Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 15 / 5

Let $M$ be a compact Riemannian manifold without boundary. Let $\delta =d^{\ast }$ be the formal $L^2$ adjoint operator of the exterior derivative, and let ${\Delta }_H=d\delta +\delta d$ be the Hodge Laplacian. Write ${\mathcal{H}}^r(M)=\ker {\Delta }_H\cap {\Omega }^r(M)$ for the smooth harmonic differential forms. The Hodge decomposition theorem states that this space is finite-dimensional andan $L^2$-orthogonal direct sum. Thus every smooth $r$-form is uniquely the sum of a harmonic, an exact and a coexact form; its exact and coexact components are unique, though the primitives used to write those components need not be unique. AlsoThe isomorphism sends a harmonic form to its de Rham cohomology class. For example, a closed coexact component $\delta \gamma$ must vanish because $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$; an exact harmonic form also vanishes by orthogonality. These facts explain uniqueness of the harmonic representative of a cohomology class.

On an oriented $n$-manifold the codifferential is $\delta =(-1{)}^{n(r+1)+1}\ast d\ast$ on $r$-forms. The general formulation above also holds on nonorientable Riemannian manifolds: use the globally defined Riemannian volume density for the $L^2$ inner product and its formal adjoint, rather than require a global ordinary volume form. The closed-manifold hypothesis is essential unless suitable boundary conditions are added.

Use the sign convention in the supplied scalar Laplace-Beltrami operator identity, and let $X={\omega }^{♯}$ be the metric-dual vector field. Integrate that Bochner formula for one-forms over the closed Riemannian manifold, with its volume density $d{V}_g$. The integral of a scalar Laplacian is zero: on an oriented manifold this is the Riemannian divergence theorem applied to its gradient, and the same density formula is valid without orientability. HenceBoth terms are nonnegative, because $\mathrm{Ric}\ge 0$. The smooth nonnegative integrand must therefore vanish everywhere: a strictly positive value would remain positive on an open neighborhood of positive volume and make its integral positive. In particular $|\nabla \omega {|}^2=0$ at every point, provingThus harmonic one-forms are parallel under nonnegative Ricci curvature. The same argument also gives $\mathrm{Ric}(X,X)=0$ pointwise. No orientation or connectedness assumption is needed for this step.

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.