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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 $k \ne i$.

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

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

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

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