Arbitrary-Hilbert Fredholm determinant from a separable reducing support

By [A1] and the nuclear characterization [A2], a nuclear representation exists. Fix any such representation and let $M$ be the closed complex linear span in the statement. The argument below applies to every representation; this initial choice only constructs one support.

Let $w_{2j-2}=u_j$ and $w_{2j-1}=v_j$. Let $\mathbb{Q}+i\mathbb{Q}$ denote the Gaussian rationals. The finite sums ${\sum }_{k<r}{c}_k{w}_{n_k}$ with $c_k\in \mathbb{Q}+i\mathbb{Q}$ and $n_k\in \mathbb{N}$ form a subset $D$ of $M$. This set is at most countable: fix a bijection $\eta :\mathbb{N}\to \mathbb{Q}$ from [A7] and a bijection $\pi :\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ from [A10]. Encode each tuple $(n_k,a_k,b_k)\in {\mathbb{N}}^3$ by $\pi (n_k,\pi (a_k,b_k))$, where $c_k=\eta (a_k)+i\eta (b_k)$. The injective finite-sequence code in [A7] therefore codes every finite list of such tuples by a natural number. Decode each valid sequence code as the corresponding sum and send a natural number that is not a valid code to $0$. This defines a surjection $\mathbb{N}\to D$; $D$ is nonempty because it contains the empty sum $0$. Thus [A10] makes $D$ at most countable.

Write $T_{n}x={\sum }_{j\le n}⟨x,u_j⟩v_j$ for the nuclear partial sums. For each $x\in H$, $T_{n}x\in M$, and $T_{n}x\to Tx$ by [A2]; operator-norm convergence implies pointwise convergence by the bound in [A4]. Since $M$ is closed, $Tx\in M$. For each $y\in H$, the series $w_y:={\sum }_{j\ge 1}⟨y,v_j⟩u_j$ converges absolutely in $H$ by [A16] and therefore converges in $H$ by [A18], because $H$ is Banach. Its partial sums lie in $M$, so $w_y\in M$. Conjugate-linearity in the second argument, the adjoint identity [A3], and inner-product continuity from [A16] give $⟨x,w_y⟩={\lim }_n{\sum }_{j\le n}⟨x,u_j⟩⟨v_j,y⟩={\lim }_n⟨T_{n}x,y⟩=⟨Tx,y⟩\mathrm{.}$ Uniqueness of the adjoint in [A3] yields $T^{\ast }y=w_y$, so $T^{\ast }(H)\subseteq M$. If $x\in M^{\perp }$, then $⟨x,u_j⟩=⟨x,v_j⟩=0$ for every $j$, so the nuclear series for both $Tx$ and $T^{\ast }x$ vanish. Consequently $M$ and $M^{\perp }$ are invariant under both $T$ and $T^{\ast }$; therefore $M$ reduces $T$. The decomposition [A5] now gives $T=S\oplus 0$ on $H=M\oplus M^{\perp }$, where $S=T{\mathrm{\mid }}_M$.

The set $D$ is dense in $M$. Given $x\in M$ and $\epsilon >0$, [A20] gives a point $y$ of the span of the listed vectors with $\mathrm{\parallel }x-y\mathrm{\parallel }<\epsilon \mathrm{/}2$; by [A19], write it as a finite complex linear combination $y={\sum }_{k<r}{c}_k{w}_{n_k}$. By density of $\mathbb{Q}$ in $\mathbb{R}$ and the coordinate/modulus formula in [A8], each $c_k$ can be approximated by $q_k\in \mathbb{Q}+i\mathbb{Q}$ closely enough that ${\sum }_{k<r}\mathrm{\mid }{c}_k-q_k\mathrm{\mid }\mathrm{\parallel }{w}_{n_k}\mathrm{\parallel }<\epsilon \mathrm{/}2$; if a listed vector is zero its summand is already zero. Then $d={\sum }_{k<r}{q}_k{w}_{n_k}\in D$ and $\mathrm{\parallel }x-d\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }+{\sum }_{k<r}\mathrm{\mid }{c}_k-q_k\mathrm{\mid }\mathrm{\parallel }{w}_{n_k}\mathrm{\parallel }<\epsilon \mathrm{.}$ Consequently $D$ is a countable dense subset of $M$, so $M$ is separable by [A9]. This also covers an empty sequence, a finite list, and $M=\{0\}$.

More generally, call a closed separable subspace $N\subseteq H$ a support for $T$ when $T(H)\subseteq N$ and $T(N^{\perp })=\{0\}$. By [A5], $H=N\oplus N^{\perp }$; hence $T=S_N\oplus 0$ for $S_N:=T{\mathrm{\mid }}_N$. The adjoint identity [A3] gives $T^{\ast }=S_N^{\ast }\oplus 0$, so $N$ is reducing in the standard sense (invariant under both $T$ and $T^{\ast }$). Let ${\iota }_N:N\hookrightarrow H$ be inclusion and $P_N$ the projection of [A5]. Then $S_N=P_{N}T{\iota }_N$: on $N$, $T$ takes values in $N$ and $P_N$ is the identity. The inclusion is bounded with its inherited norm, and [A5] makes $P_N$ bounded. Thus [A6] shows that $S_N$ is trace class. This applies to the representation support $M$ of step 2.2 and to every support used below.

