes 1) u_m\| + \| u_m^* (x_j\otimes 1) u_m - a_j \| < \varepsilon.
\]
Moreover, from $z\otimes 1 \in_{\varepsilon/2} C^*(y_0,\ldots,y_n)$ it follows that
\[
u_m^*(z\otimes 1)u_m \in_{\varepsilon/2} C^*(u_m^* y_0 u_m,\ldots, u_m^* y_n u_m),
\]
and thus
\[
c\in_\varepsilon C^*(u_m^* y_0 u_m,\ldots, u_m^* y_n u_m),
\]
which shows that $u_m^*y_0u_m,\ldots,u_m^*y_nu_m$ have the desired properties.
\end{proof}

\begin{prp}
\label{prp:UHFabs}
Let $A$ be a separable, unital \ca{} of real rank zero that tensorially absorbs a UHF-algebra of infinite type.
Then $\gr(A)\leq 1$.
\end{prp}
\begin{proof}
Let $D$ be UHF-algebra of infinite type such that $A$ is $D$-stable.
Since $A$ is unital, we have $\gr(A)=\grPre(A)$.
To verify condition~(3) of \autoref{prp:genDabsorbing}, let $a,b\in A_\sa$, $\varepsilon>0$ and $c\in A$.
We need to find $x,y\in (A\otimes D)_\sa$ such that
\[
\| x - (a\otimes 1) \| < \varepsilon, \quad
\| y - (b\otimes 1) \| < \varepsilon, \andSep
c\otimes 1 \in_\varepsilon C^*(x,y).
\]

Since $A$ has real rank zero, we may assume that $a$ is invertible and that its spectrum $\sigma(a)$ is finite, so that there exist $\lambda_1,\ldots,\lambda_n\in\RR\setminus\{0\}$ and pairwise orthogonal projections $p_1,\ldots,p_n\in A$ that sum to $1_A$ such that $a=\sum_{j=1}^n \lambda_j p_j$.
Choose $\mu>0$ such that $\mu$ is strictly smaller than the distance between any two values in $\sigma(a)\cup\{0\}$.
We may assume that $\mu<\varepsilon$.

Choose $d\geq 5$ such that $D$ admits a unital embedding $M_d\subseteq D$.
Using that $c$ can be written as a linear combination of four positive, invertible elements, we can choose positive, invertible elements $c_2,c_3\ldots,c_d\in A$ such that
\[
\|c_j\| < \frac{\varepsilon}{d}, \ \text{ for } j=2,\ldots,d, \andSep
c \in C^*(c_2,\ldots,c_d).
\]
Let $(e_{j,k})_{j,k=1,\ldots,d}$ be matrix units for $M_d$.
We define $x,y\in A\otimes M_{d}$ as
\begin{align*}
x := a\otimes 1 + \sum_{j=1}^d \frac{j}{d}\mu e_{jj}, \andSep
y := b\otimes 1 + \sum_{j=2}^d c_j\otimes \left( e_{1j} + e_{j1} \right) .
\end{align*}

As matrices, these elements look as follows:
\[
x=\left(
\begin{array}{cccccc}
a + \frac{1}{d}\mu & 0 & \cdots & 0 \\
0 & a + \frac{2}{d}\mu & & \vdots \\
\vdots & & \ddots & 0 \\
0 & \cdots & 0 & a + \mu \\
\end{array}\right), \
y=\left(
\begin{array}{cccccc}
b & c_2 & c_3 & \cdots & c_d \\
c_2 & b & 0 & \cdots & 0 \\
c_3 & 0 & b & & \vdots \\
\vdots & \vdots & & \ddots & 0 \\
c_d & 0 & \cdots & 0 & b \\
\end{array}\right), \
\]

Then
\[
\| x - (a\otimes 1) \|
= \left\| \sum_{j=1}^d \frac{j}{d}\mu e_{jj} \right\|
= \mu < \varepsilon,
\]
and
\begin{align*}
\| y - (b\otimes 1) \|
= \left\| \sum_{j=2}^d c_j\otimes \left( e_{1j} + e_{j1} \right) \right\|
\leq \sum_{j=2}^d \|c_j\| < \varepsilon.
\end{align*}

Set $B:=C^*(x,y)\subseteq A\otimes M_{d}\subseteq A\otimes D$.
We will verify that $z\otimes 1 \in B$.
The $j$-th element on the diagonal of $x$ is $a+\tfrac{j}{d}\mu$, whose spectrum is
\[
\sigma(a+\frac{j}{d}\mu)
= \left\{ \lambda_1+\frac{j}{d}\mu,\ldots,\lambda_n+\frac{j}{d}\mu \right\}.
\]

Given distinct $j,k\in\{1,\ldots,d\}$, it follows from the choice of $\mu$ that the spectra of $a+\tfrac{j}{d}\mu$ and $a+\tfrac{k}{d}\mu$ are finite disjoint sets not containing $0$.
For $j\in\{1,\ldots,d\}$, let $f_j\colon\RR\to[0,1]$ be a continuous functions that takes the value $1$ on $\sigma(a+\tfrac{j}{d}\mu)$, and that takes the value $0$ on $\{0\}\cup\bigcup_{k\neq j}\sigma(a+\frac{k}{d}\mu)$.
Then
\[
1\otimes e_{jj} = f_j(x) \in B \subseteq A\otimes M_d.
\]
Thus, $B$ contains the diagonal matrix units of $M_d$.

To show that $B$ also contains the other matrix units, we follow ideas of Olsen and Zame from \cite{OlsZam76CaSingleGen}.
Given $k\in\{2,\ldots,d\}$, we have
\[
c_k\otimes e_{1k} = (1\otimes e_{11}) y (1\otimes e_{kk}) \in B.
\]
Then
\[
c_k^2 \otimes e_{11} = (c_j\otimes e_{1k})(c_j\otimes e_{1k})^* \in B.
\]

Since $c_k$ is positive and invertible, we have $c_k^{-1}\in C^*(c_k^2)\subseteq A$, and it follows that $c_k^{-1}\otimes e_{11}\in B$.
Hence
\begin{align*}
1\otimes e_{1k} =
(c_k^{-1}\otimes e_{11})(c_k\otimes e_{1k}) \in B.
\end{align*}

It follows that $1\otimes M_d\subseteq B$.
For each $k\in\{2,\ldots,d\}$, we deduce that
\[
c_k\otimes 1
= \sum_{j=1}^d (1\otimes e_{j1})(c_k\otimes e_{1k})(1\otimes e_{kj}) \in B.
\]
Since $c\in C^*(c_2,\ldots,c_d)$, we get $c\otimes 1\in B\subseteq A\otimes D$, as desired.
\end{proof}

We use $M_{2^\infty}$ to denote the UHF-algebra of type $2^\infty$, and similarly for $M_{3^\infty}$.

\begin{prp}
\label{prp:gr8}
Let $A$ be a unital, separable \ca{} of real rank zero.
Then $\gr(A\otimes\mathcal{Z})\leq 8$.
\end{prp}
\begin{proof}
Let $Z_{2^\infty,3^\infty}$ be the generalized dimension-drop algebra given by
\[
Z_{2^\infty,3^\infty} = \big\{ f\in C([0,1],M_{2^\infty}\otimes M_{3^\infty}) : f(0)\in M_{2^\infty}\otimes 1, f(1)\in 1\otimes M_{3^\infty} \big\}.
\]
By \cite[Theorem~3.4]{RorWin10ZRevisited}, $\mathcal{Z}$ is an inductive limit of a sequence of \ca{s} each isomorphic to $Z_{2^\infty,3^\infty}$.
Hence, $A\otimes\mathcal{Z}$ is isomorphic to an inductive limit of \ca{s} isomorphic to $A\otimes Z_{2^\infty,3^\infty}$.
Using that the generator rank behaves well with respect to inductive limits (\autoref{prp:permanence}), we get
\[
\gr(A\otimes\mathcal{Z})
\leq \liminf_{n\to\infty} \gr(A\otimes Z_{2^\infty,3^\infty})
= \gr(A\otimes Z_{2^\infty,3^\infty}).
\]

It thus suffices to verify $\gr(A\otimes Z_{2^\infty,3^\infty})\leq 8$.
Set
\[
I := \big\{ f\in Z_{2^\infty,3^\infty} : f(0)=f(1)=0 \big\}.
\]
Then $I$ is a closed, two-sided ideal in $Z_{2^\infty,3^\infty}$ and $Z_{2^\infty,3^\infty}/I\cong M_{2^\infty}\oplus M_{3^\infty}$.
Since $Z_{2^\infty,3^\infty}$ is nuclear, we obtain a short exact sequence
\[
0 \to A\otimes I \to A\otimes Z_{2^\infty,3^\infty} \to A\otimes(M_{2^\infty}\oplus M_{3^\infty}) \to 0.
\]
Note that $A\otimes I$ is isomorphic to a closed, two-sided ideal in $A\otimes M_{6^\infty}\otimes C([0,1])$.

