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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.

Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.