Coulomb gauges from small critical curvature on four-balls

A constructed radial gauge. Put $M(s,x)={\sum }_i{x}_i{A}_i(sx)$ and define $P(s,x)$ by the convergent iterated-integral series $I+{\sum }_{k\ge 1}(-1{)}^k{\int }_{0<t_k<\cdots <t_1<s}M(t_1,x)\cdots M(t_k,x)d{t}_k\cdots d{t}_1$. On the compact parameter set every term has norm at most $L^k\mathrm{/}k!$; any fixed number $N$ of spatial derivatives is bounded by $C_N{k}^N{L}_N^k\mathrm{/}k!$. Uniform convergence of these derivative series, followed by the coordinate-segment fundamental theorem, identifies them as the derivatives of the sum. The series also gives the integral equation $P(s,x)=I-{\int }_0^{s}M(t,x)P(t,x)dt$, which also gives smoothness in $s$. It satisfies ${\mathrm{\partial }}_{s}P=-MP$, $P(0,x)=I$. Since $M$ is skew-Hermitian and trace zero, differentiation gives ${\mathrm{\partial }}_s(P^{\ast }P)=0$; thus $P$ is unitary, and the two-by-two determinant derivative gives ${\mathrm{\partial }}_s\det P=-\mathrm{tr}(M)\det P=0$, so $P\in SU(2)$. Uniqueness of the integral equation follows by iterating its difference estimate $k$ times and taking $k\to \mathrm{\infty }$ using $L^k\mathrm{/}k!\to 0$. Set $U(x)=P(1,x)$. Rescaling the integration parameters gives $U(sx)=P(s,x)$, so $x\cdot dU=-(x\cdot A)U$. Therefore $C=A^U$ is smooth and $x\cdot C=0$.

Gauge algebra and the scaling path. The matrix identity $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $F_{A^g}=g^{-1}{F}_{A}g$ and $(A^g{)}^h=A^{gh}$ by expansion and cancellation. It applies to smooth gauges first; [F4]–[F5] pass it to $W^{2,3}$ gauges and their $W^{1,3}$ connection forms. Unitarity preserves pointwise curvature norms. Define $A_t(x)=tC(tx)$ for $0\le t\le 1$. It is a smooth path of smooth zero-normal forms, starts at zero, and ends at $C$. Its curvature is $F_{A_t}(x)=t^2{F}_C(tx)$, hence $\mathrm{\parallel }{F}_{A_t}{\mathrm{\parallel }}_2=\mathrm{\parallel }{F}_C{\mathrm{\parallel }}_{L^2(B_t)}\le \mathrm{\parallel }{F}_A{\mathrm{\parallel }}_2$ for $t>0$, and $\mathrm{\parallel }{F}_{A_t}{\mathrm{\parallel }}_3=t^{2\mathrm{/}3}\mathrm{\parallel }{F}_C{\mathrm{\parallel }}_{L^3(B_t)}\le \mathrm{\parallel }{F}_C{\mathrm{\parallel }}_3$. The same inequalities at $t=0$ are immediate. There are finite constants $L,X$ for which $\mathrm{\parallel }{A}_t-A_s{\mathrm{\parallel }}_{W^{1,3}}\le L\mathrm{\mid }t-s\mathrm{\mid }$ and $1+{\sup }_t(\mathrm{\parallel }{A}_t{\mathrm{\parallel }}_{W^{1,3}}+\mathrm{\parallel }{F}_{A_t}{\mathrm{\parallel }}_3)\le X$: differentiate $tC(tx)$ and $t^{2}DC(tx)$ in $t$, using bounded derivatives of the fixed smooth $C$ through order two.

