Principal rays $r,r^{\prime}$ are _elementarily parallel_ , if there is a
chamber containing them both in which $r,r^{\prime}$ are parallel with respect
to the Euclidean metric $|\cdot,\cdot|$. This generates an equivalence
relation on the set of principal rays, and we say that $r,r^{\prime}$ are
_parallel_ if they are in the same equivalence class. A class of parallel rays
determines a point $\zeta$ in the visual boundary
$\partial_{\infty}\mathbf{X}$. We call such a point $\zeta$ _principal_.
Analogously we define _antiprincipal_ points in $\partial_{\infty}\mathbf{X}$.

######  Lemma 8.1.

Let $\zeta\in\partial_{\infty}\mathbf{X}$ be the principal (resp.
antiprincipal) point that is the class of the principal (resp. antiprincipal)
rays in $\mathbf{E}^{+}_{\mathrm{id}}$. Then the stabilizer
$\mathrm{Stab}(\zeta)<\mathrm{Tame}(\mathbbm{k}^{3})$ equals $C$ (resp.
$B^{\prime}$).

We need a preparatory result.

######  Lemma 8.2.

Let $r$ be a principal (resp. antiprincipal) ray in
$\mathbf{E}^{+}_{\mathrm{id}}$, and let $r^{\prime}$ be any geodesic ray in
$\mathbf{X}$ asymptotic to $r$. Then $r^{\prime}$ has a subray that is
principal (resp. antiprincipal) and parallel to $r$.

###### Proof.

Suppose first that $r\colon[0,\infty)\to\mathbf{E}^{+}_{\mathrm{id}}$ is
principal. Let $d>0$ be such that
$d_{\mathbf{X}}\big{(}r(t),r^{\prime}(t)\big{)}\leq d$ for all $t\geq 0$.
After passing to subrays of $r$ and $r^{\prime}$, we can assume that
$\rho_{+}(r)$ is at distance $>d$ from the set $\\{\alpha_{2}=\alpha_{3}\\}$
in the metric $|\cdot,\cdot|$. Let
$\overline{x}_{0}=r(0),\overline{x}_{1},\ldots,\overline{x}_{k}=r^{\prime}(0)$
be points guaranteed by Lemma 3.3(iii) such that for each $i=0,\ldots,k-1$
both $\overline{x}_{i}$ and $\overline{x}_{i+1}$ lie in a common chamber
$\mathbf{E}_{f_{i}}^{+}$, and
$\sum_{i=0}^{k-1}\big{|}\rho_{+}(\overline{x}_{i}),\rho_{+}(\overline{x}_{i+1})\big{|}=d$.
Thus all $\rho_{+}(\overline{x}_{i})$ lie outside
$\\{\alpha_{2}=\alpha_{3}\\}$. Hence, by Remark 3.8, the principal rays from
$\overline{x}_{i}$ in $\mathbf{E}_{f_{i}}^{+}$ and $\mathbf{E}_{f_{i-1}}^{+}$
coincide. Consequently $r$ and $r^{\prime}$ are parallel, as desired. If $r$
is antiprincipal, the argument is analogous. ∎

###### Proof of Lemma 8.1.

Recall from Section 3.2 that the group $C$ is generated by $K$ and $H$.
Applying Corollary 3.7, we obtain the following. If $f\in K$, then
$\mathbf{E}^{+}_{\mathrm{id}}$ and $\mathbf{E}^{+}_{f}$ share a $1$-principal
ray. If $f\in H$, then $\mathbf{E}^{+}_{\mathrm{id}}$ and $\mathbf{E}^{+}_{f}$
share an $a$-principal ray for $a$ sufficiently large (and thus they have
parallel $1$-principal rays). In particular, $\mathrm{Stab}(\zeta)$ contains
$C$. Conversely, by Proposition 3.4, if for
$f\in\mathrm{Tame}(\mathbbm{k}^{3})$ the chambers
$\mathbf{E}^{+}_{\mathrm{id}}$ and $\mathbf{E}^{+}_{f}$ share a principal ray,
then $f$ belongs to $H$ or $K$. Thus to show that for
$f\in\mathrm{Stab}(\zeta)$ we have $f\in C$ it remains to justify that
$\mathbf{E}^{+}_{\mathrm{id}}$ and $\mathbf{E}^{+}_{f}$ have parallel
principal rays. This follows from applying Lemma 8.2 to $r^{\prime}=f(r)$. The
proof of the second assertion is analogous. ∎

###  8.2. H-generic rays

######  Construction 8.3.

  1. (1)

Let $f$ be an element of $B^{\prime}$ not conjugate into $B$ or $H$. Then
there is an isometrically embedded Euclidean half-plane $\mathbf{H}$ in
$\mathbf{X}$, invariant under $f$. Indeed, by Corollary 3.7, for
$g,g^{\prime}\in B^{\prime}$ with $g^{-1}g^{\prime}\in B\setminus H$, the
chambers $\mathbf{E}^{+}_{g},\mathbf{E}^{+}_{g^{\prime}}$ intersect along a
$1$-antiprincipal ray, and for $g^{-1}g^{\prime}\in H\setminus B$ they
intersect along a region whose boundary contains an $a$-antiprincipal ray for
some $a\geq 2$. After possibly conjugating $f$, we can assume that its normal
form in $B^{\prime}=B\ast_{B\cap H}H$ is $h_{1}b_{1}\cdots h_{n}b_{n}$. For
$i=1,\ldots,n$, let $r^{\prime}_{i},r_{i}$ be the above antiprincipal rays for
$g=h_{1}\cdots b_{i-1},g^{\prime}=gh_{i}$ and $g=h_{1}\cdots
h_{i},g^{\prime}=gb_{i}$, respectively. Choose a geodesic $\sigma$ in
$(\nabla,|\cdot,\cdot|)$ perpendicular to the line $\alpha_{2}=\alpha_{3}$
that intersects the images under $\rho_{+}$ of all $r_{i},r_{i}^{\prime}$. Let
$r_{0}=f^{-1}(r_{n})$. For $i=1,\ldots,n$, consider the geodesic $\sigma_{i}$
in $\mathbf{E}^{+}_{g}$ for $g=h_{1}\cdots b_{i-1}$ (respectively,
$\sigma_{i}^{\prime}$ in $\mathbf{E}^{+}_{g^{\prime}}$ for
$g^{\prime}=gh_{i}$) that joins $r_{i-1}$ to $r^{\prime}_{i}$ (respectively,
$r^{\prime}_{i}$ to $r_{i}$) and maps into $\sigma$ under $\rho_{+}$. Then the
union of the concatenation
$\sigma_{1}\cdot\sigma_{1}^{\prime}\cdots\sigma_{n}\cdot\sigma_{n}^{\prime}$
and its translates under $\langle f\rangle$ is an axis of $f$ that bounds an
isometrically embedded Euclidean half-plane $\mathbf{H}$ in $\mathbf{X}$,
which is the union of the half-strips bounded by
$\sigma_{i},\sigma_{i}^{\prime}$ and subrays of $r_{i},r_{i}^{\prime}$ (Remark
6.6(i)).

  2. (2)

We have a following similar construction of an invariant convex subset
$\mathbf{H}$ for an element $f\in C$ that is not conjugate into $K$ or $H$. By
Corollary 3.7, for $g,g^{\prime}\in C$ with $g^{-1}g^{\prime}\in K\setminus
H$, the chambers $\mathbf{E}^{+}_{g},\mathbf{E}^{+}_{g^{\prime}}$ intersect
along a $1$-principal ray, and for $g^{-1}g^{\prime}\in H\setminus K$ they
intersect along a region whose boundary contains a curve projecting under
$\rho_{+}$ inside the (possibly non-principal) admissible line
$\alpha_{1}=m_{2}\alpha_{2}+m_{3}\alpha_{3}$ for some $m_{2}\geq
1,m_{2}+m_{3}\geq 2$. After possibly conjugating $f$, we can assume that its
normal form in $C=K\ast_{K\cap H}H$ is $h_{1}k_{1}\cdots h_{n}k_{n}$. For
$i=1,\ldots,n$, let $c_{i},r_{i}$ be the above curves and principal rays for
$g=h_{1}\cdots k_{i-1},g^{\prime}=gh_{i}$ and $g=h_{1}\cdots
h_{i},g^{\prime}=gk_{i}$, respectively. Choose a geodesic $\sigma$ in
$(\nabla,|\cdot,\cdot|)$ perpendicular to the line $\alpha_{1}=\alpha_{2}$
that intersects the images under $\rho_{+}$ of all $c_{i},r_{i}$. Let
$r_{0}=f^{-1}(r_{n})$. For $i=1,\ldots,n$, consider the geodesic $\sigma_{i}$
in $\mathbf{E}^{+}_{g}$ for $g=h_{1}\cdots k_{i-1}$ (respectively,
$\sigma_{i}^{\prime}$ in $\mathbf{E}^{+}_{g^{\prime}}$ for
$g^{\prime}=gh_{i}$) that joins $r_{i-1}$ to $c_{i}$ (respectively, $c_{i}$ to
$r_{i}$) and maps into $\sigma$ under $\rho_{+}$. Let $R_{i},R_{i}^{\prime}$
be the Euclidean half-strips bounded by $\sigma_{i},\sigma_{i}^{\prime}$ and
the principal rays issuing from their endpoints. Let
$R^{\prime}_{0}=f^{-1}(R^{\prime}_{n})$. Note that for $i=1,\ldots,n,$ the
half-strips $R_{i-1}^{\prime},R_{i}$ intersect along the $1$-principal rays by
which they are bounded, and $R_{i},R^{\prime}_{i}$ intersect along the regions
bounded by the other principal rays and the curves inside $c_{i}$. Thus the
union of all $R_{i},R_{i}^{\prime}$ and their translates under $\langle
f\rangle$ form a subset $\mathbf{H}$ that is, by Remark 6.6(iii), locally
convex, and hence convex in $\mathbf{X}$.

We have the following immediate consequence of Construction 8.3.

######  Corollary 8.4.

  1. (1)

Let $f$ be an element of $B^{\prime}$ not conjugate into $B$ or $H$. Then $f$
is loxodromic not of rank $1$. Furthermore, the translation length $|f|$ is
the sum of the distances from $r_{i-1}$ to $r^{\prime}_{i}$, and from
$r^{\prime}_{i}$ to $r_{i}$, for $i=1,\ldots,n$, in the notation of
Construction 8.3(1).

  2. (2)

Let $f$ be an element of $C$ not conjugate into $K$ or $H$. Then $f$ is
loxodromic not of rank $1$, if all $c_{i}$ are principal rays, or parabolic,
otherwise. Furthermore, $|f|$ is the sum of the distances (that might be zero)
from $r_{i-1}$ to $c_{i}$, and from $c_{i}$ to $r_{i}$, for $i=1,\ldots,n$, in
the notation of Construction 8.3(2).

######  Example 8.5.

The map $f=(x_{2},x_{1}+x_{2}x_{3},x_{3})\in C$ is parabolic of translation
length $0$. The map $f=(x_{2},x_{1}+x^{2}_{2}x_{3},x_{3})\in C$ is parabolic
with positive translation length.

######  Definition 8.6.

Let $f$ be an element of $B^{\prime}$ not conjugate into $B$ or $H$ or an
element of $C$ not conjugate into $K$ or $H$, with positive translation
length. Let $\xi\in\partial_{\infty}\mathbf{X}$ be a limit point of $f$. A
geodesic ray $r\colon J\to\mathbf{H}$ representing $\xi$, for $\mathbf{H}$ as
in Construction 8.3, is _$\mathbf{H}$ -generic_ if it is disjoint from the
vertex set $\mathbf{X}^{0}$ and transverse to the $1$-skeleton
$\mathbf{X}^{1}$. The consecutive points $l_{k}\in J$ at which $r$ passes from
one chamber $\mathbf{E}^{+}_{g}$ to another are called _transition points_.

######  Remark 8.7.

Let $f$ be an element of $B^{\prime}$ not conjugate into $B$ or $H$ or an
element of $C$ not conjugate into $K$ or $H$, with positive translation
length, and let $\mathbf{H}$ be as in Construction 8.3. Let
$\zeta\in\partial_{\infty}\mathbf{X}$ be the antiprincipal or principal point
fixed by $B^{\prime}$ or $C$, respectively. Let
$\xi\in\partial_{\infty}\mathbf{X}$ be a limit point of $f$. Note that since
$\mathbf{H}$ is convex and $f$-invariant, each geodesic ray starting in
$\mathbf{H}$ and representing $\xi$ is contained in $\mathbf{H}$. By [CL]*Thm
1.1 (see also [D]*thm 6.1), we have that $\xi$ is at Tits distance $\pi$ in
$\partial_{\infty}\mathbf{H}\subset\partial_{\infty}\mathbf{X}$ from the other
limit point of $f$, and so $\angle(\xi,\zeta)=\frac{\pi}{2}$. In particular
$\xi\neq\zeta$.

  1. (i)

An $\mathbf{H}$-generic ray exists. Indeed, choose a ray $s$ representing
$\zeta$ contained in $\mathbf{H}$ and disjoint from all the curves $c_{i}$.
Let $\mathcal{R}$ be the set of geodesic rays $r$ representing $\xi$ starting
at the points of $s$. Note that distinct rays $r,r^{\prime}\in\mathcal{R}$ are
disjoint, because they are contained in a subspace of $\mathbf{H}$ with no
edges of degree $3$, and because $r\subset r^{\prime}$ would imply
$\xi=\zeta$. Since there are uncountably many points in $s$, there exists
$r\in\mathcal{R}$ disjoint from the vertices of $\mathbf{X}$ in $\mathbf{H}$.
Similarly, since the edges of $\mathbf{X}$ have geodesic curvature of constant
sign in each $2$-cell, if two rays in $\mathcal{R}$ are tangent to a given
edge of $\mathbf{X}$ intersecting $\mathbf{H}$, then we have one of the
configurations in Figure 21. The bottom configuration is not allowed since the
subspace bounded by the two rays and the edge, which is CAT(0) by the same
argument as in Definition 6.2, would have two asymptotic geodesic rays
starting at the same point. Consequently, at most two rays in $\mathcal{R}$
are tangent to a given edge of $\mathbf{X}$. Thus, there exists a ray in
$\mathcal{R}$ transverse to $\mathbf{X}^{1}$ and disjoint from
$\mathbf{X}^{0}$.

  2. (ii)

By [BH]*II.9.8(2), the Alexandrov angle at $r(t)$ between $r$ and the geodesic
ray representing $\zeta$ converges to $\frac{\pi}{2}$ and so the same holds at
the intersection with the curves $c_{i}$.

  3. (iii)

Consequently, using [BH]*II.9.3, if at least one $c_{i}$ projects to a non-
principal admissible line in $\nabla^{+}$, then $\rho_{+}(r)$ leaves every
compact set of $\nabla^{+}$.

  4. (iv)

Conversely, suppose that all $c_{i}$ project to principal lines in
$\nabla^{+}$. Then $\mathbf{H}$ is isometric to the Euclidean half-plane and
all above intersection angles are $\frac{\pi}{2}$. Consequently, we have
$f(r)\subset r$.

