Rough-coefficient covariant Green inverse and Coulomb splitting

Fix the smooth reference connection $A_0$ from the configuration data and set $h=A-A_0\in H^k(T^{\ast }M\otimes \mathrm{ad}P)$. In a local orthonormal frame, $d_A=d_{A_0}+[h,\cdot ]$ on zero-forms and $d_A^{\ast }=d_{A_0}^{\ast }-{\iota }_h$ on one-forms, where ${\iota }_h(a)=g^{ij}[h_i,a_j]$. Therefore $K_A:={\mathrm{\Delta }}_A-{\mathrm{\Delta }}_{A_0}$ is a finite sum of terms of the forms $({\mathrm{\nabla }}_{A_0}h)\xi$, $h{\mathrm{\nabla }}_{A_0}\xi$, and $h^2\xi$. These terms define a bounded map $H^{k+1}\to H^{k-1}$. They are compact: $({\mathrm{\nabla }}_{A_0}h)\xi$ factors through the compact inclusion $H^{k+1}\hookrightarrow H^{k-1}$ followed by multiplication in the algebra $H^{k-1}$; $h{\mathrm{\nabla }}_{A_0}\xi$ and $h^2\xi$ first map boundedly into $H^k$ and then use the compact inclusion $H^k\hookrightarrow H^{k-1}$. Here $k-1>d\mathrm{/}2$, so the needed multiplication bounds follow from [F8].

Apply [F3] with $r=k-1\ge 0$ to ${\mathrm{\Delta }}_{A_0}:H^{k+1}\to H^{k-1}$. Its kernel is finite dimensional, and the Hodge theorem gives the topological direct decomposition $H^{k-1}=\ker {\mathrm{\Delta }}_{A_0}\oplus {\mathrm{\Delta }}_{A_0}(H^{k+1})$. Thus its range is closed with codimension equal to the kernel dimension, so ${\mathrm{\Delta }}_{A_0}$ is Fredholm of index zero. By [F6] and the compactness of $K_A$ from step 1.1, ${\mathrm{\Delta }}_A={\mathrm{\Delta }}_{A_0}+K_A$ is Fredholm of index zero.

The smooth reference estimate [F4] and ${\mathrm{\Delta }}_{A_0}={\mathrm{\Delta }}_A-K_A$ give $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^{k+1}}\le C(\mathrm{\parallel }{\mathrm{\Delta }}_A\xi {\mathrm{\parallel }}_{H^{k-1}}+\mathrm{\parallel }{K}_A\xi {\mathrm{\parallel }}_{H^{k-1}}+\mathrm{\parallel }\xi {\mathrm{\parallel }}_{L^2})$. The three terms in $K_A$ from step 1.1 satisfy $\mathrm{\parallel }{K}_A\xi {\mathrm{\parallel }}_{H^{k-1}}\le C_A\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^k}$. The Sobolev interpolation estimate $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^k}\le \epsilon \mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^{k+1}}+C_{\epsilon }\mathrm{\parallel }\xi {\mathrm{\parallel }}_{L^2}$ follows by finite-atlas localization and the estimate in [F5]. Choose $\epsilon$ so that the resulting $H^{k+1}$ term is absorbed into the left side. This yields $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^{k+1}}\le C_A^{\mathrm{\prime }}(\mathrm{\parallel }{\mathrm{\Delta }}_A\xi {\mathrm{\parallel }}_{H^{k-1}}+\mathrm{\parallel }\xi {\mathrm{\parallel }}_{L^2})$.

If ${\mathrm{\Delta }}_A\xi =0$ for $\xi \in H^{k+1}(\mathrm{ad}P)$, the formal-adjoint identity, first for smooth approximants and then by Sobolev continuity, gives $0=⟨{\mathrm{\Delta }}_A\xi ,\xi {⟩}_{L^2}=\mathrm{\parallel }{d}_A\xi {\mathrm{\parallel }}_{L^2}^2$. Hence $d_A\xi =0$. Parallel transport makes $\xi$ determined by $\xi (p_m)$, and that value commutes with holonomy. By [F2], $\exp (t\xi (p_m))$ lies in the finite group $Z_N$ for every real $t$, because $\xi (p_m)$ commutes with holonomy. The path is continuous and equals the identity at $t=0$, so it is constant; differentiating at zero gives $\xi (p_m)=0$. Parallel transport then gives $\xi =0$.