Since real rank zero is preserved by passing to matrix algebras and inductive limits, we get $\rr(A\otimes M_{k^\infty})=0$ for $k=2,3,6$.
By \autoref{prp:UHFabs}, we obtain $\gr(A\otimes M_{k^\infty})\leq 1$ for $k=2,3,6$.
Using that the generator rank does not increase when passing to closed, two-sided ideals (\autoref{prp:permanence}) at the first step, and \autoref{prp:tensI} for $A\otimes M_{6^\infty}$ at the second step, we get
\[
\gr(A\otimes I)\leq \gr\big( A\otimes M_{6^\infty}\otimes C([0,1]) \big) \leq 6.
\]

By \autoref{prp:gr_sum_rr0}, we have
\[
\gr\big( A\otimes(M_{2^\infty}\oplus M_{3^\infty}) \big)
= \max\big\{ \gr(A\otimes M_{2^\infty}), \gr(A\otimes M_{3^\infty}) \big\}
\leq 1.
\]

Applying the estimate for the generator rank of an extension (\autoref{prp:permanence}), we get
\[
\gr( A\otimes Z_{2^\infty,3^\infty} )
\leq \gr(A\otimes I)+\gr(A\otimes(M_{2^\infty}\oplus M_{3^\infty}) + 1
\leq 8,
\]
as desired.
\end{proof}

\section{Establishing generator rank one}
\label{sec:reduceGr}

In this section we prove our main result:
separable, $\mathcal{Z}$-stable \ca{s} of real rank zero have generator rank one;
see \autoref{prp:mainThm}.
We deduce some interesting corollaries, most importantly that every classifiable, simple, nuclear \ca{} has generator rank one;
see \autoref{prp:classifiable-gr1}.

Recall that the dimension-drop algebra $Z_{2,3}$ is defined as
\[
Z_{2,3} := \big\{ f\in C([0,1],M_2\otimes M_3 : f(0)\in M_2\otimes 1, f(1)\in 1\otimes M_3 \big\}.
\]
Below, we always view $Z_{2,3}$ as the subalgebra of $C([0,1),M_6)$ given by the next result.

\begin{lma}
The dimension-drop algebra $Z_{2,3}$ is isomorphic to the subalgebra of $C([0,1],M_6)$ consisting of the continuous functions $f\colon [0,1]\to M_6$ that satisfy
\[
f(0)=\left(
\begin{array}{cccccc}
\alpha_{11}\!\!\! & \alpha_{12}\!\!\! \\
\alpha_{21}\!\!\! & \alpha_{22}\!\!\! \\
& & \alpha_{11}\!\!\! & \alpha_{12}\!\!\! \\
& & \alpha_{21}\!\!\! & \alpha_{22}\!\!\! \\
& & & & \alpha_{11}\!\!\! & \alpha_{12} \\
& & & & \alpha_{21}\!\!\! & \alpha_{22} \\
\end{array}\right),\
f(1)=\left(
\begin{array}{cccccc}
\beta_{11} \!\!\! & \beta_{12} \!\!\! & \beta_{13} \!\!\! \\
\beta_{21} \!\!\! & \beta_{22} \!\!\! & \beta_{23} \!\!\! \\
\beta_{31} \!\!\! & \beta_{32} \!\!\! & \beta_{33} \!\!\! \\
& & & \beta_{33} \!\!\! & \beta_{31} \!\!\! & \beta_{32} \\
& & & \beta_{13} \!\!\! & \beta_{11} \!\!\! & \beta_{12} \\
& & & \beta_{23} \!\!\! & \beta_{21} \!\!\! & \beta_{22} \\
\end{array}\right),
\]
for some $\alpha_{jk},\beta_{jk}\in\CC$.
\end{lma}
\begin{proof}
Using an identification of $M_2\otimes M_3$ with $M_6$, we naturally view $Z_{2,3}$ as a subalgebra of $C([0,1],M_6)$.
Let $u\in M_6$ be the permutation matrix given as:
\[
u :=
\left(
\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 \\
\end{array}\right).
\]
Let $t\mapsto v_t$ be a continuous path of unitaries in $M_6$ with $v_0=1$ and $v_1=u$.
Then $v$ is a unitary in $C([0,1],M_6)$ that conjguates $Z_{2,3}\subseteq C([0,1],M_6)$ onto the subalgebra  of functions described in the statement.
\end{proof}

\begin{lma}
\label{prp:construction}
Let $A$ be a unital, separable \ca{} of real rank zero, and let $n\in\NN$ such that $\gr(A)\leq n+2$.
Let $x_0,\ldots,x_{n+1}\in A_\sa$, $\varepsilon>0$ and $z\in A$.
Then there exist $y_0,\ldots,y_{n+1}\in (A\otimes Z_{2,3})_\sa$ such that
\[
\left\| y_j - (x_j\otimes 1) \right\| < \varepsilon\ \text{ for } j=0,\ldots,n+1, \andSep
z\otimes 1 \in_\varepsilon C^*(y_0,\ldots,y_{n+1}).
\]
\end{lma}
\begin{proof}
Consider the $n+3$ elements $\tfrac{\varepsilon}{2},x_0,\ldots,x_{n+1}\in A_\sa$.
Using that $A$ is unital, separable with $\gr(A)\leq n+2$, apply \autoref{prp:grCharSep} to obtain $a,x_0',\ldots,x_{n+1}'\in A_\sa$ such that
\[
\left\| a-\frac{\varepsilon}{2} \right\| < \frac{\varepsilon}{2}, \
\left\| x_j' - x_j \right\| < \varepsilon\ \text{ for } j=0,\ldots,n+1, \ \text{and}\
A=C^*(a,x_0',\ldots,x_{n+1}').
\]
Note that $a$ is positive and invertible.
To simplify notation, set $b:=x_0'$ and $c:=x_1'$.
Choose a polynomial $p$ in noncommuting variables such that
\[
\left\| z-p(a,b,c,x_2',\ldots,x_{n+1}') \right\| < \varepsilon.
\]
Since $p$ is continuous as a map $A^{n+3}_\sa\to A$, we can choose $\delta>0$ such that every $\bar{c}\in A_\sa$ with $\|\bar{c}-c\|<\delta$ satisfies
\[
\left\| z-p(a,b,\bar{c},x_2',\ldots,x_{n+1}') \right\| < \varepsilon.
\]

Since $A$ has real rank zero, we can choose an invertible element $c'\in A_\sa$ with finite spectrum $\sigma(c')$ and such that $\|c'-c\|<\tfrac{\delta}{2}$.
Then $c'=\sum_{j\in J}\lambda_j q_j$ for some finite set $J$, some pairwise disjoint real numbers $\lambda_j$ (the eigenvalues of $c'$), and some pairwise orthogonal projections $q_j$ that sum to $1_A$.

Choose $\mu>0$ such that any two points in $\{0\}\cup\{\lambda_j:j\in J\}$ have distance strictly larger than $4\mu$.
We may assume that $2\mu<\varepsilon$ and $2\mu<\tfrac{\delta}{2}$.

