A Neumann Coulomb-gauge correction by contraction

Sobolev matrix algebra and boundary approximation. By [F3]–[F4] and the weak product rule obtained by smooth approximation, $\mathrm{\parallel }vw{\mathrm{\parallel }}_{W^{2,3}}\le K_A\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{2,3}}\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{2,3}}$: the second-derivative cross terms obey $\mathrm{\parallel }DvDw{\mathrm{\parallel }}_3\le \mathrm{\parallel }Dv{\mathrm{\parallel }}_6\mathrm{\parallel }Dw{\mathrm{\parallel }}_6$, and [F3] controls both $L^6$ norms. Similarly $\mathrm{\parallel }cv{\mathrm{\parallel }}_{W^{1,3}}+\mathrm{\parallel }vc{\mathrm{\parallel }}_{W^{1,3}}\le K_A\mathrm{\parallel }c{\mathrm{\parallel }}_{W^{1,3}}\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{2,3}}$. Passing the derivative test identities to $L^3$ limits proves completeness of the ambient Sobolev spaces from [F4]. The power series for $e^u$ therefore converges in $W^{2,3}$, locally uniformly in that norm, and defines a continuously differentiable composition there; termwise differentiation is justified by the convergent exponential majorants. For $u\in \mathfrak{s}\mathfrak{u}(2)$ pointwise, $u^2=-s^{2}I$ for a real $s$, so $e^u=\cos sI+(\sin s\mathrm{/}s)u$ (with the continuous value at zero) is unitary with determinant one. Polynomial Neumann approximants to $u\in X$ exponentiate to smooth $SU(2)$ maps with zero normal derivative; continuity gives $e^u\in {\mathcal{G}}_N$. Products of maps in ${\mathcal{G}}_N$ remain in it, and inverse is matrix adjoint and hence continuous. Gauge transformation by such maps sends $Y$ to $Y$, by approximating both inputs with smooth zero-normal inputs and using the above product estimates. Also $du\in Y$ for $u\in X$.

The exact nonlinear remainder. Put $h=e^u$ and define $R(u)=h^{-1}dh-du+h^{-1}(a+\lambda )h-a$. By step 1.1, $R(u)\in Y$ whenever $u\in X$. Let $E(u)=e^{-u}d(e^u)-du$. Differentiating the convergent matrix series, or integrating the series termwise, gives $e^{-u}d(e^u)={\int }_0^1{e}^{-tu}(du)e^{tu}dt$, hence $E(u)=A(u)du$ where $A(0)=0$ and $A$ and its first two matrix derivatives are uniformly bounded for $\mathrm{\mid }u\mathrm{\mid }\le 1$. More precisely $\mathrm{\mid }A(u)\mathrm{\mid }\le C\mathrm{\mid }u\mathrm{\mid }$ there. Thus $R(u)=E(u)+(e^{-u}a{e}^u-a)+e^{-u}\lambda e^u$, with $R(0)=\lambda$.

The quadratic differential term. Suppose $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}},\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{2,3}}\le \eta$ with $\eta$ small enough that [F3] gives $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\le 1$. The formula in step 2.1 gives $\mathrm{\mid }{d}^{\ast }E(u)\mathrm{\mid }\le C(\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }{D}^{2}u\mathrm{\mid }+\mathrm{\mid }Du{\mathrm{\mid }}^2)$. Subtracting the two formulas, using bounded first and second derivatives of $A$, and applying [F3]–[F4] gives $\mathrm{\parallel }{d}^{\ast }E(u){\mathrm{\parallel }}_3\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}^2$ and $\mathrm{\parallel }{d}^{\ast }(E(u)-E(v)){\mathrm{\parallel }}_3\le C\eta \mathrm{\parallel }u-v{\mathrm{\parallel }}_{W^{2,3}}$. Indeed terms with two first derivatives use $L^6\cdot L^6\subset L^3$, while a second derivative is paired with an $L^{\mathrm{\infty }}$ factor; differences of coefficients are bounded by $C\mathrm{\mid }u-v\mathrm{\mid }$, and the resulting terms use the same two estimates. These formulas hold for the Sobolev classes by the smooth approximation and product continuity of step 1.1.

The critical Coulomb coefficient and the perturbation. In $d^{\ast }(e^{-u}a{e}^u-a)$ the term $e^{-u}(d^{\ast }a)e^u-d^{\ast }a$ vanishes. Only derivatives falling on the exponentials remain, so their pointwise norm is at most $C\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }Du\mathrm{\mid }$ for $\mathrm{\mid }u\mathrm{\mid }\le 1$. Subtracting at $u,v$ gives at most $C\mathrm{\mid }a\mathrm{\mid }(\mathrm{\mid }D(u-v)\mathrm{\mid }+\mathrm{\mid }u-v\mathrm{\mid }(\mathrm{\mid }Du\mathrm{\mid }+\mathrm{\mid }Dv\mathrm{\mid }))$. By $L^4\cdot L^{12}\subset L^3$, this gives a difference bound $C\mathrm{\parallel }a{\mathrm{\parallel }}_4\mathrm{\parallel }u-v{\mathrm{\parallel }}_{W^{2,3}}$ when $\eta \le 1$. The contracted derivative of $e^{-u}\lambda e^u$ consists of $e^{-u}(d^{\ast }\lambda )e^u$ and terms bounded by $C\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }Du\mathrm{\mid }$. The same subtraction, [F3] and Holder give a difference bound $C\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{W^{1,3}}\mathrm{\parallel }u-v{\mathrm{\parallel }}_{W^{2,3}}$. Thus, after choosing one universal $C_0$, the preceding estimates prove $\mathrm{\parallel }{d}^{\ast }(R(u)-R(v)){\mathrm{\parallel }}_3\le C_0(\eta +\mathrm{\parallel }a{\mathrm{\parallel }}_4+\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{W^{1,3}})\mathrm{\parallel }u-v{\mathrm{\parallel }}_{W^{2,3}}$. The cancellation of $d^{\ast }a$ is why no higher norm of $a$ occurs in this bound.

