Kulkarni–Nomizu product, trace-free Ricci tensor, and Weyl curvature

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

For symmetric covariant two-tensors $h$ and $k$, their Kulkarni–Nomizu product is the covariant four-tensor

$\begin{array}{rl}{\displaystyle (h\odot k)(X,Y,Z,T)=}& {\displaystyle h(X,T)k(Y,Z)+h(Y,Z)k(X,T)}\\ {\displaystyle }& {\displaystyle -h(X,Z)k(Y,T)-h(Y,T)k(X,Z)\mathrm{.}}\end{array}$

This fixes the order and sign convention used below. In particular,

$(g\odot g)(X,Y,Z,T)=2(g(X,T)g(Y,Z)-g(X,Z)g(Y,T)),$

so the constant-sectional-curvature-$K$ tensor is $(K\mathrm{/}2)(g\odot g)$.

On an $n$-dimensional Riemannian manifold with $n\ge 1$, define the trace-free Ricci tensor by

${\mathrm{Ric}}_0:=\mathrm{Ric}-\frac{S}{n}g\mathrm{.}$

Indeed ${\mathrm{tr}}_g{\mathrm{Ric}}_0=S-(S\mathrm{/}n)n=0$. In dimension zero, where division by $n$ has no meaning, set ${\mathrm{Ric}}_0=0$; this is the unique covariant two-tensor on every zero-dimensional tangent space.

For $n\ge 3$, define the Weyl curvature tensor by

$W:=\mathrm{Rm}-\frac{1}{n-2}({\mathrm{Ric}}_0\odot g)-\frac{S}{2n(n-1)}(g\odot g)\mathrm{.}$

The next proposition proves that this is an algebraic curvature tensor with vanishing Ricci contraction and that it is the unique trace-free summand in the Ricci decomposition. Weyl curvature is not defined by this formula in dimensions zero, one, or two; the separate low-dimensional curvature formulas are also proved there.

All constructions are pointwise tensor operations on smooth tensors and so are smooth, including at boundary points. On an empty manifold of an allowed fixed dimension they give the corresponding unique empty tensor fields. No basis is chosen, and the finite contractions make no further family choice beyond the stated inherited assumption. Metric degeneracy is excluded by the Riemannian hypothesis.