  5. (v)

By (ii), the distances in $J$ between the consecutive transition points of $r$
are uniformly bounded from above.

Figure 21. The solid curve $e$ represents an edge to which two distinct rays
in $\mathcal{R}$ are tangent.

######  Lemma 8.8.

Let $\widetilde{\phi}_{n}\colon D_{n}\to\mathbf{X}$ be a sequence of reduced
relative disc diagrams. Suppose that
$\phi_{n}=\rho_{+}\circ\widetilde{\phi}_{n}$ have the limit half-plane diagram
$\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$ with folding locus $L$.
Furthermore, assume that $\phi^{\prime}$ is bordered by an
$\mathbf{H}$-generic ray $r\colon J\to\mathbf{X}$. Suppose that the sum of the
curvatures of the vertices of $J\subset\partial D^{\prime}$ is
$\geq-\varepsilon_{0}$ from Corollary 7.6.

Then the transition points $l_{k}$ are contained in single edges of $L$ at
angle to $J$ converging to $\frac{\pi}{2}$. See Figure 22.

Furthermore, for any $l\neq l_{k}$ in $L\cap J$, the angles between the edges
of $L$ at $l$ and $J$ are $<\frac{\varepsilon_{0}}{2}$.

Figure 22.

###### Proof.

Let $l=l_{k}$ be a transition point. Then $l\in L$. Moreover, since $x=r(l)$
is not a vertex of $\mathbf{X}$, there is an edge $e$ in $\mathbf{X}$
containing $x$ in its interior. Since $e$ lies in a translate of
$c_{i},r_{i}$, or $r_{i}^{\prime}$, the Alexandrov angles between $r$ and such
$e$ converge with $k$ to $\frac{\pi}{2}$ by Remark 8.7(ii).

Let $e_{1},e_{2}$ be the connected components of $e\setminus x$. If the point
$l$ is not contained in a single edge $f$ of $L$, since all such edges map
under $\widetilde{\phi}_{n}$ into $e_{1}$ or $e_{2}$, we have that the limit
inferior (as $k\to\infty$) of the curvature of such $l$ is $\leq-\pi$, which
is a contradiction. Similarly, the angles between $J$ and $f$ have to coincide
with the Alexandrov angles between $r$ and $e$.

For the last assertion, let $x=r(l)$, where $l$ is not a transition point. If
$l\in L$, then $l$ is contained in two edges $f_{1},f_{2}$ mapping under
$\widetilde{\phi}_{n}$ into $e_{1}$ and $e_{2}$. The angles that $f_{i}$ make
with $J$ are equal, so if they are $\geq\frac{\varepsilon_{0}}{2}$, then,
since the angle between $f_{1}$ and $f_{2}$ is $\pi$, this contradicts the
assumption that the curvature of $l$ is $>-\varepsilon_{0}$. See Figure 23. ∎

Figure 23.

######  Proposition 8.9.

Under the assumptions of Lemma 8.8, suppose additionally that the sum of the
curvatures of the interior edges and the interior vertices of $D^{\prime}$ as
well as of the vertices of $J$ is $\geq-\varepsilon_{0}$.

  1. (1)

Suppose that $l_{k}$ are contained in connected components $L_{k}$ of $L$ that
are asymptotic geodesic rays. Then there is $k$ and a subray
$L^{\prime}_{k}\subset L_{k}$ such that

     * •

$\phi^{\prime}_{|L^{\prime}_{k}}$ is an embedding into an admissible line
diverging from $\\{\alpha_{2}=\alpha_{3}\\}$ (resp.
$\\{\alpha_{1}=\alpha_{2}$}), and

     * •

for each $l^{\prime}_{k}\in L^{\prime}_{k}$, for $n$ sufficiently large, the
class in $\partial_{\infty}\mathbf{X}$ of the unique principal (resp.
antiprincipal) ray in $\mathbf{X}$ starting at
$\widetilde{\phi}_{n}(l_{k}^{\prime})$ is fixed by $f$ from Definition 8.6.

  2. (2)

Alternatively, suppose that $L_{k}$ are geodesic segments, whose length
becomes arbitrarily large when $k\to\infty$, ending at points
$v_{k}\in\partial D^{\prime}$ where $v_{k}v_{k+1}\subset\partial D^{\prime}$
is geodesic and
$\angle_{v_{k}}(L_{k},v_{k+1})+\angle_{v_{k+1}}(L_{k+1},v_{k})\geq\pi$. Then
there is $k$ such that for all $m\geq k$ we have that

     * •

$\phi^{\prime}(v_{m})$ lies outside $\\{\alpha_{2}=\alpha_{3}\\}$ (resp.
$\\{\alpha_{1}=\alpha_{2}\\}$), and

     * •

for $n$ sufficiently large, the class in $\partial_{\infty}\mathbf{X}$ of the
unique principal (resp. antiprincipal) ray in $\mathbf{X}$ starting at
$\widetilde{\phi}_{n}(v_{m})$ is fixed by $f$.

###### Proof.

We focus on the assertion (1). Suppose first that $f\in C$ and there exists
$c_{i}$ in Construction 8.3 that projects to a non-principal admissible line
in $\nabla^{+}$. By Remark 8.7(iii), the projection $\rho_{+}(r)$ leaves every
compact set $K$ of $\nabla^{+}$. Choose $K$ to contain $c\setminus c^{\prime}$
for $c=\rho_{+}(c_{i})$ and $c^{\prime}$ as in Remark 7.7(iii). Let
$c_{i}^{\prime}$ be the preimage in $c_{i}$ of $c^{\prime}$ under $\rho_{+}$.
Then we can consider $c_{k}^{\prime},c_{m}^{\prime}$ defined as two
consecutive translates of $c_{i}^{\prime}$ intersected by $r$ in points
$r(l_{k}),r(l_{m})$. Let $c^{\prime\prime}_{k},c^{\prime\prime}_{m}$ be the
unbounded connected components of $c^{\prime}_{k}\setminus
r(l_{k}),c^{\prime}_{m}\setminus r(l_{m})$. By [BH]*II.9.3 applied in the
subspace of $\mathbf{H}$ bounded by
$c^{\prime\prime}_{k},c^{\prime\prime}_{m}$ and $r(l_{k})r(l_{m})$, we then
have
$\angle_{r(l_{m})}\big{(}c^{\prime\prime}_{m},r(l_{k})\big{)}+\angle_{r(l_{k})}\big{(}c^{\prime\prime}_{k},r(l_{m})\big{)}<\pi$.

Similarly, since $D^{\prime}$ is $\mathrm{CAT}(0)$ and $L_{k},L_{m}$ are
asymptotic, we have
$\angle_{l_{m}}(L_{m},l_{k})+\angle_{l_{k}}(L_{k},l_{m})<\pi$. Consequently,
without loss of generality the first edge of $L_{k}$ is mapped under each
$\widetilde{\phi}_{n}$ to $c^{\prime\prime}_{k}$. By Remark 7.7(iv), we have
$\widetilde{\phi}_{n}(L_{k})\subset c^{\prime\prime}_{k}$, as desired.

Second, suppose that all $c_{i}$ in Construction 8.3 project to principal
lines in $\nabla^{+}$. Then we have $f(r)\subset r$ by Remark 8.7(iv). Note
that since $J$ intersects all $L_{k}$ perpendicularly, and all $L_{k}$ are
asymptotic, by [BH]*II.9.3(2), the regions between consecutive $L_{k}$ are
Euclidean half-strips. Cutting $D$ along $L_{1}$, we can assume that
$D^{\prime}=D$ is a quadrant in $\mathbb{R}^{2}$ with one boundary ray $J$,
and $\phi^{\prime}=\phi$. By Remark 6.6(i), we can assume that all
$\widetilde{\phi}_{n}$ are embeddings. Let $f_{D}\colon D\to D$ be the
translation that acts on $J$ as $f$ does on $r$.

We apply Remark 7.3 with $D^{+}=D^{-}=D,\phi^{+}=\phi,\phi^{-}=f\circ\phi$ and
$J^{+}=f_{D}(J),J^{-}=J$. By Remark 6.6(ii), we have that $R^{\pm}$ are convex
in $D$. Note that $R^{+}=f_{D}(R^{-})$.

The path $P^{+}$ is asymptotic to a line in $D$ determining a direction in
$\partial_{\infty}D$ at angle $\theta$ from the direction of $J$. Consider
first the case where $\theta=0$. Then there is an infinite subcurve
$\gamma\subset P^{+}$ with unit tangent directions mapped under
$\phi^{\angle}$ (see Definition 7.8) arbitrarily close in $S^{\angle}$ to the
midpoint of $S^{\angle}$.

We claim that a neighbourhood $N$ of $\gamma$ in $\overline{D^{+}\setminus
R^{+}}$ can be identified with a subset of $\widehat{D}=\widehat{D}^{\prime}$
on which $\phi^{+},\widehat{\phi}$ coincide. If $\gamma=J^{+}$, then the claim
follows from the last paragraph of Remark 7.3. Otherwise, $\gamma$ maps to the
$1$-skeleton of $\widehat{D}$ under $\overline{D^{+}\setminus
R^{+}}\subset\widehat{D}$, so $N$ can be taken as a union of cells of
$\widehat{D}$ contained entirely in $D^{+}$ and the claim follows from the
second to last paragraph of Remark 7.3.

Figure 24.

We keep the notation $\gamma,N$ for their image in $\widehat{D}$. Since the
function $\phi^{\angle}$ is invariant under translation, and applying the
claim to the neighbourhood $f^{-1}_{D}(N)$ of $f^{-1}_{D}(\gamma)$ in
$\overline{D^{-}\setminus R^{-}}$, we have that the symmetry of $N\cup
f^{-1}_{D}(N)\subset\widehat{D}$ interchanging $N$ and $f^{-1}_{D}(N)$
preserves $\widehat{\phi}^{\angle}$. See Figure 24. Since the unit tangent
directions of $\gamma$ are not mapped under $\phi^{\angle}$ to $\partial
S^{\angle}$, we conclude that $\gamma$ is contained in the folding locus
$\widehat{L}$ of $\widehat{D}$. This contradicts Lemma 7.10.

Second, consider the case where $\theta>0$. We treat $L_{k}$ as rays
$L_{k}\colon[0,\infty)\to D$ parametrised by arc-length. By Lemma 7.9(ii),
there is $M$ such that each $\phi\big{(}L_{k}[M,\infty)\big{)}$ is contained
in a principal line. Furthermore, since $l_{k}$ are at distance uniformly
bounded from above (Remark 8.7(v)), we can assume that the geodesic path
$L_{k}(M)L_{k+1}(M)$ in $D$ maps under $\phi$ outside
$\\{\alpha_{2}=\alpha_{3}\\}$ (resp. $\\{\alpha_{1}=\alpha_{2}\\}$).

Since $\theta>0$, there are $m$ and $n$ with $l^{\prime}_{m}=L_{m}(M)$
contained in $R^{+}_{n}$. Then for
$l^{\prime}_{k}=f^{-1}_{D}(l^{\prime}_{m})$, and
$x_{k}=\widetilde{\phi}_{n}(l^{\prime}_{k}),x_{m}=\widetilde{\phi}_{n}(l^{\prime}_{m})$,
we have $f(x_{k})=x_{m}$. Consequently, $f$ maps the unique principal (resp.
antiprincipal) ray starting at $x_{k}$ to the one starting at $x_{m}$. These
rays are parallel in $\mathbf{X}$ since $l^{\prime}_{k},l^{\prime}_{m}$ are
connected by a path in $D_{n}$ mapping under $\phi_{n}$ to a piecewise
geodesic path outside $\\{\alpha_{2}=\alpha_{3}\\}$ (resp.
$\\{\alpha_{1}=\alpha_{2}\\}$), and by Remark 3.8.

The proof of assertion (2) is identical, with the following modifications. In
the beginning of the proof, to justify
$\angle_{l_{m}}(L_{m},l_{k})+\angle_{l_{k}}(L_{k},l_{m})<\pi$, we use the
inequality in the hypothesis. Later, $D$ might be no longer a quadrant but a
possibly smaller subset of $\mathbb{R}^{2}$. The last difference is that now
$L_{k}$ are bounded, but using the assumption that $L_{k}$ become longer than
$M$ when $k\to\infty$, we can still find $k$ such that for each $m\geq k,$ for
sufficiently large $n$, we have that $l^{\prime}_{m}=L_{m}(M)$ belongs to
$R^{+}_{n}$. ∎

##  9\. Classification

In this section we finish the classification of parabolic isometries and
loxodromic isometries not of rank 1, for the action of
$\mathrm{Tame}(\mathbbm{k}^{3})$ on $\mathbf{X}$. In particular, as a converse
to Corollary 8.4, we show that such isometries always belong to the conjugates
of the groups $B^{\prime},C$ introduced in Section 3.2.

###  9.1. Parabolic isometries

######  Theorem 9.1.

If $f\in\mathrm{Tame}(\mathbbm{k}^{3})$ is parabolic, then $f$ belongs to a
conjugate of $C$. Furthermore, such a conjugate is unique.

###### Proof.

Let $\mathbf{X}_{l}\subset\mathbf{X}$ be the subspace of points $x$ with
$d_{\mathbf{X}}\big{(}x,f(x)\big{)}\leq l$. We first prove that for each
compact set $K\subset\nabla^{+}$ there is $l>|f|$ with
$\rho_{+}(\mathbf{X}_{l})\cap K=\emptyset$. Indeed, otherwise, there are
$x_{n}\in\mathbf{X}_{|f|+\frac{1}{n}}$ with $\rho_{+}(x_{n})\in K$. After
passing to a subsequence, we can assume $\rho_{+}(x_{n})\to[\alpha]\in K$.
Choose $\epsilon$ so that each $[\alpha^{\prime}]\in\nabla_{+}$ with
$|[\alpha^{\prime}],[\alpha]|<\epsilon$ lies in the star of the cell
containing $[\alpha]$ in its interior. Choose $n$ such that
$\frac{1}{n}<\frac{\epsilon}{3}$ and
$|\rho_{+}(x_{n}),[\alpha]|<\frac{\epsilon}{3}$. Then there is
$y\in\mathbf{X}$ with $\rho_{+}(y)=[\alpha]$ and
$|x_{n},y|<\frac{\epsilon}{3}$. Since $f$ does not have a fixed point and acts
without inversions on $\mathbf{X}$, we have $f(y)\neq y$ and
$|y,f(y)|\geq\epsilon$. This contradicts
$|y,f(y)|\leq|y,x_{n}|+|x_{n},f(x_{n})|+|f(x_{n}),f(y)|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}$.

Let $\nabla^{+}_{l}\subset\nabla^{+}$ be the set of points at distance $\leq
l$ from $\partial\nabla^{+}$. We claim that
$\rho_{+}(\mathbf{X}_{l})\subset\nabla^{+}_{l}$. Indeed, for each
$x\in\mathbf{X}_{l}$, by Lemma 3.3(iii), there is a chain of length $\leq l$
from $x$ to $f(x)$. If $|\rho_{+}(x),\partial\nabla^{+}|>l$, then all the
points in that chain map under $\rho_{+}$ outside $\partial\nabla^{+}$.
Consequently, by Corollary 3.7, the chambers containing the points of that
chain have nonempty intersection, which is fixed pointwise by $f$,
contradiction.

