Proof. If \(TK=Y\) with \(K\) compact, take \(R=\max_{p\in K}d(y,p)\). Every \(tp\) is within \(R\) of \(ty\). For \(L=d(sy,ty)\), divide the geodesic into \(N=\max(1,\lceil L\rceil)\le L+1\) pieces, each of length at most one. Choose nearby orbit points at the division points, keeping the prescribed endpoint labels \(s,t\). Each successive group increment \(q\) satisfies \(d(y,qy)\le2R+1\). There are finitely many such \(q\): the compact ball \(\overline B(y,2R+1)\) contains both \(y\) and \(qy\), so properness applies. The largest word length of these increments, enlarged to at least one, can be used for \(C\). Summing their word lengths proves (19). ◻

Proof. Two incident side reflections generate the reflection group of three lines spaced at angles \(\pi/3\) about their intersection. Their two length-three braid words have the same linear part. The translation part of \(uvu\) is \[e_{D_u}+e_{r_uD_v}+e_{r_ur_vD_u} =e_{D_u}+e_{D_{uv}}+e_{D_v}.\] For \(vuv\), the three terms occur in the reverse order. Thus the translation parts also agree, proving the affine braid relations.

Each generator or inverse moves \(0\) a vector of norm one, so a word of length \(k\) moves \(0\) a distance at most \(k\). For the linear permutation part \(r_s\) of \(s\), the cocycle identity gives \[V(t)-V(s)=r_s V(s^{-1}t).\] Taking norms proves (20).

Reflection in \(D_u\) fixes its own basis vector, giving the translation vector \(2e_{D_u}\) for \(u^2\). Also \((uv)^3=(uvu)^2\). The affine map \(uvu\) has translation vector equal to the sum of the three mirror vectors, and its linear part preserves that sum and has square one. Squaring proves the second formula in (21). The third mirror at a vertex of an equilateral triangle is parallel to the opposite side and distinct from it. The three opposite directions are different. This proves the last assertion. ◻

Proof. For integral \((i,j)\), the pure translation formulas give \[V((U^iH^j)^m)=m(iL_u+jL_{uv}).\] Use (20) and (19), divide by \(m\), and let \(m\to\infty\). It follows that \[\|iL_u+jL_{uv}\|_1\le C\|(i,j)\|_{U,H}.\] The same inequality holds on rational vectors by homogeneity and on real vectors by continuity. The two displayed translation vectors are linearly independent: the extra-mirror coordinate first determines \(j\), and the \(D_u\)-coordinate then determines \(i\). The stable form is therefore positive definite.

The central pairing formula (12), with \(r=3\), gives \[\langle U,H\rangle=\frac{\ell(H)^2}{3}.\] Hence \[ \cos\theta=\frac{\langle U,H\rangle}{\ell(U)\ell(H)} =\frac{\ell(H)}{3\ell(U)}>\frac12. \tag{24}\] Positive definiteness ensures \(\cos\theta<1\). This proves the angle bounds.

Along a ray for an element \(W\), exponent \(t\ge0\) has distance \(t\ell(W)\) from the origin in either adjacent sector. Thus the identifications preserve both exponents and metrics. The resulting cone is parametrized by a radius \(r\ge0\) and a cyclic angle of period \(\Omega\), with all angle values identified at \(r=0\). Its part \(0\le r\le1\) is the asserted disk.

The formulas for \(v_*\) agree on every shared ray. If a vector in its image has a nonzero extra-mirror coordinate, that coordinate specifies \((u,v)\) and \(\mu\). The excess over \(2\mu\) on one of the original two side coordinates specifies the remaining exponent and which of the two sectors is used. If there is no excess, the point is on the shared \(H\)-ray. With no extra-mirror coefficient, the vector is on one squared-generator ray or is zero. These descriptions distinguish points precisely up to the stated ray identifications. They prove injectivity, including at the origin. ◻

Proof. Triangulate the entire exponent quadrant by squares of side \(1/n\), each split along a diagonal, and assign the vertex \((i/n,j/n)\) the point \(U^iH^j y\). Use constant-speed geodesics on edges. On each triangle, cone the opposite edge to its remaining vertex using geodesics. This gives a continuous map: all restrictions to common edges are the same geodesic parametrization. Geodesic convexity bounds the distance of every image point from a vertex by the maximum of the corresponding vertex distances. Commutation bounds those distances uniformly by the displacements of \(U,H\). Restricting to the radius-one sector therefore gives (26). The restrictions on shared rays agree, so the six maps glue. There are only six sectors, allowing a common \(C_0\).

