# Generalized Hodge Decomposition on Manifolds with Boundary

This question is motivated by the problem of finding heat kernels to use for the renormalization of quantum field theories on manifolds with boundary.

If $$(\mathscr{E}, Q)$$ is an elliptic complex on a closed manifold $$M$$, then if one chooses a metric $$(\cdot,\cdot)$$ for $$\mathscr{E}$$, one can define the formal adjoint $$Q^*$$ to $$Q$$ and the cohomological degree-zero operator $$QQ^*+Q^*Q$$ is elliptic. Its kernel is precisely the cohomology of $$\mathscr{E}$$. Moreover, there is a decomposition

$$\mathscr{E}= \ker(Q)\cap\ker(Q^*)\oplus \text{Im}Q\oplus \text{Im}Q^*.$$

On a compact manifold with boundary, there is a similar decomposition (see Günter Schwarz: Hodge Decomposition - A Method for Solving Boundary Value Problems) for differential forms: a $$k$$ form can be written uniquely as the sum of a harmonic $$k$$-form (one which is both $$d$$- and $$\delta$$-closed), the boundary of a form with vanishing tangential component on $$\partial M$$, and the coboundary of a form with vanishing normal component on the boundary.

Is there a similar statement which applies for a general elliptic complex on $$M$$? I would be happy enough to know this for an elliptic complex which is isomorphic to one of the form $$(\mathscr{E’}\otimes \Omega^\bullet_{[0,\epsilon)},Q’\otimes 1+1\otimes d)$$ near $$\partial M$$ (after a choice of collar neighborhood for $$\partial M$$), and the metric on $$\mathscr{E}$$ is a product metric under this identification. Here, $$(\mathscr{E}’, Q’)$$ is an elliptic complex on $$\partial M$$. In this situation, it still makes sense to define the tangential and normal parts of an element $$e\in \mathscr{E}$$ near $$\partial M$$. So, more precisely, my question is: in the situation just described, can one write

$$\mathscr{E} = \ker Q\cap \ker Q^*\oplus Q\mathscr{T}^{k-1}\oplus Q^* \mathscr{N}^{k+1},$$

where $$\mathscr{T}^{k-1}$$ is the space of elements of degree $$k-1$$ in $$\mathscr{E}$$ with vanishing tangential component on the boundary and $$\mathscr{N}^{k+1}$$ is the space of elements of degree $$k+1$$ in $$\mathscr{E}$$ with vanishing normal component on the boundary?

The answer to this question as asked is no. However, you generally obtain something similar.

Consider $$D = Q + Q^*$$. By standard arguments, $$\ker(D) = \ker(Q)\cap \ker(Q^*).$$ (Of course $$D\Phi = 0$$ means that $$D^2\Phi=0$$. But $$D^2\Phi = 0$$ implies $$0 = \langle \Phi, (Q^*Q+QQ^*) \Phi \rangle = \|Q\Phi\|^2 + \|Q^*\Phi\|^2,$$ hence $$Q\Phi = Q^*\Phi = 0$$. In other words, $$\ker(D) \subseteq \ker (D^2) \subseteq \ker(Q)\cap \ker(Q^*).$$ The other direction is trivial.)

Now we have the integration by parts rule $$\int_X\Bigl( \langle D\Phi, \Psi\rangle - \langle \Phi, D \Psi \rangle \Bigr) = \int_{\partial X} \langle \Phi|_{\partial X}, \sigma(\nu)\Psi|_{\partial X}\rangle,$$ where $$\sigma$$ is the principal symbol of $$D$$, $$\nu$$ is the normal vector to the boundary (choose the one that makes the sign correct :P). This shows that a smooth $$D\Psi$$ is orthogonal to $$\ker(D)$$ if and only if $$\sigma(\nu)\Psi|_{\partial X} \perp \{\Phi|_{\partial X} \mid D\Phi = 0\}. \qquad (*)$$ Hence we obtain the splitting $$\mathscr{E} = \ker(D) \oplus D\mathscr{R},$$ where $$\mathscr{R}$$ is the space of $$\Psi$$ satisfying the orthogonality requirement $$(*)$$. If we further take into account the grading and figure out what the integration by parts formula gives us, we get $$\mathscr{E}^k = \mathscr{E}^k \cap \ker(D) \oplus Q \mathscr{T}^{k-1} \oplus Q^* \mathscr{N}^{k+1},$$ where \begin{aligned} \mathscr{T}^{k-1} &= \{ \Psi \in \mathscr{E}^{k-1} \mid \rho(\nu)\Psi|_{\partial X}, \perp L^{k}\},& \quad L^{k} &= \{\Phi|_{\partial X} \mid Q^*\Phi = 0, \Phi \in \mathscr{E}^k\}\\ \mathscr{N}^{k+1} &= \{ \Psi \in \mathscr{E}^{k+1} \mid \rho^*(\nu)\Psi|_{\partial X} \perp M^{k}\},& \quad M^{k} &= \{\Phi|_{\partial X} \mid Q\Phi = 0, \Phi \in \mathscr{E}^k\}.\end{aligned} Here $$\rho$$, $$\rho^*$$ are the principal symbols of $$Q$$ and $$Q^*$$, respectively, so that $$\sigma = \rho + \rho^*$$.

In your specific example, it so happens that $$L^k = M^k = \Gamma(X, \Lambda^k T^*X)$$, while $$\rho(\nu)$$ and $$\rho^*(\nu)$$ are wedging with, respectively insertion of the normal vector. Therefore $$\mathscr{T}$$ and $$\mathscr{N}$$ have the description you gave.

As an example where this is not the case, take $$0 \longrightarrow \mathscr{S}^+ \stackrel{Q}{\longrightarrow} \mathscr{S}^- \longrightarrow 0,$$ where $$\mathscr{S} = \mathscr{S}^+ \oplus \mathscr{S}^-$$ are the smooth sections of the spinor bundle over an even-dimensional spin manifold, and $$Q= D^+$$ is (half of) the Dirac operator. In this case, $$Q$$ and $$Q^* = D^-$$ are elliptic, and $$L^-$$, $$M^+$$ are not everything. In fact, it is a theorem that $$\mathscr{S}^+|_{\partial X} = \rho^*(\nu) L^- \oplus M^+, \qquad \text{and} \qquad \mathscr{S}^-|_{\partial X} = L^- \oplus \rho(\nu) M^+,$$ where each of the sums is direct. (Here I wrote $$\pm$$ instead of $$0$$, $$1$$ for the grading.)

• Thank you for your answer, Matthias! What is the standard argument that shows that $\ker(D)=\ker(Q)\cap \ker(Q^*)$? The only one I can think of uses the formal self-adjointness of $D$, which fails to hold in this case. – Eugene Rabinovich Apr 24 '19 at 21:00
• It should be noted that the argument works as written for first order $Q$. If $Q$ is a differential operator of higher order, then the Green formula (aka integration by parts formula) just preceding $(*)$ needs to be adjusted and the change propagated through to all the orthogonality conditions. – Igor Khavkine May 25 '19 at 8:19
• Thanks Igor, that's completely right. – Matthias Ludewig May 28 '19 at 2:19