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!