Now, let $K$ be the $4|f|$ neighbourhood of $[1,1,1]$ in $\nabla^{+}$, and
choose $2|f|>l>|f|$ with $\rho_{+}(\mathbf{X}_{l})\cap K=\emptyset$. Note that
$K$ separates $\nabla^{+}_{l}$ into two connected components, which are
contained in the $l$-neighbourhoods of the sets $\\{\alpha_{1}=\alpha_{2}\\}$,
or $\\{\alpha_{2}=\alpha_{3}\\}$, respectively. Since $\mathbf{X}_{l}$ is
convex, hence connected, it is mapped under $\rho_{+}$ into one of these two
connected components.

Consider first the former possibility. Let $x,f(x)\in\mathbf{X}_{l}$ and
assume without loss of generality $x\in\mathbf{E}^{+}_{\mathrm{id}}$. Then the
minimal length chain from $x$ to $f(x)$ in $\mathbf{X}_{l}$ maps under
$\rho_{+}$ outside the set $\\{\alpha_{2}=\alpha_{3}\\}$. Then by Remark 3.8,
the principal rays in $\mathbf{E}^{+}_{\mathrm{id}},\mathbf{E}^{+}_{f}$
starting at $x$ and at $f(x)$ are parallel. By Lemma 8.1, we obtain $f\in C$,
as desired. If $\mathbf{X}_{l}$ maps under $\rho_{+}$ to the other connected
component of $\nabla^{+}_{l}\setminus K$, we analogously obtain $f\in
B^{\prime}$. However, this contradicts Corollary 8.4(1).

For the remaining assertion, consider parabolic $f\in C\cap gCg^{-1}$ for some
$g\in\mathrm{Tame}(\mathbbm{k}^{3})$. Let $r\colon[0,\infty)\to\mathbf{X}$ be
the geodesic ray starting at $r(0)=x\in\mathbf{X}_{l}$ that represents the
same point $\zeta$ in $\partial_{\infty}\mathbf{X}$ as the principal rays in
$\mathbf{E}^{+}_{g}$. Since $f\in gCg^{-1}$, we have $f(\zeta)=\zeta$.
Consequently, for each $t\geq 0$, we have we have
$d_{\mathbf{X}}\big{(}r(t),f(r(t))\big{)}\leq
d_{\mathbf{X}}\big{(}r(0),f(r(0))\big{)}\leq l$, and so
$r[0,\infty)\subset\mathbf{X}_{l}$. By Lemma 8.2, we have that a subray of $r$
is parallel to a principal ray in $\mathbf{E}^{+}_{g}$. Consequently, there is
a chamber $\mathbf{E}^{+}_{h}$ with $h\in C$ with $h$ simultaneously in $gC$,
and so $g\in C$. ∎

######  Proposition 9.2.

Let $f\in C$ be parabolic with translation length zero and let
$\zeta\in\partial_{\infty}\mathbf{X}$ be the class of the principal rays in
$\mathbf{E}^{+}_{\mathrm{id}}$. Then $\zeta$ is the positive (and the
negative) limit point of $f$.

In the language of Section 2, Proposition 9.2 states that $f$ is not vile.

In the proof we will use the following. Let $a>0$. An _$a$ -sector_ is the
$\mathrm{CAT}(0)$ space obtained by glueing two Euclidean half-strips
$[0,\infty)\times[0,R]$ along copies of the set of points $(t,y)$ defined by
$y\geq f(t)$ for some convex function $f\colon[0,\infty)\to[0,R]$ satisfying
$\frac{1}{c}{e^{-at}}\leq f\leq ce^{-at}$ for some $c>0$. We call the union of
the two copies of the ray $[0,\infty)\times\\{0\\}$ the _frontier_ of the
sector. An _$a$ -quasicusp_ is the locally $\mathrm{CAT}(0)$ space glued of
finitely many $a$-sectors by cyclically identifying them along their
frontiers.

######  Lemma 9.3.

The universal cover $\mathbf{H}$ of an $a$-quasicusp is quasi-isometric with a
horodisc $\mathbf{H}^{\prime}$ in the rescaled hyperbolic plane
$\frac{1}{a}\mathbb{H}^{2}$. Furthermore, the quasi-isometry conjugates the
deck transformation group of $\mathbf{H}$ with a group of parabolic isometries
of $\mathbf{H}^{\prime}$.

Note that we could replace $\frac{1}{a}\mathbb{H}^{2}$ by $\mathbb{H}^{2}$ in
the statement by composing with the obvious quasi-isometry between
$\frac{1}{a}\mathbb{H}^{2}$ and $\mathbb{H}^{2}$, but a constant $a$ will
appear naturally in the proof of Proposition 9.2.

###### Proof.

Let $f^{\prime}$ be a parabolic isometry preserving $\mathbf{H}^{\prime}$. Let
$r$ be the image of a geodesic ray in $\mathbf{H}^{\prime}$ starting on the
horosphere bounding $\mathbf{H}^{\prime}$. It suffices to show that the union
of $f^{\prime n}(r)$, over all $n\in\mathbb{Z}$, with the metric induced from
$\mathbf{H}^{\prime}$, is quasi-isometric with the union of all the lifts to
$\mathbf{H}$ of all $[0,\infty)\times\\{0\\}$ and all $\\{(t,f(t))\\}$ in all
the sectors of a quasicusp, with the metric induced from $\mathbf{H}$. To do
that, it suffices to show that this metric on $r\cup f^{\prime}(r)$ is
bilipschitz to the metric on the union of $[0,\infty)\times\\{0\\}$ and
$\\{(t,f(t))\\}$ in $[0,\infty)\times[0,R]$.

To do that, note first that the maps $t\to(t,0),t\to(t,f(t))$ from
$[0,\infty)$ to their image in $[0,\infty)\times[0,R]$ are bilipschitz (the
first one is in fact an isometric embedding). Furthermore, the distance
between $(t^{\prime},0)$ and $(t,f(t))$ is bounded from above by
$|t^{\prime}-t|+f(t)$ and from below by both $|t^{\prime}-t|$ and $f(t)$.
Consequently, up to a multiplicative constant, this distance equals
$|t-t^{\prime}|+e^{-at}$. The same formula holds in
$\mathbf{H}^{\prime}\subset\frac{1}{a}\mathbb{H}^{2}$ for the distances
between the points on $r$ and $f^{\prime}(r)$. ∎

###### Proof of Proposition 9.2.

Let $\mathbf{H}$ be the $f$-invariant convex subset of $\mathbf{X}$ described
in Construction 8.3(2). We claim that $\mathbf{H}$ is the universal cover of a
quasicusp with $\langle f\rangle$ the deck transformation group. Since $|f|=0$
(Corollary 8.4(2)), for each $i=1,\ldots,n$, the curve $c_{i}$ bounding the
intersection $R_{i}\cap R_{i}^{\prime}\subset\mathbf{H}$ maps under $\rho_{+}$
to the admissible line with equation $\alpha_{1}=\alpha_{2}+m\alpha_{3}$ for
some integer $m\geq 1$. Passing to the Euclidean coordinates on the plane
$\beta_{1}+\beta_{2}+\beta_{3}=0$ for $\beta_{i}=\log\alpha_{i}$ and
$\alpha_{1}\alpha_{2}\alpha_{3}=1$ (see Section 3.1), this means

$e^{\beta_{1}}=e^{\beta_{2}}+me^{\beta_{3}}=e^{\beta_{2}}+me^{-(\beta_{1}+\beta_{2})}.$

On that plane we have orthogonal coordinates

$y=\frac{\beta_{1}-\beta_{2}}{\sqrt{2}},t=\frac{\sqrt{3}(\beta_{1}+\beta_{2})}{\sqrt{2}}.$

The equation above has then the form

$e^{\frac{\sqrt{\frac{2}{3}}t+\sqrt{2}y}{2}}=e^{\frac{\sqrt{\frac{2}{3}}t-\sqrt{2}y}{2}}+me^{-\sqrt{\frac{2}{3}}t},$

i.e. $2\sinh{\frac{y}{\sqrt{2}}}=me^{-\sqrt{\frac{3}{2}}t}$. For $y$ close to
$0$, up to a multiplicative constant, $\sinh{\frac{y}{\sqrt{2}}}$ equals $y$,
and so $y=y(t)$ has the required asymptotics for an $a$-sector with
$a=\sqrt{\frac{3}{2}}$. This justifies the claim.

By Lemma 9.3, we have an equivariant quasi-isometry
$\mathbf{H}\to\mathbf{H}^{\prime}$ for $\mathbf{H}^{\prime}$ a horodisc in a
hyperbolic plane $\frac{1}{a}\mathbb{H}^{2}$ preserved by a parabolic isometry
$f^{\prime}$. The horodisc $\mathbf{H}^{\prime}$ is Gromov-hyperbolic and for
any point $x^{\prime}\in\mathbf{H}^{\prime}$, for $n\to\pm\infty$, the
sequence $f^{\prime n}(x^{\prime})$ converges w.r.t. the Gromov product to the
unique point in the Gromov boundary of $\mathbf{H}^{\prime}$. Consequently,
$\mathbf{H}$ is Gromov-hyperbolic and for any point $x\in\mathbf{H}$, for
$n\to\pm\infty$, the sequence $f^{n}(x)$ converges w.r.t. the Gromov product
to the unique point in the Gromov boundary of $\mathbf{H}$. Since $\mathbf{H}$
is proper, by [BH]*III.H.3.7(2) and 3.17(6), for $n\to\pm\infty$, the sequence
$f^{n}(x)$ converges in the cone topology to $\zeta$. ∎

###  9.2. Loxodromic isometries not of rank 1

######  Theorem 9.4.

Let $f\in\mathrm{Tame}(\mathbbm{k}^{3})$ be loxodromic not of rank $1$. Then
$f$ belongs to a conjugate of $C$ or $B^{\prime}$. Furthermore, there is at
most one such a conjugate of $C$ and at most one such a conjugate of
$B^{\prime}$.

######  Example 9.5.

The map $f=(x_{2},x_{1}+x_{2}^{2},x_{3})$ belongs to both $C$ and
$gB^{\prime}g^{-1}$ for $g=(x_{3},x_{2},x_{1})$.

To prove the uniqueness in Theorem 9.4, it will be convenient to use the
following, whose proof we postpone to Section 11.

######  Lemma 9.6.

Let $\xi\neq\eta\in\partial_{\infty}\mathbf{X}$ be principal and not far. Then
$\angle(\xi,\eta)=\frac{2\pi}{3}$, the midpoint of the geodesic $\xi\eta$ in
$\partial_{\infty}\mathbf{X}$ is antiprincipal, and there is a chamber
containing geodesic rays representing $\xi$ and that midpoint. The same
statement holds if we interchange the words ‘principal’ and ‘antiprincipal’.

Note that since $\partial_{\infty}\mathbf{X}$ is $\mathrm{CAT}(1)$, the
geodesic $\xi\eta$ above is unique.

######  Remark 9.7.

$\partial_{\infty}\mathbf{X}$ has geometric dimension 1 (see [K]), which can
be justified by the following argument proposed by Pierre-Emmanuel Caprace.
Namely, $\partial_{\infty}\mathbf{X}$ embeds in the space of directions of any
asymptotic cone of $\mathbf{X}$, where the sequence of basepoints is constant.
Since $\mathbf{X}$ has geometric dimension 2, we have that the asymptotic cone
has also geometric dimension $2$ [LY]*Lem 11.1, and so
$\partial_{\infty}\mathbf{X}$ has geometric dimension $1$.

Consequently, $\partial_{\infty}\mathbf{X}$ is locally an $\mathbb{R}$-tree,
see e.g. the proof of [OP]*Lem 6.3.

###### Proof of Theorem 9.4.

Let $\gamma$ be an axis of $f$ in $\mathbf{X}$. If $f$ is not of rank $1$,
then $\gamma$ is not contracting. Let $p_{0}\in\gamma$ and for
$k\in\mathbb{Z}$ let $p_{k}=f^{k}(p_{0})$. By Lemma 2.8 applied with
$\beta=\gamma$ and $\mathcal{P}=\\{p_{k}\\}$, for each $n>0$, there is a
$(p_{k},\frac{1}{n},n)$-quadrilateral with a side in $\gamma$. Thus its
translate under $f^{-k}$ is a $(p_{0},\frac{1}{n},n)$-quadrilateral
$\square_{n}$ with a side in $\gamma$. By Lemma 2.2, we can assume that
$\square_{n}$ is embedded.

Let $\overline{\phi}_{n}\colon\overline{D}_{n}\to\mathbf{X}$ be a reduced
relative disc diagram with boundary $\square_{n}$ guaranteed by Lemma 6.3.
Note that $\overline{\phi}_{n}$ is an embedding by Theorem 6.10. Then for each
subcomplex $D_{n}\subset\overline{D}_{n}$ with
$I_{n}=D_{n}\cap\partial\overline{D}_{n},$ the restriction
$\widetilde{\phi}_{n}$ of $\overline{\phi}_{n}$ to $D_{n}$ is a disc diagram
in $\mathbf{X}$ relative to $I_{n}$.

Since $\nabla^{+}$ is locally finite and since $\overline{\phi}_{n}$ are
reduced disc diagrams, by Theorem 6.9 there is a uniform bound on the degree
of a vertex in $\overline{D}_{n}$ at a given distance from
$q_{n}=\overline{\phi}_{n}^{-1}(p_{0})$. Furthermore, since $\square_{n}$ is a
$(p_{0},\frac{1}{n},n)$-quadrilateral, we can assume that the radius $n$
closed ball in $\overline{D}_{n}$ centred at $q_{n}$ is disjoint from
$\partial\overline{D}_{n}\setminus\overline{\phi}_{n}^{-1}(\gamma)$. Thus we
can find $D_{n}$ as above so that $I_{n}=\widetilde{\phi}^{-1}_{n}(\gamma)$,
the pairs $(D_{n},I_{n})$ converge to a half-plane diagram $(D,\partial D)$,
and all $q_{n}$ coincide.

Furthermore, we can assume that we have the limit $\phi\colon D\to\nabla^{+}$
of $\rho_{+}\circ\widetilde{\phi}_{n}$ bordered by two rays with union
$\gamma$. (Here $D^{*}=D^{\prime}=D$.) Let $\gamma_{n}=\phi_{n}(I_{n})$. Note
that since $\square_{n}$ were $(p_{0},\frac{1}{n},n)$-quadrilaterals, by
Theorem 6.9 we have that $D$, with the degenerate piecewise smooth Euclidean
metric pulled back under $\phi$ from $\nabla^{+}$, is isometric to the
Euclidean half-plane. Let $f_{D}\colon D\to D$ be the translation that
preserves $\partial D$ and acts on it as $f$ does on $\gamma$.

We apply a variant of Remark 7.3 with
$D^{+}=D^{-}=D,\phi^{+}=\phi,\phi^{-}=f\circ\phi$, identity frillings, and
$J^{+}=J^{-}=\partial D$, which are lines instead of rays. By Remark 6.6(ii),
we have that $R^{\pm}$ are convex in $D^{\pm}$. Thus $R^{+}=R^{-}$ is a
Euclidean strip (possibly $R^{\pm}=D^{\pm}$).

