# What is the adjoint of the curl operator?

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?

• 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? Sep 14, 2023 at 23:37
• Have you thought about interpreting this in terms of $1$-forms and the exterior derivative operator to $2$-forms? Sep 15, 2023 at 0:13
• Yes, the spaces defined are for vector fields, not scalar functions. I have edited my question to make this more clear. Sep 15, 2023 at 14:29
• I don't have any knowledge of differential forms, so unfortunately I can't understand your suggestion. In any case, I appreciate your input. Sep 15, 2023 at 14:31

This is an exercise in integration by parts. Since the Levi-Civita symbol satisfies $$\epsilon_{ijk} = -\epsilon_{kji}$$,
$$\sum_i\int v_i(\nabla\times u)_i dx= \sum_{i,j,k}\int v_i \epsilon_{ijk}\partial_j u_k dx = \sum_{i,j,k}\int \epsilon_{kji} \partial_j v_i u_k dx = \sum_k \int u_k (\nabla\times v)_k dx$$
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.