Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 4 / d / Solution

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.