Consider first the case where $R^{\pm}\neq D^{\pm}$. Then $\widehat{D}$ from
Remark 7.3 becomes a combinatorial complex isometric to the Euclidean plane
and $\widehat{\phi}\colon\widehat{D}\to\nabla^{+}$ is combinatorial and
$\mathbf{X}$-reduced. Let $L^{\pm}\subset
D^{\pm},\widehat{L}\subset\widehat{D}$ denote the folding loci. We then have
$\widehat{L}\cap(D^{\pm}\setminus R^{\pm})=L^{\pm}\setminus R^{\pm}$. Observe
that $\widehat{L}\cap(D^{+}\setminus R^{+})$ is send to
$\widehat{L}\cap(D^{-}\setminus R^{-})$ by the glide reflection
$f_{\widehat{D}}$ of $\widehat{D}$ that is the composition of $f_{D}$ with the
reflection in $\mathrm{fr}\,R^{+}=\mathrm{fr}\,R^{-}$.

All the edges of $\widehat{L}$ are oriented, since $\widehat{D}$ is locally
Euclidean. By Remark 7.7(ii), for each oriented edge $e$ of $\widehat{L}$, the
entire oriented geodesic ray in $\widehat{D}$ starting at $e$ lies in
$\widehat{L}$. By Lemma 7.9(i), such a geodesic ray cannot be parallel to
$\mathrm{fr}\,R^{+}$, in view of the fact that $\rho_{+}(\gamma)$ is bounded.
Furthermore, by Corollary 7.6, none of such geodesics can intersect, except
when one is contained in another. Thus by the observation above on
$f_{\widehat{D}}$, we have that $\widehat{L}$ has no vertices $v$ of degree
$\geq 3$, since otherwise some of the six oriented geodesics issuing from $v$
and $f_{\widehat{D}}(v)$ would intersect.

If a connected component of $\widehat{L}$ was a line $l$ intersecting
$\mathrm{fr}\,R^{+}$, then either $l$ would intersect $f_{\widehat{D}}(l)$,
which is a contradiction, or $l$ would be perpendicular to
$\mathrm{fr}\,R^{+}$. Then the function
$\widehat{\phi}^{\angle}\colon\mathbb{S}\to S^{\angle}$ as in Definition 7.8
would simultaneously assign to the direction of $l$, which is the opposite of
the direction of $f_{\widehat{D}}(l)$, the principal and the antiprincipal
point, contradiction. Consequently, $\widehat{L}$ is empty, which contradicts
the last assertion of Lemma 7.9.

Second, consider the case where $R^{+}=D$. Then $f_{D}$ preserves the folding
locus $L$ of $D$. Thus, as in the previous case, by Remark 7.7(ii) and
Corollary 7.6, all the vertices of order $\geq 3$ in $L$ are at a uniform
distance from $\partial D$. Since $L$ is again nonempty, and contains no line
parallel to $\partial D$, there is a half-plane $D_{0}\subseteq D$ such that
$D_{0}\cap L$ is a family of parallel oriented geodesic rays
$(L_{k})_{k=-\infty}^{\infty}$ starting or ending at $\partial D_{0}$. By
Lemma 7.9(ii), there is $M$ such that each $\phi(L_{k}[M,\infty))$ is
contained in a principal line. Choose $m$ and $n$ with $L_{m}(M)$ contained in
$R^{+}_{n}$. Then for $L_{k}(M)=f^{-1}_{D}(L_{m}(M))$, and
$x_{k}=\widetilde{\phi}_{n}(L_{k}(M)),x_{m}=\widetilde{\phi}_{n}(L_{m}(M))$,
we have $f(x_{k})=x_{m}$. Consequently, $f$ maps the unique (see Remark 3.8)
principal or antiprincipal ray starting at $x_{k}$ to the one starting at
$x_{m}$. These rays are asymptotic in $\mathbf{X}$ since $L_{k}$ and $L_{m}$
are asymptotic in $D$. By Lemma 8.1, $f$ lies in a conjugate of $C$ or
$B^{\prime}$.

For the uniqueness of the conjugate of $B^{\prime}$, suppose $f\in
B^{\prime}\cap gB^{\prime}g^{-1}$ with $g\notin B^{\prime}$ and let $\zeta\neq
g\zeta$ be the antiprincipal points represented by the antiprincipal rays in
$\mathbf{E}^{+}_{\mathrm{id}},\mathbf{E}^{+}_{g}$. Let $\xi,\eta$ be the limit
points of $f$. By Construction 8.3(1), it follows that $\zeta,g\zeta$ are the
midpoints of length $\pi$ paths
$\omega_{\mathrm{id}}\neq\omega_{g}\subset\partial_{\infty}\mathbf{X}$ between
$\xi$ and $\eta$. Since $\partial_{\infty}\mathbf{X}$ is $\mathrm{CAT}(1)$ and
is locally an $\mathbb{R}$-tree (Remark 9.7), we have that
$\omega=\omega_{\mathrm{id}}\cdot\omega_{g}$ is a circle of length $2\pi$
isometrically embedded in $\partial_{\infty}\mathbf{X}$. By Lemma 6.7, we
obtain a flat $F\subset\mathbf{X}$ with $\partial_{\infty}F=\omega$. Thus
$\zeta$ and $g\zeta$ are at distance $\pi$ but they are not far. This
contradicts Lemma 9.6.

The proof of the uniqueness of the conjugate of $C$ is analogous. ∎

##  10\. $(\xi,\eta)$-diagrams

We will be using the following diagrams to describe pairs of points in
$\partial_{\infty}\mathbf{X}$ that are not far (see Definition 2.3).

######  Construction 10.1.

Let $\xi\neq\eta\in\partial_{\infty}\mathbf{X}$ be not far, and let
$x\in\mathbf{X}$ be such that $\angle_{x}(\xi,\eta)>0$. We construct a
following $\mathbf{X}$-reduced relative half-plane diagram
$\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$, which we call a _$(\xi,\eta)$
-diagram_.

Consider $u_{n},v_{n}\in\mathbf{X}\cup\partial_{\infty}\mathbf{X}$, with
$u_{n}\to\xi,v_{n}\to\eta$, and at most one of each $u_{n},v_{n}$ in
$\partial_{\infty}\mathbf{X}$, such that either:

  1. (i)

for each closed metric ball $P$ in $\mathbf{X}$, the geodesics $u_{n}v_{n}$
are disjoint from $P$ for $n$ large enough, or

  2. (ii)

there is a closed metric ball $P$ in $\mathbf{X}$ containing a point $p_{n}$
on each $u_{n}v_{n}$, such that there is a
$(p_{n},\frac{1}{n},n)$-quadrilateral $\square_{n}$ with a side
$\bar{u}_{n}\bar{v}_{n}$ in $u_{n}v_{n}$.

In case (i), by [BH]*II.9.8(4), if $\angle(\xi,\eta)<\pi$, then we can assume
that $u_{n},v_{n}$ lie on the geodesic rays from $x$ to $\xi,\eta$,
respectively. Let $\bar{u}_{n}\in xu_{n}\cap u_{n}v_{n},\bar{v}_{n}\in
xv_{n}\cap u_{n}v_{n}$ be closest possible to $x$. Let $\triangle_{n}$ be the
geodesic triangle $x\bar{u}_{n}\bar{v}_{n}$, which is an embedded closed path.
Let $\overline{\phi}_{n}\colon\overline{D}_{n}\to\mathbf{X}$ be a reduced
relative disc diagram with boundary $\triangle_{n}$ guaranteed by Lemma 6.3.
Note that $\overline{\phi}_{n}$ is an embedding by Theorem 6.10. Then for each
subcomplex $D_{n}\subset\overline{D}_{n}$ with
$I_{n}=D_{n}\cap\partial\overline{D}_{n},$ the restriction
$\widetilde{\phi}_{n}$ of $\overline{\phi}_{n}$ to $D_{n}$ is a disc diagram
in $\mathbf{X}$ relative to $I_{n}$.

Since $\nabla^{+}$ is locally finite and since $\overline{\phi}_{n}$ are
reduced disc diagrams, by Theorem 6.9 there is a uniform bound on the degree
of a vertex in $\overline{D}_{n}$ at a given distance from
$q_{n}=\overline{\phi}_{n}^{-1}(x)$. Furthermore, by the hypothesis in case
(i), we can assume that the radius $n$ closed ball in $\overline{D}_{n}$
centred at $q_{n}$ is disjoint from
$\overline{\phi}_{n}^{-1}(\bar{u}_{n}\bar{v}_{n})$. Thus, after possibly
passing to a subsequence, we can find $D_{n}$ as above so that $(D_{n},I_{n})$
converge to a half-plane diagram $(D,\partial D)$ and all $q_{n}$ coincide
with a point $q\in\partial D$. Furthermore, we can assume Construction
6.13(ii). By Remark 7.1, where $I_{a},I_{b}$ are the connected components of
$\partial D\setminus q$ and $I_{c}$ is empty, we also have Construction
6.13(iii). This gives the frilling $D^{\prime}$ of $D$, and the limit map
$\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$ of
$\phi_{n}=\rho_{+}\circ\widetilde{\phi}_{n}$ that is an $\mathbf{X}$-reduced
relative half-plane diagram. Note that by [BH]*II.9.8(4) we have that
$I_{a},I_{b}$ represent points in $\partial_{\infty}D$ at angle
$\angle(\xi,\eta)$.

In case (ii), by Lemma 2.2, we can assume that each $\square_{n}$ is an
embedded closed path. Let
$\overline{\phi}_{n}\colon\overline{D}_{n}\to\mathbf{X}$ be reduced relative
disc diagrams with boundary $\square_{n}$ guaranteed by Lemma 6.3. Again,
$\overline{\phi}_{n}$ is an embedding by Theorem 6.10. All $\rho_{+}(p_{n})$
lie in a finite set of cells of $\nabla^{+}$. Thus, after possibly passing to
a subsequence, we have $D_{n}\subset\overline{D}_{n}$ with
$I_{n}=D_{n}\cap\partial\overline{D}_{n}\subset\overline{\phi}_{n}^{-1}(\bar{u}_{n}\bar{v}_{n})$
such that $(D_{n},I_{n})$ converge to a half-plane diagram $(D,\partial D)$.
Moreover, all $q_{n}=\overline{\phi}^{-1}(p_{n})$ coincide with a point
$q\in\partial D$. Furthermore, denoting
$\widetilde{\phi}_{n}=\overline{\phi}_{|D_{n}}$, we have the frilling
$D^{\prime}$ of $D$, and the limit map $\phi^{\prime}\colon
D^{\prime}\to\nabla^{+}$ of $\phi_{n}=\rho_{+}\circ\widetilde{\phi}_{n}$ that
is an $\mathbf{X}$-reduced relative half-plane diagram. Since $\square_{n}$
were $(p_{n},\frac{1}{n},n)$-quadrilaterals, $D^{\prime}$ is isometric to the
Euclidean half-plane in $\mathbb{R}^{2}$.

######  Lemma 10.2.

Let $\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$ be a $(\xi,\eta)$-diagram.
If $\xi$ is principal or antiprincipal, then there is
$J^{\prime}\subset\partial D^{\prime}$ with $\phi^{\prime}_{|J^{\prime}}$ an
embedding into a principal line.

Note that if $\phi^{\prime}$ is bordered by a geodesic ray representing $\xi$,
which happens for example for $\angle(\xi,\eta)<\pi$, then Lemma 10.2 follows
immediately from Lemma 8.2.

###### Proof.

By Lemma 8.2, there is a point $x^{\prime}$ on the geodesic ray from $x$ to
$\xi$ such that the geodesic ray $r$ from $x^{\prime}$ to $\xi$ is principal
or antiprincipal. Let $J_{\xi}\subset\partial D$ be the ray starting with $q$
with $J_{\xi}\cap I_{n}$ mapping into $x\bar{u}_{n}$ or $p_{n}\bar{u}_{n}$
under $\widetilde{\phi}_{n}$.

In case (i) of Construction 10.1, since $u_{n}\to\xi$, the restrictions of
$\widetilde{\phi}_{n}$ to some $J\subseteq J_{\xi}$ converge to $r$ (in
$C^{1}$ piecewise, up to a parametrisation). Let $l$ be the principal line
containing $\rho_{+}(r)$.

In case (ii) of Construction 10.1, since $p_{n}\in P$, after possibly passing
to a subray of $r$, each geodesic ray from $p_{n}$ to $\xi$ contains a subray
$r_{n}$ parallel to $r$ with $\rho_{+}(r_{n})$ in a uniform neighbourhood of
$\rho_{+}(r)$. Furthermore, since $u_{n}\to\xi$, the restrictions of
$\widetilde{\phi}_{n}$ to each compact subset of some subray $J\subset
J_{\xi}$, become arbitrarily close to $r_{n}$. Since
$\rho_{+}\circ\widetilde{\phi}_{n}$ have a limit, we conclude that
$\rho_{+}(r_{n})$ and $\rho_{+}\circ\widetilde{\phi}_{n}(J)$ converge into
some principal line $l$.

In both cases, $\phi(J)$ lies in $l$, and so does $\phi^{\prime}(J^{\prime})$
for the image $J^{\prime}$ of $J$ in $\partial D^{\prime}$ by Remark 7.1. ∎

######  Lemma 10.3.

Let $\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$ be a $(\xi,\eta)$-diagram
from Construction 10.1(i), and assume
$\angle_{x}(\xi,\eta)>\angle(\xi,\eta)-\varepsilon$, for some $\varepsilon>0$.
Suppose furthermore $u_{n},v_{n}\in\mathbf{X}$ and
$\angle_{x}(\xi,u_{n}),\angle_{x}(\eta,v_{n})<\varepsilon$. Then for any
$D_{n}$ (and hence for $D^{\prime}$) the sum of the curvatures of vertices and
edges in the interior of $D_{n}$ and in the interior of $I_{n}\setminus q_{n}$
is $\geq-3\varepsilon$. Furthermore, the angle at $q_{n}$ is
$\leq\angle(\xi,\eta)$.

###### Proof.

Suppose first that $\angle(\xi,\eta)<\pi$, in which case we assumed that
$\bar{u}_{n},\bar{v}_{n}$ lie on the geodesic rays from $x$ to $\xi,\eta$,
respectively. By Lemma 2.4, we observe
$\angle_{\bar{u}_{n}}(x,\bar{v}_{n})+\angle_{\bar{v}_{n}}(x,\bar{u}_{n})\geq\pi-\angle(\xi,\eta)$.
Thus by Theorem 6.9, the sum of the curvatures in question is
$\geq-\varepsilon$. The assertion on the angle at $q_{n}$ also follows from
the above observation and Theorem 6.9.

Second, suppose $\angle(\xi,\eta)=\pi$, and so
$\angle_{x}(\xi,\eta)>\pi-\varepsilon$. Thus
$\angle_{x}(\bar{u}_{n},\bar{v}_{n})>\pi-3\varepsilon.$ Then again by Theorem
6.9, the sum of the curvatures in question is $\geq-3\varepsilon$. ∎

We need a similar result for $\xi=\eta$. We start with the following.