For one commuting pair put \[F(i,j)=d_Y(y,U^iH^j y)\qquad((i,j)\in\mathbb Z^2).\] The triangle inequality and commutation give \[|F(i,j)-F(i',j')| \le d(y,Uy)|i-i'|+d(y,Hy)|j-j'|.\] For a fixed rational vector \(q=(a/m,b/m)\), compare \(\lfloor nq\rfloor\) with \(\lfloor n/m\rfloor(a,b)\); their difference stays bounded. The definition of stable length consequently gives \[\frac{F(\lfloor nq\rfloor)}n \longrightarrow\frac{\ell(U^aH^b)}m=\|q\|_{U,H}.\] This includes negative \(a,b\). On a bounded real region, choose a finite rational net of arbitrarily small mesh. The common Lipschitz bound above, with a rounding error \(O(1/n)\), and continuity of the stable norm upgrade this convergence to uniform convergence on the region.

For \(p,p'\) in a sector, the difference between their floored exponent vectors differs from the floor of \(n(p'-p)\) by a bounded integer vector. The orbit distance between their labels is \(F\) of that difference. The uniform result and (26) prove (27). Taking one point to be \(0\) gives the radial assertion. The other assertion of (28) follows directly from (21); its error is in fact \(O(1/n)\).

Finally, compactness and injectivity give \[\eta=\min_{p\in A_1,\ q\in A_2}\|v_*(p)-v_*(q)\|_1>0.\] By (28), the norm of \(V(t_n(p))-V(t_n(q))\) is at least \(\eta n/2\), uniformly for these pairs and large \(n\). Combining (20), (19), and (26) gives \[d_Y(f_n(p),f_n(q)) \ge\frac{\eta n}{2C}-1-2C_0.\] Decreasing the positive coefficient proves (29). ◻

Proof. For any two points of \(\partial P\), choose a shorter angular arc between them. Its angle \(\phi\) is at most \(\Omega/2<\pi\). Unfold its sectors to a Euclidean wedge. The wedge intersected with the unit disk is convex, so the chord between the points stays within it, as in Figure 1. Its length is \[2\sin(\phi/2)\le2\sin(\Omega/4).\] Each sector appears at most once in this development, except that a sector containing both endpoints may appear in two end pieces. Each such piece is convex, so its intersection with the chord is a segment. The chord therefore splits into at most seven single-sector pieces. Apply the uniform estimate (27) to each piece. Uniformly for \(p,p'\in\partial P\), this gives \[ d_Y(f_n(p),f_n(p')) \le2\sin(\Omega/4)n+o(n). \tag{30}\]

Fix \(p_0\in\partial P\) and cone the boundary map to \(f_n(p_0)\) by geodesics. Their continuous dependence makes this a continuous map of a parameter disk. If \(q\in[A,B]\), the two triangle inequalities give \[d(y,q)\ge \max\{d(y,A)-d(A,q),\,d(y,B)-d(q,B)\} \ge\frac{d(y,A)+d(y,B)-d(A,B)}2.\] Here both endpoints have distance \(n+o(n)\) from \(y\), uniformly, by (28); (30) bounds their distance. The entire coning segment therefore stays at distance at least \[\bigl(1-\sin(\Omega/4)\bigr)n-o(n)\] from \(y\). Take \(\epsilon=\tfrac12(1-\sin(\Omega/4))>0\). ◻

Proof. The first two relations give \(h=x_1x_2\) and \(x_3=x_2^{-1}x_1x_2\). Substituting these in the third gives \(x_1x_2x_1=x_2x_1x_2\). Conversely, that braid relation and these definitions of \(x_3,h\) give all three relations (31). Since the added generators are private to their blocks, eliminating them independently recovers the defining presentation of \(T\). ◻

Proof. Fix a Cayley vertex \(v\). For the relation \(x_ix_{i+1}=h\), its initial, middle, and terminal corners occur at \(v\) in exactly the translates \(g=v\), \(g=vx_i^{-1}\), and \(g=vh^{-1}\), respectively. Their link edges are exactly \[ \{x_i^+,h^+\},\qquad \{x_i^-,x_{i+1}^+\},\qquad \{x_{i+1}^-,h^-\}. \tag{33}\] Thus one block is a theta graph with three internally disjoint paths of length three from \(h^+\) to \(h^-\): \[\begin{gathered} h^+\; -\; x_1^+\; -\; x_3^-\; -\; h^-,\\ h^+\; -\; x_2^+\; -\; x_1^-\; -\; h^-,\\ h^+\; -\; x_3^+\; -\; x_2^-\; -\; h^-. \end{gathered}\] Its nine link edges are distinct. The three blocks meet only at the signed directions of \(a,b,c\). Across different blocks the only link edges whose two endpoints are shared directions are \[\{a^-,b^+\},\qquad \{b^-,c^+\},\qquad \{c^-,a^+\}.\] They too are distinct. This verifies both the complete link description and the absence of parallel edges; possible coincidences of group elements do not identify labelled edge directions.