Next, we define auxiliary functions $\alpha^{(k)}\colon[0,1]\to\RR$ for $k=1,\ldots,6$ by
\begin{align*}
\alpha^{(1)}(t):= 1+t(1-t), \quad
\alpha^{(5)}(t):= 1, \quad
\alpha^{(3)}(t):=1-t,
\end{align*}
and
\[
\alpha^{(6)}(t):=-\alpha^{(1)}(t), \quad
\alpha^{(2)}(t):=-\alpha^{(5)}(t), \quad
\alpha^{(4)}(t):=-\alpha^{(3)}(t), \quad
\]
for $t\in[0,1]$.
The functions are shown in the following picture:
\[
\makebox{
\begin{tikzpicture}[xscale=1.5, yscale=0.8]
\draw [help lines, ->] (-0.1,0) -- (1.2,0);
\draw [help lines, ->] (0,-1.5) -- (0,1.5);
\draw [thick, domain=0:1, samples=500] plot (\x, {1+\x-\x^2});
\draw [thick] (0,1) -- (1,1);
\draw [thick] (0,1) -- (1,0);
\draw [thick] (0,-1) -- (1,0);
\draw [thick] (0,-1) -- (1,-1);
\draw [thick, domain=0:1, samples=500] plot (\x, {-1-\x+\x^2});
\node [left] at (0,1) {$1$};
\node [left] at (0,-1) {$-1$};
\draw [->]  (1.1,1.2) -- (0.85,1.2);
\node [right] at (1.3,1.3) {$\alpha^{(1)}$};
\draw [->]  (1.3,0.5) -- (0.75,0.95);
\node [right] at (1.3,0.5) {$\alpha^{(5)}$};
\draw [->] (-0.15,0.4) -- (0.35,0.55);
\node [left] at (-0.1,0.4) {$\alpha^{(3)}$};
\draw [->] (-0.15,-0.4) -- (0.35,-0.55);
\node [left] at (-0.1,-0.4) {$\alpha^{(4)}$};
\draw [->]  (1.3,-0.5) -- (0.75,-0.95);
\node [right] at (1.3,-0.5) {$\alpha^{(2)}$};
\draw [->]  (1.1,-1.2) -- (0.85,-1.2);
\node [right] at (1.3,-1.3) {$\alpha^{(6)}$};
\end{tikzpicture}
}
\]

We note the following properties:
\begin{enumerate}
\item[(a)]
$\alpha^{(k)}$ is continuous with $\|\alpha^{(k)}\|_\infty<2$, for $k=1,\ldots,6$;
\item[(b)]
$\alpha^{(1)}(0)=\alpha^{(5)}(0)=\alpha^{(3)}(0)$ and $\alpha^{(4)}(0)=\alpha^{(2)}(0)=\alpha^{(6)}(0)$;
\item[(c)]
$\alpha^{(1)}(1)=\alpha^{(5)}(1)$, $\alpha^{(3)}(1)=\alpha^{(4)}(1)$, and $\alpha^{(2)}(1)=\alpha^{(6)}(1)$;
\item[(d)]
$\alpha^{(6)}(t)<\alpha^{(2)}(t)<\alpha^{(4)}(t)<\alpha^{(3)}(t)<\alpha^{(5)}(t)<\alpha^{(1)}(t)$ for each $t\in(0,1)$.
\end{enumerate}

For each $k=1,\ldots,6$, we define $f_{kk}\colon[0,1]\to A$ by
\begin{align*}
f_{kk}(t) := \sum_{j\in J} \big(\lambda_j+\mu \alpha^{(k)}(t)\big) q_j
\end{align*}
for $t\in[0,1]$.
We let $e_{kl}\in M_6$ denote the matrix units, for $k,l=1,\ldots,6$.
Then define $f,g\colon[0,1]\to A\otimes M_6$ by
\begin{align*}
f(t) &:= \sum_{k=1}^6 f_{kk}(t)\otimes e_{kk}, \\
g(t) &:= b\otimes 1
+ a\otimes(e_{12}+e_{21}+e_{56}+e_{65})
+ \mu t (e_{23}+e_{32}+e_{46}+e_{64}) \\
&\quad + (1-t)a(e_{34}+e_{43}).
\end{align*}
for $t\in[0,1]$.

This means, that $f(t)$ and $g(t)$ have the following matrix form:
\begin{align*}
f(t) &:=\left( \begin{array}{ccc}
f_{11} \\
& \ddots \\
& & f_{66}\\
\end{array} \right),\
g(t) &:=\left( \begin{array}{cc|cc|cc}
b & a & & & \\
a & b & \mu t & & \\
\hline
& \mu t & b & (1-t)a & & \\
& & (1-t)a & b & & \mu t \\
\hline
& & & & b & a \\
& & & \mu t & a & b\\
\end{array} \right).
\end{align*}

We view elements in $A\otimes Z_{2,3}$ as continuous functions $[0,1]\to A\otimes M_6$.
By~(a), $f$ is continuous.
Using~(b) and~(c), we deduce that $f$ belongs to $A\otimes Z_{2,3}$.
We also have $g\in A\otimes Z_{2,3}$.
Set
\[
B := C^*(f,g,x_2'\otimes 1,\ldots,x_{n+1}'\otimes 1)\subseteq A\otimes Z_{2,3}.
\]
For each $t\in[0,1]$, we let $B(t)\subseteq A\otimes M_6$ be the image of $B$ under the evaluation map $A\otimes Z_{2,3}\to A\otimes M_6$, $h\mapsto h(t)$.
We use $e_{kl}$ to denote the matrix units in $M_6$.

\emph{Claim~1a: Let $t\in(0,1)$. Then $1\otimes M_6\subseteq B(t)$.}
To prove the claim, note that the spectrum of $f_{kk}(t)$ is
\[
\sigma(f_{kk}(t)) = \big\{ \lambda_j+\alpha_j^{(k)}(t) : j\in J \big\},
\]
for $k=1,\ldots,6$.
Using~(d), we obtain that $\sigma(f_{kk}(t))$ and $\sigma(f_{ll}(t))$ are disjoint whenever $k\neq l$.

Given $k\in\{1,\ldots,6\}$, let $h_k\colon\RR\to[0,1]$ be a continuous function that takes the value $1$ on $\sigma(f_{kk}(t))$, and that takes the value $0$ on $\{0\}\cup\bigcup_{l\neq k}\sigma(f_{ll}(t))$.
Then $1\otimes e_{kk}=h_k(f(t))\in B(t)$.
Thus, $B(t)$ contains the diagonal matrix units of $1\otimes M_6$.

It follows that $B(t)$ contains
\[
1\otimes e_{23}
= \frac{1}{\mu t} (1\otimes e_{22})g(t)(1\otimes e_{33}), \andSep
1\otimes e_{46}
= \frac{1}{\mu t} (1\otimes e_{44})g(t)(1\otimes e_{66}).
\]

We have $a\otimes e_{12} = (1\otimes e_{11})g(t)(1\otimes e_{22}) \in B(t)$, and thus
\[
a^2\otimes e_{22} = (a\otimes e_{12})^*(a\otimes e_{12}) \in B(t).
\]
As in the proof of \autoref{prp:UHFabs}, we use that $a$ is positive and invertible, to deduce $a^{-1}\otimes e_{22}\in B(t)$, and thus
\[
1\otimes e_{12} = (a\otimes e_{12})(a^{-1}\otimes e_{22})\in B(t).
\]
Analogously, we get that $B(t)$ contains $1\otimes e_{34}$ and $1\otimes e_{56}$.
It follows that $B(t)$ contains $1\otimes e_{kl}$ for every $k,l\in\{1,\ldots,6\}$, and so $1\otimes M_6\subseteq B(t)$.
Similarly, one proves:

\emph{Claim~1b: We have
\[
1\otimes(e_{kl}+e_{k+2,l+2}+e_{k+4,l+4})\in B(0), \text{ for } k,l\in\{1,2\}.
\]
}

\emph{Claim~1c: We have $1\otimes(e_{33}+e_{44})\in B(1)$ and
\[
1\otimes(e_{kl}+e_{k+4,l+4})\in B(1), \text{ for } k,l\in\{1,2\}.
\]
and
\[
1\otimes(e_{l3}+e_{l+4,4}), 1\otimes(e_{3l}+e_{4,l+4}), \in B(1), \text{ for } l\in\{1,2\}.
\]
}

\emph{Claim~2: Let $t\in[0,1]$. Then $z\otimes 1_{M_6} \in_\varepsilon B(t)$.}
First, we assume that $t\in(0,1)$.
By Claim~1a, we have $1\otimes e_{11},1\otimes e_{21}\in B(t)$, and thus
\[
a\otimes e_{11} = (1\otimes e_{11})g(t)(1\otimes e_{21})\in B(t).
\]
Analogously, we obtain
\[
b\otimes e_{11},\ f_{11}(t)\otimes e_{11},\ x_2'\otimes e_{11},\ldots,x_{n+1}'\otimes e_{11}\in B(t).
\]
We have
\[
\| f_{11}(t)-c \|
\leq \| f_{11}(t)-c' \| + \| c'-c \|
< 2\mu +\frac{\delta}{2}
\leq \delta.
\]
By choice of $\delta$, we get
\[
\left\| z-p(a,b,f_{11}(t),x_2',\ldots,x_{n+1}') \right\| < \varepsilon.
\]
and thus
\[
z\otimes e_{11} \in_\varepsilon B(t).
\]
It follows that $z\otimes e_{kk} \in_\varepsilon B(t)$ for each $k$, and consequently $z\otimes 1 \in_\varepsilon B(t)$.

