Weak topology on CAT(0) spaces

Define the topology $\mathcal{T}_{\Delta}$ as follows. We say that a subset $A\subset X$ is $\mathcal{T}_{\Delta}$-closed, if, for any bounded sequence $x_{n}\in A$ weakly converging to a point $x\in X$, we have $x\in A$.

By definition, the empty set and the whole set are $\mathcal{T}_{\Delta}$-closed. Moreover, any intersection of $\mathcal{T}_{\Delta}$-closed subsets $A_{\alpha}$ is $\mathcal{T}_{\Delta}$-closed.

Finally, if $A_{1},...,A_{m}$ are $\mathcal{T}_{\Delta}$-closed and $(x_{n})$ is a bounded sequence in $A_{1}\cup....\cup A_{m}$ weakly converging to $x$, then we find a subsequence of $(x_{n})$ contained in one of the $A_{i}$. This subsequence also weakly converges to $x$, therefore $x\in A_{i}$. Hence $A_{1}\cup...\cup A_{m}$ is $\mathcal{T}_{\Delta}$-closed.

Altogether, this shows that the family of all $\mathcal{T}_{\Delta}$-closed sets is the family of closed sets of a topology, which we will denote by $\mathcal{T}_{\Delta}$.

We claim that a sequence $(x_{n})$ in $X$ converges to a point $x$ with respect to $\mathcal{T}_{\Delta}$ if and only if $(x_{n})$ is bounded and converges to $x$ weakly.

Firstly, let $(x_{n})$ be bounded and weakly converge to $x$. If $(x_{n})$ does not $\mathcal{T}_{\Delta}$-converge to $x$, we would find a $\mathcal{T}_{\Delta}$-open subset $U$ containing $x$ and a subsequence $(x_{m_{n}})$ contained in the complement $A\mathrel{:=}X\setminus U$. However, $(x_{m_{n}})$ also converges to $x$ weakly, hence, by the definition of $\mathcal{T}_{\Delta}$-closed subsets, we infer $x\in A$, a contradiction.

On the other hand, let a sequence $(x_{n})$ converge in the $\mathcal{T}_{\Delta}$-topology to $x$. If $(x_{n})$ is not bounded, we could find a subsequence $(x_{m_{n}})$ such that $d(x_{1},x_{m_{n}})\to\infty$. Then the countable set $\{x_{m_{n}}\}$ is $\mathcal{T}_{\Delta}$-closed. Hence, $(x_{m_{n}})$ does not $\mathcal{T}_{\Delta}$-converge to $x$. Therefore, $(x_{n})$ must be bounded.

Assume that $x_{n}$ does not converge weakly to $x$. Then we find a subsequence $(x_{m_{n}})$ of $(x_{n})$ which converges weakly to some point $y\neq x$. Moreover, deleting finitely many elements from the sequence, we may assume that $x_{m_{n}}$ is not equal to $x$ for all $n$. Then the union $A$ of all $x_{m_{n}}$ and the point $y$ is $\mathcal{T}_{\Delta}$-closed. Thus, the complement of $A$ is a $\mathcal{T}_{\Delta}$-open neighborhood of $x$, which does not contain all but finitely many elements of the sequence $(x_{n})$. This contradiction proves that $(x_{n})$ weakly converges to $x$ and finishes the proof of the claim.

The claim and the definition of $\mathcal{T}_{\Delta}$ imply that a subset $A$ of $X$ is $\mathcal{T}_{\Delta}$-closed if every $\mathcal{T}_{\Delta}$-limit $x\in X$ of a sequence of points in $A$ is contained in $A$. This means that $\mathcal{T}_{\Delta}$ is sequential.

We have verified the required properties of $\mathcal{T}_{\Delta}$. Let $\mathcal{T}$ be another sequential topology on $X$, for which a sequence $(x_{n})$ converges to $x$ if and only if $(x_{n})$ is bounded and weakly converges to $x$. Then, for $\mathcal{T}$ and $\mathcal{T}_{\Delta}$ the convergence of sequences coincide. Since both topologies are sequential, this implies that the properties of being closed with respect to $\mathcal{T}$ and $\mathcal{T}_{\Delta}$ coincide. Hence, $\mathcal{T}=\mathcal{T}_{\Delta}$. ∎

Consider the set

which has appeared above. As explained in the proof of Proposition 1.3 above, the set $A$ is $\mathcal{T}_{\Delta}$-closed.

