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$\DeclareMathOperator\div{div}\DeclareMathOperator\curl{curl}$Consider a div–curl system \begin{align*} Lu &= (\div(u), \curl(u)) \text{ in } \Omega \subset M, \text{ a 3-manifold}, \\ u &= 0 \text{ on } \partial \Omega. \end{align*} Here, $u$ can be a $p$-covariant symmetric tensor field, $p \geq 0$. These are defined to be $(p-1)$-covariant and $p$-covariant, respectively by: \begin{align*} \div(u)_{a_1 \cdots a_{p-1}} &= \nabla^a u_{a_1 \cdots a_{p - 1} a} \\ \curl(u)_{a_1 \cdots a_{p}} &= {\epsilon}_{a_1}^{\ \ \ \ ab}\nabla_a u_{b a_2 \cdots a_p}. \end{align*}

The operator $L = (\div, \curl)$ is a first-order elliptic operator. Is it known whether or not this operator is injective in various natural settings (e.g. on a compact Riemannian manifold with or without boundary, on an unbounded complete Riemannian manifold, weighted/unweighted Sobolev spaces…)?

In, for instance, flat space $\Omega \subset \mathbb{R}^3$ bounded, $L$ is injective on the domain $H^k_0(\Omega)$ (trace-zero vector-valued functions in $H^k(\Omega)$), this follows from the vector calculus identity $\nabla \times (\nabla \times v) = \nabla(\nabla \cdot v) - \Delta v$. If $\div(v) = 0$ and $\curl(v) = 0$ with $v \in H^k_0(\Omega)$, then $\Delta v = 0$, and this implies $v = 0$ since the Laplacian is injective in this setting.

However, in a curved space this vector calculus identity gives only \begin{align*} \operatorname{curl}^2 (v)_a = \nabla_a \operatorname{div}(v) + R_{ae}v^e - \Delta v_a \end{align*} and the resulting equation obtained from the divergence and curl vanishing is \begin{align*} \Delta v_a - R_{ae} v^e = 0, \end{align*} and I am not sure of the injectivity of the operator $-\Delta + \operatorname{Ric}$.

Edit: I think for vector fields on Riemannian manifolds you have the same result if the topology is trivial: by the Poincaré Lemma, if the curl of $v$ is zero, the it is equal to the gradient of a potential function $\phi$; the divergence being zero then gives $\Delta \phi = 0$. Assuming that the spectrum of $-\Delta$ is positive (a result known for $M$ a compact manifold; I'm unsure if it holds for bounded domains), this would give the result. However, 1) I'm unsure, as just mentioned, of this positivity of the spectrum of $-\Delta$ for non-compact manifolds/manifolds with boundary; and 2) the question is still unresolved for higher-rank tensor fields.

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  • $\begingroup$ Does the Hodge decomposition of forms answer your question? It sounds like you're taking part in the time-honored tradition of accidentally rediscovering de Rham cohomology! $\endgroup$ Commented Oct 23, 2023 at 15:51
  • $\begingroup$ @DanielShapero - I think Hodge decomposition is sort of what I want, but I'd like some analogous statement for (not necessarily alternating) tensor fields. Are you aware of any such extensions? $\endgroup$
    – Chris
    Commented Oct 23, 2023 at 22:07
  • $\begingroup$ Hm there are other complexes besides the de Rham complex. For example there's a cochain complex associated to the linear elasticity problem, for which one of the spaces consists of symmetric tensors; see this paper. $\endgroup$ Commented Oct 23, 2023 at 23:08
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    $\begingroup$ For compact manifolds, you already have a fairly trivial obstruction: the Riemannian metric itself is both curl-free and divergence free. (Interestingly this is ruled out if you look at manifolds with boundary or non-compact manifolds, the latter using decay at infinity.). This extends to all $p$ even. $\endgroup$ Commented Dec 15, 2023 at 1:33
  • $\begingroup$ There are some sufficient conditions for injectivity following from the Bochner argument. For example, if the manifold is conformally flat (vanishing Weyl curvature; this condition can be replaced also by a sign condition on the Weyl curvature that is harder to state) and has non-negative Ricci curvature, then any element in the kernel must be parallel; if in addition Ricci is strictly positive at at least one point, then the the kernel can contain only those tensors built from symmetric products of the metric with itself. $\endgroup$ Commented Dec 15, 2023 at 2:27

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