It remains to bound a simple circuit. A circuit staying within one block has length at least six. Otherwise split it into maximal nonempty paths within individual blocks, and let \(k\) be their number. Their endpoints are the places where the circuit changes blocks. If \(k=2\), both changes occur in the intersection of the same two blocks. Their endpoints must be the two distinct signs of the one shared generator. Within either block their distance is three, so the circuit has length at least six.

If \(k=3\), the circuit visits all three blocks, changing at one signed direction of each of \(a,b,c\). In a block the two relevant directions are those of \(x_1,x_2\). Equal signs have distance at least two; opposite signs have distance at least three except for \(x_1^-,x_2^+\), which are adjacent. When all three signs agree, the length is therefore at least \(2+2+2\). Otherwise there are two changes of sign in the cyclic order \(a,b,c,a\): one from minus to plus, and one from plus to minus. At most one of the corresponding two block paths can use the exceptional adjacency. The lower bound is then \(1+3+2=6\).

Finally, if \(k\ge4\), a block path of length one must be one of the three displayed edges between shared directions. Each shared direction is incident to at most one such edge, so two length-one paths cannot be consecutive on the simple circuit. At most \(\lfloor k/2\rfloor\) paths have length one and the others have length at least two. The total length is at least \(2k-\lfloor k/2\rfloor\ge6\). ◻

Proof. Encoding the sphere. In each noncollapsed source triangle draw a central node with three legs ending at its side midpoints. A triangle collapsed to an edge has exactly two nonconstant sides. Join their midpoints by an arc; both sides map to the same labelled target edge. A triangle collapsed to a vertex needs no arc. Joining across source sides produces an embedded trivalent graph, together with possible closed curves containing no nodes. Discard these curves. Each remaining node records a particular occurrence of a target triangle, with its orientation sign, and each graph edge pairs two sides mapping to the same labelled target edge.

We must retain endpoint information before using this graph as a diagram. Give it a thin regular neighbourhood, consisting of node disks and ribbons along the edges. Along either side of a ribbon the actual group label of the corresponding endpoint of its target edge stays constant. Indeed, passing through an edge-collapsed triangle connects the two sides at the same endpoint of that edge. At a node the two ribbon sides bounding one sector carry the same group label, namely the corresponding corner of that target triangle. It follows that every boundary circle of the regular neighbourhood carries one group-vertex label throughout.

For each connected component containing nodes, cap its neighbourhood by a disk on every boundary circle and use that constant label at the corresponding dual vertex. This gives a labelled triangular cell decomposition of a sphere: graph nodes correspond to its triangular faces, graph edges to its edges, and neighbourhood boundary circles to its vertices. Here we use the elementary finite planar-topology fact that a polygonal arc properly embedded in a disk cuts it into two disks. To see how it applies, thicken a spanning tree of the connected planar graph to a disk. Add the other embedded edges one at a time as bands. Each runs through a complementary disk and divides it into two disks. Capping the remaining boundary circles therefore gives a sphere. This also proves the usual Euler formula for the resulting cell decomposition. It need not be a strict simplicial complex; incidences, rather than distinct pairs of vertices, are counted.

The labelling rules exclude certain possible degeneracies. A graph edge cannot join two legs at the same node: these legs represent different target sides, whereas a ribbon has one fixed target edge. No two corners of a face can become the same dual vertex, since their actual target vertex labels are distinct. Also a bridge of the graph would have both ribbon sides on the same neighbourhood boundary circle. Constancy around that circle would equate the distinct endpoints of its target edge, so a correctly labelled graph here has no bridge. Multiple edges between different nodes are harmless.

Deleting folds. A fold is a pair of faces sharing an edge and mapping to the same target triangle across the same target side. Their signs are opposite. Such a pair can be deleted directly in the planar graph. Take a small disk around the chosen connecting edge and its two nodes, cutting the other four legs close to the nodes. The chosen edge and its endpoint disks form a disk even if the same two nodes have additional connecting edges: the middle portions of those additional edges lie outside this disk. If the remaining target sides are called \(i,j\), the exposed legs occur in cyclic order \(i,j,j,i\), up to reversal and cyclic permutation. Join \(i\) to \(i\) and \(j\) to \(j\) inside the disk without crossing and delete the two nodes. The endpoint labels match on each new ribbon side, including at the common target corner between the two pairings. This preserves all the label rules, planarity, and the sum of signs for each target triangle. It may separate components or produce circles without nodes; discard such circles and cap the neighbourhoods of the remaining components as above. Figure 2 illustrates the reconnection and its endpoint labels.