Next, we consider the case $t=0$.
Set $\tilde{e}_{kl}:=e_{kl}+e_{k+2,l+2}+e_{k+4,l+4}\in M_6$ for $k,l\in\{1,2\}$.
By Claim~1b, we have $1\otimes\tilde{e}_{kl}\in B(0)$ for each $k,l$, and thus
\[
a\otimes \tilde{e}_{11}
= (1\otimes\tilde{e}_{11}) g(0) (1\otimes\tilde{e}_{21}) \in B(0).
\]
Analogously, we obtain
\[
b\otimes \tilde{e}_{11},\ f_{11}(0)\otimes \tilde{e}_{11},\ x_2'\otimes \tilde{e}_{11},\ldots,x_{n+1}'\otimes \tilde{e}_{11}\in B(0).
\]
Arguing as in the proof of Claim~2a, we get $z\otimes\tilde{e}_{11}\in_\varepsilon B(0)$.
It follows that $z\otimes\tilde{e}_{22}\in_\varepsilon B(0)$, and consequently $z\otimes 1 \in_\varepsilon B(0)$.

Similarly, one proves $z\otimes 1 \in_\varepsilon B(1)$.

\emph{Claim~3: Let $s,t\in[0,1]$ with $s\neq t$. Then there exists $d\in B$ such that $d(s)=0$ and $d(t)=1$.}
We first consider the case $s<t$.
By choice of $\mu$, the intervals $[\lambda_j-2\mu,\lambda_j+2\mu]$ are pairwise disjoint for $j\in J$.
We may therefore choose a continuous function $h\colon\RR\to[0,1]$ that takes the value $0$ on
\[
S:=\bigcup_{j\in J} \big( [\lambda_j-2\mu,\lambda_j-(1-s)\mu] \cup [\lambda_j+(1-s)\mu,\lambda_j+2\mu]).
\]
and that takes the value $1$ on
\[
T:=\bigcup_{j\in J} [\lambda_j-(1-t)\mu,\lambda_j+(1-t)\mu],
\]
Note that $T$ consists of the real numbers that have distance at most $(1-t)\mu$ to some $\lambda_j$.
For the purposes below, one could consider $S$ as the real numbers that have distance at least $(1-s)\mu$ to each $\lambda_j$.

Set $c:=h(f)\in B$, the element obtained by applying functional calculus for $h$ to $f$.
Since $f=\diag(f_{11},\ldots,f_{66})$, we have $c=\diag(h(f_{11}),\ldots,h(f_{66})$.

Let $r\in[0,1]$.
We have
\[
\sigma(f_{kk}(r))=\{\lambda_j+\mu\alpha^{(k)}(r):j\in J\}.
\]
If $k\in\{1,5,2,6\}$, then $|\alpha^{(k)}(r)|\geq 1$, and thus
\[
|\lambda_j-(\lambda_j+\mu\alpha^{(k)}(r))|\geq \mu \geq (1-s)\mu,
\]
whence
\[
\sigma(f_{kk}(r))\subseteq S.
\]
It follows that
\[
h(f_{11})=h(f_{55})=h(f_{22})=h(f_{66})=0.
\]
For $k\in\{3,4\}$, we have
\[
|\lambda_j-(\lambda_j+\mu\alpha^{(k)}(s))| = \mu|1-s|, \andSep
|\lambda_j-(\lambda_j+\mu\alpha^{(k)}(t))| = \mu|1-t|,
\]
whence
\[
\sigma(f_{kk}(s))\subseteq S, \andSep
\sigma(f_{kk}(t))\subseteq T.
\]
It follows that
\[
h(f_{33})(s)=h(f_{44})(s)=0, \andSep
h(f_{33})(t)=h(f_{44})(t)=1.
\]
In conclusion, we have
\[
c(s)=0, \andSep
c(t)=1\otimes(e_{33}+e_{44}).
\]

If $t<1$, then by Claim~1a there exist $b_{kl}\in B$ such that $b_{kl}(t)=1\otimes e_{kl}$ for $k,l\in\{1,\ldots,6\}$.
Then the following element has the desired properties:
\[
d := c + b_{13}cb_{31} + b_{23}cb_{32} + b_{53}cb_{35} + b_{63}cb_{36}.
\]

If $t=1$, then by Claim~1c there exist $b_{k3},b_{3l}\in B$ such that $b_{k3}(1)=1\otimes (e_{k3}+e_{k+4,4})$ and $b_{3l}(1)=1\otimes (e_{3l}+e_{4,l+4})$ for $k,l\in\{1,2\}$.
Then the following element has the desired properties:
\[
d := c + b_{13}cb_{31} + b_{23}cb_{32}.
\]

Next, let us indicate how to proceed in the case $s>t$.
Let $h\colon\RR\to[0,1]$ be a continuous function that takes the value $0$ on
\[
S:=\bigcup_{j\in J} \big( [\lambda_j-2\mu, \lambda_j-\mu] \cup [\lambda_j-(1-s)\mu,\lambda_j+(1-s)\mu] \cup [\lambda_j+\mu,\lambda_j+2\mu]).
\]
and that takes the value $1$ on
\[
T:=\bigcup_{j\in J} \{\lambda_j-(1-t)\mu,\lambda_j+(1-t)\mu\}.
\]
The element $c:=h(f)\in B$ satisfies $c(s)=0$ and $c(t)\neq 0$.
Similar as in the case $s<t$, one then constructs $d\in B$ such that $d(s)=0$ and $d(t)=1$, which proves the claim.

Claim~3 verifies condition~(b) in \autoref{prp:genCenter}, whence $B$ contains $C([0,1])$.
Thus, $B$ is a $C([0,1])$-subalgebra of $A\otimes Z_{2,3}$.
If $\pi_t$ denotes the quotient map from $A\otimes Z_{2,3}$ to the fiber at $t$, then Claim~2 shows that $\pi_t(z\otimes 1)\in_\varepsilon\pi_t(B)=B(t)$ for every $t\in[0,1]$.
By \cite[Lemma~2.1]{Dad09CtsFieldsOverFD}, we get $z\otimes 1\in_\varepsilon B$, which proves the result.
\end{proof}

\begin{thm}
\label{prp:mainThm}
Let $A$ be a separable, $\mathcal{Z}$-stable \ca{} of real rank zero.
Then~$A$ has generator rank one.
In particular, a generic element of $A$ is a generator.
\end{thm}
\begin{proof}
We first prove the theorem under the additional assumption that $A$ is unital.
In this case, we have $\gr(A)\leq 8$ by \autoref{prp:gr8}.
Next, we successively reduce the upper bound for $\gr(A)$.

\emph{Claim:

Let $n\in\NN$ such that $\gr(A)\leq n+2$.
Then $\gr(A)\leq n+1$.
}
To prove the claim, we verify condition~(3) of \autoref{prp:genDabsorbing}.
Let $x_0,\ldots,x_{n+1}\in A_\sa$, $\varepsilon>0$ and $z\in A$.
We need to find $y_0,\ldots,y_{n+1}\in(A\otimes\mathcal{Z})_\sa$ such that
\[
\left\| y_j - (x_j\otimes 1) \right\| < \varepsilon\ \text{ for } j=0,\ldots,n+1, \andSep
z\otimes 1 \in_\varepsilon C^*(y_0,\ldots,y_{n+1}).
\]
By identifying $Z_{2,3}$ with a unital sub-\ca{} of $\mathcal{Z}$, elements $y_0,\ldots,y_{n+1}$ with the desired properties are provided by \autoref{prp:construction}.

Applying the claim seven times, we obtain that $\gr(A)\leq 1$.