Let $N_1,N_2$ be two reducing supports. If either is zero, the closure of their sum is the other support and is separable. Otherwise choose nonempty dense sets $X_i\subseteq N_i$ by [A9]. They are countable, so by the surjection criterion in [A10] fix surjections $s_i:\mathbb{N}\to X_i$. Using the bijection $\pi$ from [A10] to reindex pairs, the map $(m,n)\mapsto s_1(m)+s_2(n)$ has countable image $X_1+X_2=\{x_1+x_2:x_i\in X_i\}$. It is dense in $N_1+N_2$: for $x_i\in N_i$ and any $\epsilon >0$, choose $x_i^{\mathrm{\prime }}\in X_i$ with $\mathrm{\parallel }{x}_i-x_i^{\mathrm{\prime }}\mathrm{\parallel }<\epsilon \mathrm{/}2$, so $\mathrm{\parallel }(x_1+x_2)-(x_1^{\mathrm{\prime }}+x_2^{\mathrm{\prime }})\mathrm{\parallel }<\epsilon$. Its closure $L:=\overline{N_1+N_2}$ is therefore separable. Also $T(H)\subseteq N_1\subseteq L$ and $L^{\perp }\subseteq N_1^{\perp }$, so $T(L^{\perp })=\{0\}$. Hence $L$ is a reducing support.

For any reducing support $N$, the decomposition $H=N\oplus N^{\perp }$ gives, for every $\lambda \mathrm{\ne }0$ and integer $k\ge 1$, $(T-\lambda I{)}^k=(S_N-\lambda I_N{)}^k\oplus (-\lambda {)}^k{I}_{N^{\perp }}\mathrm{.}$ Since $(-\lambda {)}^k\mathrm{\ne }0$, $\ker (T-\lambda I{)}^k=\ker (S_N-\lambda I_N{)}^k\oplus \{0\}\mathrm{.}$ Thus $T$ and $S_N$ have the same nonzero eigenvalues and the same stabilized generalized eigenspaces and algebraic multiplicities. Both are trace class and therefore compact; their nonzero spectral values are eigenvalues by [A11], and the multiplicity is the dimension of the stabilized kernel.

Let $N_1,N_2$ be arbitrary reducing supports and let $L=\overline{N_1+N_2}$ from step 4.1. Each of $S_{N_1}$, $S_{N_2}$ and $S_L$ is trace class by step 3.2 and acts on a separable Hilbert space. Step 4.2 gives the same nonzero eigenvalue list, including algebraic multiplicity, for all three restrictions. The local product theorem [A13] therefore gives $D_{S_{N_1}}(z)=D_{S_L}(z)=D_{S_{N_2}}(z)(z\in \mathbb{C})\mathrm{.}$ Taking $N_1,N_2$ to be supports arising from any two nuclear representations proves representation independence as well as independence from every separable reducing support.

Define $D_H(I+zT):=D_{S_M}(z)$ using any support $M$. Step 5.1 makes this well-defined. By [A12], it is entire and equals $1$ at $z=0$. By [A13], $D_{S_M}(z)={\prod }_j(1+z{\lambda }_j(S_M))$ locally uniformly. Step 4.2 identifies the nonzero eigenvalues and their algebraic multiplicities with those of $T$, so this is the asserted locally uniform product for $T$. If the eigenvalue list is empty, the product is $1$; finite lists are finite products, and the formula also holds for $H=\{0\}$ and $T=0$.

Suppose $T$ has finite rank and write $R=\mathrm{ran}T$. Then $R\subseteq M$ by step 2.2, $R$ is finite dimensional, and $T(R)\subseteq R$. Apply the finite-rank clause of [A12] to $S_M$ and the invariant subspace $R\subseteq M$ to obtain $D_H(I+zT)=D_{S_M}(z)={\det }_R(I_R+z(T{\mathrm{\mid }}_R))\mathrm{.}$ If $R=\{0\}$, this is $1=1$ by [A12] and the zero-dimensional determinant convention.

Let $E\subseteq H$ be any finite-dimensional subspace containing $R$. It is $T$-invariant because $T(E)\subseteq R\subseteq E$. For $R\mathrm{\ne }\{0\}$, extend a basis of $R$ to a basis of $E$. Relative to the resulting decomposition $E=R\oplus F$, the matrix of $T{\mathrm{\mid }}_E$ has block form $\left(\begin{array}{cc}A& B\\ 0& 0\end{array}\right),$ where $A$ is the matrix of $T{\mathrm{\mid }}_R$. Thus the matrix of $I_E+z(T{\mathrm{\mid }}_E)$ is $\left(\begin{array}{cc}{I}_R+zA& zB\\ 0& I_F\end{array}\right)\mathrm{.}$ In any nonzero term of the Leibniz formula [A15], each of the $\dim R$ columns from $R$ must use an $R$ row because the lower-left block is zero. Since there are exactly $\dim R$ such rows, they are all occupied, so no $F$ column can use an $R$ row. Each $F$ column must then use its matching identity entry in the $I_F$ block, and the remaining permutation sum is the Leibniz determinant of $I_R+zA$. Hence ${\det }_E(I_E+z(T{\mathrm{\mid }}_E))={\det }_R(I_R+z(T{\mathrm{\mid }}_R)),$ which with step 7.1 proves the formula for every such $E$. When $R=0$, both operators are identities and both determinants are $1$. Basis independence of these operator determinants is [A14].

At $z=0$, [A12] gives determinant one and the finite-dimensional formula is the determinant of the identity. The empty nonzero-eigenvalue list has empty product one by step 6.1; a zero operator and a zero-dimensional $H$ are included there. A one-dimensional nonzero range is covered by step 7.1, where the finite-dimensional determinant is the corresponding single linear factor. The proof uses AC exactly as [A1] states: it supplies the Countable Choice needed to obtain the nuclear representation and the Hilbert projection, and AC-qualified spectral multiplicities; the explicit countable coding and the later comparison of supports make no further selections. The conclusion is a direct equality, not a biconditional. [A1, A11, A12, step 2.1, step 4.2, step 6.1, step 7.1] \qed