The Euler contradiction. If nodes remain and no fold is available, choose a component and its dual spherical cell decomposition. The successive corners at each dual vertex give a closed walk in the link of its actual target vertex. A backtrack in this walk would use the very same target corner on both sides of a shared edge. The exact corner description (33) then identifies the same translated target triangle on both sides, which is a fold. These must be two distinct source faces: one face cannot use two different corners at this dual vertex, nor can two of its distinct sides be joined by the same ribbon. Thus a one-face incidence cannot evade the deletion rule.

Every link walk is consequently nonbacktracking. Any finite nonempty closed nonbacktracking walk contains a circuit, so Lemma 15 makes its length at least six. Let \(F,E,V\) count faces, edges, and vertices of this dual component, with all incidences counted. We obtain \[3F=2E,\qquad 2E\ge6V,\qquad V-E+F=2.\] The first two imply \(V-E+F\le0\), a contradiction. Thus whenever nodes remain, a fold can be removed. Each removal decreases their number by two, so the process terminates with no nodes. All signed target-cell counts have been preserved and are then zero. ◻

Proof. Uniform boundary types. Choose \(R<\infty\) such that every point of \(Y\) is within \(R\) of \(Ty\). For example, if \(Y=TK\) with \(K\) compact, take \(R=\max_{z\in K}d_Y(y,z)\). By Lemma 9, choose an integer \(M\) such that \[ d_Y(sy,ty)\le2R+1\quad\Longrightarrow\quad d_T(s,t)\le M. \tag{35}\] Set \(D=\max\{d_Y(y,ay),d_Y(y,by),d_Y(y,cy)\}\).

By uniform continuity, first refine the given parameter triangulation, still respecting \(B\), until every face image has diameter at most one. Give each vertex \(v\) a group label \(s_v\) with \(d_Y(s_vy,f(v))\le R\). For each unoriented edge, choose once and for all a path between its two labels of length at most \(M\) in the Cayley graph on \(a,b,c\); the opposite orientation uses the reverse of this same path. This is possible by (35). Add three constant steps at each end of every chosen path. Subdivide the parameter edge accordingly, once for both adjacent faces. Its length is at most \(M+6\), and every face boundary is a labelled null loop of length at most \(3(M+6)\).

Relative to the label of an original vertex of a face, only finitely many such boundary types are possible: their lengths are bounded and their letters belong to \(\{a^{\pm1},b^{\pm1},c^{\pm1},1\}\). We now construct one finite filling of each type, keeping its parameter boundary fixed. This construction also explains why an ordinary parameter disk, with permissible collapsed triangles, can always be used.

Filling the boundary loops. Reserve the first three constant steps at the chosen original base vertex, and leave them unchanged. Ignoring constants, the remaining boundary word represents the identity in the presentation of Lemma 14. By the definition of a presented group, it lies in the normal closure of the relators in the free group. Hence it can be reduced to the empty word by finitely many free insertions or deletions of \(qq^{-1}\), insertions or deletions of a translated triangular relator loop or its reverse, and constant-step moves. Here each individual move takes place in one target edge, triangle, or vertex. More explicitly, an insertion of a conjugate \(w r w^{-1}\) is achieved by inserting the journey \(ww^{-1}\) through successive free moves and then inserting the triangular loop \(r\) at its tip. Free cancellations account for equality in the free group. This proves the claimed finite reduction without assuming a disk-diagram theorem. All moves are performed outside the three reserved steps, so every intermediate cyclic parameter path has at least three edges.

Realize two successive cycles as the disjoint boundary circles of a triangulated annulus. Along unchanged edges use strips whose two ends have identical target labels on the two circles; triangulate each strip rectangle into two edge-collapsed or vertex-collapsed triangles. For a nonempty substituted arc on each circle, the intervening region is a polygon whose vertex labels lie in the one target simplex used by the move, so it can be triangulated with the corresponding simplex maps.