If $A$ is nonunital, we use that $A$ is separable and has real rank zero to choose an increasing approximate unit $(p_n)_n$ of projections in $A$;
see \cite[Proposition~2.9]{BroPed91CAlgRR0}.
For each~$n$, we consider the unital corner $A_n:=p_nAp_n$.
By \cite[Corollary~2.8]{BroPed91CAlgRR0}, real rank zero passes to hereditary sub-\ca{s}.
By \cite[Corollary~3.1]{TomWin07ssa}, $\mathcal{Z}$-stability passes to hereditary sub-\ca{s}.
Thus, $A_n$ is a unital, separable, $\mathcal{Z}$-stable \ca{} of real rank zero, and thus $\gr(A_n)\leq 1$.
By \autoref{prp:permanence}, we get
\[
\gr(A)\leq\liminf_n \gr(A_n) = 1.
\]
Since $A$ is noncommutative (if nonzero), we have $\gr(A)\neq 0$ by \autoref{prp:gr0}, and so $\gr(A)=1$.
Since $A$ is separable and has real rank zero, $\gr(A)=1$ means that generators in $A$ are a dense $G_\delta$-subset;
see \autoref{rmk:gr1}
\end{proof}

\begin{rmk}
Using the methods developed in \cite[Section~4]{Thi12arX:GenRnk}, one can remove the assumption of separability in \autoref{prp:mainThm}:
Every $\mathcal{Z}$-stable \ca{} of real rank zero has generator rank one.
They key point is that for every $\mathcal{Z}$-stable \ca{} $A$ and every separable sub-\ca{} $B_0\subseteq A$, there exists a separable, $\mathcal{Z}$-stable sub-\ca{} $B\subseteq A$ with $B_0\subseteq B$.
Similar methods are used in the proof of \autoref{prp:nuclPI} below.
\end{rmk}

For the definition and the basic properties of pure infiniteness for nonsimple \ca{s}, we refer to \cite{KirRor00PureInf}.

\begin{cor}
\label{prp:nuclPI}
Every nuclear, purely infinite \ca{} of real rank zero has generator rank one.
\end{cor}
\begin{proof}
Let $A$ be a nuclear, purely infinite \ca{} of real rank zero.
As in \cite[Paragraph~4.1]{Thi12arX:GenRnk}, we let $\SubSep(A)$ denote the collection of separable sub-\ca{s} of $A$; a subset $\mathcal{S}\subseteq\SubSep(A)$ is \emph{$\sigma$-complete} if $\overline{\bigcup\mathcal{T}}\in\mathcal{S}$ for every countable, directed subset $\mathcal{T}\subseteq\mathcal{S}$;
a subset $\mathcal{S}\subseteq\SubSep(A)$ is \emph{cofinal} if for every $B_0\in\SubSep(A)$ there is $B\in\mathcal{S}$ with $B_0\subseteq B$.

We let $\mathcal{S}_{\mathrm{nuc}}$, $\mathcal{S}_{\mathrm{pi}}$ and $\mathcal{S}_{\mathrm{rr0}}$ denote the sets of separable sub-\ca{s} of $A$ that are nuclear, purely infinite and have real rank zero, respectively.
It follows from Paragraph~II.9.6.5 and Proposition~IV.3.1.9 in \cite{Bla06OpAlgs} that $\mathcal{S}_{\mathrm{nuc}}$ is $\sigma$-complete and cofinal.
Using Proposition~4.18 and Corollary~4.22 in \cite{KirRor00PureInf}, it follows that $\mathcal{S}_{\mathrm{pi}}$ is $\sigma$-complete and cofinal.
Lastly, it was noted at the end of \cite[Paragraph~4.1]{Thi12arX:GenRnk} that real rank zero satisfies the `L\"{o}wenheim-Skolem condition', which means that $\mathcal{S}_{\mathrm{rr0}}$ is $\sigma$-complete and cofinal.
Set
\[
\mathcal{S}:= \mathcal{S}_{\mathrm{nuc}}\cap\mathcal{S}_{\mathrm{pi}}\cap\mathcal{S}_{\mathrm{rr0}}.
\]
It is well-known that the intersection of countably many $\sigma$-complete, cofinal subsets is again $\sigma$-complete and cofinal, whence $\mathcal{S}$ is $\sigma$-complete and cofinal.

Let $B\in\mathcal{S}$.
Then $B$ is a separable, nuclear, purely infinite \ca{} of real rank zero.
It follows from \cite[Theorem~9.1]{KirRor02InfNonSimpleCalgAbsOInfty} that $B$ is $\mathcal{O}_\infty$-stable, and thus $\mathcal{Z}$-stable.
By \autoref{prp:mainThm}, we have $\gr(B)=1$.
Since $A$ is the inductive limit of the system $\mathcal{S}$ (indexed over itself), we obtain $\gr(A)\leq 1$ by \autoref{prp:permanence}.
Since purely infinite \ca{s} are by definition noncommutative, we we have $\gr(A)\neq 0$ by \autoref{prp:gr0}, and so $\gr(A)=1$.
\end{proof}

Recall that a \emph{Kirchberg algebra} is a separable, simple, nuclear, purely infinite \ca.
By Zhang's theorem, \cite[Proposition~V.3.2.12, p.454]{Bla06OpAlgs}, every simple, purely infinite \ca{} has real rank zero.
Thus, Kirchberg algebras have real rank zero.
Applying \autoref{prp:nuclPI}, we obtain:

\begin{cor}
\label{prp:KirchbergAlg}
Every Kirchberg algebra has generator rank one.
\end{cor}

Let us say that a simple, nuclear \ca{} is \emph{classifiable} if it is unital, separable, $\mathcal{Z}$-stable and satisfies the Universal Coefficient Theorem (UCT).
By the recent breakthrough in the Elliott classification program, \cite{GonLinNiu15arX:mainClassification, EllGonLinNiu15arX:classFinDR2, TikWhiWin17QDNuclear}, two simple, nuclear, classifiable \ca{s} are isomorphic if and only if their Elliott invariants ($K$-theoretic and tracial data) are isomorphic.

\begin{cor}
\label{prp:classifiable-gr1}
Let $A$ be a unital, separable, simple, nuclear, $\mathcal{Z}$-stable \ca{} satisfying the UCT.
Then $A$ has generator rank one.
In particular, a generic element in $A$ is a generator.
\end{cor}
\begin{proof}
By \cite[Theorem~4.1.10]{Ror02Classification}, $A$ is either stably finite or purely infinite.
In the second case, $A$ is a Kirchberg algebra and we obtain $\gr(A)=1$ by \autoref{prp:KirchbergAlg}.
(The purely infinite case does not require the UCT.)

In the first case, it follows from \cite[Theorem~6.2(iii)]{TikWhiWin17QDNuclear} that $A$ is an approximately subhomogeneous (ASH) algebra.
By \cite[Theorem~5.10]{Thi20arX:grSubhom}, every $\mathcal{Z}$-stable ASH-algebra has generator rank one.
Thus, we have $\gr(A)=1$ in either case.

Since $A$ is unital and separable, $\gr(A)=1$ means that generators in $A$ are a dense $G_\delta$-subset;
see \autoref{rmk:gr1}.
\end{proof}

\begin{rmks}
(1)
It seems likely that the proof of the main theorem can be generalized to show the following:
If $A$ is a unital, separable, $\mathcal{Z}$-stable \ca{} such that $A\otimes M_{2^\infty}$ and $A\otimes M_{3^\infty}$ have real rank zero, then $A$ has generator rank one.

(2)
Let $A$ be a unital, separable, $\mathcal{Z}$-stable \ca{}.
By \cite[Theorem~3.8]{ThiWin14GenZStableCa}, $A$ is singly generated.
Our results show that under additional assumptions, $A$ even contains a dense set of generators.
It is reasonable to expect that every $\mathcal{Z}$-stable \ca{} has generator rank one.
However, by \cite[Proposition~3.10]{Thi12arX:GenRnk}, the real rank is a lower bound for the generator rank, and it is not known that every $\mathcal{Z}$-stable \ca{} has real rank at most one.

