Betti-number vanishing under sharp curvature-operator positivity

Prove that if a closed $n$-dimensional Riemannian manifold has curvature operator that is $\frac{p(n-p)}{2}$-positive, then its $p$-th Betti number vanishes for every $2\leq p\leq n-2$.

Background

The sharp curvature criterion in Theorem \ref{SharpnessCriterion} converts an operator inequality for corrected action operators into vanishing results for harmonic forms. If the higher-dimensional Thorpe correction conjecture holds, the resulting threshold for the curvature operator would be p(n−p)2\frac{p(n-p)}{2}.

The proposed statement is unresolved in general but is proved in dimension six for the relevant third Betti number by Theorem \ref{TheoremThirdBettiNumber}. For orientable manifolds, Poincare duality would then imply the real homology-sphere conclusion under (n−2)(n-2)-positive curvature operator.

References

Together with Theorem \ref{SharpnessCriterion}, Conjecture \ref{ConjectureControllingLichnerowicz} would imply \begin{conjecture} \label{RiemannianConjecture} Let $(M,g)$ be a closed $n$-dimensional Riemannian manifold. If the curvature operator of $(M,g)$ is $\frac{p(n-p)}{2}$-positive, then the $p$-th Betti number vanishes, $b_p(M)=0,$ for $2 \leq p \leq n-2.$

In particular, if $(M,g)$ is orientable with $(n-2)$-positive curvature operator, then $M$ is a real homology sphere. \end{conjecture}

— A Thorpe Trick for the Bochner Technique  (2609.11653 - Wink, 10 Sep 2026) in Conjecture \ref{RiemannianConjecture} in the Introduction