We are going to prove that $\frac{1}{4}\cdot p$ is contained in the $\mathcal{T}_{co}$ closure of $A$. Assuming the contrary, we find finitely many convex, metrically closed subsets $C_{1},...,C_{n}$ in $X$ which cover $A$ and do not contain $\frac{1}{4}\cdot p$.

For any $b\in B$, consider the set $E^{b}$ of points in $E$ which are at distance $\frac{\pi}{4}$ from $b$. Then a counting argument implies that at least one of the sets $C_{i}$ contains at least 2 points in any of the sets $E^{b_{1}},E^{b_{2}}$, for different $b_{1},b_{2}\in B$. Then this convex set $C_{i}$ contains the origin $o$ (as the midpoint of a point in $E^{b_{1}}$ and $E^{b_{2}}$), the points $\frac{1}{\sqrt{2}}\cdot b_{i}$ and therefore their midpoint $\frac{1}{2}\cdot p$. Hence, $C_{i}$ also contains the whole geodesic $[o,\frac{1}{2}\cdot p]$ and, therefore, $\frac{1}{4}\cdot p\in C_{i}$, in contradiction to our assumption.

Thus, the set $A$ is not $\mathcal{T}_{co}$-closed, finishing the proof. ∎

Replacing the net by a subnet we may assume that the net $r_{\alpha}\mathrel{:=}d(x_{\alpha},x)$ of real numbers converges to some $r\geq 0$. If $r=0$, we find some $x_{\alpha_{i}}$ such that $\lim_{i\to\infty}r_{\alpha_{i}}=0$. Thus, the sequence $x_{\alpha_{i}}$ converges to $x$ in the metric topology, and, therefore, also weakly. Thus, we may assume $r>0$ and, after rescaling, $r=1$.

We choose inductively $\alpha_{k}\in I$, for $k=1,2,...$, starting with an arbitrary $\alpha_{1}$. Let the elements $\alpha_{1}\leq...\leq\alpha_{k}$ in $I$ be already chosen.

For any non-empty subset $S\subset\{1,...,k\}$, denote by $m_{S}$ the unique circumcenter $m_{S}$ of the finite set $\{x_{\alpha_{i}},i\in S\}$. Since the net $(x_{\alpha})$ converges weakly to $x$ and $(r_{\alpha})$ converges to $1$, we find some $\alpha_{k+1}\geq\alpha_{k}$ with the two following properties, for any $\alpha\geq\alpha_{k+1}$:

1) $|r_{\alpha}-1|\leq 2^{-k-1}$.

2) For all nonempty $S\subset\{1,...,k\}$ the projection $Proj_{c}(x_{\alpha})$ of $x_{\alpha}$ onto the geodesic $c=[xm_{S}]$ has distance at most $2^{-k-1}$ from $x$.

Note that any subsequence of the sequence $(x_{\alpha_{i}})$ has also the properties (1) and (2). We claim that the so-defined sequence $(x_{\alpha_{i}})$ converges to $x$ weakly. The proof of the claim relies only on the strict convexity of the squared distance functions and is rather straightforward. For the convenience of the reader, we present the somewhat lengthy details.

Assuming the contrary and replacing the sequence by a subsequence we may assume that the sequence converges weakly to a point $z\neq x$. Set $\delta\mathrel{:=}d(z,x)$. Choosing yet another subsequence we may assume that $s_{\alpha_{i}}\mathrel{:=}d(x_{\alpha_{i}},z)$ converge to some $s\geq 0$, for $i\to\infty$.

We set $\varepsilon\leq\frac{\delta^{2}}{10}$ and find some $i_{0}$ such that $(1-2^{-i_{0}-1})^{2}>1-\varepsilon$ and such that, for all $i\geq i_{0}$,

Using the weak convergence of $(x_{\alpha_{i}})$ to $z$ and $\mathrm{CAT}(0)$ comparison, we may assume in addition, that for all $i\geq i_{0}$

For $j=1,2...$ we consider the point $p_{j}\mathrel{:=}x_{\alpha_{i_{0}+j}}$. By above, the circumradius $t$ of the countable set $\{p_{j}\}$ satisfies

Denote by $0\leq t_{k}\leq t$ the circumradius of the set $\{p_{1},...,p_{k}\}$. We claim that there exists some positive $\rho>0$, such that $t^{2}_{k+1}-t^{2}_{k}>\rho$ for all $k\geq 1$. Since the sequence $(t_{k})$ is bounded above by $t$, this would provide a contradiction and finish the proof.