Note that every unital, separable, \emph{simple}, $\mathcal{Z}$-stable \ca{} has real rank at most one:
It is either purely infinite and then has real rank zero;
or it is stably finite and thus has stable rank one by \cite[Theorem~6.7]{Ror04StableRealRankZ}, which entails real rank at most one.
Therefore, the following question has no obvious obstruction:
\end{rmks}

\begin{qst}
Does every unital, separable, \emph{simple}, $\mathcal{Z}$-stable \ca{} have generator rank one?
\end{qst}

\providecommand{\href}[2]{#2}

config: arxiv


TITLE: Books on complex analysis for self learning that includes the Riemann zeta function?
QUESTION [2 upvotes]: I am searching for an introductory book in the field of complex analysis for self learning, that would contain the following:

Analytic number theory : the connection between complex analysis and
number theory and most importantly the Riemann hypothesis
Some proofs : I prefer to look behind the proofs of the theorem that I study before accepting it fully
The book must be suitable for self study, so it would be appreciated if it is clear enough and contains everything needed
The book must contain the necessary parts of complex analysis that are used in physics, given that I'm interested in physics too
I would like the algebraic approach of complex analysis but it would be good if the book contained the geometric beauty of complex numbers too like fractals (Julia set, Mandelbrot set)

I will use complex analysis later in two mathematical fields: Partial Differential Equations and Number Theory, so I am searching for a book that covers the fundamentals of complex analysis on an undergraduate level that will make me comfortable in studying PDE and Number Theory later on.
My background in mathematics:
ODE - Real analysis - Multivariable analysis - Vector Analysis - Linear algebra (matrices and vector space)

REPLY [3 votes]: All your requirements (except "geometric beauty") are satisfied by the book of
John Stalker,
Complex Analysis, Springer 1998.
It is written by a mathematical physicist who loves formulas and number theory but thinks and explicitly says that geometry is useless in the subject:-)
(I did not understand your requirement about
"some proofs". Books in Complex Analysis, like other mathematics textbooks usually contain complete proofs of all statements). For fractals and "geometric beauty" in general, I recommend another book, Milnor, Dynamics in
one complex variable.


TITLE: Find $w^2$ and compute a quadratic equation satisfied by $w$
QUESTION [1 upvotes]: I'm given $$\zeta_7 = e^{\frac{2\pi i}{7}} = \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$$
and I know $$\sum_{k=0}^{6}\zeta_7^k=0 $$
I'm then given $$w=\zeta_7+\zeta_7^2+\zeta_7^4$$
and that $\zeta_7^7 = 1$.
I'm asked to find $w^2$ and I found that that's $\zeta^2_7+2\zeta_7^3+\zeta_7^4+2\zeta_7^5+2\zeta_7^6+\zeta_7^8$ so $w^2=\zeta_7^3+\zeta_7^5+\zeta_7^6-1$ and if we add w, we get $\zeta_7^0+\zeta_7+\zeta_7^2+\zeta_7^3+\zeta_7^4+\zeta_7^5+\zeta_7^6-2\zeta_7^0-w$ and that, I think, is just $-2-w$
So, if $w^2=-2-w$ then $w^2+w+2=0$ and w solves it, so that's the quadratic, right?

REPLY [1 votes]: Here's a slightly simpler approach:
$$w^2=(\zeta_7+\zeta_7^2+\zeta_7^4)^2=\zeta_7^2+\zeta_7^4+\zeta_7^8+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6,$$
so
$$w^2+w=2\zeta_7^2+2\zeta_7^4+2\zeta_7^1+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6=2(0-\zeta_7^0)=2(-1),$$
hence
$w^2+w+2=0$.


TITLE: Is this formula correct for the stirling numbers of the second kind?
QUESTION [1 upvotes]: If $S \left( n , k \right)$ is the number of partitions with $k$ blocks we can make from a set with $n$ elements, is the formula $$ S \left( n + 1, k \right) = \sum_{i = 0} ^{n} \binom{n}{i} S \left( i, k - 1 \right)$$ correct?
I ask because I have to prove the right-hand side counts the same thing as the left-hand side, but my strategy is not working and I think it is correct.

REPLY [1 votes]: Yes, it’s correct.
HINT: Use the fact that $$\binom{n}iS(i,k-1)=\binom{n}{n-i}S(i,k-1)\;.$$ Think about first picking the block that will contain $n+1$, then splitting up the rest of $[n]$.


TITLE: Do the Lienard Wiechert Potentials satisfy the wave equation?
QUESTION [0 upvotes]: The Lienard Wiechert potential (leaving out the vector potential for simplicity),
$$\left.\phi(\vec{r},t)=\frac{e}{4\pi\epsilon_0R\,(1-\hat{n}\cdot\vec{\beta})} \right|_{t'=t_{\rm ret}},$$
where the quantities should be evaluated in the retarded time given by $t_{\rm ret}=t-R/c$, should satisfy the wave equation. The wave equation for the potential is derived from Maxwell's equations in the Lorenz gauge and given by
$$\left(\nabla^2-\frac{1}{c^2}\partial_t^2\right)\phi(\vec{r},t)=-\frac{\rho}{\epsilon_0}, $$
where $\rho$ is the charge density and should be
$\rho=e\,\delta(\vec{r}-\vec{r}')$ for the point charge. Plugging $\phi(\vec{r},t)$ into the wave equation we need to compute the second-order spatial and time derivatives with respect to the observer. To assist with the derivation, I have computed the following spatial derivatives in cartesian coordinates ($i=x,y,$ or $z$),
$$\partial_i R=n_i, $$
$$ \partial_i \frac{1}{R} = -\frac{n_i}{R^2},$$
$$\partial_i n_i = \frac{1-n_i^2}{R},$$
$$\partial_i (1-\hat{n}\cdot\vec{\beta}) = \frac{n_i(\hat{n}\cdot\vec{\beta})-\beta_i}{R},$$
and the following observer time derivatives (taking mind to switch from $\partial_t$ to $\partial_{t'}$ using $dt=dt'(1-\hat{n}\cdot\vec{\beta})$ when necessary),
$$\partial_t R=-c\frac{\hat{n}\cdot\vec{\beta}}{(1-\hat{n}\cdot\vec{\beta})}, $$
$$ \partial_t \frac{1}{R} = c\frac{\hat{n}\cdot\vec{\beta}}{R^2(1-\hat{n}\cdot\vec{\beta})},$$
$$\partial_t n_i = \frac{c}{R(1-\hat{n}\cdot\vec{\beta})}\left({n_i(\hat{n}\cdot\vec{\beta})-\beta_i }\right),$$
$$\partial_t (1-\hat{n}\cdot\vec{\beta}) = \frac{c\left({ \vec{\beta}\cdot\vec{\beta}-(\hat{n}\cdot\vec{\beta})^2 }\right)}{R\left({1-\hat{n}\cdot\vec{\beta}}\right)}.$$
Now, when I compute the second-order derivatives I get:
$$\nabla^2\phi(\vec{r},t) = -\frac{e}{2\pi\epsilon_oR^3(1-\hat{n}\cdot\vec{\beta})^3}\left({ 1-(\hat{n}-\vec{\beta})\cdot(\hat{n}-\vec{\beta}) }\right)$$
$$\frac{1}{c^2}\partial_t^2 = \frac{e}{4\pi\epsilon_oR^3(1-\hat{n}\cdot\vec\beta)^7}
\left({(1-\hat{n}\cdot\vec\beta)^3 \left({(\hat{n}\cdot\vec\beta)^2-\vec\beta\cdot\vec\beta +2(\hat{n}\cdot \vec\beta)(\hat{n}\cdot\vec\beta-\vec\beta\cdot\vec\beta)}\right) + (\hat{n}\cdot\vec\beta-\vec\beta\cdot\vec\beta)((\hat{n}\cdot\vec\beta)^2-\vec\beta\cdot\vec\beta) }\right) $$
which I dont see any way that their difference will be equal to $-\rho/\epsilon_0$ for a point charge as dictated by the wave equation. Am I doing something wrong here?

