Fredholm determinant of a trace-class operator

Assume the Axiom of Choice (The Axiom of Choice). Let $H$ be a complex Hilbert space and let $T\in {\mathcal{S}}_1(H)$ (Trace class operator). Choose a nuclear representation $Tx={\sum }_{j\ge 1}⟨x,u_j⟩v_j,{\sum }_{j\ge 1}\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }<\mathrm{\infty },$ and put $M=\stackrel{}{\mathrm{span}}\{u_j,v_j:j\ge 1\}$ and $S=T{\mathrm{\mid }}_M$. The Fredholm determinant of $I+zT$ is

$$\underset{H}{\det }(I+zT):=D_S(z),$$

where $D_S$ is the locally constructed separable determinant of Local separable trace-class determinant construction, identified across supports by Arbitrary-Hilbert Fredholm determinant from a separable reducing support. If $H=\{0\}$, this definition gives ${\det }_H(I+zT)=1$.

The well-definedness argument below proves that this restriction preserves the trace, trace norm, nonzero singular values, and nonzero generalized-eigenvalue data, and that the resulting determinant is independent of the nuclear representation and separable reducing support.

Nuclear representations exist by Nuclear series characterizes trace norm. Finite rational-complex linear combinations of the vectors $u_j,v_j$ form a countable dense subset of $M$, so $M$ is separable. If $x\in M^{\perp }$, every coefficient in the nuclear series vanishes and $Tx=0$; if $x\in M$, every partial sum and hence $Tx$ lies in the closed space $M$. Thus, using Orthogonal decomposition by a closed subspace,

$$H=M\oplus M^{\perp },T=S\oplus 0.$$

The restriction $S$ is bounded and compact. Indeed, a bounded sequence in $M$ is bounded in $H$. Since $T$ is compact, its images have a norm-convergent subsequence by Sequential characterization of compact operators; the limit lies in the closed space $M$. The converse direction of that same characterization makes $S:M\to M$ compact. Full AC supplies its DC hypothesis. The same nuclear series, now regarded inside $M$, therefore makes $S$ trace class by Nuclear series characterizes trace norm. The nuclear trace formula in Trace is absolutely convergent and basis independent gives ${\mathrm{tr}}_M(S)={\mathrm{tr}}_H(T)$.

The block identity gives $T^{\ast }T=S^{\ast }S\oplus 0$. Hence $\mathrm{\mid }S\mathrm{\mid }\oplus 0$ is a compact positive square root of $T^{\ast }T$, and uniqueness in Positive square root of a compact positive operator gives $\mathrm{\mid }T\mathrm{\mid }=\mathrm{\mid }S\mathrm{\mid }\oplus 0$. Therefore $S$ and $T$ have the same nonzero singular values, with multiplicities, and $\mathrm{\parallel }S{\mathrm{\parallel }}_1=\mathrm{\parallel }T{\mathrm{\parallel }}_1$.

For every $\lambda \mathrm{\ne }0$ and $r\ge 1$,

$$(T-\lambda I{)}^r=(S-\lambda I_M{)}^r\oplus (-\lambda {)}^r{I}_{M^{\perp }}\mathrm{.}$$

Consequently all generalized $\lambda$-eigenvectors lie in $M$, and $S$ and $T$ have identical nonzero eigenvalues, generalized kernels, stabilization indices, and algebraic multiplicities. The locally proved product formula and support comparison in Arbitrary-Hilbert Fredholm determinant from a separable reducing support therefore make $D_S$ independent of the chosen nuclear representation and of every separable closed reducing support on whose orthogonal complement $T$ is zero. No arbitrary invariant subspace is asserted to reduce $T$.