Local separable trace-class determinant construction

For $n=0$, ${\mathrm{\Lambda }}^{0}T=I_{\mathbb{C}}$, so $c_0=1$. For $n\ge 1$, [A2] makes ${\mathrm{\Lambda }}^{n}T$ trace class and [A3] gives $\mathrm{\mid }{c}_n\mathrm{\mid }\le \mathrm{\parallel }{\mathrm{\Lambda }}^{n}T{\mathrm{\parallel }}_1\le \mathrm{\parallel }T{\mathrm{\parallel }}_1^n\mathrm{/}n!$. Hence for each real $R\ge 0$, ${\sum }_{n\ge 0}\mathrm{\mid }{c}_n\mathrm{\mid }{R}^n\le {\sum }_{n\ge 0}\frac{(R\mathrm{\parallel }T{\mathrm{\parallel }}_1{)}^n}{n!}<\mathrm{\infty }$ by [A4]. Since this holds at every radius, the complex power series has radius $+\mathrm{\infty }$ by [A22]. The separability hypothesis is the dense-subset condition [A20] and is retained, though this estimate uses only trace-class membership.

The finite-dimensional subspace $E$ is closed by [A9], so [A11] supplies its orthogonal projection $P_E$. It is bounded, linear and self-adjoint by [A24]. Its defining decomposition also gives $P_E^2=P_E$. Choose an orthonormal basis $e_0,\dots ,e_{d-1}$ of $E$ by [A13], where $d=\dim E$ and the list is empty if $d=0$. For each $n\ge 0$, put ${\mathcal{H}}_n:={\mathrm{\Lambda }}^{n}K$, $G_n:={\mathrm{\Lambda }}^{n}E\subseteq {\mathcal{H}}_n$, and $Q_n:={\mathrm{\Lambda }}^n{P}_E$. The increasing wedges $e_{i_1}\wedge \cdots \wedge e_{i_n}$ with $i_1<\cdots <i_n$ are orthonormal by the Gram formula [A1]. They span $G_n$: every algebraic tensor in $E^{\otimes n}$ expands in the basis tensors from $(e_j)$, and antisymmetrizing sends a repeated-index tensor to zero and every other one to a multiple of an increasing wedge; the algebraic tensors are dense and the projection range is closed. Thus this is a finite orthonormal basis of $G_n$, empty for $n>d$; for $n=0$, $G_0=\mathbb{C}$ with basis $1$. It follows that $G_n$ is closed in ${\mathcal{H}}_n$ by [A9]. By functoriality in [A1], $Q_n^2=Q_n$. For $n=0$, [A1] gives $Q_0=I_{\mathbb{C}}$ and $G_0=\mathbb{C}$, so $Q_0$ is the orthogonal projection onto $G_0$. For $n\ge 1$, on decomposable wedges the Gram identity and self-adjointness of $P_E$ give $⟨Q_n(x_1\wedge \cdots \wedge x_n),y_1\wedge \cdots \wedge y_n⟩=\det [⟨P_E{x}_i,y_j⟩]=\det [⟨x_i,P_E{y}_j⟩]=⟨x_1\wedge \cdots \wedge x_n,Q_n(y_1\wedge \cdots \wedge y_n)⟩\mathrm{.}$ Density of decomposable wedges makes $Q_n$ self-adjoint. It maps decomposable wedges into $G_n$ and fixes every decomposable wedge in $G_n$; continuity and density therefore show that its range is exactly $G_n$. Thus $Q_n$ is the orthogonal projection onto $G_n$.

Let $d:=\dim E$ and use the basis $(e_j)$ chosen in step 1.2. By [A10], $A:=F{\mathrm{\mid }}_E$ is an endomorphism; for $x\in E$, [A23] gives $\mathrm{\parallel }Ax\mathrm{\parallel }=\mathrm{\parallel }Fx\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, so $A$ is bounded. Write its matrix as $(a_{ij})$, so $A{e}_j={\sum }_{i<d}{a}_{ij}{e}_i$ by [A15]. For each $0\le n\le d$, the increasing wedges $e_I:=e_{i_1}\wedge \cdots \wedge e_{i_n}$, with $I=(i_1<\cdots <i_n)$, form the orthonormal basis of $G_n$ established in step 1.2. Expanding ${\mathrm{\Lambda }}^{n}A(e_I)=A{e}_{i_1}\wedge \cdots \wedge A{e}_{i_n}$ by multilinearity and antisymmetry shows that its diagonal coefficient at $e_I$ is the principal minor $\det A[I,I]$. Thus [A14] yields $\mathrm{tr}({\mathrm{\Lambda }}^{n}A)={\sum }_{I\subseteq \{0,\dots ,d-1\},\mathrm{\mid }I\mathrm{\mid }=n}\det A[I,I],$ with the $n=0$ term equal to $1$; for $n>d$, $G_n=\{0\}$ and the trace is $0$. When $d=0$ this says the sole coefficient is $1$ in degree zero and all positive-degree coefficients vanish.

