# An Einstein manifold has constant scalar curvature.

I know this is called Schur's lemma. But I cannot find a proof. All references available to me either does not give a proof, or says that it is similar to the lemma for sectional curvature, making use of the second Bianchi identity. But in the sectional curvature case, we can express explicitly $R_{ijkl}$ in $g_{ij}$ (in a unit orthogonal basis), while here we only have a sum $R_{ikjk}$. And I don't know how to manipulate the Bianchi identity to make the covariant derivative index uniform. Can anyone give some help?

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try any text on general relativity. The fact that the Einstein tensor has divergence 0 is one of the first calculations you find there. –  user8268 Sep 25 '13 at 14:45
Well, the Einstein condition says $R_{ab} = k g_{ab}$. Taking trace we have $R = R^a_a = k g^a_a = k n$ where $n$ is the dimension of the manifold. I suppose you had something else in mind? –  Marek Sep 25 '13 at 15:24
Given the reference to Schur's Lemma I'm guessing you're allowing the proportionality factor $k$ to vary on the manifold? Einstein manifolds have $k$ constant, in which case the proof is very easy as pointed out by Marek. –  Anthony Carapetis Sep 25 '13 at 15:51

There seems to be some disagreement in the literature about whether we assume $\lambda$ is constant from the outset when we write $\text{Ric}_{ij} = \lambda g_{ij}$. Here is a hint to a proof that $\lambda$ is constant using the classical second Bianchi identity $$R_{ijk\ell;m}+R_{ij\ell m;k}+R_{ijmk;\ell}=0\,.$$ Working with $g_{ij}(p) = \delta_{ij}$, consider $\sum\limits_{i,k,\ell} \delta_{k\ell} R_{iki\ell}$.
It is weird that I didn't notice that $R_{ijjk,i}$ and $R_{ijki,j}$ are the same if we are to contract $i,j$. –  Ash GX Oct 8 '13 at 6:08
$\DeclareMathOperator{\Rc}{Rc}\DeclareMathOperator{\Scal}{Scal}\DeclareMathOperator{\dv}{div}\DeclareMathOperator{\tr}{tr}$I believe what you want a proof of is this:
If a connected Riemannian $n$-manifold, $n>2$ satisfies $\Rc = f\ g$ for some $f \in C^\infty(M,\mathbb R)$, then $f$ is constant; so in particular $M$ is Einstein with factor $f$ and thus has constant scalar curvature $nf$.
Start with the twice-contracted second Bianchi identity $d \Scal = 2 \dv \Rc$. In this case this is $$d (f \tr g) = 2 \dv (f g).$$ The LHS is simply $n\ df$, while the RHS is $$2 g^{ij}\nabla_i(f g_{jk})\ dx^k= 2 g^{ij} g_{jk} \nabla_i f\ dx^k = 2\ df.$$ Thus we have $2\ df = n\ df$, so $n > 2$ means that $df=0$. Since $M$ is connected this implies $f$ is constant.