Ricci curvature

For $X,Y\in T_{p}M$, the Ricci curvature is

${\mathrm{Ric}}_p(X,Y):=\mathrm{tr}(Z⟼R_p(Z,X)Y)\mathrm{.}$

The trace is the basis-independent trace of this endomorphism of $T_{p}M$; no orthonormal basis is part of the definition. Equivalently, Ricci contracts the first input of the curvature endomorphism with its output. The next lemma proves smoothness and symmetry and derives the orthonormal-basis formula.

On a zero-dimensional tangent space the endomorphism and its empty trace are zero. On an empty manifold this is the unique empty two-tensor, and the same pointwise definition applies in dimension one and at a boundary point. No choice of bases over the points of $M$ is made.