For precision, an empty arc requires a fan, not an identification of parameter vertices. Suppose an arc with successive vertices \(p_0,\ldots,p_k\) on one circle is deleted at the vertex \(q\) of the other. The region between these arcs has the two distinct radial edges \(qp_0\) and \(qp_k\), together with the fan triangles \(qp_{j-1}p_j\) for \(1\le j\le k\). Although \(p_0,p_k,q\) have the same group label, \(p_0\) and \(p_k\) are distinct parameter vertices. The unchanged strips attach along the two radial edges. At \(q\) the preceding strip, the fan, and the following strip have a single interval as their link, so this construction has no pinched parameter point. Reversing the construction handles insertion at an empty arc. The reserved constant steps ensure a nonempty unchanged part around the other side of the annulus. Thus all these pieces form an actual annulus with its two boundary circles unchanged. Since all vertices of the fan map to the simplex used by the move, its triangles are legal, including when some or all of their labels agree.

As a concrete case, deleting \(xx^{-1}\) from a five-edge cycle consisting of that excursion and three constant edges gives a three-edge cycle. The connecting annulus has two fan triangles collapsed to the \(x\)-edge and three unchanged strips. Its inner circle remains a circle, despite all its labels being equal; see Figure 3.

Stack the finitely many annuli and cap the final three-step constant cycle by a cone mapping to its vertex. The result is a finite triangulated disk with the original subdivided boundary. Identify this parameter disk with the original face by a boundary-preserving homeomorphism. Such an identification is elementary: realize the stacked annuli as successive collars inside a disk and the final cone as its central disk; a prescribed parametrization of the outer circle extends radially. The triangulation and its affine simplex maps are understood in these abstract triangle coordinates; the homeomorphism need not be affine in any pre-existing coordinates on the parameter surface. All added triangles are therefore contained in the original face. Gluing these constructions along the already agreed edge subdivisions gives a triangulated sphere and preserves \(B\) as a subcomplex.

The uniform estimate. It remains to verify the uniform estimate at every parameter point. Fix all the above fillings for the finite set of boundary types. Normalize their original base vertex to \(1\in T\), and let \(L\) be the largest original word length \(d_T(1,s)\) of any vertex label appearing in any one of these fillings. This number is finite, including for private generators in the triangular presentation, since they are words in \(a,b,c\). It is chosen once and does not depend on \(f\). In a translated filling based at an original vertex \(v_0\) labelled \(s_0\), every label \(s\) satisfies \(d_T(s_0,s)\le L\). Every point \(p\) of that original face then satisfies \[\begin{split} d_Y(sy,f(p)) &\le d_Y(sy,s_0y)+d_Y(s_0y,f(v_0)) +d_Y(f(v_0),f(p))\\ &\le DL+R+1. \end{split}\] Each final triangle lies within that face, so taking \(C_1=DL+R+1\) proves (34). ◻

Proof. Fix \(p\in B\) and any vertex label \(s\) of a triangle containing \(p\). The bounds (39) and (26) imply \[d_Y(sy,t_n(p)y)\le C_1+C_0.\] By (19) and (20), there is a constant \(C_2\), independent of \(n,p,s\), such that \[\|V(s)-V(t_n(p))\|_1\le C_2.\] The second convergence in (28) has error \(O(1/n)\), so \[\left\|\frac{V(s)}n-v_*(p)\right\|_1=O(1/n)\] uniformly for every such vertex label. Apply \(J\) and (38). Since \(q_n(p)\) is a convex combination of those vertex values, it satisfies the same uniform error bound from \(\kappa(p)\). This is precisely where the all-points form of (39) is used. ◻

Proof. By (29), the \(f_n\)-images of \(B'\) and \(P\setminus\operatorname{int}B\) are separated by a positive multiple of \(n\). On the second disk, Lemma 13 gives distance at least \(\epsilon n\) from \(y\). On \(B'\), (36) and (28) give distance at most \((\epsilon/3)n+o(n)\) from \(y\). Thus the images of \(B'\) and the entire complement of \(\operatorname{int}B\) in the sphere have mutual distance tending to infinity, uniformly.

Let \(\tau\) be a contributing labelled target triangle. One of its occurrences contains \(p\in B'\) with \(q_n(p)=w\). Fix any one of the three group vertex labels \(s\) of \(\tau\). Every other occurrence of this same target cell has the same vertex label \(s\), including the full translate information. If some occurrence met the complement of \(\operatorname{int}B\), take a point \(p'\) in that intersection. The all-points bound (39) would give \[d_Y(f_n(p),f_n(p'))\le d_Y(f_n(p),sy)+d_Y(sy,f_n(p'))\le2C_1,\] contradicting the uniform separation for large \(n\). This also excludes an occurrence touching \(\partial B\). ◻

Proof of Proposition 8. Choose \(n\) large enough for the constructions and Lemma 19. Orient every labelled target triangle of \(\mathcal K\) once and for all. Its three group vertices have fixed projected values \(JV(s)/n\), so its affine planar image is fixed. If it covers \(w\), its determinant has a fixed nonzero sign \(\eta_\tau\in\{1,-1\}\). The covering sign of each occurrence is \(\eta_\tau\) times that occurrence’s orientation sign relative to the chosen orientation of the target.

