Totally geodesic spacelike surfaces in de Sitter space

Question

Let $S^2_1$ be the de Sitter space $$\{x:\langle x,x\rangle=x_0^2-x_1^2-x_2^2-x_3^2=-1\}$$ in Minkowski space $\mathbb {R}^3_1$ with $ds^2=dx_0^2-dx_1^2-dx_2^2-dx_3^2$. Is it true that for each pair of distinct points $x,y \in\ S^2_1$ there exists a totally geodesic spacelike surface in $S^2_1$ separating them?

Answer 1

So, topologically, $dS_3 \simeq \mathbb{R} \times S^2$ and is given by the hyperboloid of one sheet in $\mathbb{R}^{3,1}$ as you've written. The isometries of this surface are just the isometries of the ambient space which fix the origin $O$ of $\mathbb{R}^{3,1}$. And the geodesics on this surface are precisely the intersections, in the ambient space, between the surface and planes through $O$. Thus, for example, the $S^2$ at $x^0 = 0$ is a totally geodesic surface.

If two points on $dS_3$ are timelike- or null-separated, then I think the answer to your question is obvious. So let's consider when two points are spacelike-separated. Call the points $A$ and $B$, and consider the plane $OAB$.

Since $A$ and $B$ are spacelike-separated, there is an isometry of $\mathbb{R}^{3,1}$ which boosts and rotates such that the plane $OAB$ is mapped to $OA'B'$, where $A'$ and $B'$ lie in the $S^2$ at $x^0 = 0$.

Now we act with one more isometry. Take the midpoint $C'$ of the line $A'B'$, and consider the ray $OC'$. Now act with a boost in $\mathbb{R}^{3,1}$ which leaves $OC'$ invariant. This moves $A' \to A''$ and $B' \to B''$, where now, say, $A''$ has coordinate $x^0 > 0$ and $B''$ has coordinate $x^0 < 0$.

Now it is simple to see a totally geodesic spacelike surface which separates $A''$ and $B''$: it is given by the $S^2$ at $x^0 = 0$. Since we have obtained this result acting only with isometries, it must be true that any spacelike-separated points can be separated by a totally geodesic spacelike surface.