######  Lemma 10.4.

Let $X$ be a complete $\mathrm{CAT}(0)$ space, and let $d>0$. Then there
exists a function $\psi\colon[0,\infty)\to\mathbb{R}_{>0}$ with
$\lim_{t\to\infty}\psi(t)=0$, such that for any disjoint geodesic rays
$r,r^{\prime}$ representing the same point $\xi\in\partial_{\infty}X$, with
$d_{X}\big{(}r(0),r^{\prime}(0)\big{)}\leq d$, the projection $p(t)$ of $r(t)$
to $r^{\prime}[0,\infty)$ satisfies

$\Big{|}\angle_{r(t)}\big{(}r(0),p(t)\big{)}-\frac{\pi}{2}\Big{|}\leq\psi(t).$

###### Proof.

Without loss of generality, we can assume
$b_{\xi}(r(t))=b_{\xi}(r^{\prime}(t))=-t$ for a Busemann function $b_{\xi}$.
Since the horoball $b_{\xi}\leq-t$ is a limit of balls, we have the inequality
$\angle_{r(t)}\big{(}\xi,r^{\prime}(t)\big{)}\leq\frac{\pi}{2}$. Consequently,
$\alpha(t)=\angle_{r(t)}\big{(}r(0),r^{\prime}(t)\big{)}\geq\frac{\pi}{2}$,
and analogously
$\beta(t)=\angle_{r^{\prime}(t)}\big{(}r^{\prime}(0),r(t)\big{)}\geq\frac{\pi}{2}$.

Let $x$ be the last point on the geodesic $r(0)r^{\prime}(0)$ whose projection
to the geodesic $r(t)r^{\prime}(t)$ equals $r(t)$. Then we have
$\angle_{r(t)}\big{(}r^{\prime}(t),x\big{)}=\frac{\pi}{2}$. Considering the
triangle $r(0)r(t)x$, by [BH]*II.1.7(5) we have

$d\geq d_{X}\big{(}r(0),r^{\prime}(0)\big{)}\geq d_{X}\big{(}r(0),x\big{)}\geq
t\sin\big{(}\alpha(t)-\tfrac{\pi}{2}\big{)}.$

Consequently, there is a function $\psi$ with $\lim_{t\to\infty}\psi(t)=0$
satisfying for all $r,r^{\prime}$ the bounds
$\frac{\psi(t)}{2}\geq\alpha(t)-\frac{\pi}{2}$, and analogously
$\frac{\psi(t)}{2}\geq\beta(t)-\frac{\pi}{2}$. Since the sum of the Alexandrov
angles in the triangle $r(t)r^{\prime}(t)p(t)$ is $\leq\pi$, we obtain
$\angle_{r(t)}\big{(}r^{\prime}(t),p(t)\big{)}\leq\frac{\psi(t)}{2}$.
Consequently,

$\Big{|}\angle_{r(t)}\big{(}r(0),p(t)\big{)}-\frac{\pi}{2}\Big{|}\leq\Big{|}\angle_{r(t)}\big{(}r(0),r^{\prime}(t)\big{)}-\frac{\pi}{2}\Big{|}+\Big{|}\angle_{r(t)}\big{(}r^{\prime}(t),p(t)\big{)}\Big{|}\leq\psi(t).\qed$

######  Corollary 10.5.

Let $r\colon[0,\infty)\to X$ be a geodesic ray in a complete $\mathrm{CAT}(0)$
space $X$. Then for any $\varepsilon>0,d>0$ and any non-increasing function
$\delta\colon[0,\infty)\to\mathbb{R}_{>0}$, there is $T>0$ such that

  * •

for any geodesic ray $r^{\prime}\colon[0,\infty)\to X$ asymptotic to $r$
satisfying

    * –

$d_{X}\big{(}r(0),r^{\prime}(0)\big{)}\leq d,$ and

    * –

$d_{X}\big{(}r(t),r^{\prime}[0,\infty)\big{)}\geq\delta(t)$ for each
$t\in[0,\infty),$ and

  * •

for any $t\in(T,\infty)$, for any
$0<\varepsilon^{\prime}<\min\\{\delta(t)^{4},d,\varepsilon\\}$, and for any
point $s\in X$ at distance $\leq\varepsilon^{\prime}$ from $r^{\prime}(t+4d)$
such that the projections $s(t),s(T)$ of $r(t),r(T)$ onto $sr^{\prime}(0)$ are
distinct,

we have that the Alexandrov angles of the geodesic quadrilateral
$s(t)s(T)r(T)r(t)$ are $\geq\frac{\pi}{2}$ at $s(t)$ and $s(T)$ and
$\geq\frac{\pi}{2}-9\sqrt{d}\sqrt[4]{\varepsilon}$ at $r(t)$ and $r(T)$.

In particular, by Theorem 6.9, for any reduced relative disc diagram
$\phi\colon D\to X$ with boundary path $s(t)s(T)r(T)r(t)$, the angles in $D$
at the preimages under $\phi$ of $s(t),s(T),r(T),r(t)$ are nearly
$\frac{\pi}{2}$.

Figure 25. Proof of Corollary 10.5. Projections are denoted by arrows.

###### Proof.

Choose $T$ so that the function $\psi$ from Lemma 10.4 satisfies
$\psi[T,\infty)<\sqrt{d}\sqrt[4]{\varepsilon}$ (this choice does not matter
until the last line of the proof).

Let $p(t),p^{\prime}(t)$ be the projections of $r(t),s(t)$ to $r^{\prime}$,
and let $d(t)=d_{X}\big{(}r(t),p(t)\big{)}$. See Figure 25. Since $p(t)$ is at
distance $\leq\varepsilon^{\prime}$ from a point on $r^{\prime}(0)s$, and
$s(t)$ is the projection of $r(t)$ onto $r^{\prime}(0)s$, we have
$d_{X}\big{(}r(t),s(t)\big{)}\leq d(t)+\varepsilon^{\prime}$. In particular,
we have $d_{X}\big{(}r(t),s(t)\big{)}<2d$, and so
$d_{X}\big{(}r^{\prime}(t),s(t)\big{)}<3d$, which implies that $s(t)$ is
distinct from $s$. Analogously, $s(T)$ is distinct from $s$. Similarly, taking
$T>3d$ we have that $s(t)$ and $s(T)$ are distinct from $r^{\prime}(0)$. Then
$\angle_{s(T)}\big{(}s(t),r(T)\big{)}\geq\frac{\pi}{2}$ and
$\angle_{s(t)}\big{(}s(T),r(t)\big{)}\geq\frac{\pi}{2}$.

Furthermore, we have $d_{X}\big{(}r(t),p^{\prime}(t)\big{)}\leq
d(t)+2\varepsilon^{\prime}$ and so, considering the triangle
$r(t)p(t)p^{\prime}(t)$, we obtain
$d_{X}\big{(}p(t),p^{\prime}(t)\big{)}\leq\sqrt{4d(t)\varepsilon^{\prime}+4\varepsilon^{\prime
2}}$. Using the assumptions $d(t)\leq d$ and $\varepsilon^{\prime}<d$, we get
$d_{X}\big{(}p(t),s(t)\big{)}\leq\varepsilon^{\prime}+\sqrt{4d\varepsilon^{\prime}+4\varepsilon^{\prime
2}}<4\sqrt{d\varepsilon^{\prime}}$. Analogously, the projection $p(T)$ of
$r(T)$ onto $r^{\prime}$ is at distance $<4\sqrt{d\varepsilon^{\prime}}$ from
$s(T)$. Consequently, by [BH]*II.1.7(5) we have that
$\sin\angle_{r(t)}\big{(}p(t),s(t)\big{)}$ and
$\sin\angle_{r(T)}\big{(}p(T),s(T)\big{)}$ are
$<4\sqrt{d\varepsilon^{\prime}}/\delta(t)$. Since $\theta\leq 2\sin\theta$ for
$\theta\in[0,\frac{\pi}{2}]$, and
$\sqrt{\varepsilon^{\prime}}/\delta(t)<\sqrt[4]{\varepsilon^{\prime}}$, these
Alexandrov angles are $<8\sqrt{d}\sqrt[4]{\varepsilon}$. By the definition of
$\psi$, the differences between $\frac{\pi}{2}$ and
$\angle_{r(t)}\big{(}p(t),r(T)\big{)}$ and
$\angle_{r(T)}\big{(}p(T),r(t)\big{)}$ are $\leq\sqrt{d}\sqrt[4]{\varepsilon}$
and the corollary follows. ∎

##  11\. Limit points that are not far

In this section we describe particular pairs of limit points that are not far.

###  11.1. Angles between principal (and antiprincipal) points in
$\partial_{\infty}\mathbf{X}$

We first prepare the proof of Lemma 9.6.

######  Lemma 11.1.

Let $\phi\colon D\to\nabla^{+}$ be an $\mathbf{X}$-reduced half-plane diagram
with $\partial D=I_{a}\cdot I_{b}\cdot I_{c}$, where $I_{a},I_{c}$ are
geodesic rays, and $I_{b}$ is a possibly trivial geodesic segment. Suppose
that the sum of the curvatures of vertices and edges in the interior of $D$
and in the interior of $I_{a},I_{b},I_{c}$ is
$\geq-\varepsilon\geq-\varepsilon_{0}$ from Corollary 7.6. Let $\theta$ be the
sum of the curvatures of $I_{a}\cap I_{b}$ and $I_{b}\cap I_{c}$.

Then for each compact subcomplex $F\subset D$, there is a compact subcomplex
$F\subseteq F^{\prime}\subset D$, such that for any point $y_{a}\in
I_{a}\setminus F^{\prime}$ and a unit vector $u_{a}$ tangent to $I_{a}$ at
$y_{a}$, there exists a point $y_{c}\in I_{c}\setminus F$ such that the unit
vector $u_{c}$ tangent to $I_{c}$ at $y_{c}$, directed towards the same end of
$\partial D$ as $u_{a}$, satisfies the following. Namely, there is an embedded
path of length in $[\theta-3\varepsilon,\theta+3\varepsilon]$ in the unit
circle $S$, whose endpoints project to
$\phi^{\angle}_{y_{a}}(u_{a}),\phi^{\angle}_{y_{c}}(u_{c})\in S^{\angle}$.

###### Proof.

Choose $F^{\prime}$ to be the minimal subcomplex of $D$ such that

  * •

$D\setminus F^{\prime}$ is simply connected, and

  * •

$F^{\prime}$ contains $F$, and

  * •

$F^{\prime}$ contains all the non-oriented components of the folding locus $L$
of $D$ with

    * –

both ends on $\partial D$, and

    * –

starting at $I_{b}$ or intersecting $F$.

Let $u_{a}$ be a unit vector tangent to $I_{a}$ at a point $y_{a}$ outside
$F^{\prime}$. Then there is a piecewise geodesic path $\beta$ in $D\setminus
D^{0}$ from $y_{a}$ to a point $y_{c}\in I_{c}$ that intersects each non-
oriented component $\mathcal{L}$ of $L$ at most once. Furthermore, by our
choice of $F^{\prime}$, we can assume that for each such $\mathcal{L}$, there
is a connected component of $\mathcal{L}\setminus\beta$ that does not end on
$\partial D$. Let $u_{c}$ be the unit vector tangent to $I_{c}$ at $y_{c}$,
directed towards the same end of $\partial D$ as $u_{a}$.

We consider the non-oriented edges $f$ of $L$ intersected by $\beta$ at points
$z$, and we denote by $\vec{f}$ a unit tangent vector to $f$ at $z$ that is
not directed towards $\partial D$. By Lemma 7.10, the total angle of all
$\phi_{z}^{\angle}(\vec{f})$ from the principal direction is
$\leq\varepsilon$.

Let $D_{\beta}$ be the disc diagram that is the bounded connected component of
$D\setminus\beta$. The sum of the curvatures outside $\beta$ in $D_{\beta}$
lies in $[\theta-\varepsilon,\theta]$. By Theorem 6.9, the sum of the
curvatures along $\beta$ (including its endpoints) lies in
$[2\pi-\theta,2\pi-\theta+\varepsilon]$. Thus, if we parallel transport
$u_{a}$ along $\beta$, we obtain a vector $u$ in the obvious extension of
$S_{y_{c}}$ to the unit circle, at angle from $u_{c}$ in
$[\theta-\varepsilon,\theta]$. Furthermore, $\phi^{\angle}_{y_{c}}(u)$ and
$\phi^{\angle}_{y_{a}}(u_{a})$ differ by an angle $\leq 2\varepsilon$ since
$\phi_{z}^{\angle}(\vec{f})$ were at total angle $\leq\varepsilon$ from
$\partial S^{\angle}$. ∎

###### Proof of Lemma 9.6.

Suppose $\angle(\xi,\eta)\neq\frac{2\pi}{3}$. Choose
$\varepsilon\leq\varepsilon_{0}$ from Corollary 7.6 such that the interval
$[\angle(\xi,\eta)-4\varepsilon,\angle(\xi,\eta)+4\varepsilon]$ does not
contain $0$ and does not contain $\frac{2\pi}{3}$. Choose $x$ so that
$\angle_{x}(\xi,\eta)>\angle(\xi,\eta)-\frac{\varepsilon}{3}$. Let
$\phi^{\prime}\colon D^{\prime}\to\nabla^{+}$ be a $(\xi,\eta)$-diagram for
that $x$.

Suppose first that we are in case (ii) of Construction 10.1, where
$D^{\prime}$ is isometric to the Euclidean half-plane in $\mathbb{R}^{2}$.
Consider a unit vector $u$ tangent to $\partial D^{\prime}$. Since $\xi,\eta$
are both principal, by Lemma 10.2 the function
$\phi^{\prime\angle}\colon\mathbb{S}\to S^{\angle}$ as in Definition 7.8
simultaneously assigns to $u$ the principal and the antiprincipal point,
contradiction.

Now assume that we are in case (i) of Construction 10.1. Let $q^{\prime}$ be
the image of $q$ in $D^{\prime}$. By Lemma 10.3, the sum of the curvatures of
vertices and edges in the interior of $D^{\prime}$ and in the interior of
$\partial D^{\prime}\setminus q^{\prime}$ is $\geq-\varepsilon$. Let
$J_{\xi},J_{\eta}\subset\partial D^{\prime}$ be the rays $J^{\prime}$ from
Lemma 10.2 applied to $\xi$ and $\eta$. Then the closure $F$ of $\partial
D^{\prime}\setminus(J_{\xi}\cup J_{\eta})$ is compact. Choose
$F^{\prime}\supseteq F$ provided by Lemma 11.1, and let $y_{a}\in
J_{\xi}\setminus F^{\prime}$ with $\phi_{y_{a}}^{\prime\angle}(u_{a})$ the
principal point. Let $u_{c}\in S_{y_{c}}$ be the vector guaranteed by Lemma
11.1. By Lemma 10.3, we have
$\theta\in[\pi-\angle(\xi,\eta),\pi-\angle(\xi,\eta)+\varepsilon]$, which is
at distance $>3\varepsilon$ from $\frac{\pi}{3}$ and from $\pi$. Since
$\phi_{y_{c}}^{\prime\angle}(u_{c})$ is the antiprincipal point, there is no
path in $S$ of length in $[\theta-3\varepsilon,\theta+3\varepsilon]$ whose
endpoints project to
$\phi^{\prime\angle}_{y_{a}}(u_{a}),\phi^{\prime\angle}_{y_{c}}(u_{c})$. This
contradicts the conclusion of Lemma 11.1.

