CAT Comparison, Link Criteria, and Local Globalization — Examples

(i). Every interval $I$ is a convex subset of $\mathbb{R}$; for $x<y$ in $I$ the interval $[x,y]\subseteq I$ with its usual parametrization is a geodesic segment from $x$ to $y$ by [F1], and it is the only one, since a distance-preserving map into $\mathbb{R}$ from an interval is determined by its values at the endpoints and is monotone. A geodesic triangle with vertices in $I$ has its three vertices in a common interval and the sum of two of its side lengths equal to the third, so it is degenerate and by [F5] it is congruent to its Euclidean comparison triangle; hence the CAT(0) inequality holds with equality.

Reduced paths compute the metric. Subdivide edges at the finitely many points under discussion. Subdivision preserves connectedness and cannot create a cycle: a cycle in the subdivided graph would traverse each inserted degree-two vertex straight through and collapse to a cycle in $\mathrm{\Gamma }$. Thus the subdivided graph is still a finite tree, and any two of its vertices $x,y$ have a unique edge path $P$. Traverse $P$ at unit speed, with length $D$ equal to the sum of its edge lengths. Define $f:X_{\mathrm{\Gamma }}\to [0,D]$ to be distance along $P$ on $P$, and constant on each branch attached to $P$. Each component off $P$ attaches at exactly one vertex; two attachment vertices would create a second path between them and hence a cycle. Consequently $f$ is well defined, continuous, and $1$-Lipschitz on every edge. For any chain, summing edgewise inequalities gives $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }=D\le \mathrm{\ell }(\text{chain})$. The path $P$ gives the reverse bound, so $d(x,y)=D$. Taking $x,y$ on a single edge also proves that edge's metric is its interval metric.

Geodesics and uniqueness. Apply step 1.2 after subdividing at any two points $x,y$. Unit-speed traversal of their reduced path is distance preserving on every subinterval, again by the reduced-path formula, so is a geodesic. If $w$ lies off that path, its unique attachment point $v$ satisfies $d(x,w)+d(w,y)=d(x,y)+2d(v,w)>d(x,y)$. Every point of any minimizing segment must instead satisfy equality in this sum. Thus the segment lies on the reduced path, where its distance from $x$ fixes its position, proving uniqueness. The geodesic is a path of length $d(x,y)$, so the space is a length space. Its finite union of compact interval cells is compact: each cell inclusion is $1$-Lipschitz for the chain metric, so the preimages of any open cover have finite subcovers; taking their finite union covers the entire finite gluing. Completeness is [F3].

The median. Subdivide at $x,y,z$. The paths from $x$ to $y$ and from $x$ to $z$ have a common initial path: if they separated and later met, their two portions between the first separation and reunion would contradict unique paths in the tree. Let $o$ be the last vertex of that common initial path. The remaining paths from $o$ to $y$ and from $o$ to $z$ have no vertex in common except $o$, so their concatenation is the unique path from $y$ to $z$. Hence the triple intersection of the three paths is exactly $\{o\}$, including the cases of repeated points, and each side is the union of the corresponding two arms. Every point of an arm lies on its two incident sides.

(ii)(b). Fix a geodesic triangle with vertices $x,y,z$, let $o$ be its median from step 3.1 and put $L:=d(y,z)>0$ if $y\mathrm{\ne }z$ (if two vertices coincide, uniqueness from step 2.1 makes the two nonconstant sides coincide, so [F5] applies). Parametrize the geodesic $[y,z]$ by $\gamma :[0,1]\to X_{\mathrm{\Gamma }}$ with $\gamma (t_o)=o$, so that $d(x,\gamma (t))=h+L\mathrm{\mid }t-t_o\mathrm{\mid }$ with $h:=d(x,o)$, $d(x,y)=h+L{t}_o$ and $d(x,z)=h+L(1-t_o)$. Substituting into the squared right-hand side of [F4], the difference $(1-t)d(x,y{)}^2+td(x,z{)}^2-t(1-t)L^2-d(x,\gamma (t){)}^2$ equals $4hLt(1-t_o)\ge 0$ for $t\le t_o$ and $4hL(1-t)t_o\ge 0$ for $t\ge t_o$; this is a direct expansion, and the two cases are interchanged by $t\leftrightarrow 1-t$, $y\leftrightarrow z$. By [F4] the hinged inequality at the vertex $x$ holds; the same computation with $x$ replaced by $y$ and by $z$, using the median description of step 3.1, gives the hinged inequality at the other two vertices.

