Proving that some subset of $ \mathbb{H}^{2} \times \mathbb{H}^{2} $ is connected.

Question

Since we know that the hyperbolic plane $\mathbb{H^{2}}$ is metric space with metric $ d_{H}(z,w) = 2 \tanh^{-1} \frac{ \mid z - w \mid}{ \mid z - \overline{w} \mid }$, where $ z, w \in \mathbb{H^{2}}$. Now we define the distance function on $\mathbb{H^{2}} \times \mathbb{H^{2}} $ given by $$ \gamma(z,w) = ( (d_{H}(z_{1},w_{1}))^{2} + (d_{H} z_{2},w_{2}))^{2})^{\frac{1}{2}} $$ where $$ z = ( z_{1}, z_{2}) , w = (w_{1}, w_{2}) \in \mathbb{H^{2}} \times \mathbb{H^{2}}.$$

Now consider the set $ E (z,w) = \lbrace x \in \mathbb{H^{2}} \times \mathbb{H^{2}} \mid \gamma( x, z) = \gamma ( x, w) \rbrace $, where $ x = ( x_{1}, x_{2} )$. I want to prove that $E(z,w)$ is connected set. if $ z_{1} = w_{1}$ or $z_{2} = w_{2}$, in that case $E (z,w)$ is simply the product of $ \mathbb{H^{2}}$ with a straight line and hence connected. But if $z_{i} \neq w_{i}$ for $ i = 1 , 2$, In that case why it is connected? Can some help in carrying out the proof?

Answer 1

Even more is true. Suppose that $M$ is a simply connected complete Riemannian manifold of nonpositive curvature (a Hadamard manifold). (Your space is Hadamard since it is the product of two hyperbolic spaces.) I will denote $d$ the Riemannian distance function on $M$ and $xy$ the geodesic segment connecting $x$ to $y$. For $x, y\in M$ define the bisector $$ B(x,y)=\{z\in M: d(z,x)=d(z,y)\}, $$ open half-spaces $$ V(x,y)=\{z\in M: d(x,z)< d(z,y)\} $$ and closed half-spaces $$ \bar{V}(x,y)=\{z\in M: d(x,z)\le d(z,y)\}. $$ It is clear that open half-spaces are open subsets of $M$ and closed subspaces are closed subsets.

Theorem. For any pair of points $x, y\in X$ the bisector $B(x,y)$ is connected. Furthermore, $B(x,y)$ is acyclic: $\tilde{H}^i(B(x,y))=0$ for all $i\ge 0$.

Proof. I first establish some basic properties of half-spaces and bisectors. Recall that a subset $A\subset M$ is star-like with respect to $x\in A$, if $xa\subset A$ for all $a\in A$. Since geodesic segments $xa$ depend continuously on $a$, each star-like subset is contractible.

Lemma 1. $V(x,y), \bar{V}(x,y)$ are star-like with respect to $x$. Moreover, for each $a\in B(x,y)$, $xa\cap B(x,y)=\{a\}$.

Proof. Follows from the triangle inequality. qed

For each pair $x, y$ define the following subset $A(x,y)\subset \bar{V}(x,y)$: $$ A(x,y)=\bigcup_{z\in B(x,y)} zx - \{x\}.$$

For each $a\in M -\{x\}$ define the geodesic ray $r_a: [0,\infty)\to M$ with $r_a(0)=x$, $a\in R_a=r_a([0,\infty))$. Then $A(x,y)$ is the set of points $a\in V(x,y)-\{x\}$ for which $R_a\cap B(x,y)$ is nonempty and, hence, is a singleton $\{b\}$. Set $f_{x,y}(a):=b$.

Lemma 2. The map $f_{x,y}: A(x,y)\to B(x,y)$ is surjective and continuous.

Proof. Surjectivity is clear; continuity follows from continuous dependence of $r_a$ on $a$. qed

Thus, $f_{x,y}$ is a retraction of $A(x,y)$ to $B(x,y)$. Moreover, this map is a deformation retraction, a homotopy-equivalence $A(x,y)\to B(x,y)$. This follows from continuity of the map $a\mapsto d(a, f_{x,y}(a))$, $a\in A(x,y)$.

