Einstein manifold

Remarks

Example: the unit sphere $S^n \subset \R^{n+1}$ with $n \ge 2$ has sectional curvature $1$, so $\operatorname{Ric} = (n-1) g$, and it is Einstein with $\lambda = n - 1$.

Non-example: the product $S^2 \times S^3$ of unit spheres has $\operatorname{Ric} = g$ on vectors tangent to $S^2$ and $\operatorname{Ric} = 2g$ on vectors tangent to $S^3$. No single constant $\lambda$ works, so it is not Einstein.

Taking the trace gives $\lambda = R/n$, where $R$ is the scalar curvature and $n = \dim M$, so an Einstein metric has constant scalar curvature. If $M$ is connected, $n \ge 3$ and $\operatorname{Ric} = f g$ for a smooth function $f$, the contracted Bianchi identity forces $f$ to be constant. In dimension $2$ every metric satisfies $\operatorname{Ric} = K g$ with $K$ the Gaussian curvature, so a surface is Einstein exactly when its curvature is constant. The same definition is used for pseudo-Riemannian metrics.