Finite stabilization defines the analytic Fredholm determinant line

Since $D_{b_0}$ is Fredholm, its cokernel is finite-dimensional. Choose representatives in $F$ for a basis of $\mathrm{coker}{D}_{b_0}$ and let $W$ be their span. Then $W$ satisfies [A2], so the quotient map $q_W:F\to F\mathrm{/}W$ makes $q_W{D}_{b_0}$ surjective. The map $D_{b_0}:K_W(b_0)\to W$ has kernel $\ker D_{b_0}$ and finite-dimensional image $W\cap \mathrm{im}{D}_{b_0}$, hence $K_W(b_0)$ is finite-dimensional.

We record the bounded-inverse fact needed below. Let $T:X_0\to Y_0$ be a bounded surjection between complete normed spaces (Banach spaces) and let $B_{X_0}(0,r)=\{x\in X_0:\mathrm{\parallel }x\mathrm{\parallel }<r\}$. Since $Y_0={\bigcup }_{n\ge 1}\overline{T(B_{X_0}(0,n))}$, the Baire property of the complete metric space $Y_0$ implies that one of these closed sets has nonempty interior. The Baire property follows by contradiction: if the closed sets in a countable cover all had empty interior, choose nested nonempty closed balls with radii tending to zero, the $n$th ball lying outside the $n$th closed set; completeness gives a point in all balls and outside the whole cover. Taking differences in the interior ball and rescaling gives some $\delta >0$ with $B_{Y_0}(0,\delta )\subseteq \overline{T(B_{X_0}(0,1))}$. For $\mathrm{\parallel }y\mathrm{\parallel }<\delta \mathrm{/}2$, recursively choose $\mathrm{\parallel }{x}_j\mathrm{\parallel }\le 2^{-j}$ so that $\mathrm{\parallel }y-{\sum }_{i=1}^{j}T{x}_i\mathrm{\parallel }<\delta 2^{-j-1}$; each choice is possible by applying this density statement to the rescaled current residual. The series $x={\sum }_j{x}_j$ converges, satisfies $Tx=y$, and has $\mathrm{\parallel }x\mathrm{\parallel }\le 1$. If $T$ is also bijective, scaling this conclusion shows $\mathrm{\parallel }{T}^{-1}y\mathrm{\parallel }\le 4\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}\delta$, so $T^{-1}$ is bounded.

Choose the orthogonal complement $X=\ker (q_W{D}_{b_0}{)}^{\perp }$ in the Hilbert space $E$. Since $W$ is finite-dimensional and closed, $F\mathrm{/}W$ is isometric to the Hilbert space $W^{\perp }$; $X$ is also a Hilbert space. The restriction $q_W{D}_{b_0}{\mathrm{\mid }}_X:X\to F\mathrm{/}W$ is bounded and bijective, because $q_W{D}_{b_0}$ is surjective and its kernel is $K_W(b_0)$. By step 1.2 its inverse is bounded. Shrink the neighborhood of $b_0$ so that $q_W{D}_b{\mathrm{\mid }}_X=(q_W{D}_{b_0}{\mathrm{\mid }}_X)(I+R_b)$ with $\mathrm{\parallel }{R}_b\mathrm{\parallel }\le r<1$. The space of bounded operators on $X$ is complete in operator norm: an operator-norm Cauchy sequence converges on each vector in the complete space $X$, and the pointwise limit is its operator-norm limit. Thus ${\sum }_{m=0}^{\mathrm{\infty }}(-R_b{)}^m$ converges in operator norm, since its norm series is bounded by ${\sum }_{m=0}^{\mathrm{\infty }}{r}^m$; multiplying its partial sums by $I+R_b$ and taking the limit gives the identity. The series converges locally uniformly in $b$, so the inverse of $q_W{D}_b{\mathrm{\mid }}_X$ is continuous, respectively holomorphic. Hence $q_W{D}_b$ is surjective.

For every nearby $b$, $0\to \ker D_b\to K_W(b)\stackrel{D_b}{\to }W\to \mathrm{coker}{D}_b\to 0$ is exact: the image of $D_b{\mathrm{\mid }}_{K_W(b)}$ is $W\cap \mathrm{im}{D}_b$, and $W\to \mathrm{coker}{D}_b$ is surjective because $q_W{D}_b$ is surjective. Thus $K_W(b)$ is finite-dimensional. Decompose $E=K_W(b_0)\oplus X$ as in step 2.1. Every $K_W(b)$ is the graph over $K_W(b_0)$ of $-(q_W{D}_b{\mathrm{\mid }}_X{)}^{-1}{q}_W{D}_b{\mathrm{\mid }}_{K_W(b_0)}$, so these kernels form a vector bundle of constant finite rank, continuously or holomorphically according to the family. Taking dimensions in the exact sequence gives $\mathrm{ind}{D}_b=\dim K_W(b)-\dim W$, so the index is locally constant.