Since ${\mathrm{\Delta }}_A$ is Fredholm of index zero and its kernel is zero by step 3.1, its cokernel has dimension zero. Its range is closed by the Fredholm property, so ${\mathrm{\Delta }}_A$ is onto and therefore bijective.

If no constant bounded $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^{k+1}}$ by $\mathrm{\parallel }{\mathrm{\Delta }}_A\xi {\mathrm{\parallel }}_{H^{k-1}}$, choose $\mathrm{\parallel }{\xi }_j{\mathrm{\parallel }}_{H^{k+1}}=1$ with ${\mathrm{\Delta }}_A{\xi }_j\to 0$. By [F4], a subsequence converges in $L^2$; applying the estimate of step 2.2 to differences shows it is Cauchy in $H^{k+1}$. Its limit has $H^{k+1}$ norm one and lies in $\ker {\mathrm{\Delta }}_A$, contradicting step 3.1. Thus $\mathrm{\parallel }\xi {\mathrm{\parallel }}_{H^{k+1}}\le C_A^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }{\mathrm{\Delta }}_A\xi {\mathrm{\parallel }}_{H^{k-1}}$. Together with bijectivity this proves the inverse $G_A$ is bounded.

The operator-valued map $A\mapsto {\mathrm{\Delta }}_A\in \mathcal{B}(H^{k+1},H^{k-1})$ is a continuous polynomial of degree at most two in the local coefficients of $A$, by step 1.1 and the product estimates [F8]. Near a fixed irreducible $A$, write ${\mathrm{\Delta }}_{A+h}={\mathrm{\Delta }}_A+E(h)$; for small $h$, [F7] gives $({\mathrm{\Delta }}_{A+h}{)}^{-1}=(I+G_{A}E(h){)}^{-1}{G}_A={\sum }_{n=0}^{\mathrm{\infty }}(-G_{A}E(h){)}^n{G}_A$. On a neighborhood where $\mathrm{\parallel }{G}_{A}E(h)\mathrm{\parallel }\le \rho <1$, the Neumann series converges in operator norm. Its $m$-th derivative is bounded termwise by a constant times $n^m{\rho }^{n-m}$ for $n\ge m$, whose series converges; hence all derivatives converge uniformly on a smaller neighborhood. This proves that inversion and $A\mapsto G_A$ are smooth there.

Define ${\mathrm{\Pi }}_A=d_A{G}_A{d}_A^{\ast }$. The Sobolev orders give a bounded map $H^k({\mathrm{\Omega }}^1(\mathrm{ad}P))\to H^k({\mathrm{\Omega }}^1(\mathrm{ad}P))$. Since ${\mathrm{\Delta }}_A{G}_A=I$ and $G_A{\mathrm{\Delta }}_A=I$, one has ${\mathrm{\Pi }}_A{d}_A\xi =d_A\xi$ and $d_A^{\ast }(I-{\mathrm{\Pi }}_A)a=0$ for every $\xi \in H^{k+1}$ and $a\in H^k$. Thus ${\mathrm{\Pi }}_A$ is a bounded projection with range $\mathrm{im}{d}_A$, while $I-{\mathrm{\Pi }}_A$ has range $\ker d_A^{\ast }$. Their ranges intersect trivially: if $d_A\xi \in \ker d_A^{\ast }$, then ${\mathrm{\Delta }}_A\xi =0$, so step 3.1 gives $\xi =0$. This proves the direct-sum decomposition and projection formula. Smooth dependence of $G_A$, $d_A$, and $d_A^{\ast }$ from step 5.1 makes ${\mathrm{\Pi }}_A$ smooth in $A$.