(ii)(b), conclusion. Every geodesic triangle in $X_{\mathrm{\Gamma }}$ has all three hinged inequalities at its vertices, and by [F4] (the vertex-opposite-side criterion) it satisfies the CAT(0) inequality for all pairs of its points; hence $X_{\mathrm{\Gamma }}$ is CAT(0), and the Bruhat–Tits inequality is its case $t=1\mathrm{/}2$.

(i) and the metric properties are the case $\mathrm{\ell }=2\pi$ of [F1]: for two points at distance $\theta <\pi$ the shorter arc, parametrized proportionally, is a geodesic segment, and any geodesic between them has length $\theta$ and is monotone along the circle, hence is that arc; so it is unique.

(ii). Let three points have pairwise distances $a,b,c$ with $a+b+c<2\pi$. If vertices repeat, the two nonzero sides coincide by step 1.1 and realize a degenerate comparison. Otherwise order them cyclically on the circle, writing the three gaps as $g_1,g_2,g_3>0$ with $g_1+g_2+g_3=2\pi$. If all $g_i<\pi$ then the pairwise distances are $g_1,g_2,g_3$ and $a+b+c=2\pi$, contrary to hypothesis; so some gap, say $g_3\ge \pi$, and the complementary arc of length $s:=2\pi -g_3\le \pi$ contains all three points; the two remaining gaps satisfy $g_1+g_2=s$, and the pairwise distances are $g_1,g_2$ and $g_1+g_2=s$, so $a+b+c=2s$ and $s<\pi$. All three points then lie in an arc of length $s<\pi$, on which the circle metric is the interval metric, so the triangle is degenerate and isometric to a triangle on a great arc of $S^2$ of the same length; by [F3] and the comparison-triangle uniqueness of Comparison triangles in the Euclidean plane and the round sphere, model spaces, and CAT(0) and CAT(1) consequences clause (ii) it realizes its spherical comparison triangle isometrically, and the CAT(1) inequality holds with equality.

(iii). For the three equally spaced points the gaps are $2\pi \mathrm{/}3$ each, the pairwise distances are $2\pi \mathrm{/}3$ and the perimeter is exactly $2\pi$, so the hypothesis "perimeter $<2\pi$" of the CAT(1) definition is not met and the triple is not tested; step 2.1 shows that every tested triangle is degenerate and satisfies the inequality with equality, so no triangle of perimeter $<2\pi$ witnesses a failure.

Conclusion. By [F2] the space $S_{2\pi }^1$ is CAT(1), in agreement with steps 2.1 and 3.1, and by [F1] it is compact, complete, geodesic and contains itself as an isometrically embedded circle of length $2\pi$.

The metric, compactness, completeness and geodesic character are clause (vi) of [F1]. For the failure, the three points $x=0$, $y=\mathrm{\ell }\mathrm{/}3$, $z=2\mathrm{\ell }\mathrm{/}3$ of $S_{\mathrm{\ell }}^1$ have pairwise distances $\mathrm{\ell }\mathrm{/}3$ by [F5], so the geodesic triangle they determine has perimeter $\mathrm{\ell }<2\pi$ and all its sides are $<\pi$; its comparison triangle in $S^2$ has three equal sides $\mathrm{\ell }\mathrm{/}3$ by the comparison-triangle uniqueness of Comparison triangles in the Euclidean plane and the round sphere, model spaces, and CAT(0) and CAT(1) consequences clause (ii).

