Conditional nodal gluing regularity for Pardon implicit atlases

Since $\mathrm{\parallel }{1}_Y-D_p{T}_p\mathrm{\parallel }\le \delta <1$, the Neumann series $(D_p{T}_p{)}^{-1}={\sum }_{m\ge 0}(1_Y-D_p{T}_p{)}^m$ converges in operator norm and has norm at most $(1-\delta {)}^{-1}$. Define $Q_p=T_p(D_p{T}_p{)}^{-1}$. Then $D_p{Q}_p=1_Y$ and $\mathrm{\parallel }{Q}_p\mathrm{\parallel }\le C_0$. Hence $B=\ker D_p\oplus \mathrm{im}{Q}_p$: for every $x$, write $x=(x-Q_p{D}_{p}x)+Q_p{D}_{p}x$, with the first summand in $\ker D_p$; if $Q_{p}y\in \ker D_p$, then $y=D_p{Q}_{p}y=0$. By [F3], the Fredholm index is $\dim \ker D_p-\dim \mathrm{coker}{D}_p$; the right inverse makes the cokernel zero, so the index equals $\dim \ker D_p=d$.

For $k\in \ker D_p$ with $\mathrm{\parallel }k\mathrm{\parallel }\le \rho$, the fundamental theorem of calculus along $tk$ and $D_{p}k=0$ give $\mathrm{\parallel }F(p,k)\mathrm{\parallel }\le \eta +L{\rho }^2\mathrm{/}2$. For $y\in Y$ with $\mathrm{\parallel }y\mathrm{\parallel }\le r$, set $G_{p,k}(y)=y-F(p,k+Q_{p}y)$. The derivative bound and $\mathrm{\parallel }{Q}_p\mathrm{\parallel }\le C_0$ give $\mathrm{\parallel }{D}_y{G}_{p,k}(y)\mathrm{\parallel }=\mathrm{\parallel }{1}_Y-D_{x}F(p,k+Q_{p}y)Q_p\mathrm{\parallel }\le C_{0}L(\rho +C_{0}r)\le 1\mathrm{/}4$, since $D_p{Q}_p=1_Y$.

Also $\mathrm{\parallel }{G}_{p,k}(0)\mathrm{\parallel }=\mathrm{\parallel }F(p,k)\mathrm{\parallel }\le r\mathrm{/}2$. For $\mathrm{\parallel }y\mathrm{\parallel }\le r$, step 2.1 gives $\mathrm{\parallel }{G}_{p,k}(y)\mathrm{\parallel }\le \mathrm{\parallel }{G}_{p,k}(0)\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}4\le 3r\mathrm{/}4$, so the closed radius-$r$ ball is mapped into itself. It is nonempty because it contains $0$. It is complete: a Cauchy sequence $y_j$ in this ball converges to some $y\in Y$ by the Banach completeness in [F3], and $\mathrm{\mid }\mathrm{\parallel }y\mathrm{\parallel }-\mathrm{\parallel }{y}_j\mathrm{\parallel }\mathrm{\mid }\le \mathrm{\parallel }y-y_j\mathrm{\parallel }\to 0$ gives $\mathrm{\parallel }y\mathrm{\parallel }\le r$, so the limit remains in the ball. The derivative bound makes $G_{p,k}$ a contraction there with constant at most $1\mathrm{/}4$. By [F3] it has a unique fixed point $y(p,k)$ in that ball, equivalently $F(p,k+Q_{p}y(p,k))=0$.

