Is the intersection of two distinct sufficiently small metric spheres always empty, a point or a metric sphere of lower dimension?

Question

Let $(X,d)$ be an $n$-dimensional $(n< \infty)$ complete geodesic metric space, where any two points in $X$ are joined by a unique shortest geodesic. Let $S$ be a sufficiently small metric $(n-1)$-sphere in $X$. Let $\epsilon>0$ be the radius of $S$. Pick a point $p \in S$. Find a metric $(n-1)$-sphere $S'$ of radius $0<r < \epsilon/2$ around $p$. A $k$-dimensional metric sphere $S^k$ in $(X,d)$ of radius $r$ centered at $p$ is given by a $k$-dimensional subset of $(X,d)$: $S^k \subset \{x\in X|d(p,x) = r\}$. Here dimension I'm referring to covering dimension.

Is the intersection $S \cap S'$ always an $(n-2)$-dimensional metric sphere?

Answer 1

Let $C_\alpha$ denote the $n$-dimensional closed Euclidean solid cone with the cone angle $\alpha\in (0, \pi)$. Let $X_\alpha$ be the metric space obtained by gluing two copies $C^\pm_\alpha$ of $C_\alpha$ at their tips, and equipped with the natural path-metric. Let $o\in X_\alpha$ denote the common tip of the cones. I will use it as the center of the first sphere $S=S(o,1)$; I will use $\epsilon=1$. For every point $x\in S$, the sphere $S'=S(x,r)$ is $n-1$-dimensional. However, for each $r\in (0,1)$, for all sufficiently small $\alpha$, for every $x\in S\subset X_\alpha$, the intersection $S'\cap S$ is empty, hence, has dimension $-1$, not $n-2$. By modifying this construction one can build uniquely geodesic spaces where spheres as in your question are $n-1$-dimensional but have arbitrary dimension of their intersection, between $-1$ and $n-1$.

Edit. Here is a generalization, to ensure that the spheres $S, S'$ are connected (and locally path-connected) and the intersection $S\cap S'$ is nonempty.

Let $Y$ be a closed solid cone in the Euclidean space $E^k, 1\le k<n$, with the tip $o$. I will glue $Y$ to the space $X_\alpha$ as above so that $o$ is identified with the common tip $o\in X$ and two boundary rays of $Y$ are identified with geodesic rays in the two cones $C^\pm_\alpha \subset X$. Let $Z$ denote the resulting path-metric space. It is easy to check (say, using Reshetnyak gluing theorem) that $Z$ is a $CAT(0)$ space, hence, is uniquely geodesic.

Moreover, both spheres $S, S'\subset Z$ (defined as before) are connected and locally path-connected. Furthermore, for every $r$, there is $\alpha$ such that $S\cap S'$ is $(k-2)$-dimensional. (Note that $k-2<n-2$.)

By working a bit harder, one can modify this construction so that spheres are still connected and locally path-connected, while $S\cap S'$ is $(n-1)$-dimensional, where $n$ is the dimension of the ambient space.

Answer 2

Even for generic Riemannian metrics, the answer to the title question is no.

Consider the metric $ds^2=dx^2 + (2-x^2)dy^2+ dz^2$.

The geodesic equations for this metric are \begin{align} x'' &= -x y'^2 \\ y'' &= 2x x' y'/(2-x^2) \\ z'' &= 0 \end{align}

The geodesic that begins at $(-1,0,0)$ with initial time-derivative $(1.00,1,0)$ arrives at $(0,0.54,0)$ after distance $1.21$.

The geodesic that begins at $(-1,0,0)$ with initial time-derivative $(1.47,0,1)$ arrives at $(0,0,0.68)$ after distance $1.21$.

So the spheres of radius $1.21$ from $(\pm1,0,0)$ intersect in a topological circle that includes the four points $(0,\pm 0.54,0)$ and $(0,0,\pm 0.68)$.

The first two points are diametrically opposite on that topological circle, and they are $0.76$ units from the origin.

The second two points are also diametrically opposite on that topological circle, and they are $0.68$ units from the origin.

So the topological circle of intersection is not a metric circle, and the same would be true even for the intersections of spheres with smaller radii.