Assume $(M,g)$ is a smooth Riemannian manifold, $p\in M$ and $\{e_i\}\subset T_pM$ is orthogonal basis (may not normal). If I have $$ K_{ij}=-1,~~~\forall~ i,j=1,...,n; i\ne j $$ where $K_{ij}$ is the sectional curvature of plane generated by $e_i,e_j$. Then whether I can get $$ K(\sigma)=-1,~~~\forall \sigma\subset T_pM $$ where $\sigma$ is 2-dim plane contain origin.
This problem is from the 3-section of 8-chapter of do Carmo's Riemannian Gemetry. I feel his way is a little troublesome. Therefore, I want to get the constant curvature from the $K_{ij}=-1$. But I don't know whether this idea is right and how to prove it.
PS: I can't understand the Laz's explain. Who can talk it detail ? Thanks very much.
In the section 3 of chapter 8 of do Carmo's Riemannian Geometry, do Carmo give an example of a space of constant curvature $-1$. Let $$ H^n =\{(x_1,...,x_n)\in \mathbb R^n; x_n >0\},~~~ g_{ij} (x_1,...,x_n) = \frac{\delta_{ij}}{x_n^2} $$ Let $f=\ln x_n$, then $$ x_n^2 R_{ijij} = -\sum_l f_l^2 + f_i^2 +f_j^2+f_{ii}+f_{jj} $$ where $f_i =\partial_i f$. Besides, if all four indices are distinct, we have $$ R_{ijkl}=0 $$ And if any three indices are distinct, we have $$ R_{ijk}^i = -f_kf_j -f_{kj}, ~~~ R_{ijk}^j = f_i f_k +f_{ki},~~~ R_{ijk}^k = 0 $$ Then we have
But in my view, do Carmo only prove $$ K_{ij} = -1 $$ If I want to get $K(\sigma)=-1$ for all $\sigma \subset T_pM$, I must to show $R_{ijkl}=0$ which have not the form $R_{ijij}, R_{ijji}$. do Carmo's proof seemly be not complete, since he do not show $R_{ijjj}=0, (i\ne j)$. In fact, I am not sure how to understand "all four indices are distinct", whether it contains $R_{1333}$.
Since I feel it's too much trouble and I feel my original question maybe right, I ask whether $$ K_{ij}=-1,~~~\forall~ i,j=1,...,n; i\ne j \Rightarrow K(\sigma)=-1,~~~\forall \sigma\subset T_pM $$ If so, we don't need to calculate all $R_{ijkl}$.
