A topological PEC zero mode on a spherical shell

The function $-1\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ is smooth on the shell and $\mathrm{\nabla }(-1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })=x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^3=u$. Therefore $\mathrm{curl}u=0$. Direct differentiation gives $\mathrm{div}u={\sum }_{i=1}^3(\mathrm{\mid }x{\mathrm{\mid }}^{-3}-3x_i^2\mathrm{\mid }x{\mathrm{\mid }}^{-5})=0$. For $0<\delta <1\mathrm{/}8$, choose a smooth radial cutoff ${\chi }_{\delta }(r)$ that vanishes near $r=1,2$ and equals one on $[1+2\delta ,2-2\delta ]$. The field $u_{\delta }(x)={\chi }_{\delta }(\mathrm{\mid }x\mathrm{\mid })u(x)$ is smooth and compactly supported in $\mathrm{\Omega }$. Both $\mathrm{\nabla }{\chi }_{\delta }$ and $u$ are radial, so $\mathrm{curl}{u}_{\delta }=0$. Since $\mathrm{\mid }u\mathrm{\mid }\le 1$ on the shell and $u-u_{\delta }$ is supported in a layer of volume $O(\delta )$, $\mathrm{\parallel }u-u_{\delta }{\mathrm{\parallel }}_{H(\mathrm{curl})}\to 0$. Thus [F1] gives $u\in H_0(\mathrm{curl})$ and $u\in K$.

If $q\in H_0^1(\mathrm{\Omega })$, choose $q_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with $q_j\to q$ in $H^1$ by [F3]. For each $j$, integration by parts and $\mathrm{div}u=0$ give $(u,\mathrm{\nabla }{q}_j{)}_{L^2}=-(\mathrm{div}u,q_j{)}_{L^2}=0$. Passing to the limit shows $(u,\mathrm{\nabla }q{)}_{L^2}=0$. Hence this zero mode survives the zero-boundary gradient gauge.

Suppose for contradiction that $u=\mathrm{\nabla }q$ for some $q\in H_0^1(\mathrm{\Omega })$. For $1<r<2$ define the spherical mean $M(r):=\frac{1}{4\pi }{\int }_{S^2}q(r\omega )d\omega$. For $q_j$ from step 2.1, define $M_j$ by the same formula. Polar coordinates, Jensen's inequality and [F2] show $\mathrm{\parallel }{M}_j-M_k{\mathrm{\parallel }}_{H^1(1,2)}\le C\mathrm{\parallel }{q}_j-q_k{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$: the derivative is the spherical average of $\mathrm{\nabla }(q_j-q_k)(r\omega )\cdot \omega$, and $1\le r^2\le 4$. Thus $M_j\to M$ in $H^1(1,2)$. Each $M_j$ is smooth and compactly supported in $(1,2)$, so [F3] gives $M\in H_0^1(1,2)$. For a smooth $g$ on $[1,2]$, choose $s$ with $\mathrm{\mid }g(s)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(1,2)}$; then $\mathrm{\mid }g(r)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }{g}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\le \sqrt{2}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^1}$ for every $r$. Applying this estimate to $M_j-M_k$ makes $(M_j)$ uniformly Cauchy. Its continuous limit has value zero at both endpoints, and it represents $M$ because $M_j\to M$ in $L^2$. Thus $M(1)=M(2)=0$.

For the smooth $q_j$, differentiation under the spherical integral gives $M_j^{\mathrm{\prime }}(r)=(4\pi {)}^{-1}{\int }_{S^2}\mathrm{\nabla }{q}_j(r\omega )\cdot \omega d\omega$. Passing to the $H^1$ limit and using $\mathrm{\nabla }q=u$ yields $M^{\mathrm{\prime }}(r)=(4\pi {)}^{-1}{\int }_{S^2}{r}^{-2}d\omega =r^{-2}$ in distributions. Therefore $M(r)=c-r^{-1}$ for some constant $c$. The endpoint values from step 3.1 would force $c=1$ and $c=1\mathrm{/}2$ simultaneously, a contradiction. Thus $u\mathrm{\notin }\mathrm{\nabla }{H}_0^1(\mathrm{\Omega })$.

The field is a nonzero member of the full curl kernel by step 1.1, so it is excluded from $X=K^{\perp }$; by [F4] it is a zero eigenfield of the unquotiented pencil. Step 4.1 shows why the gradient-only gauge would fail on this shell. It is not a positive eigenvalue of the contractible-cavity approximation theorem.