Let $a:=\mathrm{\ell }\mathrm{/}6$ be the midpoint of the shorter arc $[x,y]$, so that $d_{\mathrm{\ell }}(a,z)=\mathrm{\ell }\mathrm{/}2$ by [F5]. The comparison point $\bar{a}$ of $a$ is the midpoint of the corresponding side of the spherical comparison triangle, and by the midpoint identity [F2], applied with $c=\mathrm{\ell }\mathrm{/}3<\pi$, its distance from the opposite vertex $\bar{z}$ satisfies $\cos d_S(\bar{a},\bar{z})={\displaystyle \frac{\cos (\mathrm{\ell }\mathrm{/}3)+\cos (\mathrm{\ell }\mathrm{/}3)}{2\cos (\mathrm{\ell }\mathrm{/}6)}}={\displaystyle \frac{\cos (\mathrm{\ell }\mathrm{/}3)}{\cos (\mathrm{\ell }\mathrm{/}6)}}$, the denominator being positive because $0<\mathrm{\ell }\mathrm{/}6<\pi \mathrm{/}2$.

The comparison distance is strictly less than $\mathrm{\ell }\mathrm{/}2$. Indeed the addition formula [F3] gives $2\cos (\mathrm{\ell }\mathrm{/}2)\cos (\mathrm{\ell }\mathrm{/}6)=\cos (2\mathrm{\ell }\mathrm{/}3)+\cos (\mathrm{\ell }\mathrm{/}3)$, and $\cos (2\mathrm{\ell }\mathrm{/}3)-\cos (\mathrm{\ell }\mathrm{/}3)=-2\sin (\mathrm{\ell }\mathrm{/}2)\sin (\mathrm{\ell }\mathrm{/}6)<0$, since both sine arguments lie in $(0,\pi )$; hence $2\cos (\mathrm{\ell }\mathrm{/}2)\cos (\mathrm{\ell }\mathrm{/}6)<2\cos (\mathrm{\ell }\mathrm{/}3)$, that is $\cos (\mathrm{\ell }\mathrm{/}2)<\cos (\mathrm{\ell }\mathrm{/}3)\mathrm{/}\cos (\mathrm{\ell }\mathrm{/}6)$ after dividing by the positive number $2\cos (\mathrm{\ell }\mathrm{/}6)$. Since both $\mathrm{\ell }\mathrm{/}2$ and $\arccos (\cos (\mathrm{\ell }\mathrm{/}3)\mathrm{/}\cos (\mathrm{\ell }\mathrm{/}6))$ lie in $[0,\pi ]$ and $\cos$ is strictly decreasing there, $\arccos (\cos (\mathrm{\ell }\mathrm{/}3)\mathrm{/}\cos (\mathrm{\ell }\mathrm{/}6))<\mathrm{\ell }\mathrm{/}2$.

An ambient space cannot escape the failure. Let $X$ be a metric space and let $\phi :S_{\mathrm{\ell }}^1\to X$ be an isometric embedding of a circle of length $\mathrm{\ell }<2\pi$; the image $Z:=\phi (S_{\mathrm{\ell }}^1)$ with the induced metric is isometric to $S_{\mathrm{\ell }}^1$, hence geodesic, and every geodesic triangle of $Z$ of perimeter $<2\pi$ is a geodesic triangle of $X$ with the same side lengths and comparison distances, so if $X$ were CAT(1) then $Z$ would be CAT(1) by [F4]; but $Z$, being isometric to $S_{\mathrm{\ell }}^1$, is not CAT(1) by step 2.1, a contradiction. Hence a metric space containing an isometrically embedded circle of length $\mathrm{\ell }<2\pi$ is not CAT(1).

(i) is the first part of [F1]: $S_{\mathrm{\ell }}^1$ is compact, complete and geodesic, hence a length space, and it is locally isometric to $\mathbb{R}$.

Small balls are intervals. Let $x\in S_{\mathrm{\ell }}^1$, $0<r<\mathrm{\ell }\mathrm{/}4$ and $t$ any lift of $x$. For $s,s^{\mathrm{\prime }}\in [t-r,t+r]$ we have $\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }\le 2r<\mathrm{\ell }\mathrm{/}2$, so $d_{\mathrm{\ell }}(p(s),p(s^{\mathrm{\prime }}))=\min \{\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid },\mathrm{\ell }-\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }\}=\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }$; hence $p$ restricts to a distance-preserving bijection of the interval $[t-r,t+r]$ onto the closed ball $\bar{B}(x,r)$.

