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By the Hodge decomposition theorem, the first Betti number is $b_1(M)=\dim {\mathcal{H}}^1(M)$. Part (b) makes every element of this space a parallel differential form. For any fixed $p\in M$, consider the linear evaluation mapIt is injective. A parallel form with ${\omega }_p=0$ remains zero under parallel transport along any path from $p$; connectedness of the smooth manifold makes every point reachable by such a path. Equivalently, the derivative of the squared norm of a parallel form is zero in every direction, so its norm is constant on the connected manifold. Either argument gives $\omega =0$ globally. Since $\dim T_p^{\ast }M=n$, the dimension bound for harmonic one-forms under nonnegative Ricci curvature gives

For an explicit obstruction in dimension three, take $N={\Sigma }_2\times S^1$, where ${\Sigma }_2$ is a closed orientable surface of genus two. It is a compact connected smooth three-manifold. A cell decomposition of ${\Sigma }_2$ has one vertex, four one-cells and one two-cell attached by a product of commutators. Its cellular boundary into the one-cells is zero, because each generator has total exponent zero in that attaching word; the boundary of every one-cell is also zero. Therefore $H^1({\Sigma }_2;\mathbb{R})={\mathbb{R}}^4$. The Künneth theorem in degree one and $H^1(S^1;\mathbb{R})=\mathbb{R}$ giveConsequentlyThis is the Betti-number obstruction to nonnegative Ricci curvature. It rules out every Riemannian metric on that underlying manifold, not merely a particular product metric.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.

The Bochner formula for one-forms, with the positive Hodge Laplacian, isHere $\Delta =d\delta +\delta d$ acts on one-forms. The covariant derivative $\nabla \theta$ is a section of $T^{\ast }M\otimes T^{\ast }M$, and its formal-adjoint composition, the rough Laplacian, isThis expression is independent of the local orthonormal frame. The Ricci term is the zero-order bundle endomorphism specified byThus the identity separates the differential energy of the form from the curvature contribution; the sign of the Ricci term matches positive sectional curvature on the sphere.

Let $\theta$ be a harmonic one-form on a compact connected manifold with nonnegative Ricci curvature. Pair the formula with $\theta$ and integrate:Both integrands are nonnegative and continuous, so each vanishes everywhere. In particular $\nabla \theta =0$, proving that harmonic one-forms are parallel under nonnegative Ricci curvature. This integration uses the Riemannian volume density and formal adjoints, so orientation is not necessary for the final conclusion.

Fix $p\in M$. The evaluation mapis injective: a parallel form with zero value at $p$ stays zero under parallel transport along every path, and connectedness makes every point reachable from $p$. ThereforeThis is the dimension bound for harmonic one-forms under nonnegative Ricci curvature. The bound is sharp: on a flat $n$-torus the $n$ constant coordinate one-forms are parallel and harmonic. If Ricci curvature is positive definite at even one point, the same integral and parallelism argument force every harmonic one-form to vanish.