By [A5], the series converges uniformly on every closed disk of finite radius, and so locally uniformly on $\mathbb{C}$. Its infinite radius from step 1.1 and [A6] make its sum holomorphic on all of $\mathbb{C}$; therefore $D_T$ is entire by [A7].

The constant coefficient in step 1.1 gives $D_T(0)=1$. The derivative formula in [A6] gives $D_T^{\mathrm{\prime }}(0)=c_1=\mathrm{tr}({\mathrm{\Lambda }}^{1}T)=\mathrm{tr}(T)$ by [A1].

Let $F$ be bounded and finite rank. By [A8], it is trace class, so the series defining $D_F$ is well-defined and the entire-function conclusion of steps 1.1 and 2.2 applies. Its restriction $A:=F{\mathrm{\mid }}_E$ is the bounded endomorphism established in step 2.1.

Since $\mathrm{ran}F\subseteq E$, the wedge action in [A1] gives $\mathrm{ran}({\mathrm{\Lambda }}^{n}F)\subseteq G_n$, hence $Q_n{\mathrm{\Lambda }}^{n}F={\mathrm{\Lambda }}^{n}F$. Each ${\mathrm{\Lambda }}^{n}F$ is trace class by [A2] for $n\ge 1$; for $n=0$ it is the finite-rank identity on $\mathbb{C}$, trace class by [A8]. Cyclicity [A12] therefore gives $\mathrm{tr}({\mathrm{\Lambda }}^{n}F)=\mathrm{tr}(({\mathrm{\Lambda }}^{n}F)Q_n)\mathrm{.}$

Let ${\iota }_n:G_n\hookrightarrow {\mathcal{H}}_n$ be inclusion and let $A_n:={\mathrm{\Lambda }}^n(F{\mathrm{\mid }}_E)$. Use the finite orthonormal basis of $G_n$ from step 1.2. Because $Q_n$ is its orthogonal projection, $Q_{n}x={\sum }_j⟨x,g_j⟩g_j$. On $G_n$, functoriality gives $({\mathrm{\Lambda }}^{n}F){\iota }_n={\iota }_n{A}_n$, and therefore $(({\mathrm{\Lambda }}^{n}F)Q_n)x={\sum }_j⟨x,g_j⟩{\iota }_n(A_n{g}_j)\mathrm{.}$ This is a finite nuclear representation. By [A25] its trace is $\mathrm{tr}(({\mathrm{\Lambda }}^{n}F)Q_n)={\sum }_j⟨{\iota }_n(A_n{g}_j),g_j⟩={\mathrm{tr}}_{G_n}(A_n),$ where the last equality is the finite-dimensional trace formula [A14]. This includes $G_n=\{0\}$, when both traces are zero. Together with step 4.1, it proves $\mathrm{tr}({\mathrm{\Lambda }}^{n}F)=\mathrm{tr}({\mathrm{\Lambda }}^n(F{\mathrm{\mid }}_E))$ for every $n\ge 0$.

In the basis $(e_j)$, [A16] identifies ${\det }_E(I_E+zA)$ with the determinant of the matrix $({\delta }_{ij}+z{a}_{ij})$. Expanding its Leibniz formula [A17] and choosing the $zA$ entry in precisely the columns indexed by a subset $I$ forces the permutation to fix every column outside $I$; the remaining signed sum is $z^{\mathrm{\mid }I\mathrm{\mid }}\det A[I,I]$. Grouping by $\mathrm{\mid }I\mathrm{\mid }=n$ and using step 2.1 gives ${\det }_E(I_E+zA)={\sum }_{n=0}^d{z}^n\mathrm{tr}({\mathrm{\Lambda }}^{n}A)\mathrm{.}$ By step 5.1, these coefficients equal $\mathrm{tr}({\mathrm{\Lambda }}^{n}F)$, and they vanish for $n>d$. Therefore the right side is exactly the series defining $D_F(z)$, proving the finite-rank identity for every $z\in \mathbb{C}$. If $d=0$, step 2.1 makes this identity $1=1$. The proof permits any finite-dimensional invariant $E$ containing $\mathrm{ran}F$; it never assumes that $E$ reduces $F$.

The empty exterior degree and $z=0$ are covered by steps 1.1 and 2.2. If $F=0$, then every positive-degree exterior power vanishes and the determinant is $1$. A zero-dimensional $E$ is handled in step 6.1. When $\dim E=1$, step 2.1 gives coefficients $1$ and $\mathrm{tr}(F{\mathrm{\mid }}_E)$ and all higher coefficients vanish, matching the linear determinant; for every $n>d$, $G_n=\{0\}$ and the trace coefficient vanishes. Countable Choice is the exact declared assumption [A21]; the only bases chosen locally are finite-dimensional orthonormal bases [A13], and no full AC or DC is used.