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?


1 Answer 1


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.)

  • $\begingroup$ 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. $\endgroup$ Apr 24, 2019 at 21:00
  • $\begingroup$ 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. $\endgroup$ May 25, 2019 at 8:19
  • $\begingroup$ Thanks Igor, that's completely right. $\endgroup$ May 28, 2019 at 2:19

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