An interval is CAT(0). Let $V=[a,b]\subset \mathbb{R}$ with the induced metric; it is geodesic, and a geodesic triangle with vertices $\alpha \le \beta \le \gamma$ in $V$ has its three sides contained in $[\alpha ,\gamma ]$ and side lengths $\beta -\alpha ,\gamma -\beta ,\gamma -\alpha$, so its Euclidean comparison triangle is the degenerate segment $[\bar{\alpha },\bar{\gamma }]$ of length $\gamma -\alpha$ with the three comparison vertices at positions $\alpha ,\beta ,\gamma$; a point of the triangle lies on some side and its comparison point has the same position in $[\bar{\alpha },\bar{\gamma }]$, so all distances between points of the triangle equal their comparison distances and the CAT(0) inequality holds with equality. Hence $V$ is CAT(0).

Conclusion of (i). By steps 1.2 and 1.3 each point of $S_{\mathrm{\ell }}^1$ has a closed ball of radius $r<\mathrm{\ell }\mathrm{/}4$ that is isometric, for the induced metric, to an interval, and intervals are CAT(0); so $S_{\mathrm{\ell }}^1$ is locally CAT(0).

(ii) $p$ is a covering map. The quotient map $p$ is continuous and surjective, and for every $x$ and $0<r<\mathrm{\ell }\mathrm{/}4$ the preimage $p^{-1}(B(x,r))$ is the disjoint union of the open intervals $(t_k-r,t_k+r)$ about the lifts $t_k=t+k\mathrm{\ell }$ of $x$, since two such intervals meet only if their centres differ by less than $2r<\mathrm{\ell }$; by step 1.2 each of them is mapped isometrically onto $B(x,r)$. Hence every ball of radius $<\mathrm{\ell }\mathrm{/}4$ is evenly covered and $p$ is a covering map.

(iii) the witness triple. Let $a:=0$, $b:=\mathrm{\ell }\mathrm{/}3$, $c:=2\mathrm{\ell }\mathrm{/}3$ and $m:=\mathrm{\ell }\mathrm{/}6$ in $S_{\mathrm{\ell }}^1$. The three pairwise distances are $d_{\mathrm{\ell }}(a,b)=d_{\mathrm{\ell }}(b,c)=d_{\mathrm{\ell }}(a,c)=\mathrm{\ell }\mathrm{/}3$, since $\mathrm{\mid }0-\mathrm{\ell }\mathrm{/}3\mathrm{\mid }=\mathrm{\mid }\mathrm{\ell }\mathrm{/}3-2\mathrm{\ell }\mathrm{/}3\mathrm{\mid }=\mathrm{\ell }\mathrm{/}3$ and $d_{\mathrm{\ell }}(a,c)=\min \{2\mathrm{\ell }\mathrm{/}3,\mathrm{\ell }-2\mathrm{\ell }\mathrm{/}3\}=\mathrm{\ell }\mathrm{/}3$; moreover $m$ lies on the geodesic $[a,b]$ given by the arc from $0$ to $\mathrm{\ell }\mathrm{/}3$, and $d_{\mathrm{\ell }}(m,c)=\min \{\mathrm{\ell }\mathrm{/}2,\mathrm{\ell }-\mathrm{\ell }\mathrm{/}2\}=\mathrm{\ell }\mathrm{/}2$, so there is a geodesic triangle of $S_{\mathrm{\ell }}^1$ whose comparison is tested.

(ii) The generator is not nullhomotopic. The loop $\alpha (t)=p(t)$, $t\in [0,\mathrm{\ell }]$, lifts from $0$ to $\tilde{\alpha }(t)=t$ and ends at $\mathrm{\ell }$, whereas the based constant loop lifts to a path ending at $0$. Thus [F4] excludes a based nullhomotopy, giving a nontrivial class in ${\pi }_1(S_{\mathrm{\ell }}^1,0)$. To exclude a free nullhomotopy as well, suppose $K:[0,\mathrm{\ell }]\times [0,1]\to S_{\mathrm{\ell }}^1$ deforms $\alpha$ through loops to a constant loop; then $K(0,s)=K(\mathrm{\ell },s)$. Lift $K$ with initial lift $t\mapsto t$. The two paths $s\mapsto \tilde{K}(\mathrm{\ell },s)$ and $s\mapsto \tilde{K}(0,s)+\mathrm{\ell }$ lift the same path and both start at $\mathrm{\ell }$, hence coincide by [F4]. At $s=1$ the lifted constant loop is constant by path-lifting uniqueness in [F4]: the constant path at its initial lift is another lift of the same constant loop. Their difference is then both $\mathrm{\ell }$ and $0$, impossible. The circle is path-connected by its arcs and is not simply connected.

