Ricci decomposition of the Riemann tensor in dimension at least three

Directly exchanging the four inputs in [F1]'s formula shows that $h\odot g$ has both pair skews, pair interchange, and the cyclic Bianchi identity whenever $h$ is symmetric. Hence it is an algebraic curvature tensor. For such a tensor $F$, define its Ricci contraction by $c(F)(X,Y)={\sum }_{i}F(e_i,X,Y,e_i)$ in an orthonormal basis; [F3] makes this intrinsic.

Substitution into [F1]'s four-term formula gives $c(h\odot g)=(n-2)h+({\mathrm{tr}}_{g}h)g$: the four sums are respectively $({\mathrm{tr}}_{g}h)g$, $nh$, $-h$, and $-h$. In particular, $c(g\odot g)=2(n-1)g$.

By [F4]–[F5], $c(\mathrm{Rm})=\mathrm{Ric}$ and ${\mathrm{tr}}_g\mathrm{Ric}=S$, so ${\mathrm{tr}}_g{\mathrm{Ric}}_0=0$. Applying step 2.1 to [F1]'s formula for $W$ gives $c(W)=\mathrm{Ric}-{\mathrm{Ric}}_0-(S\mathrm{/}n)g=0$. The defining equation for $W$, rearranged, is the displayed decomposition.

More generally, suppose $\mathrm{Rm}=W^{\mathrm{\prime }}+h_0\odot g+a(g\odot g)$ with $c(W^{\mathrm{\prime }})=0$ and ${\mathrm{tr}}_g{h}_0=0$. Step 2.1 gives $\mathrm{Ric}=(n-2)h_0+2a(n-1)g$. Taking the metric trace yields $S=2an(n-1)$; subtracting $(S\mathrm{/}n)g$ then gives ${\mathrm{Ric}}_0=(n-2)h_0$. Thus $a=S\mathrm{/}(2n(n-1))$, $h_0={\mathrm{Ric}}_0\mathrm{/}(n-2)$, and the residual $W^{\mathrm{\prime }}$ equals $W$.

Let $n=3$ and choose an orthonormal basis $(e_1,e_2,e_3)$ using [F6]. Put $A_{ij}=W(e_i,e_j,e_j,e_i)$ for $i<j$. The three diagonal equations $c(W)(e_i,e_i)=0$ are $A_{12}+A_{13}=0$, $A_{12}+A_{23}=0$, and $A_{13}+A_{23}=0$, so all $A_{ij}$ vanish. Each off-diagonal equation $c(W)(e_i,e_j)=0$ has only the term indexed by the remaining basis vector, because the other two terms vanish by a pair skew; hence all three off-diagonal components of the symmetric bilinear form induced by $W$ on ${\mathrm{\Lambda }}^2{T}_{p}M$ also vanish. The pair symmetries in [F2] say these six entries determine $W$, so $W=0$.

Let $n=2$ and choose an orthonormal basis $(e_1,e_2)$. If $K=\mathrm{Rm}(e_1,e_2,e_2,e_1)$, then [F2] and [F4] give $\mathrm{Ric}(e_1,e_1)=\mathrm{Ric}(e_2,e_2)=K$ and $\mathrm{Ric}(e_1,e_2)=0$; hence [F5] gives $S=2K$. The tensors $\mathrm{Rm}$ and $(S\mathrm{/}4)(g\odot g)$ both have the algebraic curvature symmetries, and their single component on the one-dimensional space ${\mathrm{\Lambda }}^2{T}_{p}M$ is $K$, because $(g\odot g)(e_1,e_2,e_2,e_1)=2$. Therefore they are equal.