Uniform critical improvement in any admissible gauge. Choose $\rho >0$ no larger than ${\rho }_c,{\rho }_2,{\rho }_3$ in [F1]–[F2]. Call $t$ admissible if some $g\in {\mathcal{G}}_N$ has $a=A_t^g\in Y_3$, $d^{\ast }a=0$ and $\mathrm{\parallel }a{\mathrm{\parallel }}_4\le \rho$. All such forms belong to $Y_2$ by finite-volume norm comparison of their smooth normal approximants. By [F1], curvature covariance and [F4], there is a fixed $K_0$ such that $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,2}}\le C_2\mathrm{\parallel }{F}_A{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }a{\mathrm{\parallel }}_4\le K_0\mathrm{\parallel }{F}_A{\mathrm{\parallel }}_2$. Also $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,3}}\le C_3\mathrm{\parallel }{F}_{A_t}{\mathrm{\parallel }}_3\le C_{3}X$. Choose the theorem's fixed $\epsilon >0$ so that $K_0\epsilon \le \rho \mathrm{/}4$. Consequently every admissible gauge has the improved margin $\mathrm{\parallel }a{\mathrm{\parallel }}_4\le \rho \mathrm{/}4$, and its higher connection norm is bounded by a constant depending only on $X$. These estimates require only the given curvature bound, once admissibility holds.

A uniform bound on the transported perturbation. In an admissible gauge, $dg=ga-A_{t}g$ holds as a weak matrix identity. Unitarity and [F4] give $\mathrm{\parallel }dg{\mathrm{\parallel }}_{12}\le C(\mathrm{\parallel }a{\mathrm{\parallel }}_{12}+\mathrm{\parallel }{A}_t{\mathrm{\parallel }}_{12})\le C_X$, with $C_X$ independent of $t$ and of the chosen admissible gauge. For another parameter $s$, set $\lambda =g^{-1}(A_s-A_t)g$. Product differentiation, $g^{-1}=g^{\ast }$, Holder and $W^{1,3}\hookrightarrow L^4$ give $\mathrm{\parallel }\lambda {\mathrm{\parallel }}_{W^{1,3}}\le C(1+\mathrm{\parallel }dg{\mathrm{\parallel }}_{12})\mathrm{\parallel }{A}_s-A_t{\mathrm{\parallel }}_{W^{1,3}}\le C_{X}L\mathrm{\mid }s-t\mathrm{\mid }$. Moreover $\lambda \in Y_3$: approximate $g$ by the defining smooth Neumann gauges of ${\mathcal{G}}_N$; conjugating the smooth zero-normal form $A_s-A_t$ by these gauges preserves zero normal component, and [F5] gives convergence in $W^{1,3}$.

A uniform admissibility step. Let $K_4$ be the fixed embedding norm from $W^{1,3}$ to $L^4$. Using the uniform bound for $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,3}}$ in step 3.1, choose ${\delta }_X>0$ below the perturbation threshold of [F2] and so small that $K_4{C}_c(1+C_{3}X){\delta }_X\le \rho \mathrm{/}4$. Choose $\kappa >0$ so that $C_{X}L\kappa <{\delta }_X$, replacing $L$ by $1+L$ in this choice if necessary. Whenever $t$ is admissible and $\mathrm{\mid }s-t\mathrm{\mid }<\kappa$, step 4.1 allows [F2] to correct $a+\lambda$ by $h\in {\mathcal{G}}_N$. The result is $b=(a+\lambda {)}^h=A_s^{gh}\in Y_3$, Coulomb, and $\mathrm{\parallel }b-a{\mathrm{\parallel }}_4\le \rho \mathrm{/}4$. The group ${\mathcal{G}}_N$ is closed under multiplication by the Sobolev product and approximation argument in [F2], so $gh\in {\mathcal{G}}_N$. Step 3.1 gives $\mathrm{\parallel }b{\mathrm{\parallel }}_4\le \rho \mathrm{/}2<\rho$. Thus $s$ is admissible, with the same step length $\kappa$ for every admissible $t$.

Finite construction reaches the original connection. At $t=0$, take $g=I$ and $a=0$, so admissibility holds. Choose a positive integer $N$ with $1\mathrm{/}N<\kappa$ and successively apply step 5.1 to $t_j=j\mathrm{/}N$, $j=0,\dots ,N$. This finite induction constructs an admissible $G\in {\mathcal{G}}_N$ at $t=1$, with $a=C^G$ Coulomb and in $Y_3$. Take the total gauge $g=UG$. Step 2.1 gives $a=A^g$, and step 3.1 gives $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,2}}+\mathrm{\parallel }a{\mathrm{\parallel }}_4\le (C_2+K_0)\mathrm{\parallel }{F}_A{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,3}}\le C_3\mathrm{\parallel }{F}_A{\mathrm{\parallel }}_3$. In particular the construction obtains a gauge from small curvature alone. It used no assumed small potential norm of the original connection: the admissible small potentials were constructed from the zero starting connection by finitely many proved correction steps.

