Scalar curvature

This item assumes ${\mathrm{A}\mathrm{C}}_{\omega }$, namely countable choice. In the propagated dependency chain, that assumption is required through Ricci curvature is symmetric and basis independent; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

Use the inverse metric to raise the first covariant index of $\mathrm{Ric}$, obtaining the endomorphism ${\mathrm{Ric}}^{\mathrm{♯}}:T_{p}M\to T_{p}M$ characterized by

$g_p({\mathrm{Ric}}^{\mathrm{♯}}X,Y)={\mathrm{Ric}}_p(X,Y)\mathrm{.}$

The scalar curvature is the smooth function

$S(p):=\mathrm{tr}({\mathrm{Ric}}_p^{\mathrm{♯}})={\mathrm{tr}}_g{\mathrm{Ric}}_p\mathrm{.}$

The musical isomorphism makes ${\mathrm{Ric}}^{\mathrm{♯}}$ well-defined and smooth, and tensor contraction makes its trace intrinsic. In any orthonormal basis $(e_1,\dots ,e_n)$ of $T_{p}M$, the metric dual basis is $(g_p(e_i,\cdot ))$, so

$S(p)={\sum }_{i=1}^n{\mathrm{Ric}}_p(e_i,e_i)\mathrm{.}$

Thus the displayed sum is independent of the orthonormal basis. In dimension zero it is the empty sum $0$; on an empty manifold it is the unique empty smooth function. The definition is unchanged in dimension one or at boundary points. Positive definiteness excludes a degenerate metric, and fixing a single finite basis at an arbitrary point requires no global choice.