# Proving a manifold with certain homogeneous property is Einstein

Let $$M$$ be a Riemannian manifold such that for all pairs of points $$(p, q), (r, s)$$ satisfying $$d(p, q) = d(r, s)$$, there exists an isometry $$f: M \to M$$ which takes $$p$$ to $$r$$ and $$q$$ to $$s$$. Prove that $$M$$ is an Einstein manifold.

My proof: let $$p, q$$ be arbitrary points in $$M$$ and $$v, w$$ unit vectors in $$T_p M$$ and $$T_q M$$, respectively. Take totally normal balls of radius $$\varepsilon$$ around $$p$$ and $$q$$ and choose points $$r, s$$ on the geodesics $$\gamma_v, \gamma_w$$ (i.e, the only geodesics starting at $$p$$ and $$q$$ with initial tangent vectors $$v$$ and $$w$$, respectively) such that $$d(p,r)=d(q,s) = \frac{\varepsilon}{2}$$. Then there exists an isometry $$f$$ which takes $$p$$ to $$q$$ and $$r$$ to $$s$$. Since isometries take geodesics to geodesics (and preserve lengths) and $$\gamma_v, \gamma_w$$ are the only geodesics connecting $$p$$ to $$r$$ and $$q$$ to $$s$$ with length $$\frac{\varepsilon}{2}$$, it follows that $$f \circ \gamma_v = \gamma_w$$, and hence that $$\mathrm{d}f_p(v) = w$$. Since the Ricci tensor is invariant under isometries, this proves that the Ricci tensor is constant on the unit bundle of $$M$$ and it follows easily (from polarization, for instance) that $$M$$ is Einstein.

Is this alright, do you see any mistakes in my argument or anything that could be improved? I`d appreciate any help. Thanks in advance!

EDIT: Following Kajelad's comment, I thought it would be good to add a minor change to the argument. Now I think it's entirely correct.

• Looks good to me. Commented Jun 24, 2021 at 4:21
• Looks good to me as well, save for a very minor nitpick: $\gamma_v$ and $\gamma_w$ are not necessarily the only geodesics joining their respective endpoints, but they are the only such geodesics with length $\varepsilon/2$. Commented Jun 24, 2021 at 4:54
• Is this your own proof really? Commented Jun 24, 2021 at 4:58
• Thanks everyone! And to answer @CFG, no, it's not entirely mine. I asked for hints to solve the problem to the T.A of the Riemannian Geometry course I'm taking, and he gave hints along the line my proof follows (to try and show arbitrary unit vectors could be reached using the isometries in the hypothesis) and a colleague suggested using geodesics, but everything else I thought on my own. Commented Jun 24, 2021 at 5:01
• @MatheusAndrade: Thanks for comment. I love the proof idea you wrote. Commented Jun 24, 2021 at 8:12