Lemma 3. For all $x, y\in M$, $A(x,y)$ is open in $\bar{V}(x,y)$.

Proof. Take $a\in A(x,y)$; then the geodesic ray $r=r_a: [0,\infty)\to M$ satisfies $R_a\cap B(x,y)\ne\emptyset$. Therefore, there exists $t_0$ such that $r(t_0)\in V(y,x)$. Consider a sequence $a_i\to a$, $a_i\in V(x,y)$. I claim that $a_i$ belongs to $A(x,y)$ for large $i$; this will imply that $a$ is in the interior of $A(x,y)$ and, hence, $A(x,y)$ is open. For each $a_i$ define the geodesic ray $r_i:=r_{a_i}$; then $\lim_{i\to\infty} r_i=r$. Hence $$ c_i=r_i(t_0)\to c=r_(t_0)\in V(y,x).$$ Thus for all sufficiently large $i$, $c_i\in V(y,x)$. By the intermediate value theorem, the segment $xc_i$ has nonempty intersection with $B(x,y)$ for all large $i$. \qed

Define the following open subsets of $M$: $$ U(x,y)= V(x,y) \cup A(y,x) $$ $$ W(x,y)= U(x,y)\cap U(y,x). $$ The subset $U(x,y)$ deformation retracts to $\bar{V}(x,y)$ by the map which is the identity on $V(x,y)$ and equals $f_{y,x}$ on $A(y,x)$. In particular, $U(x,y)$ is homotopy-equivalent to $\bar{V}(x,y)$ and, hence, is contractible. Similarly, we have a deformation retraction $W(x,y)\to B(x,y)$ defined using the maps $f_{x,y}, f_{y,x}$. Hence, $W(x,y)$ is homotopy-equivalent to $B(x,y)$. Lastly, note that $$ M= U(x,y) \cup U(y,x). $$

Now, we can prove acyclicity of $B(x,y)$. By the above considerations, it suffices to prove acyclicity of $W(x,y)$. We have the Mayer-Vietoris sequence (of reduced homology groups) associated with the above open cover of $M$: $$ ... \to 0= \tilde{H}_{i+1}(M) \to \tilde{H}_i(W(x,y))\to \tilde{H}_i(U(x,y))\oplus \tilde{H}_i(U(y,x))=0 \to ... $$ since $M$ and $U(x,y), U(y,x)$ are all contractible. Therefore, $\tilde{H}_i(W(x,y))=0$ for all $i$ and, hence, $B(x,y)$ is acyclic. qed

Remark. Here is what I do not know:

1. Is $B(x,y)$ contractible? (Edit: actually, yes.)

2. Is it a (smooth) submanifold? (Edit: Actually, yes.)

3. Homeomorphic to $R^{n-1}$ where $n=dim(M)$?

Edit. $B(x,y)$ is indeed a smooth submanifold which means that Lemma 1 suffices for the proof of the theorem as one can use the tubular neighborhood theorem instead of the explicit construction of the neighborhood $W(x,y)$ for $B(x,y)$. The fact that $B(x,y)$ is smooth can be seen by considering the function $h(z)= d^2(z,x)- d^2(z,y)$. By working harder one can see that this function is onto ${\mathbb R}$ and that every gradient line of $h$ intersects every level set of $h$ at least (and, hence, exactly) once. This implies that $M$ is diffeomorphic to $B(x,y)\times {\mathbb R}$: Send $z\in M$ to the pair $(z_0, h(z))$, where $z_0\in B(x,y)$ is the intersection point of the gradient line of $h$ through $z$ with the bisector $B(x,y)$. It then follows that $B(x,y)$ is contractible. However, a priori, it is not homeomorphic to $R^{n-1}$ as the product of the Whitehead manifold $W^3$ with $R$ is diffeomorphic to $R^4$, but $W^3$ is not homeo to $R^3$.