Consequently, we have $\angle(\xi,\eta)=\frac{2\pi}{3}$. Let $\zeta$ be the
midpoint of the geodesic $\xi\eta$ in $\partial_{\infty}\mathbf{X}$. Let
$\mathbf{X}_{\xi},\mathbf{X}_{\eta}$ be the connected components of the
preimage of $\nabla^{+}\setminus\\{\alpha_{2}=\alpha_{3}\\}$ under $\rho_{+}$,
containing the interior of some (hence all by Lemma 8.2) principal ray
$r_{\xi},r_{\eta}\colon[0,\infty)\to\mathbf{X}$ representing $\xi,\eta$,
respectively. By [BH]*II.9.10 (applied with $\frac{a^{\prime}}{a}=2$), the
geodesic ray $r_{\zeta}$ representing $\zeta$ starting at $r_{\xi}(1)$
satisfies

$d_{\mathbf{X}}\big{(}r_{\xi}(2t),r_{\zeta}(t)\big{)}/t\nearrow\sqrt{3}.$

Consequently, $\rho_{+}(r_{\zeta})$ is disjoint from
$\\{\alpha_{2}=\alpha_{3}\\}$ and so $r_{\zeta}\subset\mathbf{X}_{\xi}$.
Furthermore, since $\sqrt{3}<2,$ for any compact subset $K\subset\nabla^{+}$
we have that $\rho_{+}\big{(}r_{\zeta}(t)\big{)}$ is disjoint from $K$ for $t$
sufficiently large. We have analogous properties for the geodesic ray
$r^{\prime}_{\zeta}$ representing $\zeta$ starting at $r_{\eta}(1)$.

Since $\xi\neq\eta$, we have that $\mathbf{X}_{\xi},\mathbf{X}_{\eta}$ are
disjoint. Since $r_{\zeta},r^{\prime}_{\zeta}$ are asymptotic, we have that
$\rho_{+}(r_{\zeta})$ is contained in the set $\\{\alpha_{2}\leq
m\alpha_{3}\\}$, for some $m\geq 2$. We set $K$ above to
$\\{\alpha_{1}=\alpha_{2}\leq m\alpha_{3}\\}$, and we conclude that a subray
$r_{\zeta}[T,\infty)$ of $r_{\zeta}$ is contained in the preimage
$\mathbf{X}^{\prime}_{\xi}\subset\mathbf{X}_{\xi}$ under $\rho_{+}$ of
$\\{\alpha_{3}<\alpha_{2}\leq m\alpha_{3}\\}\setminus K$. Each connected
component of $\mathbf{X}^{\prime}_{\xi}$ is quasi-isometric to the connected
component of the preimage under $\rho_{+}$ of
$\\{\alpha_{2}=m\alpha_{3}\\}\setminus K$, which is a tree with only one
infinite path and that path is an antiprincipal ray. Consequently,
$r_{\zeta}[T,\infty)$ is antiprincipal, and it lies in the same chamber as the
principal ray representing $\xi$ starting at $r_{\zeta}(T)$, as desired.

The last assertion of the lemma follows from interchanging in the proof the
words ‘principal’ and ‘antiprincipal’, defining $K=\\{\alpha_{1}\leq
m\alpha_{2}=m\alpha_{3}\\}$ and replacing $\\{\alpha_{3}<\alpha_{2}\leq
m\alpha_{3}\\}\setminus K$ by $\\{\alpha_{2}<\alpha_{1}\leq
m\alpha_{2}\\}\setminus K$. ∎

The following application of Lemma 9.6 produces distinct principal points in
$\partial_{\infty}\mathbf{X}$ at angle $\neq\frac{2\pi}{3}$ (and hence far).

######  Lemma 11.2.

Let $g\in C$ be parabolic and let $\zeta$ be the class of the principal rays
in $\mathbf{E}^{+}_{\mathrm{id}}$. Let $\eta\in\partial_{\infty}\mathbf{X}$ be
principal with $\angle(\zeta,\eta)=\frac{2\pi}{3}$. Then
$\angle\big{(}\eta,g(\eta)\big{)}\neq\frac{2\pi}{3}$ or
$\angle\big{(}\eta,g^{2}(\eta)\big{)}\neq\frac{2\pi}{3}$.

###### Proof.

Suppose by contradiction
$\angle\big{(}\eta,g(\eta)\big{)}=\angle\big{(}\eta,g^{2}(\eta)\big{)}=\frac{2\pi}{3}$.
Consider the unique geodesics
$\omega_{1},\omega_{2},\omega_{3}\subset\partial_{\infty}\mathbf{X}$ joining
$\zeta,\eta,$ and $g(\eta)$. The midpoints $\xi_{i}$ of $\omega_{i}$ are
antiprincipal by Lemma 9.6. If $\omega_{1}\cdot\omega_{2}$ is not a locally
embedded path, then $\angle(\xi_{1},\xi_{2})<\frac{2\pi}{3}$, and so by Lemma
9.6 we have $\xi_{1}=\xi_{2}=\xi_{3}$. Consequently, $g$ fixes the entire
geodesic $\zeta\xi_{1}$, which by Lemma 9.6 constitutes the boundary of a
chamber. Thus by Lemma 8.1, we have that $g$ lies in a conjugate of $C\cap
B^{\prime}=H$. By Proposition 3.4, $g$ is elliptic, which is a contradiction.

Since $\partial_{\infty}\mathbf{X}$ is $\mathrm{CAT}(1)$ and is locally an
$\mathbb{R}$-tree (Remark 9.7), we have that
$\omega=\omega_{1}\cdot\omega_{2}\cdot\omega_{3}$ is isometrically embedded
and thus by Lemma 6.7 there is a flat $F\subset\mathbf{X}$ with
$\partial_{\infty}F=\omega$. Since $g^{2}$ is also not elliptic, we have
analogously a flat $F^{\prime}$ with $\partial_{\infty}F^{\prime}$ containing
$\zeta,\eta,g^{2}(\eta)$.

We will now show that $F\cap F^{\prime}$ is nonempty. Choose $x\in
F,x^{\prime}\in F^{\prime}$ and let $a=d_{\mathbf{X}}(x,x^{\prime})$. Let
$r_{\zeta},r_{\eta}\colon[0,\infty)\to
F,r^{\prime}_{\zeta},r^{\prime}_{\eta}\colon[0,\infty)\to F^{\prime}$ be the
geodesic rays representing $\zeta,\eta$ starting at $x,x^{\prime}$. Let $y$ be
a point in the sector bounded by $r_{\zeta},r_{\eta}$ in $F$ and at distance
$>a$ from $r_{\zeta}\cup r_{\eta}$. We can assume that $y$ lies in the
interior of a $2$-cell. Let $R>d_{\mathbf{X}}(x,y)+a$, and let $\beta\subset
F$ be the arc of the circle centred at $x$ of radius $R$ from $r_{\zeta}(R)$
to $r_{\eta}(R)$. Let $\beta^{\prime}\subset F^{\prime}$ be the arc of the
circle centred at $x^{\prime}$ of radius $R$ from $r^{\prime}_{\zeta}(R)$ to
$r^{\prime}_{\eta}(R)$. Then the corresponding points of the closed paths
$\alpha=r_{\zeta}[0,R]\cdot\beta\cdot r_{\eta}[R,0]$ and
$\alpha^{\prime}=r^{\prime}_{\zeta}[0,R]\cdot\beta^{\prime}\cdot
r^{\prime}_{\eta}[R,0]$ are at distance $\leq a$, since they are points of the
corresponding asymptotic geodesic rays starting at $x,x^{\prime}$.
Consequently, $\alpha^{\prime}$ is homotopic to $\alpha$ in $\mathbf{X}-y$. By
projecting to a small metric ball around $y$ homeomorphic to a disc, we see
that $\alpha$ is homotopically nontrivial in $\mathbf{X}-y$. Thus the same is
true for $\alpha^{\prime}$, and so $y\in F^{\prime}$. Consequently, $F\cap
F^{\prime}$ contains the geodesic ray representing $\zeta$ starting at $y$.
Analogously, $F\cap g(F)$ and $F^{\prime}\cap g(F)$ contain geodesic rays
representing $\zeta$.

We will now prove that there is a geodesic ray $r$ representing $\zeta$ in
$F\cap F^{\prime}\cap g(F)$. For contradiction, assume that all of $V=F\cap
F^{\prime},V^{\prime}=F\cap g(F),V^{\prime\prime}=g(F)\cap F^{\prime}$ are
disjoint. Let $b_{\zeta}$ be a Busemann function at $\zeta$. Note that
$V,V^{\prime}\subset F$ are nonempty, closed, convex (Remark 6.6(i)) and
invariant under positive translation in the direction $\zeta$. Let
$c,c^{\prime}\colon\mathbb{R}\to F$ parametrise $\partial V,\partial
V^{\prime},$ in a way that for $t$ sufficiently large we have
$b_{\zeta}(c(t))=b_{\zeta}(c^{\prime}(t))=-t$. There is $t$ such that for
$T>t$ the curves $\partial V,\partial V^{\prime}$ are at $c(T),c^{\prime}(T)$
at Alexandrov angle $>\frac{\pi}{6}$ from the geodesic $c(T)c^{\prime}(T)$. We
can arrange that the same condition holds for
$c^{\prime}(T)c^{\prime\prime}(T)\subset gF$ joining $\partial V^{\prime}$
with $\partial V^{\prime\prime}$ and for $c^{\prime\prime}(T)c(T)\subset
F^{\prime}$ joining $\partial V^{\prime\prime}$ with $\partial V$. Then the
geodesic triangle $c(T)c^{\prime}(T)c^{\prime\prime}(T)$ has all Alexandrov
angles $>\frac{\pi}{3}$, which is a contradiction.

Consequently, there is a geodesic ray $r$ representing $\zeta$ in $F\cap
F^{\prime}\cap g(F)$. Note that $r$ lies in $\partial V,\partial V^{\prime},$
and $\partial V^{\prime\prime}$. Choose $y\in r$ and let $\gamma$ be the
geodesic $yg(y)$. Note that $\gamma\cap V^{\prime}=y$, since otherwise $g$
would fix the midpoint of $\gamma$ or would fix $g(\eta)$. See Figure 26.
Consequently, $g^{-1}(\gamma)\cdot\gamma$ is a local geodesic. Thus
$\bigcup_{k\in\mathbb{Z}}g^{k}(\gamma)$ is an axis of $g$, and so $g$ is
loxodromic, which is a contradiction. ∎

Figure 26.

###  11.2. Reaching far

######  Proposition 11.3.

Let $g,g^{\prime}\in\mathrm{Tame}(\mathbbm{k}^{3})$ be conjugate and parabolic
of nonzero translation length or loxodromic not of rank $1$. Suppose that for
principal points $\zeta,\zeta^{\prime}\in\partial_{\infty}\mathbf{X}$ fixed by
$g,g^{\prime},$ respectively (if these points exist), we have
$\zeta\neq\zeta^{\prime}$, and the same holds for $\zeta,\zeta^{\prime}$
antiprincipal. (Equivalently, $g,g^{\prime}$ do not belong to a common
conjugate of $C$ or $B^{\prime}$). Then any limit points $\xi$ of $g$ and
$\eta$ of $g^{\prime}$ are far.

###### Proof.

To begin with, we justify $\xi\neq\eta$. Otherwise, if, say, $g\in B^{\prime}$
and $\zeta\neq\zeta^{\prime}\in\partial_{\infty}\mathbf{X}$ are the
antiprincipal points fixed by $g,g^{\prime}$, then, by Construction 8.3(1), we
have length $\frac{\pi}{2}$ paths
$\omega,\omega^{\prime}\subset\partial_{\infty}\mathbf{X}$ between $\xi=\eta$
and $\zeta,\zeta^{\prime}$, respectively. By Lemma 9.6, we have that $\zeta$
is at distance $\frac{2\pi}{3}$ from $\zeta^{\prime}$. Since
$\partial_{\infty}\mathbf{X}$ is $\mathrm{CAT}(1)$ and is locally an
$\mathbb{R}$-tree (Remark 9.7), the geodesic $\zeta\zeta^{\prime}$ is
contained in the union of $\omega$ and $\omega^{\prime}$, and so in particular
the midpoint of $\zeta\zeta^{\prime}$, which is principal by Lemma 9.6, lies
in $\omega$ and $\omega^{\prime}$. Consequently, this midpoint is represented
by a geodesic ray in $\mathbf{H}$ from Construction 8.3. However, such a ray
does not satisfy Lemma 8.2, which is a contradiction.

The remaining proof is divided into five parts.

######  Part 1.

Defining $D^{\prime+}$ and $D^{\prime-}$.

Choose $x$ so that
$\angle_{x}(\xi,\eta)>\angle(\xi,\eta)-\frac{\varepsilon}{3}$ for
$\varepsilon<\varepsilon_{0}$ from Corollary 7.6. Let $\phi^{\prime+}\colon
D^{\prime+}\to\nabla^{+}$ be a $(\xi,\eta)$-diagram from Construction 10.1.

Denote by $q^{\prime}\in\partial D^{\prime+}$ the image of $q\in\partial
D^{+}$, and by $J^{+}_{\xi},J^{+}_{\eta}$ the closures of the appropriate
connected components of $\partial D^{\prime+}\setminus q^{\prime}$. Recall
that $D^{\prime+}$ is $\mathrm{CAT}(0)$, and $J^{+}_{\xi},J^{+}_{\eta}$ are
geodesic rays in $D^{\prime+}$ representing points in
$\partial_{\infty}D^{\prime+}$ at angle $\angle(\xi,\eta)$. Let
$\beta^{+}_{n}=\widetilde{\phi}^{+}_{n|J^{+}_{\xi}\cap I_{n}}$.

Let $r$ representing $\xi$ be an $\mathbf{H}$-generic ray guaranteed by Remark
8.7(i). Let $r^{\prime}_{n}$ representing $\xi$ be starting at $p_{n}$ (in
case (ii) of Construction 10.1) or $x$ (in case (i)). Suppose first that:

($\heartsuit$) for each $t$ there is $\delta(t)>0$ such that each
$r^{\prime}_{n}$ it at distance $\geq\delta(t)$ from $r(t)$.

Let $d$ be the maximal distance from $r(0)$ to $x$ or a point of $P$. For
$\varepsilon$ above replaced by $\varepsilon^{4}/18^{4}d^{2}$, let $T>0$ be
provided by Corollary 10.5, which applies to all $r^{\prime}_{n}$. Then for
each $t>T,$ for sufficiently small $\varepsilon^{\prime}>0$ there is $n$ with
$s$ in $\beta^{+}_{n}$ satisfying the second bullet point of Corollary 10.5.
By Lemma 6.3, we have reduced relative disc diagrams
$\widetilde{\phi}^{-}_{n}\colon D^{-}_{n}\to\mathbf{X}$ (embedded by Theorem
6.10) bounded by geodesic quadrilaterals $r(t)r(T)s(T)s(t)$ with total
curvature $\geq-\varepsilon$ outside the preimages under
$\widetilde{\phi}^{-}_{n}$ of $s(t),s(T),r(T),r(t)$. See Figure 27.