In order to prove the claim, consider the circumcenter $m_{k}$ of the subset $p_{1},...,p_{k}$. Thus, $m_{k}$ is the point at which the $2$-convex function,

assumes its unique minimum $t_{k}^{2}$. By the $2$-convexity, we deduce

On the other hand, $f(m_{k+1})\leq t^{2}_{k+1}$, hence

By construction of the sequence $x_{\alpha_{i}}$, we have

Thus, by the triangle inequality and the fact

we obtain some positive lower bound $\rho>0$ on $d^{2}(m_{k},m_{k+1})$. This finishes the proof of the claim and of the proposition. ∎

Assume the contrary. Then, replacing the net by a subnet, we find a $\mathcal{T}_{\Delta}$-open neighborhood $U$ of $x$ which does not contain any $x_{\alpha}$. Using Proposition 5.1 we find a sequence $x_{\alpha_{1}},....,x_{\alpha_{k}},...$ of elements of the net converging weakly to $x$. Then $x$ is contained in the $\mathcal{T}_{\Delta}$-closed set $X\setminus U$ which contains all elements of the sequence. This contradicts the definition of $\mathcal{T}_{\Delta}$-closed sets. ∎

Let $\mathcal{T}$ be a topology on $X$, such that a bounded net $(x_{\alpha})$ weakly converges to $x$ if and only if this net $\mathcal{T}$-converges to $x$. Since any net has at most one weak limit point and since the Hausdorff property can be recognized by the uniqueness of limit points of nets, we deduce that any bounded subset of $X$ is Hausdorff with respect to $\mathcal{T}$.

Let $A$ be a bounded subset of $X$. By definition, $A$ is $\mathcal{T}$-closed if and only if it contains all weak limit points of any net $(x_{\alpha})$ of elements in $A$. From Proposition 5.1, this happens if and only if $A$ contains all weak limit points of any sequence of elements in $A$. Thus, if and only if $A$ is $\mathcal{T}_{\Delta}$-closed. We infer, that $\mathcal{T}$ coincides with $\mathcal{T}_{\Delta}$ on bounded subsets.

Assume, on the other hand, that the weak topology $\mathcal{T}_{\Delta}$ is Hausdorff on any ball in $X$. We claim that a bounded net $(x_{\alpha})$ converges weakly to $x$ if and only if $(x_{\alpha})$ converges to $x$ with respect to $\mathcal{T}_{\Delta}$.

Due to Lemma 5.2, the only if conclusion always holds. On the other hand, assume that $(x_{\alpha})$ converges to $x$ with respect to $\mathcal{T}_{\Delta}$ but does not weakly converge to $x$. Replacing $(x_{\alpha})$ by a subnet we may assume that $(x_{\alpha})$ weakly converges to another point $y$. Due to Lemma 5.2, this implies that the net $(x_{\alpha})$ converges to the point $y$ with respect to the topology $\mathcal{T}_{\Delta}$. But this contradicts the assumption that $\mathcal{T}_{\Delta}$ is Hausdorff on the bounded ball which contains the net $(x_{\alpha})$.

This proves the ”if”-direction and finishes the proof of the theorem. ∎

We may replace $X$ by a ball $B_{r}(x)$ and assume that $X$ is bounded. The only if statement follows from Proposition 3.1.

On the other hand, assume that $\mathcal{T}_{co}$ is sequential and sequentially Hausdorff on the bounded $\mathrm{CAT}(0)$ space $X$. Due to Proposition 3.1, the identity map $Id\colon(X,\mathcal{T}_{\Delta})\to(X,\mathcal{T}_{co})$ is continuous. In order to prove that the inverse $Id\colon(X,\mathcal{T}_{co})\to(X,\mathcal{T}_{\Delta})$ is continuous, consider a $\mathcal{T}_{\Delta}$-closed subset $A$ and assume that $A$ is not $\mathcal{T}_{co}$-closed. Since $\mathcal{T}_{co}$ is sequential, we find a sequence $(x_{n})$ in $A$ which $\mathcal{T}_{co}$-converges to a point $x\in X\setminus A$. Using that $X$ is $\mathcal{T}_{\Delta}$-sequentially compact, we may replace $(x_{n})$ by a subsequence and assume that $(x_{n})$ converges to some point $y$ in $X$ with respect to $\mathcal{T}_{\Delta}$. Then, by Proposition 3.1, the sequence converges to $y$ also with respect to $\mathcal{T}_{co}$. The assumption that $\mathcal{T}_{co}$ is sequentially Hausdorff gives us $x=y$. Since $A$ is $\mathcal{T}_{\Delta}$-closed, we deduce $x=y\in A$. This contradiction implies that $Id\colon(X,\mathcal{T}_{co})\to(X,\mathcal{T}_{\Delta})$ is continuous, hence $\mathcal{T}_{\Delta}=\mathcal{T}_{co}$.