Group the nonzero winding sum on \(B\) by its contributing target triangles. By Lemma 19, every occurrence of each of these target triangles in the whole sphere lies in \(B\). Its total contribution is therefore \[\eta_\tau\, \bigl(\text{signed number of occurrences of }\tau \text{ in the sphere}\bigr)=0\] by Proposition 16. Summing gives zero, contrary to the winding number \(1\) or \(-1\). Distinct target cells remain separate in this sum even if their projected vertex data agree. The contradiction proves the proposition. ◻

Proof. For the first assertion, let \(w\) be the remaining element of \(\{a,b,c\}\). The assignment \[u\longmapsto t_1,\qquad v\longmapsto t_2, \qquad w\longmapsto t_1\] defines a homomorphism \(T\to B_3\): the braid relation for \(u,v\) is the displayed defining relation of \(B_3\), the relation for \(v,w\) is that relation with its sides interchanged, and the relation for \(w,u\) becomes \(t_1^3=t_1^3\). Its composite with the asserted inclusion is the identity of \(B_3\).

For the second assertion we construct a word invariant, retaining the entire permutation of the strands as part of its state. Place \(l\) distinct tokens, named \(1,\ldots,l\), in \(l\) ordered positions. Tokens \(1,2,3\) are called selected. A state consists of an ordered list of all \(l\) tokens and an element \(q\in B_3\). Read a signed letter \(s_i^{\varepsilon}\), \(\varepsilon\in\{1,-1\}\), as follows. Swap the tokens in positions \(i,i+1\), irrespective of the sign. If both are selected, they are consecutive in the ordered list of selected tokens; if they occupy positions \(j,j+1\) in that shorter list, replace \(q\) by \(qt_j^{\varepsilon}\). If at most one is selected, leave \(q\) unchanged. The integer \(j\) is computed before the swap, and is either \(1\) or \(2\).

We verify the defining relations as transformations of all states, not merely of the initial state. The operation for \(s_i^{-1}\) is the inverse of that for \(s_i\): a second swap restores the full token list, and, when a letter was recorded, its selected-list index is still \(j\), so that the two recorded letters cancel. For \(|i-k|\ge2\), the swaps at \(i\) and \(k\) involve disjoint pairs of positions and have the same final token list in either order. With just three selected tokens, at most one of these pairs consists of two selected tokens. Swapping the other pair does not change the selected-list index of the recorded pair, because moving a selected token past an unselected token preserves the order of the selected list. Hence the positive commutation relation is respected.

For a positive braid relation, name the three tokens initially in positions \(i,i+1,i+2\) by \(A,B,C\). The swaps in \(s_is_{i+1}s_i\) exchange, in order, the pairs \[(A,B),\ (A,C),\ (B,C),\] whereas those in \(s_{i+1}s_is_{i+1}\) exchange them in the opposite order. Both finish with the ordered triple \(C,B,A\) and fix all other positions. If at most one of \(A,B,C\) is selected, neither word records a letter. If exactly two are selected, each word records their single crossing with positive sign and the same selected-list index: the third selected token is outside these three consecutive positions and stays on the same side throughout. If all three are selected, the two records are respectively \(t_1t_2t_1\) and \(t_2t_1t_2\), which represent the same element of \(B_3\).

Thus the invertible state transformations satisfy every defining positive relation of \(B_l\). They consequently respect arbitrary signed word equalities: each defining relator acts trivially, as does its inverse or any conjugate, and free cancellations act trivially by the inverse check. This is precisely the normal-closure description of equality in the presented group. In particular, the verification applies after every signed prefix, and the equal full token lists ensure identical processing of every suffix.

Start with the token list \((1,\ldots,l)\) and \(q=1\). A word using only \(s_1^{\pm1},s_2^{\pm1}\) always swaps two of the first three tokens. Its record is exactly the same word with \(s_i\) replaced by \(t_i\), including every sign. If the original word is trivial in \(B_l\), its state transformation fixes this initial state, and its record is therefore trivial in \(B_3\). This proves injectivity. The construction asserts a state invariant; it does not assert that discarding the other tokens gives a homomorphism \(B_l\to B_3\). ◻

Proof. For \(i=1,2\), choose a set \(R_i\subset H_i\) such that every element of \(H_i\) is uniquely \(re\) with \(r\in R_i\), \(e\in E\), and with \(1\) representing \(E\). Consider formal states \[(r_1,\ldots,r_n;e),\qquad n\ge0,\] where every \(r_j\) is a nonidentity representative and their factor types alternate; for \(n=0\) only the terminal element \(e\) is present. Representatives retain their factor types as part of the state.