Higher regularity from the constructed Neumann inverse. The family gauge $G$ is in $W^{2,3}$ and obeys ${\mathrm{\partial }}_{i}G=G{a}_i-(A_1{)}_{i}G$. Contracting its derivative and using ${\sum }_i{\mathrm{\partial }}_i{a}_i=0$ gives $\mathrm{\Delta }G={\sum }_i(({\mathrm{\partial }}_{i}G)a_i-(A_1{)}_i{\mathrm{\partial }}_{i}G)-(\mathrm{div}{A}_1)G$. Initially $\mathrm{\nabla }G,a\in L^{12}$ by [F4], while $A_1$ is smooth and $G$ is bounded; hence the right side belongs to $L^6$. The form $dG$ belongs to $Y_3$ with target the real eight-dimensional matrix space, by the defining smooth Neumann approximation of $G$. The integration identity of [F1] applied to $dG$ therefore gives the zero-normal weak identity for each component of $G$. In particular $-\mathrm{\Delta }G$ has zero mean. Apply [F3] at $p=6$ to each component, after subtracting its mean, and use weak-solution uniqueness to conclude $G\in W^{2,6}$. By [F4] its first derivatives are now bounded; the displayed equation then has bounded right side, because $a=G^{-1}dG+G^{-1}{A}_{1}G$ is bounded as well. For every finite $p\ge 2$, apply [F3] again to this same zero-mean source and uniqueness to obtain $G\in W^{2,p}$. The smooth factor $U$ preserves this regularity for the total gauge $g$, and the gauge formula gives $a\in W^{1,p}$ by ordinary weak product differentiation. This proves exactly the claimed Sobolev regularity, without asserting unproved smooth boundary bootstrapping.

Given finite equivariance. Suppose the stated $\mathrm{\Gamma },\tau$ data are present. The equality $M(s,\gamma x)=d{\tau }_{\gamma }(M(s,x))$ follows from the equivariance of $A$ evaluated on the radial vector. Constant conjugation commutes with all products and integrals in step 1.1, so $U(\gamma x)={\tau }_{\gamma }(U(x))$. Hence $C$ and every $A_t$ have the same equivariance. The initial gauge $I$ is equivariant. If an admissible family gauge is equivariant, its transported perturbation in step 4.1 is equivariant, and [F2] supplies an equivariant correction. The finite induction of step 6.1 therefore produces equivariant $G$ and total $g=UG$. Only the conjugation action is used; scalar representatives of a projective unitary element act trivially, so a lift to an $SU(2)$ representation is unnecessary.

Scaling and overlap compatibility. For $B_r$, pull back the connection as $A_r(y)=rA(ry)$. Its critical curvature norm equals $\mathrm{\parallel }{F}_A{\mathrm{\parallel }}_2$. Apply the unit-ball construction and define the physical gauge by $g(x)=g_r(x\mathrm{/}r)$. The transformed one-form obeys $a_r(y)=ra(ry)$, so $\mathrm{\parallel }{a}_r{\mathrm{\parallel }}_2=r^{-1}\mathrm{\parallel }a{\mathrm{\parallel }}_2$, $\mathrm{\parallel }\mathrm{\nabla }{a}_r{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\nabla }a{\mathrm{\parallel }}_2$, $\mathrm{\parallel }{a}_r{\mathrm{\parallel }}_4=\mathrm{\parallel }a{\mathrm{\parallel }}_4$, and the derivative/curvature $L^3$ norms have common factor $r^{2\mathrm{/}3}$. This gives both stated estimates and preserves the approximation class, regularity and equivariance. Finally, on the given overlaps the continuous representatives of $g_i$ supplied by [F4] make $h_{ij}=g_i^{-1}{t}_{ij}{g}_j$ continuous and $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{2,3}$ by [F5]. Direct multiplication gives $h_{ij}{h}_{jk}=h_{ik}$ pointwise, and the gauge composition identity of step 2.1 gives $a_j=a_i^{h_{ij}}$; smooth approximation justifies the equality for the Sobolev forms. Full AC is used only through the actual analytic suppliers [F1]–[F6].