Figure 27. The images of $D^{\pm}_{n}$.

Let $I^{-}_{n}\subset\partial D^{-}_{n}$ be the union of the three sides not
mapping under $\widetilde{\phi}^{-}_{n}$ to $r(t)s(t)$. By passing to a
subsequence and subdiagrams, we can assume that $(D^{-}_{n},I^{-}_{n})$
converge to a half-plane diagram $(D^{-},\partial D^{-})$ with $\partial
D^{-}=J\cdot I_{b}\cdot I_{c}$, where the intersections of $J,I_{b},I_{c}$
with $\partial D^{-}_{n}$ map under $\widetilde{\phi}^{-}_{n}$ into
$r(t)r(T),r(T)s(T),s(T)s(t)$, respectively. Since $r(T)s(T)$ has uniformly
bounded length, the first part of the second bullet point in Remark 7.1
implies the second part. Thus $D^{-}$ has the frilling $D^{\prime-}$ with the
$\mathbf{X}$-reduced relative half-plane diagram $\phi^{\prime-}\colon
D^{\prime-}\to\nabla^{+}$ that is the limit of
$\phi^{-}_{n}=\rho_{+}\circ\widetilde{\phi}^{-}_{n}$ bordered by
$r_{0}=r_{|[T,\infty)}$, where we identify $[T,\infty)$ with $J\subset\partial
D^{\prime-}$.

######  Part 2.

Calculating the angle.

By Corollary 10.5 and Theorem 6.9, the angles of $J\cap I_{b},I_{b}\cap J_{c}$
are at distance $\leq\varepsilon$ from $\frac{\pi}{2}$. We now apply three
times Lemma 11.1, to $D^{\prime-},D^{\prime+}$ (using Lemma 10.3), and to the
analogue $D^{\prime-}(\eta)$ of $D^{\prime-}$ with $\xi$ replaced by $\eta$.
For $\varepsilon$ sufficiently small, we obtain vectors $u,u(\eta)$ tangent to
$J,J(\eta)$ at points $y,y(\eta)$, with
$\phi^{\prime-\angle}_{y}(u),\phi^{\prime-\angle}_{y}(u)(\eta)$ that are
projections of the endpoints of an embedded path in $S$ of length arbitrarily
close to $\pi-\angle(\xi,\eta)$. Furthermore, we can arrange that $y,y(\eta)$
are in arbitrary subrays of $J,J(\eta)$. Since $r_{0},r_{0}(\eta)$ are
$\mathbf{H}$-generic, this implies that
$\phi^{\prime-\angle}_{y}(u),\phi^{\prime-\angle}_{y}(u)(\eta)$ are
arbitrarily close to the midpoint of $S^{\angle}$. It follows that
$\angle(\xi,\eta)=\frac{\pi}{3},\frac{2\pi}{3},$ or $\pi$.

######  Part 3.

Curvature in $Q\subset\widehat{D}^{\prime}$.

Let $\widehat{\phi}_{n}\colon\widehat{D}_{n}\to\mathbf{X}$ be the unions of
$\widetilde{\phi}_{n}^{\pm}$ at $s(T)s(t)$, to which we apply Remark 7.3
describing their limit
$\widehat{\phi}^{\prime}\colon\widehat{D}^{\prime}\to\nabla^{+}$. In the case
where $\overline{D^{\prime-}\setminus R^{-}}$ is not homeomorphic to a half-
plane, we use the more general Remark 7.4. This covers the case of empty
$\overline{D^{\prime-}\setminus R^{-}}$, where
$\widehat{D}^{\prime}=\overline{D^{\prime+}\setminus R^{+}}$. After possibly
increasing $T$, we can assume that $R^{+}$ is disjoint from $J^{+}_{\eta}$,
and so $J^{+}_{\eta}\subset\widehat{D}^{\prime}$. Note that
$\widehat{\phi}^{\prime}$ is bordered by $r_{0}$. Observe that a geodesic ray
in $D^{\prime+}\setminus R^{+}$ asymptotic to $J^{+}_{\xi}$ in $D^{\prime+}$
is also asymptotic to $J$ in $\widehat{D}^{\prime}$. In particular, the rays
$J$ and $J^{+}_{\eta}$ represent points in
$\partial_{\infty}\widehat{D}^{\prime}$ at angle $\angle(\xi,\eta)$.

At this point, let us note that if ($\heartsuit$ ‣ 11.2) was not satisfied,
and so there is $T$ such that a subsequence of $r^{\prime}_{n}$ becomes
arbitrarily close to $r(T)$, then there are subrays of $r^{\prime}_{n}$
converging to $r_{0}=r_{|[T,\infty)}$. This implies that $\beta^{+}_{n}$ have
subpaths converging to $r_{0}$ and so $\phi^{\prime+}$ is weakly bordered by
$r_{0}$ (see Definition 7.2). After modifying $\phi^{+}_{n}$ on the cells
intersecting $J$, we have that the same $\phi^{\prime+}$ is bordered by
$r_{0}$, and we set $\widehat{\phi}^{\prime}=\phi^{\prime+}$ in the remaining
discussion (where the following claim is trivial).

We claim that the sum of the curvatures of the vertices and edges of
$\widehat{D}^{\prime}$ is finite. Indeed, if $\angle(\xi,\eta)<\pi$, then by
[BH]*II.9.8(4), the geodesics in $\widehat{D}^{\prime}$ joining the points on
$J$ and $J^{+}_{\eta}$ sufficiently far from a basepoint are disjoint from any
given compact subcomplex of $\widehat{D}^{\prime}$. Consequently, by Theorem
6.9, there is a uniform bound on the total curvature in any compact subcomplex
of $\widehat{D}^{\prime}$, as desired. For $\angle(\xi,\eta)=\pi$, one repeats
this argument after subdividing $\widehat{D}^{\prime}$ into two half-plane
diagrams along a ray representing at point in
$\partial_{\infty}\widehat{D}^{\prime}$ at distance $\frac{\pi}{2}$ from the
points represented by $J$ and $J^{+}_{\eta}$. This justifies the claim.

By the claim, there exists a compact subcomplex
$F\subset\widehat{D}^{\prime}$, such that the sum of the curvatures in
$\widehat{D}^{\prime}$ of the vertices and edges outside $F$ is
$\geq-\varepsilon_{0}$. Let $l_{k}\in J$ be the sequence of transition points
of $r_{0}$. There is $k_{0}$ such that the unbounded connected component of
$J\setminus l_{k_{0}}$ is disjoint from $F$. Applying Lemma 8.8 to a
neighbourhood of that component in $\widehat{D}^{\prime}$, we obtain that for
$k\geq k_{0}$ the points $l_{k}$ are contained in single edges of the folding
locus $\widehat{L}$ and at angle to $J$ in
$\big{(}\frac{\pi}{2}-\varepsilon_{0},\frac{\pi}{2}+\varepsilon_{0}\big{)}$
and converging to $\frac{\pi}{2}$.

After possibly increasing $k_{0}$, a geodesic $\gamma$ in $\widehat{D}$,
issuing from $l_{k_{0}}$ at angle $\frac{\pi}{2}+\varepsilon_{0}$ to the
unbounded connected component of $J\setminus l_{k_{0}}$, separates $F$ from
$l_{k_{0}+1}$. Cutting $\widehat{D}^{\prime}$ along $\gamma$, we obtain a
half-plane diagram $Q$ containing $l_{k_{0}+1}$, where the sum of all
curvatures, except the curvature of $l_{k_{0}}$, and possibly except of the
curvature of other endpoint of $\gamma$, if $\gamma$ is bounded, is
$\geq-\varepsilon_{0}$.

######  Part 4.

Folding locus.

Let $L_{k}\subset\widehat{L}$ be the connected component of $\widehat{L}$
minus the degree $>2$ vertices of $\widehat{L}$, containing $l_{k}$. For each
$k\geq k_{0}$, we have that $L_{k}\cap Q$ are geodesics in $Q$, as are $J\cap
Q$ and $\gamma$. Considering the angles that $L_{k}$ and $\gamma$ make with
$J$, we obtain that if $L_{k}\cap Q$ is bounded, then it does not end in
$J\cup\gamma$. In particular, $L_{k}\cap Q=L_{k}$.

If $L_{k}$ is contained in a non-oriented component $\mathcal{L}$ of
$\widehat{L}$, then analogously $\mathcal{L}\cap Q$ does not end in
$J\cup\gamma$, and so $\mathcal{L}\cap Q=\mathcal{L}$. Furthermore, since $J$
is not asymptotic to $J^{+}_{\eta}$, after possibly increasing $k_{0}$, we
have that if $\mathcal{L}$ has both ends in $\partial Q$, then it has
arbitrarily large length. Thus, by Lemma 7.10, for each $x\in L_{k}$ and
$\vec{f}\in S_{x}$ the direction of an edge of $L_{k}$, directed away from
$l_{k}$, we have that $\widehat{\phi}^{\prime\angle}_{x}(\vec{f})\in
S^{\angle}$ is at angle $\leq\varepsilon_{0}$ from a point in $\partial
S^{\angle}$ (same point for all $k$). The same would hold for other edges of
the folding locus starting at $J$ sufficiently far from $\gamma$, which is
excluded by the last assertion of Lemma 8.8. Consequently, after possibly
increasing $k_{0}$, we have $\widehat{L}\cap J\cap Q=\\{l_{k}\\}_{k\geq
k_{0}}$.

Figure 28.  Figure 29.

Suppose first that all $\widehat{\phi}^{\prime\angle}_{x}(\vec{f})\in
S^{\angle}$ above are at angle $\leq\varepsilon_{0}$ from the antiprincipal
point. If $L_{k}$ is bounded and ends with an interior vertex $v_{k}$ of $Q$,
then by Corollary 7.6, we have one of the configurations in Figure 28, where
the dashed edges might be oriented as indicated or non-oriented. However,
configurations (e,f) can be excluded in the following way. If the non-oriented
component $\alpha$ intersects $\partial Q$, then this contradicts the previous
paragraph. If $\alpha$ is disjoint from $\partial Q$, then cutting $\alpha$
into two and applying Lemma 7.10, we see that $\alpha$ contributes at least
$\frac{2\pi}{3}$ to the curvature of $Q$, which is a contradiction.
Analogously, the edges in configuration (d) must be oriented.

We call $v_{k}$ as in configuration (c) _descending_ , in (d)  _ascending_ ,
and in (a,b)  _minimal_. Let $\alpha^{\prime}$ be the connected component of
$\widehat{L}$ minus the degree $>2$ vertices of $\widehat{L}$, containing the
north-west edge in (a,b,c) or the south-west edge in (d). If $\alpha^{\prime}$
is non-oriented (which might happen only for $v_{k}$ descending), and ends at
$v$ in Figure 17(h), then we call $v_{k}$ _wicked_.

Suppose that $v_{k}$ is not wicked. Since $J$ and $J^{+}_{\eta}$ represent
points at $\partial_{\infty}\widehat{D}^{\prime}$ at angle
$\geq\frac{\pi}{3}>\frac{\pi}{6}$, we have that $\alpha^{\prime}$ does not end
on $\partial Q$. Consequently, $L_{k+1}$ also ends with an interior vertex
$v_{k+1}$ of $Q$. Moreover, applying Corollary 7.6 to the endpoint of
$\alpha^{\prime}$ distinct from $v_{k}$, and recalling that the edges in
configuration (d) must be oriented, we conclude that this endpoint coincides
with $v_{k+1}$, and $v_{k+1},v_{k}$ are either

  * •

ascending, ascending (see Figure 29), or

  * •

minimal, ascending (see Figure 29 for the case that uses the minimal
configuration from Figure 28(a)), or, symmetrically,

  * •

descending, descending, or

  * •

descending, minimal.

Note that for $v_{k}$ wicked we could have other combinations. However, all
$v_{k^{\prime}}$ with $k>k^{\prime}\geq k_{0}$ are then descending and not
wicked. Consequently, after possibly replacing $k_{0}$ with $k+1$, we can
assume that none of $v_{k}$ are wicked.

If the sequence $v_{k+2},v_{k+1},v_{k}$ was descending, minimal, ascending, we
would obtain a contradiction with Remark 7.7(ii) applied to the north-east
edge in Figure 28(c) and the north-west edge in Figure 28(d). Furthermore, we
cannot have an infinite sequence of ascending vertices. Consequently, after
possibly increasing $k_{0}$, all $v_{k}$ are descending, see Figure 30, left.
The remaining possibility is that all $L_{k}$ are connected components of
$\widehat{L}$, where $L_{k}$ might be unbounded (see Figure 30, centre), or
bounded (possibly non-oriented) if they end on $\partial Q$ (see Figure 30,
right).

Figure 30.  Figure 31.

Second, suppose that all $\widehat{\phi}^{\prime\angle}_{x}(\vec{f})\in
S^{\angle}$ are at angle $\leq\varepsilon_{0}$ from the principal point. If
$L_{k}$ is bounded and ends with an interior vertex $v_{k}$ of $Q$, then we
have one of the configurations in Figure 31, where the dashed edges in (a)
might be oriented as indicated or non-oriented. In (a,b,c), the oriented
geodesic $\alpha$ starting at $v_{k}$, given by Remark 7.7(ii), terminates at
$J$ at angle contradicting $\widehat{L}\cap J\cap Q=\\{l_{k}\\}_{k\geq
k_{0}}$. In (d), the non-oriented component $\alpha^{\prime}$ is disjoint from
$\partial Q$, since $J$ and $J^{+}_{\eta}$ represent points at
$\partial_{\infty}\widehat{D}^{\prime}$ at angle $>\frac{\pi}{6}$. Cutting
$\alpha^{\prime}$ into two and applying Lemma 7.10, we see that
$\alpha^{\prime}$ contributes at least $\frac{2\pi}{3}$ to the curvature of
$Q$, which is a contradiction. It follows that all $v_{k}$ are as in (e), see
Figure 32, left, or all $L_{k}$ are connected components of $\widehat{L}$, see
Figure 32, centre and right (where some $L_{k}$ are possibly non-oriented).

Figure 32.

######  Part 5.

Interaction.

To summarise all the possibilities for both $\xi$ and $\eta$, consider first
the situation where there is $k_{0}$ such that for all $k\geq k_{0}$ the
$L_{k}$, denoted from now on by $L_{k}(\xi)$, are unbounded. Since they are at
angle to $J$ converging to $\frac{\pi}{2}$ as $k\to\infty$, they are geodesic
rays representing points in $\partial_{\infty}\widehat{D}^{\prime}$ at
distance $\geq\frac{\pi}{2}$ from the point represented by $J$. Suppose that
after replacing $\xi$ by $\eta$ we reach the same conclusion for
$L_{k}(\eta)$. Then the subrays of $L_{k}(\xi)$ and $L_{k}(\eta)$ lying in
$D^{\prime+}$ are disjoint. Since they represent points in
$\partial_{\infty}D^{\prime+}$ at distance $\geq\frac{\pi}{2}$ from
$J^{+}_{\xi},J^{+}_{\eta}$, they have to be asymptotic. Proposition 8.9(1),
applied twice, implies that $g,g^{\prime}$ fix a common principal or
antiprincipal point in $\partial_{\infty}\mathbf{X}$, which is a
contradiction.

