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Let $\Omega \subseteq \mathbb{R}^3$ be a bounded, connected domain, and set $$\mathcal{V} = \{ \vec\phi = (\phi_1, \phi_2, \phi_3) \in C_c^\infty(\Omega): \nabla\cdot \vec\phi=0\}.$$ Denote $V$ to be the closure of $\mathcal{V}$ in $H^1(\Omega)$, and let $\mathbb{L}^2(\Omega)$ be the space of vector-valued functions from $\Omega$ to $\mathbb{R}^3$ with each component in $L^2(\Omega)$.

Since the curl operator $\nabla \times : V \to\mathbb{L}^2(\Omega)$ is continuous, there exists some adjoint operator $(\nabla \times)^*$ for which, given $\vec u, \vec v\in V$, $$(\nabla \times \vec u, \vec v )_{\mathbb{L}^2} = (\vec u, (\nabla \times)^* \vec v)_{\mathbb{L}^2} .$$

What is an expression for the adjoint for the curl operator?

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    $\begingroup$ What precisely do you mean by the spaces of $C^\infty$ and $L^2$ functions on $\Omega$? I presume you should be writing vector fields, not scalar functions? $\endgroup$ Sep 14, 2023 at 23:37
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    $\begingroup$ Have you thought about interpreting this in terms of $1$-forms and the exterior derivative operator to $2$-forms? $\endgroup$ Sep 15, 2023 at 0:13
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    $\begingroup$ Yes, the spaces defined are for vector fields, not scalar functions. I have edited my question to make this more clear. $\endgroup$
    – Matt E.
    Sep 15, 2023 at 14:29
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    $\begingroup$ I don't have any knowledge of differential forms, so unfortunately I can't understand your suggestion. In any case, I appreciate your input. $\endgroup$
    – Matt E.
    Sep 15, 2023 at 14:31

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This is an exercise in integration by parts. Since the Levi-Civita symbol satisfies $\epsilon_{ijk} = -\epsilon_{kji}$,

$$\sum_i\int v_i(\nabla\times u)_i dx= \sum_{i,j,k}\int v_i \epsilon_{ijk}\partial_j u_k dx = \sum_{i,j,k}\int \epsilon_{kji} \partial_j v_i u_k dx = \sum_k \int u_k (\nabla\times v)_k dx$$

Which means the curl operator is self-adjoint.

You can also think of curl as left matrix multiplication by a skew-symmetric “matrix” of partial derivatives, which picks up one minus sign when transferring onto $v$, and differentiation is skew-adjoint (by integration by parts) which picks up another minus sign, hence making curl self-adjoint.

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    $\begingroup$ Many thanks for your answer! $\endgroup$
    – Matt E.
    Sep 15, 2023 at 14:23

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