2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/225916Let $(M^n,g)$, $n \ge 4$, be a compact simply-connected Riemannian manifold with nonnegative isotropic curvature. Given $0<l\le L$, we prove that there exists $\eps = \eps (l,L,n)$ satisfying the following: If the scalar curvature $s$ of $g$ satisfies $$ l \le s \le L $$ and the Einstein tensor satisfies $$ | Ric - \frac {s}{n}g | \le \eps$$ then $M$ is diffeomorphic to a symmetric space of compact type. This is a smooth analogue of the result of S. Brendle that a compact Einstein manifold with nonnegative isotropic curvature is isometric to a locally symmetric space.5 PagesDifferential Geometry53C21text