(iii) the comparison fails. The Euclidean comparison triangle of $(a,b,c)$ is equilateral of side $\mathrm{\ell }\mathrm{/}3$, and the comparison point $\bar{m}$ of $m$ is the midpoint of the side $[\bar{a},\bar{b}]$; its distance to the opposite vertex $\bar{c}$ is the altitude $\mathrm{\ell }\sqrt{3}\mathrm{/}6$, because $(\mathrm{\ell }\sqrt{3}\mathrm{/}6{)}^2+(\mathrm{\ell }\mathrm{/}6{)}^2={\mathrm{\ell }}^2\mathrm{/}9=(\mathrm{\ell }\mathrm{/}3{)}^2$ by Pythagoras in ${\mathbb{E}}^2$. Since $\mathrm{\ell }\sqrt{3}\mathrm{/}6<\mathrm{\ell }\mathrm{/}2$ we have $d_{\mathrm{\ell }}(m,c)=\mathrm{\ell }\mathrm{/}2>d_2(\bar{m},\bar{c})$, so the CAT(0) inequality fails and $S_{\mathrm{\ell }}^1$ is not CAT(0).

The fundamental group is infinite cyclic. Parametrize based loops on $[0,1]$. Every loop $\beta$ lifts uniquely from $0$ to a path $\tilde{\beta }$ in $\mathbb{R}$, with endpoint $n\mathrm{\ell }$ for a unique integer $n$; [F4] makes $n$ invariant under based homotopy. Conversely $p((1-s)\tilde{\beta }(t)+sn\mathrm{\ell }t)$ is a based homotopy to the loop $t\mapsto p(n\mathrm{\ell }t)$, since the two endpoints of the interpolated lift stay $0,n\mathrm{\ell }$. Every integer is realized by this explicit loop. When loops of winding $n,m$ are concatenated, the lift of the second starts at $n\mathrm{\ell }$ and is its lift from $0$ translated by $n\mathrm{\ell }$, so the endpoint is $(n+m)\mathrm{\ell }$. Winding thus gives an isomorphism ${\pi }_1(S_{\mathrm{\ell }}^1,0)\cong \mathbb{Z}$, sending $\alpha$ to $1$.

(iv). Steps 1.1–1.3 and 2.1 show that $S_{\mathrm{\ell }}^1$ is a complete locally CAT(0) length space, path-connected because it is geodesic and hence connected, while step 3.1 shows that it is not simply connected and step 3.2 that it is not CAT(0); hence the simple-connectivity hypothesis of the globalization theorem [F6] cannot be dropped.

The circle is not contractible, and the theorem's clauses fail explicitly. A contraction of the circle, composed with $\alpha$, would be a free nullhomotopy of $\alpha$, excluded by step 3.1. Thus the conclusion of clause (iii) of [F6] fails, and its clause (i) fails as well: the points $0$ and $\mathrm{\ell }\mathrm{/}2$ are joined by exactly two minimizing geodesics, the two semicircular arcs of length $\mathrm{\ell }\mathrm{/}2$. Indeed lift any minimizing segment on $[0,1]$ from $0$ through $p$: on each interval chart its lift is affine with slope either $\mathrm{\ell }\mathrm{/}2$ or $-\mathrm{\ell }\mathrm{/}2$, and the slope cannot change on overlapping intervals, so the lift is precisely $t\mapsto \pm t\mathrm{\ell }\mathrm{/}2$. Thus the unique geodesic from $x_0$ to $x$ that builds the geodesic contraction is not available at $x=\mathrm{\ell }\mathrm{/}2$ and $0<t<1$.