Finally, if $\mathcal{T}_{co}$ is Hausdorff on the bounded $\mathrm{CAT}(0)$ space $X$, then the compactness of $\mathcal{T}_{\Delta}$ and the continuity of the identity map $Id\colon(X,\mathcal{T}_{\Delta})\to(X,\mathcal{T}_{co})$ imply $\mathcal{T}_{\Delta}=\mathcal{T}_{co}$. ∎

Assume first that $X$ is homeomorphic to the plane $\mathbb{R}^{2}$. Then each geodesic $\gamma\colon[a,b]\to X$ extends to a geodesic $\hat{\gamma}\colon\mathbb{R}\to X$, [BH99]. Moreover, by Jordan’s theorem, $\hat{\gamma}$ divides $X$ into two connected components both having $\hat{\gamma}$ as their boundaries. The closures of the connected components are convex. Thus, the open components are $\mathcal{T}_{co}$ open.

In order to prove that $\mathcal{T}_{co}$ is Hausdorff it suffices to find, for any pair of points $x,y$, some geodesic $\gamma\colon\mathbb{R}\to X$, such that $x$ and $y$ are in different components of $X\setminus\gamma$. We connect $x$ and $y$ by a geodesic $\eta$ and take the midpoint $m$ of $\eta$. We find two points $p^{\pm}$ sufficiently close to $m$ which lie in different components of $X\setminus\hat{\eta}$ for some extension of $\eta$ to an infinite geodesic. Then consider a geodesic $\gamma\colon\mathbb{R}\to X$ which contains $p^{+}$ and $p^{-}$. The geodesic $\gamma$ intersect $\eta$ between $x$ and $y$. We infer that $x$ and $y$ lie in different components of $X\setminus\gamma$. This finishes the proof if $X$ is homeomorphic to a plane.

Assume now that $X$ is a finite-dimensional cubical complex and choose $x,y\in X$. Taking a sufficiently fine cubical subdivision, we may assume that the diameter of all cubes is much smaller than the distance between $x$ and $y$. Then the geodesic between $x$ and $y$ intersects at least one hyperwall in $X$. Any such hyperwall is a convex subsets dividing $X$ into two connected and convex components. As above, we deduce that $x$ and $y$ are separated in $\mathcal{T}_{co}$.

Finally, let $X$ be a Riemannian manifold with pinched negative curvature and let $x,y\in X$ be arbitrary different points. Fix $r>d(x,y)$ and set $B=B_{r}(x)$. By Anderson’s construction, [And83], see also [Bor92, Theorem 2.1], we find finitely many closed convex subsets $C_{i}$ in $X\setminus B$, such that $V\mathrel{:=}X\setminus\cup_{i=1}^{m}C_{i}$ is bounded. Then $V$ is a $\mathcal{T}_{co}$-open set containing $B$ and contained in some larger closed ball $B^{\prime}$.

On the compact ball $B^{\prime}$ the topology $\mathcal{T}_{co}$ coincides with the metric topology, [Mon06, Lemma 17], hence it is Hausdorff. Thus, we find $\mathcal{T}_{co}$-open subsets $U_{1}$ and $U_{2}$ in $X$ containing $x$ and $y$, respectively, such that $U_{1}\cap U_{2}\cap B^{\prime}$ is empty. Then $U_{1}\cap U$ and $U_{2}\cap U$ are disjoint $\mathcal{T}_{co}$-neighborhoods of $x$ and $y$ in $X$. ∎

Since $X^{\infty}$ is a compact space, it is not a countable union of nowhere dense subsets, by Baire’s theorem.

Therefore, by our assumption, $X$ is not a finite union of closed convex subsets different from $X$. Thus, any finite intersection of non-empty $\mathcal{T}_{co}$-open subsets is non-empty. In particular, $\mathcal{T}_{co}$ is not Hausdorff.