Define right multiplication of a state by \(h\in H_i\) as follows. If its final representative has type \(i\), isolate the tail \(re\) with that representative \(r\); otherwise use the missing representative \(r=1\). The preceding list \(P\) is empty or ends in the other factor. Decompose the element \(reh\) of \(H_i\) uniquely as \(r'e'\). Replace the tail by \(r';e'\), omitting \(r'\) when \(r'=1\).

For fixed \(i\), these operations are a right action of \(H_i\). Indeed, after multiplying by \(h\in H_i\), the prefix \(P\) stays fixed. For a further \(k\in H_i\), the tail is rewritten from \(r'e'k=(reh)k\). If \(r'=1\) was omitted, the rule simply uses the missing representative \(1\) and rewrites \(e'k\), with exactly the same result. Uniqueness of the decomposition in \(H_i\) and associativity therefore give the operation for \(hk\). Multiplication by \(1\) fixes every state, and multiplication by \(h^{-1}\) is inverse to multiplication by \(h\).

If \(h\in E\), either factor action leaves all representatives fixed and changes only the terminal element from \(e\) to \(eh\). The two actions therefore agree on \(E\) and give a right action of the presented amalgam: all multiplication relations within the factors and all identifications of \(E\) act identically.

Every state is reached from \((\,;1)\) by reading the word \(r_1\cdots r_ne\). Conversely, applying the algorithm to an arbitrary word uses only valid equalities in the amalgam and produces a word of that form. If two state words were equal in the amalgam, their actions on \((\,;1)\) would give the same state. This proves both existence and uniqueness of these normal forms. In particular, the normal form of an element of \(H_i\) is its single representative and terminal \(E\)-element, or just its \(E\)-element; this proves the factor embeddings and the intersection assertion.

Finally, when processing an alternating product \(a_1\cdots a_n\) with every \(a_j\notin E\), at each step the terminal \(E\)-element times \(a_j\) is still outside \(E\). Since its factor type differs from that of the preceding representative, one new nonidentity representative is appended. Its normal form thus has exactly \(n\) representatives, of the indicated alternating types. It is nontrivial and cannot be the normal form of an element of a factor whose type omits any of these representatives. ◻

Proof. Every element of \(T\) commutes with \(d\) and \(e\) by the presentation. Conversely, first suppose that \(g\notin B\). Its normal form in (41) can be written \(g=wb\), where \(b\in B\) and \(w\) is a nonempty alternating product of elements outside \(T\), whose last factor lies in \(G_0\setminus T\). Indeed, absorb the terminal \(T\)-element and, if present, the last \(B\)-representative into \(b\). There is at least one \(G_0\)-representative because \(g\notin B\).

Under the projection \(B\to F(d,e)\), the element \(bdb^{-1}\) maps to a conjugate of the nonidentity free generator \(d\). Consequently \(bdb^{-1}\in B\setminus T\). The word \[gdg^{-1}=w(bdb^{-1})w^{-1}\] is reduced in the sense of Lemma 22: its middle factor lies in \(B\setminus T\), both neighbors lie in \(G_0\setminus T\), and all remaining adjacent factors alternate. It contains a \(G_0\setminus T\) factor and hence cannot belong to \(B\). In particular it cannot equal \(d\). Thus \(C_G(d)\subset B\).

It remains to compute the common centralizer within \(B\). If a reduced word \(f\in F(d,e)\) commutes with \(d\), write \(f=v d^q\), where \(q\in\mathbb Z\) and either \(v\) is empty or its last letter is \(e\) or \(e^{-1}\). This is obtained by removing the maximal terminal string of \(d\)-letters from the reduced word. Then \[fdf^{-1}=v d v^{-1}.\] If \(v\) is nonempty, this displayed word is freely reduced, has length \(2|v|+1>1\), and cannot equal \(d\). Hence \(f=d^q\). For \(q\ne0\), the word \(d^q e d^{-q}\) is freely reduced and is not \(e\), so only the identity commutes with both free generators. An element \((t,f)\in T\times F(d,e)\) therefore centralizes both \(d\) and \(e\) exactly when \(f=1\). This proves the assertion. ◻

