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Let $\Omega\subset \mathbb{R}^3$ be a Lipschitz domain and let$S_\mathrm{curl}(\Omega)$ be the largest constant such that $$\int_{\mathbb{R}^3}|\nabla\times u|^2\, dx\geq S_{\mathrm{curl}}(\Omega)\inf_{\substack{w\in W_0^6(\mathrm{curl};\mathbb{R}^3)\\ \nabla\timesw=0}}\Big(\int_{\mathbb{R}^3}|u+w|^6\,dx\Big)^{\frac13} $$ for any $u$ in$W_0^6(\mathrm{curl};\Omega)\subset W_0^6(\mathrm{curl};\mathbb{R}^3)$ where$W_0^6(\mathrm{curl};\Omega)$ is the closure of$\mathcal{C}_0^{\infty}(\Omega,\mathbb{R}^3)$ in $\{u\inL^6(\Omega,\mathbb{R}^3): \nabla\times u\in L^2(\Omega,\mathbb{R}^3)\}$ withrespect to the norm $(|u|_6^2+|\nabla\times u|_2^2)^{1/2}$. We show that$S_{\mathrm{curl}}(\Omega)$ is strictly larger than the classical Sobolevconstant $S$ in $\mathbb{R}^3$. Moreover, $S_{\mathrm{curl}}(\Omega)$ isindependent of $\Omega$ and is attained by a ground state solution to thecurl-curl problem $$ \nabla\times (\nabla\times u) = |u|^4u $$ if$\Omega=\mathbb{R}^3$. With the aid of those results, we also investigateground states of the Brezis-Nirenberg-type problem for the curl-curl operatorin a bounded domain $\Omega$ $$\nabla\times (\nabla\times u) +\lambda u =|u|^4u\quad\hbox{in }\Omega$$ with the so-called metallic boundary condition$\nu\times u=0$ on $\partial\Omega$, where $\nu$ is the exterior normal to$\partial\Omega$.

A Sobolev-type inequality for the curl operator and ground states for the curl-curl equation with critical Sobolev exponent