Sobolev and smooth irreducible gauge-orbit principal bundles

The finite center $Z_N$ acts freely on the Banach Lie group ${\mathcal{G}}_{k+1}$ by translations. Around any group element choose an open chart whose distinct $Z_N$-translates are disjoint; the quotient map restricts to a chart on each such neighborhood, with transition maps given by smooth translations. Multiplication and inversion descend smoothly, so ${\bar{\mathcal{G}}}_{k+1}$ is a Banach Lie group. Since a central gauge section acts trivially on connections, this is the structure group of the reduced action.

The tangent space of the Coulomb slice at $A$ is $\ker d_A^{\ast }$. By [F3], the bounded linear map $L_A:\ker d_A^{\ast }\oplus H^{k+1}(\mathrm{ad}P)\to H^k(T^{\ast }M\otimes \mathrm{ad}P)$, $L_A(a,\xi )=a+d_A\xi$, is an isomorphism with inverse $v\mapsto (v-d_A{G}_A{d}_A^{\ast }v,G_A{d}_A^{\ast }v)$. The derivative of $(a,\xi )\mapsto (A+a{)}^{\exp \xi }$ at $(0,0)$ is $L_A$ by [F1].

Define $F(a,\xi ):=d_A^{\ast }((A+a{)}^{\exp \xi }-A)$ for $a\in H^k(T^{\ast }M\otimes \mathrm{ad}P)$ near $0$ and $\xi \in H^{k+1}(\mathrm{ad}P)$ near $0$. This is a smooth map into $H^{k-1}(\mathrm{ad}P)$ by [F1], and $F(0,0)=0$. Its partial derivative in $\xi$ at $(0,0)$ is $d_A^{\ast }{d}_A={\mathrm{\Delta }}_A$, an isomorphism by [F3]. The implicit-function theorem [F4] therefore gives, for every connection sufficiently near $A$, a unique nearby gauge parameter moving it into the Coulomb condition $d_A^{\ast }(B^u-A)=0$. This supplies a local section of the orbit projection.

Apply the inverse-function theorem [F4] to ${\mathrm{\Phi }}_A(s,u):=s^u$ on the local Banach product $S_{A,\epsilon }\times {\bar{\mathcal{G}}}_{k+1}$ near $(A,e)$; its derivative is $L_A$ from step 1.2. Shrink $\epsilon$ so the slice remains in the open irreducible locus. At $(s,e)$, for $s=A+a$, the derivative is $L_s(b,\xi )=b+d_s\xi$ on $b\in \ker d_A^{\ast }$; this differs from $L_A$ only by $(b,\xi )\mapsto [a,\xi ]$. Its operator norm is at most $C\mathrm{\parallel }a{\mathrm{\parallel }}_{H^k}$ by [F7]. Choose $\epsilon$ so $C\epsilon \mathrm{\parallel }{L}_A^{-1}\mathrm{\parallel }<1$; then [F5] makes $L_s$ an isomorphism for every $s$ in the slice. At an arbitrary $(s,u_0)$, write $u=u_{0}v$ near $u_0$ and use $s^{u_{0}v}=(s^{u_0}{)}^v$. Applying the inverse action by $u_0$ to the target tangent and changing the gauge tangent by the invertible adjoint action identifies this derivative with $L_s$. Hence the derivative is an isomorphism at every point of the slice product, and the inverse-function theorem makes ${\mathrm{\Phi }}_A$ a local diffeomorphism throughout.

The map ${\mathrm{\Phi }}_A$ is injective after possibly shrinking the slice further. Otherwise there would be sequences $s_j,t_j\in S_{A,1\mathrm{/}j}$ and distinct pairs $(s_j,u_j)\mathrm{\ne }(t_j,v_j)$ with $s_j^{u_j}=t_j^{v_j}$. Put $w_j=u_j{v}_j^{-1}$; then $s_j^{w_j}=t_j$ and $s_j,t_j\to A$. Properness [F2] gives a subsequence of $w_j$ converging to $w$ with $A^w=A$. Freeness [F2] gives $w=e$. Hence for all sufficiently large indices, both $(s_j,w_j)$ and $(t_j,e)$ lie in the local injectivity neighborhood from step 2.1. Their equal images force $s_j=t_j$ and $w_j=e$, and freeness then gives $u_j=v_j$, contradicting distinctness.

Let $U_A:={\mathrm{\Phi }}_A(S_{A,\epsilon }\times {\bar{\mathcal{G}}}_{k+1})$. It is open because ${\mathrm{\Phi }}_A$ is a local diffeomorphism at every point of its domain, and it is exactly the saturation of the slice. By steps 3.1 and 2.1, ${\mathrm{\Phi }}_A$ is a bijective local diffeomorphism onto $U_A$, hence a diffeomorphism. If $V_A:={\pi }_k(S_{A,\epsilon })$, then ${\pi }_k^{-1}(V_A)=U_A$. For every open subset $O$ of the slice, $O\times {\bar{\mathcal{G}}}_{k+1}$ is open and its image under ${\mathrm{\Phi }}_A$ is open; thus ${\pi }_k(O)$ is open in $V_A$ by the quotient topology. Therefore the slice-to-quotient map is a smooth chart and ${\mathrm{\Phi }}_A$ is an equivariant local principal trivialization.

The orbit relation on ${\mathcal{A}}_k^{\ast }$ is closed: if $x_j^{u_j}=y_j$ and $(x_j,y_j)\to (x,y)$, properness [F2] gives a convergent subsequence $u_{j_{\mathrm{\ell }}}\to u$, and continuity of the action gives $x^u=y$. For distinct orbits, the complement of this closed relation contains their pair, so there are product neighborhoods $O_x\times O_y$ disjoint from the relation. Their saturations are open and disjoint: if the saturations intersected, some translate of a point in $O_x$ would equal a translate of a point in $O_y$, making that pair related. Their quotient images therefore separate the two orbits. This proves the quotient is Hausdorff.