A well-defined contraction, including compatibility. Define $T(u)=S(-d^{\ast }R(u))$ on the closed radius-$\eta$ ball in $X$. Step 2.1 and [F2] show $-d^{\ast }R(u)\in L_0^3$, so the Neumann solve is well defined for every iterate. Choose positive $\eta ,\rho$ so that $\eta \le 1$, the $L^{\mathrm{\infty }}$ restriction of step 3.1 holds, and $K_N{C}_0(\eta +\rho )\le 1\mathrm{/}4$. Then choose $\delta >0$ so small that $K_N{C}_0\delta \le 1\mathrm{/}4$ and $K_N\mathrm{\parallel }{d}^{\ast }\lambda {\mathrm{\parallel }}_3\le \eta \mathrm{/}4$ whenever $\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{W^{1,3}}\le \delta$; the latter follows from the fixed finite coefficient bound for $d^{\ast }$. For $\mathrm{\parallel }a{\mathrm{\parallel }}_4\le \rho$ and such $\lambda$, step 4.1 gives contraction constant at most $1\mathrm{/}2$, and $\mathrm{\parallel }T(0){\mathrm{\parallel }}_{W^{2,3}}\le \eta \mathrm{/}4$. Hence $\mathrm{\parallel }T(u){\mathrm{\parallel }}_{W^{2,3}}\le 3\eta \mathrm{/}4$ on that ball. It is a complete metric space by [F1] and the equivalence of its domain norm with $W^{2,3}$.

The correction and uniqueness. By [F4], $u_{j+1}=T(u_j)$ starting at zero converges to the unique fixed point $u$ in that ball. The equation $-\mathrm{\Delta }u=-d^{\ast }R(u)$ and $d^{\ast }du=-\mathrm{\Delta }u$ give $d^{\ast }(du+R(u))=0$. Since $d^{\ast }a=0$, the form $b=a+du+R(u)=h^{-1}dh+h^{-1}(a+\lambda )h$ is Coulomb and belongs to $Y$ by step 1.1. Conversely any $u\in X$ in the stated ball for which $b$ is Coulomb satisfies this same Neumann equation, so uniqueness of [F1] makes it a fixed point and thus this $u$. Also $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}\le \mathrm{\parallel }T(0){\mathrm{\parallel }}_{W^{2,3}}+\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}\le K_N\mathrm{\parallel }{d}^{\ast }\lambda {\mathrm{\parallel }}_3+\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}$, giving the first quantitative bound.

The correction size in the connection norm. The matrix power series and product estimates of step 1.1 imply $\mathrm{\parallel }{e}^{-u}d(e^u){\mathrm{\parallel }}_{W^{1,3}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}$, $\mathrm{\parallel }{e}^{-u}a{e}^u-a{\mathrm{\parallel }}_{W^{1,3}}\le C\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,3}}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,3}}$, and $\mathrm{\parallel }{e}^{-u}\lambda e^u{\mathrm{\parallel }}_{W^{1,3}}\le C\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{W^{1,3}}$ on the fixed small ball. For the middle bound, write $e^{-u}a{e}^u-a=(e^{-u}-1)a{e}^u+a(e^u-1)$; the terms with derivatives on $a$ have a factor bounded by $C\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$, and the other terms use $L^4\cdot L^{12}\subset L^3$. Adding the three bounds and using step 6.1 proves the claimed estimate for $b-a$, with $C$ independent of $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,3}}$.

Prescribed finite equivariance. Orthogonal pullback preserves the Laplacian, codifferential, means and all the domain norms. Constant unitary conjugation preserves matrix products, the Frobenius norm and the exponential. Therefore the transformation $u(x)\mapsto d{\tau }_{\gamma }^{-1}(u(\gamma x))$ preserves $X$ and commutes with $T$ when $a,\lambda$ have the stated equivariance; this uses the equivariant inverse of [F1]. The zero starting iterate is invariant, so every iterate and its limit are invariant. Exponentiation gives the assertion for $h$. A central unitary scalar acts trivially by conjugation, which explains why a $PU(2)$ action suffices. The existence, uniqueness, estimates and boundary compatibility are all supplied in steps 5.1–7.1; full AC is inherited as stated in [F5].