REPLY [4 votes]: Yes, the Lienard-Wiechert potential is a solution to the wave
equation with a point charge source. That's how it is derived.
However, if you want to take derivatives in order to show this
explicitly from the differential equation, you must be very
careful to not neglect any terms in the derivatives, and also
be careful to be able to extract the delta function.
For the second point,
the delta function occurs from derivatives like those in the simple
Coulomb's law static case
$\nabla^2 \frac{1}{r} = -4\pi \delta^3(\vec r)$. However, if you
were to write $\nabla^2$ in spherical coordinates it would be
$\frac{1}{r}\partial_r^2 r$ plus angular terms. Taking the derivatives
above gives
$\frac{1}{r}\partial_r^2 r \frac{1}{r} = 0$. The delta function is
hidden at the coordinate boundary at $r=0$. Exactly this case will
also occur for the
the Lienard-Wiechert potential, so it is best to isolate a
$\nabla^2 \frac{1}{R}$ term and realize this will give the delta function,
by either using the divergence theorem, smearing out the singularlity,
or changing to other coordinates.
For the derivatives, you just have to be meticulous. If you think of
a derivative as being
$\lim_{\delta\rightarrow 0} \frac{F(x+\delta)-F(x)}{\delta}$, then
when you change the observation point $x$, $y$, $z$, or $t$, the
retarded time changes. You have to take derivatives with respect to
$t_{\rm ret}$ times derivatives of $x$, $y$, $z$, or $t$ using the
chain rule.
So realizing that $R$ is a function of $t_{ret}$,
$t_{\rm ret} = t-R(t_{\rm ret})/c$
\begin{equation}
\partial_t  t_{\rm ret} = \partial_t [t-R(t_{\rm ret})/c]
= [1-\vec \beta(t_{\rm ret}) \cdot \hat R(t_{\rm ret})\partial_t t_{\rm ret}]
\end{equation}
gives
\begin{equation}
\partial_t t_{\rm ret} = \left . \frac{1}{1-\vec \beta(t')\cdot \hat R(t')}
\right |_{t'= t_{\rm ret}}
\end{equation}
similarly
\begin{equation}
\partial_x t_{\rm ret} = -\left . \frac{1}{1-\vec \beta(t') \cdot \hat R(t')}
\frac{X(t')}{cR(t')} \right |_{t'= t_{\rm ret}}
\end{equation}
with analogous expressions for $y$ and $z$. $X/R$ is the $x$ component of your $\hat n$.
Using the chain rule, your need to evaluate your derivatives as
\begin{eqnarray}
\partial_ t \left [ F \right ]_{\rm ret} &=&
\left [  \frac{1}{1-\vec \beta \cdot \hat R} \dot F \right ]_{\rm ret}
\nonumber\\
\partial_ x \left [ F \right ]_{\rm ret} &=&
\left [ \partial_x F -\frac{X}{cR} \frac{1}{1-\vec \beta \cdot \vec r} \dot F
\right ]_{\rm ret}
\end{eqnarray}
where the dot means take the partial derivative with respect to the time
argument, and I have just written the subscript ret to indicate evaulating
at the retarded tim.
If this isn't clear, just carefully take the derivatives with
the chain rule. The second terms are from the change in the retarded time.
If you do this and don't make an algebra mistake, you will find
your desired result.


TITLE: Condition on $k$ if $|z-z_1|^2+|z-z_2|^2=k$
QUESTION [0 upvotes]: How can we prove that
$$|z-z_1|^2+|z-z_2|^2=k$$
will represent a circle  if $|z_1-z_2|^2 \leq 2k$
Please give me some hints to initiate this question.

REPLY [0 votes]: Making a translation and/or a rotation if necessary, we can assume $\;a:=z_1,\,b:=z_2\in\Bbb R^\;$ (this means: to "align" the line through $\;z_1,\,z_2\;$ with the real axis by means of translation/rotation), so putting $\;z=x+iy\;$ we get
$$(x-a)^2+y^2+(x-b)^2+y^2=k\iff 2x^2+2y^2-2(a+b)x=k-(a^2+b^2)\iff$$
$$\iff\left(x-\frac{a+b}2\right)^2+y^2=\frac{k-(a^2+b^2)}2+\frac{(a+b)^2}4$$
and thus the above represents a circle (with center at $\;\left(\frac{a+b}2,\,0\right)\;$) iff the rightside is non-negative, iff
$$2(a^2+b^2)-2k\le a^2+2ab+b^2\iff (a-b)^2\le2k$$


TITLE: How to use the Mean Value Theorem or Rolle's in this word problem.
QUESTION [0 upvotes]: I'm a bit stuck on how to solve this word problem:

Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider $f(t) = g(t) - h(t)$ where g and h are the position functions of the two runners.]

I know this is the mean value theorum:

And this is Rolle's:

Here's how I started:
Assuming g(t) and h(t) are the position functions of the runners, then we know that at 2 points (the beginning and the end of the race), they are equal.
So $f(t)$ is like the difference between the two position functions. It is equal to 0 at the start and at the end.
Where do I go from here?
when I write this question out, this seems more like an application of Rolle's theorem... instead of Mean Value theorem right?
Like $f(t) = 0$ at the beginning and end
$f(t)$ is differentiable and continuous.
So there is a number c where that $f'(c) = 0$ meaning the two runners have the same speed at some point.
How do I prove this though? Is this a proof?

REPLY [2 votes]: Yes it is a proof. You are correct, this is Rolle's Theorem, which is special case of Mean Value Theorem.
So, your proof is fine.


TITLE: Space-time metric in tensor form
QUESTION [0 upvotes]: In space time metric in tensor form:
The distance is given by $$ds^2=c^2dt-dx^2-dy^2-dz^2$$
Which in tensor form is: $$ds^2=\sum_{\alpha \beta}g_{\alpha \beta}dx^\alpha dx^\beta$$
Using Einstein summation convention  we have $$ds^2=g_{\alpha \beta}dx^\alpha dx^\beta$$
I was wondering what the explict form of this looks like:
I know that:

I was wonder that $dx^\alpha$ and $dx^\beta$ look like?
All together does it look like this:

(i know the above is wrong becuase it is invalid matrix)

REPLY [0 votes]: If you use the fact that $g_{\alpha \beta} dx^{\alpha} = dx_{\beta}$ where $dx^{\alpha}$ is the column vector you have in your above expression, then $dx_{\beta}$ would be the row vector $(cdt,-dx,-dy,-dz)$. Then, the line element becomes $ds^{2} = dx_{\beta}dx^{\beta}$ which is a row vector multiplying a column vector, giving you a scalar and not a matrix.
Equivalently, you can view $g_{\mu \nu} dx^{\mu} dx^{\nu}$ as $\vec{dx}^{T} \cdot \textbf{g}\cdot \vec{dx}$ which gives a scalar and not a matrix.


TITLE: $p$-adic logarithm is a homomorphism, formal power series proof
QUESTION [9 upvotes]: Consider the $p$-adic logarithm defined by the series
$$\log (1+x) = \sum_{n\ge 1} (-1)^{n+1} \frac{x^n}{n}.$$
It converges for $|x|_p < 1$, and if $|x|_p < 1$ and $|y|_p < 1$, then we have
$$\log ((1+x)\cdot (1+y)) = \log (1+x) + \log (1+y).$$
One way to show it is to note that in the ring of formal power series $\mathbb{Q} [[X,Y]]$ (where $\log (1+X)$ is defined by the same formula) we have
$$\log ((1+X)\cdot (1+Y)) = \log (1+X) + \log (1+Y).$$
How does one see that this formal identity indeed implies the identity above?
We have to see that
$$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{n\ge 1} (-1)^{n+1}\,\left(\frac{x^n}{n} + \frac{y^n}{n}\right).$$
Let us expand the term $(x+y+xy)^n$:
$$(x+y+xy)^n = \sum_{i_1 + i_2 + i_3 = n} {n \choose i_1, i_2, i_3} \, x^{i_1}\,y^{i_2}\,(xy)^{i_3} = \sum_{i_1 + i_2 + i_3 = n} {n \choose i_1, i_2, i_3}\,x^{i_1+i_3}\,y^{i_2+i_3} = \sum_{i\ge 0} \sum_{j\ge 0} {n \choose n-j, n-i, i+j-n}\,x^i\,y^j.$$
We have then
$$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j.$$
Now the order of sums $\sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0}$ may be changed (I will go back to this point below) to obtain
$$\sum_{i\ge 0} \sum_{j\ge 0} \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j,$$
and we have to see that the numbers
$$c_{ij} = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}$$
satisfy
$$c_{ij} = \begin{cases}
(-1)^{m+1}/m, & \text{if }i = m, j = 0 \text{ or } i = 0, j = m,\\
0, & \text{otherwise}.
\end{cases}$$
But we already know that it's true thanks to the formal identity in $\mathbb{Q} [[X,Y]]$, so we are done.