Assume now that $\mathcal{T}_{co}$ is first-countable on $X$, fix an arbitrary $x\in X$ and a $\mathcal{T}_{co}$-fundamental system of its open neighborhoods $U_{1},...,U_{n},...$. By definition of $\mathcal{T}_{co}$, we may assume that each $U_{i}$ is the complement of a finite union of closed convex subsets $K_{i}^{j}$, not containing the point $x$. Hence, the union of the boundaries at infinity

is not all of $X^{\infty}$. Consider an arbitrary point $z\in X^{\infty}$ not contained in this union and a ray $\gamma$ in $X$ with endpoint $z\in X^{\infty}$, such that $x$ is not on $\gamma$. Then $X\setminus\gamma$ is a $\mathcal{T}_{co}$-open neighborhood of $x$ which does not contain any of the set $U_{i}$. This contradiction shows that $\mathcal{T}_{co}$ is not first-countable. ∎

Assume the contrary and consider any closed convex subset $A$ of $X$ such that the boundary at infinity $A^{\infty}$ of $A$ has non-empty interior in the $(n-1)$-dimensional sphere $X^{\infty}$; here $n$ is the dimension of $X$.

Thus, in the cone topology, $A^{\infty}$ has dimension $n-1$. Therefore, there are no totally geodesic symmetric spaces $Y\subsetneq X$ with $A^{\infty}\subset Y^{\infty}$. On the other hand, if $A^{\infty}=X^{\infty}$ then $A=X$. Thus, we may assume $A^{\infty}\neq X^{\infty}$. Applying [KL06, Theorem 3.1], we deduce that $A^{\infty}$ is not a sub-building of the spherical building $X^{\infty}$.

Since $A^{\infty}$ contains an open subset in the cone topology, we find a non-empty subset $O$ of $A^{\infty}$, open in the cone topology and consisting of regular points only. If, for some $p\in O$, we find an antipode $q\in A^{\infty}$ (with respect to the Tits-distance) then $A^{\infty}$ contains a spherical apartment (the boundary of a maximal flat in $X$), as the convex hull in the Tits-metric of $q$ and a Tits-ball around $p$. By [BL06, Theorem 1.1], this would imply that $A^{\infty}$ is a sub-building, in contradiction to the statements above. Thus, for no $p\in O$ and $q\in A^{\infty}$ the Tits-distance between $p$ and $q$ equals $\pi$.

We are going to construct a pair of antipodes $p\in O$ and $q\in A^{\infty}$ and achieve a contradiction. We start with an arbitrary point $p\in O$.

Let $G$ be the isometry group of $X$ (and of $X^{\infty}$) and denote by $\Delta$ the spherical Coxeter chamber $X^{\infty}/G$ of the spherical building $X^{\infty}$. Let $\mathcal{P}:X^{\infty}\to\Delta$ be the canonical projection. Denote by $I:\Delta\to\Delta$ the isometry of the Coxeter chamber induced by the action of $-Id$ on any apartment of $X^{\infty}$. The map $I$ is an involution, which is the identity map if and only if the Coxeter group $W$ of $X^{\infty}$ has a non-trivial center (note, that this is the case for all Weyl groups, which are not of type $A_{m}$, $E_{6}$ or $D_{2m+1}$, see [Hum72, p.71]).

Consider the orbit $L:=G\cdot p=\mathcal{P}^{-1}(\mathcal{P}(p))\subset X^{\infty}$. Any element $p^{\prime}\in L$ is contained in a unique Coxeter chamber $\Delta_{p^{\prime}}$. Consider the set $L^{op}_{p}$ of all elements $p^{\prime}$ in $L$ which are in an opposite Coxeter to $p$, thus such that the Coxeter chamber $\Delta_{p^{\prime}}$ through $p^{\prime}$ contains an antipode of $p$. Then $L^{op}_{p}$ is open and dense in the manifold $G\cdot p$, see, for instance, [KLP18]. Thus, we find an element $p^{\prime}\in O\cap L^{op}_{p}$.

If the isometry $I:\Delta\to\Delta$ is the identity (see the discussion above), then $p^{\prime}$ is an antipode of $p$ and we are done. If $I$ is not the identity, then looking at an apartment through $p$ and $p^{\prime}$ we deduce that the Tits-geodesic between $p$ and $p^{\prime}$ contains a point $q$ which is projected by $\mathcal{P}$ onto $I(p)$. Then $L$ contains all antipodes of $q$. By convexity, $q\in A^{\infty}$. As above, the set $L^{op}_{q}\cap O$ of elements in $O$ contained in a chamber opposite to $q$ is not empty. For any such element $p^{\prime}\in L^{op}_{q}\cap O$, the distance between $q$ and $p^{\prime}$ is $\pi$.

Thus, in both cases we have found a pair of antipodes $p\in O$ and $q\in A^{\infty}$, finishing the proof. ∎