For the smooth-space clause, fix now a smooth irreducible center $A\in {\mathcal{A}}_{\mathrm{\infty }}^{\ast }$ and use the Banach slice and trivialization established in step 4.1. Put $U_A={\mathrm{\Phi }}_A(S_{A,\epsilon }\times {\bar{\mathcal{G}}}_{k+1})$ and $V_A={\pi }_k(S_{A,\epsilon })$. Since $U_A$ is open in ${\mathcal{A}}_k^{\ast }$, $U_A\cap {\mathcal{A}}_{\mathrm{\infty }}^{\ast }$ is open in the stated $C^{\mathrm{\infty }}$ topology; set $S_A^{\mathrm{\infty }}=S_{A,\epsilon }\cap {\mathcal{A}}_{\mathrm{\infty }}^{\ast }$ and $V_A^{\mathrm{\infty }}={\pi }_{\mathrm{\infty }}(S_A^{\mathrm{\infty }})$. For a smooth $B\in U_A$, its unique reduced gauge class from step 4.1 has a representative $u\in {\mathcal{G}}_{k+1}$. It satisfies $d_A^{\ast }(B^{u^{-1}}-A)=0$. Locally in a bundle frame, write $C_j=u{B}_j{u}^{-1}-({\mathrm{\partial }}_{j}u)u^{-1}-A_j$. Substitution in $d_A^{\ast }C=0$, use of ${\mathrm{\partial }}_i(u^{-1})=-u^{-1}({\mathrm{\partial }}_{i}u)u^{-1}$, and multiplication on the right by $u$ isolate the equation $g^{ij}{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}u=R(A,B,g;u,u^{-1},du)$ on each chart. Every term in $R$ is a finite product of smooth coefficients (including first derivatives of $A,B$) and factors among $u,u^{-1},du$. To apply [F6] globally, use the matrix bundle $\mathcal{M}=P{\times }_{\mathrm{Ad}}{\mathrm{Mat}}_N(\mathbb{C})$ with its invariant Frobenius Hermitian metric, Riemannian density and smooth connection ${\mathrm{\nabla }}^0=d+[A_0,\cdot ]$. Set $L=({\mathrm{\nabla }}^0{)}^{\ast }{\mathrm{\nabla }}^0$. Integration by parts on the closed base gives its formal symmetry, and its positive-frequency principal symbol is $\mathrm{\mid }\xi {\mathrm{\mid }}_g^{2}I$; its coefficients are smooth. Locally write $L=-g^{ij}{\mathrm{\partial }}_i{\mathrm{\partial }}_j+B_0^i{\mathrm{\partial }}_i+C_0$ with smooth matrix coefficients. The local equation is consequently $Lu=\mathcal{R}$, where $\mathcal{R}=-R+B_0^i{\mathrm{\partial }}_{i}u+C_{0}u$. These expressions patch to a distributional section because they equal the global operator $Lu$ on every overlap. If $u\in H^{r+1}$ for $r\ge k$, then $u^{-1}=u^{\ast }$ and [F7] put every term of $\mathcal{R}$ in $H^r$. The base has dimension three or four, and $u\in H^1$, so all hypotheses of [F6] hold for this fixed $L$ and give $u\in H^{r+2}$. Iteration proves $u$ smooth; changing its representative by an element of the finite center preserves smoothness. Thus every smooth point of the Banach saturation has a smooth trivialization representative, and $U_A\cap {\mathcal{A}}_{\mathrm{\infty }}^{\ast }={\pi }_{\mathrm{\infty }}^{-1}(V_A^{\mathrm{\infty }})$.

Suppose smooth $B_j\to B$ in $C^{\mathrm{\infty }}$ within $U_A\cap {\mathcal{A}}_{\mathrm{\infty }}^{\ast }$. The Banach trivialization from step 4.1 gives convergence of their slice representatives in $H^k$ and of their reduced gauge classes in $H^{k+1}$. In a quotient chart around the limiting class from step 1.1, choose lifts $u_j,u$ that converge in $H^{k+1}$; this is possible because the finite center acts by disjoint translates. The global equation $L{u}_j=\mathcal{R}(A,B_j;u_j)$ for the same fixed reference operator $L$ in step 5.2 and elliptic estimates [F6] bootstrap convergence: if $u_j\to u$ in $H^{r+1}$, then the assumed $C^{\mathrm{\infty }}$ convergence gives $B_j\to B$ in $H^{r+1}$, and [F7] makes the right-hand sides converge in $H^r$. Apply the global estimate for $L$ to $u_j-u$, using also their already known $H^{k+1}$ convergence for the $L^2$ term, to get $u_j\to u$ in $H^{r+2}$. Induction gives convergence in every Sobolev order, which on compact $M$ is the $C^{\mathrm{\infty }}$ section topology. The slice representatives converge in $C^{\mathrm{\infty }}$ by their action formula. Conversely the forward action is continuous in the $C^{\mathrm{\infty }}$ topology by multiplication and differentiation of smooth sections. The saturation $U_A\cap {\mathcal{A}}_{\mathrm{\infty }}^{\ast }$ is open, and $V_A^{\mathrm{\infty }}$ is open in the quotient because the quotient map is open: the saturation of any open set is a union of its open gauge translates. The restricted map $S_A^{\mathrm{\infty }}\times {\bar{\mathcal{G}}}_{\mathrm{\infty }}\to {\pi }_{\mathrm{\infty }}^{-1}(V_A^{\mathrm{\infty }})$ is therefore a homeomorphism for the actual quotient topology, and it is equivariant. This proves the smooth-space topological principal bundle claim.