In a $C^1$ local trivialization of the kernel bundle put $x=(p,k)$ and $H(x,y)=F(p,k+Q_{p}y)$. Its derivative in $y$ is $D_{y}H=D_{x}F(p,k+Q_{p}y)Q_p=1_Y+(D_{x}F-D_p)Q_p$. The perturbation of $1_Y$ has norm at most $C_{0}L(\rho +C_{0}r)\le 1\mathrm{/}4$, so $D_{y}H$ is invertible by its Neumann series, with a uniformly bounded inverse. Fix $x$ and put $y_0=y(x)$. Comparing the two fixed-point equations with $y_0$ held fixed gives $\mathrm{\parallel }y(x^{\mathrm{\prime }})-y_0\mathrm{\parallel }\le \frac{1}{4}\mathrm{\parallel }y(x^{\mathrm{\prime }})-y_0\mathrm{\parallel }+\mathrm{\parallel }{G}_{x^{\mathrm{\prime }}}(y_0)-G_x(y_0)\mathrm{\parallel }$. Hence $\mathrm{\parallel }y(x^{\mathrm{\prime }})-y_0\mathrm{\parallel }\le \frac{4}{3}\mathrm{\parallel }{G}_{x^{\mathrm{\prime }}}(y_0)-G_x(y_0)\mathrm{\parallel }\to 0$ by continuity at the single point $(x,y_0)$. This proves continuity without a uniform derivative bound on the entire Banach ball. Continuity of $D_{x}G$ at $(x,y_0)$ now bounds it on a product neighborhood of this point. Shrink the finite-dimensional parameter neighborhood to a convex ball; the already proved continuity keeps $y(x^{\mathrm{\prime }})$ in that product neighborhood. Integrating $D_{x}G$ along the parameter segment, at the fixed value $y(x^{\mathrm{\prime }})$, and using the same contraction comparison gives a local bound $\mathrm{\parallel }y(x^{\mathrm{\prime }})-y(x^{\mathrm{\prime }\mathrm{\prime }})\mathrm{\parallel }\le C\mathrm{\parallel }{x}^{\mathrm{\prime }}-x^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }$. For an increment $h$ in the finite-dimensional $x$ variables, expand $H(x+h,y(x+h))-H(x,y(x))=0$. The preceding Lipschitz bound and Fréchet differentiability give $y(x+h)-y(x)=-(D_{y}H{)}^{-1}{D}_{x}Hh+o(\mathrm{\parallel }h\mathrm{\parallel })$. This derivative is continuous because $H$ is $C^1$, $y$ is continuous, and inversion is continuous on boundedly invertible operators. Thus $y(x)$ is $C^1$, with no implicit-function theorem invocation. Conversely, a sufficiently small zero $z$ decomposes uniquely as $z=k+Q_{p}y$ with $y=D_{p}z$ and $k=z-Q_p{D}_{p}z$; continuity of these projections makes $\mathrm{\parallel }k\mathrm{\parallel }<\rho$ and $\mathrm{\parallel }y\mathrm{\parallel }<r$ after shrinking the neighborhood. It is then the unique fixed point above. The zero set is locally a $C^1$ graph over $(p,k)$ and has dimension $\dim P+d$.

Each specified forgetting map is the coordinate projection on the graph coordinates $(p,k)$ of step 4.1. Its derivative is surjective, so it is a submersion on the zero set. The assumed equivariance makes $F^{-1}(0)$ invariant under $\mathrm{\Gamma }$. For a point with trivial stabilizer, finiteness of $\mathrm{\Gamma }$ and Hausdorffness of the zero-set chart let us choose a neighborhood disjoint from each of its nonidentity translates; its quotient map is therefore a $C^1$ chart. These charts give a $C^1$ manifold on the trivial-stabilizer quotient. On loci where source and target stabilizers are trivial, the equivariant forgetting maps descend, and the quotient charts identify them locally with the original submersions.

At a solution $z=(p,x)$ in the graph chart put $A=D_{x}F(p,x)$. By the same bound as in step 4.1, $A{Q}_p$ is invertible; hence $R=Q_p(A{Q}_p{)}^{-1}$ is a right inverse of $A$. The map $k\mapsto k-RAk$ identifies $\ker D_p$ with $\ker A$: its inverse is the projection $v\mapsto v-Q_p{D}_{p}v$ from step 1.1, as follows by writing $v=k+Q_{p}y$ and solving $Ak+A{Q}_{p}y=0$. This is a continuous family of finite-dimensional isomorphisms and equals the identity at $x=0$; continuity suffices for transporting orientations. Both $D_p$ and $A$ are surjective, so their determinant lines are the top exterior lines of their kernels, with the zero-cokernel factor canonically $\mathbb{R}$. Transport the supplied orientation of $\det (D_p)$ through this isomorphism to $\det A$. The finite-dimensional exact sequence $0⟶\ker A⟶T_{z}Z(F)\stackrel{{\mathrm{pr}}_{TP}}{\to }{T}_{p}P⟶0$ has first map $b\mapsto (0,b)$ and is surjective at the last term: for $v\in T_{p}P$, $(v,-R{D}_{\mathrm{p}\mathrm{a}\mathrm{r}}F(p,x)v)$ lies in the tangent kernel of the total derivative, where $D_{\mathrm{p}\mathrm{a}\mathrm{r}}F$ here denotes the derivative in the parameter variable. Wedge a kernel basis followed by lifts of a parameter basis, then reorder the two determinant factors with the fixed sign $(-1{)}^{d\dim P}$ to use the parameter-first convention. This identifies the oriented zero-set determinant with $\det (T_{p}P)\otimes \det A$. Since $p=(a,e)$ and $F((a,e),x)=0$, the $E$ orientation occurs precisely once in $\det (T_{p}P)=\det (T_a{P}_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}})\otimes \det E$. There is no orientation of an infinite-dimensional equation space being taken. The supplied overlap and group-action compatibility therefore orients the local zero sets consistently and descends on the free quotient. These conclusions give the stated conditional smooth-thickening regularity. An application must still verify the relative extra-section projection models and covering-group conditions in [F1]; coordinate forgetting on a solution manifold alone is not their proof. The weighted analytic estimates in [A1] remain inputs.