Second, suppose that $L_{k}(\xi)$ are bounded but end on
$\partial\widehat{D}^{\prime}$. If $L_{k}(\eta)$ are unbounded, then subrays
of $L_{k}(\eta)$ and $L_{k}(\xi)$ lying in $D^{\prime+}$ intersect, which is a
contradiction. If $L_{k}(\eta)$ are bounded connected components of the
appropriate folding locus, then, after renumbering, subrays of $L_{k}(\xi)$
and $L_{k}(\eta)$ lying in $D^{\prime+}$ coincide. This implies that
$J^{+}_{\xi}$ and $J^{+}_{\eta}$ are asymptotic, which is a contradiction.

Finally, suppose that we have the configuration from Figure 30, left. Consider
the north-east oriented geodesics $L^{\prime}_{k}\subset\widehat{L}$ starting
at $v_{k}$. In particular, if $L_{k}(\eta)$ are unbounded, then subrays of
$L_{k}(\eta)$ and $L^{\prime}_{k}$ lying in $D^{\prime+}$ intersect, which is
a contradiction. Otherwise, we have that subsegments of $L_{k}(\eta)$ coincide
with subsegments of $L^{\prime}_{k}$, since otherwise they would intersect in
$D^{\prime+}$. See Figure 33, left. The analogous remaining possibility is
described in Figure 33, right.

Figure 33.

In both of these configurations, Let $O(\xi)$ be the half-plane diagram in
$\widehat{D}^{\prime}$ bounded by $L_{k_{0}}(\xi)$, and the ray
$\alpha\subset\widehat{L}$ containing all $v_{k}$ for $k\geq k_{0}$, and let
$O(\eta)$ be defined analogously. Applying Proposition 8.9(2) to $O(\xi)$ and
$O(\eta)$, we find $k$ such that the unique antiprincipal or principal ray
starting at $v_{k}$ represents an antiprincipal or principal point in
$\partial_{\infty}\mathbf{X}$ fixed by both $g$ and $g^{\prime}$, which is a
contradiction. ∎

##  12\. Proof of the Tits alternative

We start with the following preparatory lemmas. In the applications of the
first one, the set $\partial$ will be the visual boundary
$\partial_{\infty}\mathbf{X}$, which explains the notation.

######  Lemma 12.1.

Let $G$ be a group acting on a set $\partial$. Let $Z$ be a finite subset of
$\partial$. Suppose that for each $u\in G$, the sets $Z$ and $u(Z)$ intersect.
Then there is a nonempty finite subset of the union of $u(Z)$ over $u\in G$
that is invariant under $G$.

Let $|Z|$ denote the number of elements of $Z$. Before we start the proof,
note that for $|Z|=2$, the sets $u(Z)$ have either an element in common or
take on only three values $\\{\xi,\eta\\},\\{\eta,\zeta\\},\\{\zeta,\xi\\}$.
In particular, the invariant nonempty finite set can be taken as
$\bigcap_{u\in G}u(Z)$ or $\bigcup_{u\in G}u(Z)$. For $|Z|=3$, this is no
longer true. Namely, consider the 2-dimensional simplicial complex $Y$ built
of a triangle with vertices $\xi,\eta,\zeta$, and infinitely many triangles
$T_{i}^{\zeta}$ (resp. $T_{i}^{\eta},T_{i}^{\xi}$) sharing the edge $\xi\eta$
(resp. $\xi\zeta,\eta\zeta$). Let $G$ be the automorphism group of $Y$, let
$\partial$ be the vertex set of $Y$ and let $Z$ be the vertex set of
$T_{1}^{\zeta}$. Then $\bigcap_{u\in G}u(Z)=\emptyset,\bigcup_{u\in
G}u(Z)=\partial$, and the required $G$-invariant finite set is
$\\{\xi,\eta,\zeta\\}$.

We will be applying Lemma 12.1 only with $|Z|\leq 4$.

###### Proof of Lemma 12.1.

We prove the lemma by induction on $|Z|$. If $|Z|=1$, then $Z$ is invariant
under $G$. If $|Z|>1$, then let $Z^{\prime}\subsetneq Z$ be a maximal subset
contained in infinitely many distinct translates $u(Z)$, where $u\in G$. If
$Z^{\prime}$ is empty, then the union of all $u(Z)$ over $u\in G$ is finite
and constitutes the desired finite set invariant under $G$. If $Z^{\prime}$ is
nonempty, then we claim that for each $u\in G$ the set $u(Z^{\prime})$
intersects $Z^{\prime}$, which allows to appeal to the induction and finishes
the proof.

To justify the claim, observe that by the maximality of $Z^{\prime}$, there is
a sequence $(u_{i})_{i\geq 1}$ in $G$ such that the sets $u_{i}(Z)\setminus
Z^{\prime}$ are pairwise disjoint. Consequently, the sets $uu_{i}(Z)\setminus
u(Z^{\prime})$ are pairwise disjoint. In particular, we can fix $j$ such that
$uu_{j}(Z)\setminus u(Z^{\prime})$ is disjoint from $Z^{\prime}$. Since
$uu_{j}(Z)$ intersects all $u_{i}(Z)$, and all the sets $u_{i}(Z)\setminus
Z^{\prime}$ are pairwise disjoint, we have that $uu_{j}(Z)$ intersects
$Z^{\prime}$. By the choice of $j$, we have that $u(Z^{\prime})$ intersects
$Z^{\prime}$, justifying the claim. ∎

######  Lemma 12.2.

Let $G$ be a group and let $F$ be a finite group. Suppose that for each
finitely generated subgroup $H<G$ we have a nonempty family $\mathcal{F}_{H}$
of homomorphisms from $H$ to $F$ satisfying the following. For all finitely
generated subgroups $H<H^{\prime}$ of $G$, and each
$f\in\mathcal{F}_{H^{\prime}}$, we have $f_{|H}\in\mathcal{F}_{H}$. Then there
is a homomorphism $f\colon G\to F$ such that for each finitely generated $H<G$
we have $f_{|H}\in\mathcal{F}_{H}$.

###### Proof.

The following proof was found by Marcin Sabok. Let $K=F^{G}$ be the compact
space of all the functions from $G$ to $F$. For each finitely generated
subgroup $H<G$, let $K_{H}$ be the nonempty subset of $K$ consisting of all
$f\in K$ satisfying $f_{|H}\in\mathcal{F}_{H}$. Since $F$ is finite and $H$ is
finitely generated, we have that $\mathcal{F}_{H}$ is finite, and so $K_{H}$
is closed. Note that the family $\\{K_{H}\\}$ has the finite intersection
property. Indeed, for finitely generated subgroups $H_{1},\ldots,H_{k}$ of
$G$, the group $H=\langle H_{1},\ldots,H_{k}\rangle$ is also finitely
generated, and so any element of $K_{H}$ belongs to all $K_{H_{i}}$. Since $K$
is compact, there is $f\in K$ in the intersection of $\\{K_{H}\\}$, as
desired. ∎

######  Corollary 12.3.

Let $n,m>0$ and let $G$ be a group. Suppose that each finitely generated
subgroup of $G$ contains a non-abelian free group or a solvable subgroup of
index $\leq n$ and step $\leq m$. Then $G$ satisfies the strong Tits
alternative.

###### Proof.

Suppose that $G$ does not contain a non-abelian free group. Let $F$ be the
symmetric group on $n$ elements. For each finitely generated subgroup $H<G$,
let $\mathcal{F}_{H}$ be the nonempty family of homomorphisms from $H$ to $F$
whose kernel is solvable of step $\leq m$. By Lemma 12.2, there is a
homomorphism $f\colon G\to F$ such that for each finitely generated $H<G$ we
have $f_{|H}\in\mathcal{F}_{H}$. Let $G^{\prime}$ be the kernel of $f$. Thus
each finitely generated subgroup of $G^{\prime}$ is solvable of step $\leq m$.
Consequently, $G^{\prime}$ is solvable of step $\leq m$. ∎

###### Proof of the Main Theorem..

Let $G<\mathrm{Tame}(\mathbbm{k}^{3})$. We distinguish three cases according
to the type of elements in $G$ with respect to the action of
$\mathrm{Tame}(\mathbbm{k}^{3})$ on $\mathbf{X}$.

Case 1. All elements of $G$ are elliptic.

Consider any finitely generated subgroup $H$ of $G$. By [NOP]*Cor 2.6, the
group $H$ has a global fixed point in $\mathbf{X}$. By Remark 3.6(i,ii), the
group $H$ contains a non-abelian free group or a solvable subgroup of
uniformly bounded step and index. Hence $G$ satisfies the strong Tits
alternative by Corollary 12.3.

Case 2. $G$ contains a parabolic element $g$ or a loxodromic element $g$ that
is not of rank $1$.

By Remark 3.6(iii) and Lemma 8.1, we can suppose that $G$ is not _elementary_
in the sense that it is not virtually contained in the stabiliser of a
principal or antiprincipal point in $\partial_{\infty}\mathbf{X}$. In other
words, there is no finite subset of the union of the principal and
antiprincipal points in $\partial_{\infty}\mathbf{X}$ that is invariant under
$G$.

Suppose first that $g$ has zero translation length. Let $\zeta$ be the unique
principal point in $\partial_{\infty}\mathbf{X}$ fixed by $g$ guaranteed by
Theorem 9.1. Since $G$ is not elementary, there is $h\in G$ with
$h(\zeta)\neq\zeta$. If
$\angle\big{(}\zeta,h(\zeta)\big{)}\neq\frac{2\pi}{3}$, then $\zeta$ and
$h(\zeta)$ are far by Lemma 9.6. We set $h^{\prime}=h$ in that case. If
$\angle\big{(}\zeta,h(\zeta)\big{)}=\frac{2\pi}{3}$, then by Lemma 11.2, we
have $\angle\big{(}h(\zeta),gh(\zeta)\big{)}\neq\frac{2\pi}{3}$ or
$\angle\big{(}h(\zeta),g^{2}h(\zeta)\big{)}\neq\frac{2\pi}{3}$. Furthermore,
$h(\zeta)\neq gh(\zeta),g^{2}h(\zeta)$ by the uniqueness in Theorem 9.1
applied to $g$ and $g^{2}$. Hence for $h^{\prime}=h^{-1}gh$ or
$h^{\prime}=h^{-1}g^{2}h$, we have that $\zeta$ and $h^{\prime}(\zeta)$ are
far by Lemma 9.6.

Let $g^{\prime}=h^{\prime}gh^{\prime-1}$. By Proposition 2.9 and Proposition
9.2, for sufficiently large $n$ we have that $f_{n}=(g^{\prime})^{n}g^{-n}$
has far limit points, in arbitrary neighbourhoods of
$\zeta,h^{\prime}(\zeta)$. Since $G$ is not elementary, by Lemma 12.1 there is
$u\in G$ such that $Z=\\{\zeta,h^{\prime}(\zeta)\\}$ is disjoint from
$u(Z)=\\{u(\zeta),uh^{\prime}(\zeta)\\}$. In particular, for $n$ sufficiently
large, $f_{n}$ and $uf_{n}u^{-1}$ have distinct limit points. Thus by
Corollary 2.7, there are powers of $f_{n}$ and $uf_{n}u^{-1}$ that generate a
non-abelian free group.

Second, suppose that $g$ has nonzero translation length. By Theorems 9.1 and
9.4, there is a unique principal point $\zeta$ and/or a unique antiprincipal
point $\zeta^{\prime}$ fixed by $g$ in $\partial_{\infty}\mathbf{X}$. Let
$Z\subset\partial_{\infty}\mathbf{X}$ be $\\{\zeta\\},\\{\zeta^{\prime}\\}$
(if only one of the two points exists), or $\\{\zeta,\zeta^{\prime}\\}$. Since
$G$ is not elementary, by Lemma 12.1 there is $h\in G$ with $Z\cap
h(Z)=\emptyset$. By Proposition 11.3, the limit points $\xi,\eta$ of $g$ and
$h(\xi),h(\eta)$ of $g^{\prime}=hgh^{-1}$ are far. Again, by Proposition 2.9
and Proposition 9.2, for sufficiently large $n$ we have that
$f_{n}=(g^{\prime})^{n}g^{-n}$ has far limit points, in arbitrary
neighbourhoods of $\xi,h(\xi)$. Since $G$ is not elementary, applying Lemma
12.1 to the set $Z\cup h(Z)$ in the place of $Z$, there is $u\in G$ such that
$\zeta,\zeta^{\prime},h(\zeta),h(\zeta^{\prime})$ are distinct from
$u(\zeta),u(\zeta^{\prime}),uh(\zeta),uh(\zeta^{\prime})$. By Proposition
11.3, the points $\xi,h(\xi),u(\xi),uh(\xi)$ are distinct. Thus for $n$
sufficiently large, $f_{n}$ and $uf_{n}u^{-1}$ have distinct limit points, and
we conclude as before.

Case 3. $G$ contains a loxodromic element $g$, and all its elements are
loxodromic of rank $1$ or elliptic.

If a finite index subgroup $G^{\prime}$ of $G$ fixes one of the two limit
points $\xi\in\partial_{\infty}\mathbf{X}$ of $g$, then let $b_{\xi}$ be a
Busemann function for $\xi$. The group $G^{\prime}$ acts on the level sets of
$b_{\xi}$ giving rise to a homomorphism $G^{\prime}\to\mathbb{R}$, whose
kernel $H$ preserves the level sets. By Case 1, we can assume that there is
$h\in H$ that is not elliptic. Thus $h$ is loxodromic. Let
$\sigma\colon\mathbb{R}\to\mathbf{X}$ be an axis of $h$. Since $h\in H$, the
axis $\sigma$ is contained in a level set, say $b_{\xi}=0$. For each
$s\in\mathbb{R},t\geq 0$, let $\sigma(s,t)$ be the projection of $\sigma(s)$
onto the horoball $b_{\xi}\leq-t$. Thus for each
$s,s^{\prime}\in\mathbb{R},t\geq 0$, we have
$d_{\mathbf{X}}\big{(}\sigma(s,t),\sigma(s^{\prime},t)\big{)}\leq
d_{\mathbf{X}}\big{(}\sigma(s),\sigma(s^{\prime})\big{)}$. On the other hand,
since $h$ acts on the curve $\sigma(\cdot,t)$ as a translation, and since
$\sigma$ was an axis of $h$, we have a converse inequality, and so
$d_{\mathbf{X}}\big{(}\sigma(s,t),\sigma(s^{\prime},t)\big{)}=d_{\mathbf{X}}\big{(}\sigma(s),\sigma(s^{\prime})\big{)}$.
Consequently, $\sigma$ bounds an isometrically embedded Euclidean half-plane
$\sigma(\cdot,\cdot)$, and so $h$ is not of rank $1$, which is a
contradiction.

Thus, by Lemma 12.1, there is $u\in G$ such that $g$ and $ugu^{-1}$ have
distinct limit points, and we conclude as before. ∎

## References