# Prove that an homogeneous and isotropic Riemannian manifold has constant sectional curvature

I have problems in proving that an homogeneous and isotropic Riemannian manifold has constant sectional curvature.

This is my attempt:

By definition, the manifold $M$ has constant sectional curvature $k\in \mathbb{R}$ if $\;k(\pi)=k\;$ $\forall p\in M$ and $\forall$ $2$-plane $\pi\in T_pM.$

So I have to prove these two facts:

CLAIM $1:$ If $p,q\in M$ and $\pi, \tilde \pi \in T_p M$ then $k(\pi)=k(\tilde \pi)$.

In order to prove this I use the fact that $M$ is homogeneous, so there is $F \in Iso(M)$ s.t. $F(p)=q$.

Then I can write $\pi=span\{v,w\}$ and $\tilde \pi=span\{\tilde v,\tilde w\tilde\}$ with both $\{v,w\}$ and $\{\tilde v,\tilde w\tilde\}$ orthonormal and such that $\tilde v=dF_p (v)$ and $\tilde w=dF_p (w)$.

Then since the tensor of curvature $R$ is invariant under isometries I have claim $1$.

CLAIM $2$ Let be $p\in M$ and $\pi_1, \pi_2$ $2$-planes in $T_p M$ then $k(\pi_1)=k(\pi_2)$.

I think that here I should use the fact that $M$ is isotropic.

I can write $\pi_1=span\{v_1,w_1\}$ and $\pi_2=span\{v_2,w_2\}$ with both $\{v_1,w_1\}$ and $\{v_2,w_2\}$ orthonormal. Then there are $F_1, F_2 \in Iso(M)$ s.t. $F_1(p)=F_2(p)=p$ and $(dF_1)_p(v_1)=w_1$ and $(dF_2)_p(v_2)=w_2$.

From this fact does it f0llow that $k(\pi_1)=k(\pi_2)$?

Is it correct hat I've done? How can I conclude?

Thanks for the help!

• I don't understand. What you've proven in claim 1 is that $k(dF_p \pi) = k(\pi)$. But why can you always pick an $F$ such that $dF_p \pi = \tilde \pi$? (Isn't $\mathbb{CP}^2$ an isotropic manifold? It's not a space form.) Being isotropic only lets you move one vector where you want it to go, not two. – user98602 Jul 15 '15 at 9:43
• The claim is true only in $3$ dimensions. – Jack Lee Jul 15 '15 at 11:14