Proof. Each function \(x\mapsto d_X(x,q_jx)\) is continuous and convex. For the convexity, if \(\gamma\) is the geodesic from \(x_0\) to \(x_1\) parametrized on \([0,1]\), then \(q_j\gamma\) is the geodesic from \(q_jx_0\) to \(q_jx_1\), and Lemma 2 gives \[d_X(\gamma(t),q_j\gamma(t)) \le (1-t)d_X(x_0,q_jx_0)+t d_X(x_1,q_jx_1).\] Thus \(Y_A\) is closed and convex. Its geodesic triangles and their metrics are inherited from \(X\), so it is CAT(0); each closed bounded subset of \(Y_A\) is closed and bounded in \(X\) and is therefore compact. If \(c\in C\), then \[d_X(cx,q_jcx)=d_X(cx,cq_jx)=d_X(x,q_jx),\] so \(Y_A\) is \(C\)-invariant. Compact subsets of \(Y_A\) are compact in \(X\), hence the restricted action satisfies the required compact-set properness condition.

It remains to prove cocompactness. Choose compact \(K\subset X\) with \(HK=X\), and let \[L=\{z\in X:d_X(z,K)\le A\}.\] This closed bounded neighborhood of \(K\) is compact, since \(X\) is proper. Accordingly \[\mathcal F=\{r\in H:rL\cap L\ne\varnothing\}\] is finite. For any \(h\in H\) such that \(hK\cap Y_A\ne\varnothing\), choose \(p\in K\) with \(hp\in Y_A\). For every \(j\), \[d_X(p,h^{-1}q_jhp)=d_X(hp,q_jhp)\le A.\] Both \(p\) and \(h^{-1}q_jhp\) belong to \(L\), so \(h^{-1}q_jh\in\mathcal F\). Only finitely many ordered tuples \[(h^{-1}q_1h,\ldots,h^{-1}q_mh)\] are therefore possible. Choose one representative \(h_i\) for each tuple that occurs, with \(1\le i\le N\).

If \(h\) has the same tuple as \(h_i\), then, for every \(j\), \[h^{-1}q_jh=h_i^{-1}q_jh_i \quad\Longrightarrow\quad (hh_i^{-1})q_j(hh_i^{-1})^{-1}=q_j.\] Consequently \(hh_i^{-1}\in C\), or equivalently \[h\in C h_i,\qquad h=ch_i\quad\text{for some }c\in C.\] Here \(C h_i\) is a right coset, in the standard convention. Now set \[K_C=Y_A\cap\bigcup_{i=1}^N h_iK.\] This set is compact. Given \(x\in Y_A\), write \(x=hp\) with \(p\in K\), choose its tuple representative \(h_i\), and write \(h=ch_i\) as above. Then \(c^{-1}x=h_ip\) belongs to both \(Y_A\) and \(h_iK\), so \(x\in CK_C\). Thus \(CK_C=Y_A\), proving cocompactness.

Finally, for \(y\in Y_A\), \(c\in C\), and every positive integer \(n\), invariance and the inherited metric give \[d_{Y_A}(y,c^ny)=d_X(y,c^ny).\] Dividing by \(n\) and taking limits proves equality of stable lengths. ◻

Proof of Theorem 1. Let \(M\) be the explicit finite matrix of Definition 20. Suppose, for a contradiction, that \(G=A_M\) acts properly and cocompactly by isometries on a nonempty proper CAT(0) space \(X\). No restriction is placed on the dimension of \(X\).

Choose \(x_0\in X\) and a finite number \[A\ge\max\{d_X(x_0,dx_0),d_X(x_0,ex_0)\}.\] The set \[Y=\{x\in X:d_X(x,dx)\le A,\ d_X(x,ex)\le A\}\] contains \(x_0\). By Lemmas 23 and 24, \(T\) acts geometrically on \(Y\), and its stable lengths there agree with its stable lengths in \(X\).

Sending every Artin generator of \(G\) to \(1\in\mathbb Z\) defines a homomorphism \(G\to\mathbb Z\): both sides of each defining Artin relation have the same number of letters. In particular, every squared generator maps to \(2\) and has infinite order. Lemma 3 therefore gives \(\ell_X((s_1^{uv})^2)>0\) for each \((u,v)\in\mathcal E\). The corresponding embedded \(B_{40}\) block satisfies all the relations required by Proposition 6, so that \[\frac{\ell_X((uv)^3)}{3\ell_X(u^2)} =\frac{\ell_X((uv)^3)}{3\ell_X(v^2)} >\frac12 \qquad ((u,v)\in\mathcal E).\] Equality of stable lengths transfers all three strict inequalities to the geometric \(T\)-action on \(Y\). This contradicts Proposition 8.

The \(116\)-generator Artin group \(A_M\) therefore has no proper cocompact isometric action on any nonempty proper CAT(0) space. This is the required counterexample. ◻