Harmonic one-forms are parallel under nonnegative Ricci curvature

On a closed Riemannian manifold with nonnegative Ricci curvature, integration of the Bochner-Weitzenbock formula for one-forms for a harmonic one-form givesBoth terms are nonnegative, so $\nabla \alpha =0$. Riemannian volume density makes the argument valid even without orientation.

On a compact connected Riemannian manifold without boundary with nonnegative Ricci curvature, the Bochner formula for one-forms makes every harmonic one-form parallel. Evaluation at any point is then injective: a parallel form that is zero there stays zero along every path. Hence $\dim {\mathcal{H}}^1(M)\le \dim M$. Flat tori attain equality. Positive-definite Ricci curvature at even one point instead forces ${\mathcal{H}}^1=0$, because the Bochner integral makes its Ricci contraction vanish everywhere.

On a closed connected Riemannian manifold with nonnegative Ricci curvature, harmonic one-forms are parallel under nonnegative Ricci curvature. Evaluation at one point injects their space into the cotangent space. The Hodge decomposition theorem identifies its dimension with the first Betti number, giving $b_1(M)\le \dim M$. For example, a genus-two closed orientable surface times a circle has dimension three and first Betti number five, so no Riemannian metric on that product has nonnegative Ricci curvature.