The famous maximum principle of Hamilton states the following. Let $C$ be a convex $O(n)$-invariant subset of the space of algebraic curvature operators. Then if it is invariant under the ODE $$ \dot{R} = R^2 + R^\#,$$ (where $R^\#$ is quadratic in $R$ and is defined in terms of the adjoint action), then it is invariant unter the Ricci flow, i.e. a Ricci flow whose curvature starts in $C$ will remain in $C$ for all times where it is defined.

What is known about the converse? If a set $C$ is invariant under Ricci flow, must $C$ be invariant under the ODE? If not, are there counter-examples?

My idea to prove the converse would be the following. Suppose that $C$ is Ricci-flow invariant and choose $R \in \partial C$. We want to prove that $R^2 + R^\#$ points into $C$. Now take some compact manifold $M$, a point $x \in M$ and choose a metric $g$ such that the curvature tensor $R^M(x) = R$. Now because of Ricci-flow invariance, the Ricci flow starting at $g$ remains in $C$. By the evolution equation for the curvature tensor, we obtain that $$ \Delta R + R^2 + R^\#$$ points into $M$.

Therefore the remaining question: Can we arrange the metric $g$ in such a way that at the point $x$, we have $\Delta R(x) = 0$? Is this always possible?