A constant curvature manifold is Einstein

Question

I would like to prove that a Riemannian manifold $(M,g)$ with constant curvature $k$ is Einstein, with Einstein constant $(n-1)k$.

My attempt: Let $e_1, \dots e_n$ be an orthonormal basis for $T_pM$, where $p \in M$. Then I would like to show that $Ric(e_i) = (n-1)k e_i$ for every $i = 1, \dots, n.$ $$Ric(e_i) = \sum_{j=1}^n R(e_i, e_j) e_j $$ I can see that $g(Ric(e_i), e_i) = (n-1)k$, but I don't know how to prove that $g(Ric(e_i), e_k) = 0$ for every $k \ne i$.

Answer 1

$$ (R(e_i+e_k,e_j)e_j,e_i+e_k) $$ $$= ( R(e_i,e_j)e_j,e_i ) +2( R(e_i,e_j)e_j, e_k) + (R(e_k,e_j)e_j,e_k) =2k + 2( R(e_i,e_j)e_j, e_k ) $$

Here if $i\neq k$, $$ 2(Ric(e_i),e_k)=-2(n-1)k+ \sum_{j=1}^n ( R(e_i+e_k,e_j)e_j,e_i+e_k ) $$

$$ =-2(n-1)k+ 2(n-2)k+ 2( R(e_i+e_k,e_i)e_i,e_i+e_k ) $$

since $|e_i+e_k|=\sqrt{2}$

$$ =-2k +2( R(e_k,e_i)e_i,e_k ) =0 $$

Answer 2

The easiest way I know of to see this is the following: let $v \in T_{p} M$ be an unit vector and $\{v, e_2, \cdots, e_n \}$ be an orthonormal basis of $T_{p} M$. Then:

$$\operatorname{Ric}(v, v) = \sum_{2 \leq i \leq n} K(v, e_i) = (n-1) k$$

Since a symmetric $2$-tensor is entirely determined by its image on unit vectors, we get the desired result.