Choose a basis of $\ker D_b$, extend it to a basis of $K_W(b)$ by lifts of a basis of $W\cap \mathrm{im}{D}_b$, then extend that latter basis to a basis of $W$ by lifts of a basis of $\mathrm{coker}{D}_b$. Wedge multiplication identifies $\det K_W\otimes (\det W{)}^{\ast }$ with $\det \ker D_b\otimes (\det \mathrm{coker}{D}_b{)}^{\ast }$; changes of any chosen lifts add earlier basis vectors and therefore do not change the induced line isomorphism.

For nested stabilizers $W\subseteq U$ on a common neighborhood, the map $K_U(b)⟶U\mathrm{/}W,x⟼D_{b}x+W$ is well-defined and surjective, and its kernel is $K_W(b)$. Surjectivity follows because $q_W{D}_b:E_b\to F_b\mathrm{/}W$ is surjective: for $u\in U$, choose $x$ with $D_{b}x-u\in W$; then $D_{b}x\in U$ and $D_{b}x+W=u+W$. Hence $0⟶K_W(b)⟶K_U(b)⟶U\mathrm{/}W⟶0$ is exact. The second exact sequence $0⟶W⟶U⟶U\mathrm{/}W⟶0$ uses the same quotient. The determinant isomorphisms of these finite-dimensional exact sequences, with the order and Koszul signs of [A1], cancel the common $\det (U\mathrm{/}W)$ factor and give a canonical comparison $\det K_U(b)\otimes (\det U{)}^{\ast }⟶\det K_W(b)\otimes (\det W{)}^{\ast }\mathrm{.}$ Its inverse direction is equally canonical. For arbitrary stabilizers $W_1,W_2$, their sum $U=W_1+W_2$ is a finite-dimensional stabilizer, and the two nested comparisons through $U$ identify both local lines.

In a common local trivialization, let ${\alpha }_W$ be the canonical fiber isomorphism from $\det K_W\otimes (\det W{)}^{\ast }$ to $\det (D_b)$ supplied by the exact sequence in step 4.1. For nested $W\subseteq U$, the displayed comparison in step 4.2, from the $U$ chart to the $W$ chart, is ${\alpha }_W^{-1}{\alpha }_U$. This follows by choosing bases adapted to the two short exact sequences in step 4.2 and using the same ordered lifts in the exact sequence of step 4.1; the resulting maps commute with $D_b$ and the quotient maps. Thus for any three stabilizers the transition maps compose to the identity because the intermediate ${\alpha }_W$ maps cancel. For incomparable fixed subspaces use their finite sum as in step 4.2.

It remains to compare charts from different Hilbert-bundle trivializations. In a common frame, the two stabilizers are finite-rank subbundles $W_1,W_2\subset F$: a frame change takes a constant finite-dimensional subspace to a continuous, respectively holomorphic, finite-rank subbundle. Put $K_{W_i}(b)=\{u\in E_b:D_{b}u\in W_i(b)\}$. Define $A_i:E\oplus W_i\to F$ by $A_i(u,w)=D_{b}u+w$ and $A_{12}:E\oplus W_1\oplus W_2\to F$ by $A_{12}(u,w_1,w_2)=D_{b}u+w_1+w_2$. Each $A_i$ is a bounded surjection, and $\ker A_i$ is isomorphic to $K_{W_i}(b)$ by $u\mapsto (u,-D_{b}u)$, so its kernel is finite-dimensional by step 3.1. The family $A_{12}$ is also surjective, and $0\to \ker A_1(b)\to \ker A_{12}(b)\to W_{2,b}\to 0$, with the last map $(u,w_1,w_2)\mapsto w_2$, is exact: surjectivity follows by solving $A_1(u,w_1)=-w_2$. Thus $A_{12}$ has finite-dimensional kernel. The same local splitting and graph argument as in steps 2.1 and 3.1 gives continuous, respectively holomorphic, finite-rank kernel bundles for these surjective families. Exchanging $W_1,W_2$ gives the exact sequence $0\to \ker A_2\to \ker A_{12}\to W_1\to 0$. The determinant maps from [A1] for these short exact sequences identify each stabilization chart continuously, respectively holomorphically, with the common chart $\det \ker A_{12}\otimes (\det W_1\otimes \det W_2{)}^{\ast }$. Applying the same construction to three stabilizer subbundles and the common augmentation $E\oplus W_1\oplus W_2\oplus W_3$, the composites of the exact-sequence determinant maps agree by their wedge definition and the Koszul signs in [A1]. Hence the chart comparisons obey the cocycle condition across Hilbert-bundle trivializations.