The only non-formal step in the above is changing the order of sums. Recall that in the non-archimedian case, we have
$$\sum_{i\ge 0} \sum_{j\ge 0} x_{ij} = \sum_{j\ge 0} \sum_{i\ge 0} x_{ij}$$
if $|x_{ij}| \to 0$ as $\max (i,j) \to \infty$.
In the above case, we may note that
$$\left|\sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j\right|_p \xrightarrow{\max (n,i) \to \infty} 0$$
(by the way, is it completely obvious?) so that
$$\sum_{n\ge 1} \sum_{i\ge 0} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j = \sum_{i\ge 0} \sum_{n\ge 1} \sum_{j\ge 0} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j = \sum_{i\ge 0} \sum_{j\ge 0} \sum_{n\ge 1} \frac{(-1)^{n+1}}{n}\,{n \choose n-j, n-i, i+j-n}\,x^i\,y^j$$
(we swap the two inner sums in the second equality since they are finite).

My question is the following: all these details look a bit messy. Is there a shorter justification of the transition from the formal identity to the corresponding identity with $p$-adic series?
Koblitz in his GTM 58 book says that since in the non-archimedian situation, any convergent series converges after an arbitrary reordering, we can automatically assume that we may write
$$\sum_{n\ge 1} (-1)^{n+1}\,\frac{(x+y+xy)^n}{n} = \sum_{i\ge 0}\sum_{j\ge 0} c_{ij}\,x^i\,y^j,$$
for some $c_{ij}$. Maybe I am missing something obvious, and the above change of summation order indeed doesn't require any explicit justifications?
Thank you.

REPLY [0 votes]: For a shorter proof, the main idea is simple: you evaluate the formal identity at $(X,Y) = (x,y)$ to get the special identity. Everything else is technical detail.
To address a concern in the comments, the point is that evaluation is continuous. If $\sum a_k x^k$ is a convergent power series in a topological ring $R$, then any continuous homomorphism $\varphi : R \to S$ will satisfy
$$ \varphi\left( \sum_{k=0}^{\infty} a_k x^k \right) = \varphi\left( \lim_{n \to \infty} \sum_{k=0}^{n} a_k x^k \right)
= \lim_{n \to \infty} \varphi\left( \sum_{k=0}^{n} a_k x^k \right)
\\= \lim_{n \to \infty} \sum_{k=0}^n \varphi(a_k) \varphi(x)^k
= \sum_{k=0}^\infty \varphi(a_k) \varphi(x)^k
$$
In particular, if $\log(1+z)$ is defined in $R$, then $\log(1 + \varphi(z))$ is defined in $S$ and $\varphi(\log(1+z)) = \log(1+\varphi(z))$.

The main technical obstacle is the fact that the power series $\log(1+T)$ does not have $\mathbb{Z}_p$-integral coefficients, so we can't directly invoke many of the usual facts about power series rings.
So, one must instead develop enough of the theory of convergent formal power series to show the homomorphisms involved are defined and continuous. Unfortunately, I don't recall how straightforward this is; but maybe your source already has the relevant theorems.


TITLE: In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD\times DC)$ .
QUESTION [0 upvotes]: In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD \times DC)$ .

What I Tried: Here is a picture :-

I have no idea for this problem. From the picture if $D$ lies a bit near of $B$ then I just reflect it to get $E$ near $C$ , $\Delta ADE$ is an isosceles triangle, but there's no use to it, how would that help in the original problem?
In fact, that did not look like the case when I tried it in Geogebra. $D$ lies nearly in the middle of $BC$ and, also with a bit of approximation, I got $(BD\times DC)$ to be nearly equal to $2$ , but why?
Can anyone help me with a reason here? Thank You.

REPLY [1 votes]: Draw $AH\perp BC$
$$AH^2+DH^2=1$$
$$AH^2+BH^2=AH^2+(BD+DH)^2=AH^2+BD^2+DH^2+2\times BD\times  DH=3$$
$$BD^2+2\times BD\times DH=BD(BD+2\times DH)=BD(BD+DE)=BD\times BE=BD\times DC=2$$
where $BD=EC$ and $BE=DC$


TITLE: The derivative of the normal unit vector
QUESTION [1 upvotes]: I want to derive the unit vector $N(s)=\dfrac{T'(s)}{\|T'(S)\|}$, where $T'(s)$ is the derivative of the tangent unit vector, how can I derive this vector?
Thanks!

REPLY [0 votes]: I'm not sure Omar the OP meant "differentiate" or simply "derive" as in "show that such-and-such is the case.  I assumed the former:
If the context is $\Bbb R^2$, i.e., the curves lie in the plane, then since $N(s)$ and $\kappa$ are defined via
$T'(s) = \kappa N(s), \tag{1}$
together with the requirement that $\Vert N(s) \Vert = 1$, where $\kappa$ is the curvature of the curve, it follows that
$N'(s) = -\kappa T(s);  \tag{2}$
you don't have to do any more work once you've got $T'(s)$; you're done.  A complete explanation may be found in my answer to this question.
If you're in $\Bbb R^3$, or even $\Bbb R^n$, $n > 3$ (1) still defines $\kappa$ and $N(s)$ in this case, and we can use it to write $N'(s)$ via
$T''(s) = \kappa' N(s) + \kappa N'(s), \tag{3}$
provided we can compute $\kappa'(s)$.  But
$\kappa(s) = \Vert T'(s) \Vert = \langle T'(s), T'(s) \rangle^{\frac{1}{2}}, \tag{4}$
so that
$\kappa'(s) = \frac{1}{2}\langle T'(s), T'(s) \rangle^{-\frac{1}{2}}\langle T'(s), T'(s) \rangle' = \frac{1}{2}\kappa^{-1}\langle T'(s), T'(s) \rangle'. \tag{5}$
Now
$\langle T'(s), T'(s) \rangle' = \langle T''(s), T'(s) \rangle + \langle T'(s), T''(s) \rangle = 2\langle T'(s), T''(s) \rangle, \tag{6}$
so combining this with (5) yields
$\kappa'(s) = \langle T'(s), T'(s) \rangle^{-\frac{1}{2}}\langle T'(s), T''(s) \rangle, \tag{7}$
and now using (1) and (4) we see that
$\kappa'(s) = \kappa^{-1}\langle \kappa N(s), T''(s) \rangle = \langle N(s), T''(s) \rangle. \tag{8}$
(8) may be used in (3) which, after a little re-arranging, becomes
$N'(s) = \kappa^{-1}(T''(s) -  \langle N(s), T''(s) \rangle N(s)). \tag{9}$
That's most likely how I'd do it, in any event.  A "recipe" or set of directions for obtaining $N'(s)$ based on (9) might read:
Given $T(s)$,
1.)  Differentiate $T(s)$ to obtain $T'(s) = kN$; this gives $\kappa$ and $N$;
2.)  Differentiate $T'(s)$ to obtain $T''(s)$;
3.)  From the data derived in steps (1) and (2) above, algebraically calculate, in accordance with equation (9), $N'(s) = \kappa^{-1}(T''(s) -  \langle N(s), T''(s) \rangle N(s))$.
4.)  Halt.
Of course, it is also possible to compute $N'(s)$ directly from (1) by writing
$N(s) = \kappa^{-1}T'(s), \tag{10}$
reducing it to simplest terms "algebraically", and then differentiate the resulting expression directly.  But it's not clear that it would be less work the above prescpition; indeed, my personal experience shows it would often be more.  Addionally but not extraneously, the nice thing about (9) is that it breaks out everything in geometric terms; indeed, $T''(s) -  \langle N(s), T''(s) \rangle N(s)$ is the projection of $T''(s)$ onto $N^\perp(s)$, the subspace normal to $N(s)$.  So (9) has a nice interpretation, geometrically speaking:  it says that the derivative of $N(s)$ with respect to $s$ is the component of $T''(s)$ normal to $N(s)$, divided by $\kappa$.  One can of course assign similar geometric interpretations to most of the other equations occurring in the above discussion.
Of course all of the above is subject to the usual caveats concerning the vanishing of $\kappa$; we must generally assume we are working in a neighborhood where $\kappa(s) \ne 0$, since $N(s)$ can be defined, and the ensuing divisions by $\kappa$ are permitted.
Hope this helps.  Cheerio,
and of course,
Fiat Lux!!!