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Fredholm determinant properties for trace-class operators

Statement

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and $T \in S_{1} (H)$ . The function $D_{T} (z) = \det_{H} (I + z T)$ is entire and, locally uniformly in $z$ ,

D_{T} (z) = \prod_{j} (1 + z λ_{j} (T)),

where all nonzero eigenvalues are listed with their finite algebraic multiplicities, and $\sum_{j} ∣ λ_{j} (T) ∣ \leq ∥ T ∥_{1}$ . It satisfies

D_{T} (0) = 1, D_{T}^{'} (0) = {tr}_{H} (T),

∣ D_{T} (z) ∣ \leq \prod_{j} (1 + ∣ z ∣ s_{j} (T)) \leq e^{∣ z ∣ ∥ T ∥_{1}},

and for every $ε > 0$ there is $C_{ε}$ such that $∣ D_{T} (z) ∣ \leq C_{ε} e^{ε ∣ z ∣}$ . For trace-class $A, B$ , $∣ D_{A} (z) - D_{B} (z) ∣ \leq ∣ z ∣ ∥ A - B ∥_{1} e^{1 + ∣ z ∣ ∥ A ∥_{1} + ∣ z ∣ ∥ B ∥_{1}},$ and

\underset{H}{\det} (I + A + B + A B) = \underset{H}{\det} (I + A) \underset{H}{\det} (I + B) .

Moreover $D_{T} (z) = 0$ exactly when $I + z T$ is not boundedly invertible, and the zero at $- 1 / λ$ has the algebraic multiplicity of $λ \neq 0$ .

If finite-rank $T_{n}$ converge to $T$ in trace norm, their ordinary finite-dimensional determinants converge to $D_{T}$ locally uniformly. Where $I + z T$ is invertible,

D_{T}^{'} (z) = D_{T} (z) {tr}_{H} (T (I + z T)^{- 1}) .

All assertions include $H = {0}$ , finite eigenvalue lists and the empty list.

Facts & Assumptions

Given: The Axiom of Choice, a complex Hilbert space $H$ , and the displayed trace-class operators.

[F1]

The determinant is defined through a separable reducing support; its value is independent of the support, entire and normalized, and equals the locally uniform product over the nonzero eigenvalues with algebraic multiplicity. Finite-rank values are ordinary finite-dimensional determinants (Fredholm determinant of a trace-class operator, Arbitrary-Hilbert Fredholm determinant from a separable reducing support).

[F2]

On a separable complex Hilbert space the local determinant satisfies $D_{S}^{'} (0) = tr (S)$ (Local separable trace-class determinant construction).

[F3]

On a separable complex Hilbert space the eigenvalue absolute sum is at most the trace norm, with algebraic multiplicities (Weyl product and sum inequalities for compact operators).

[F4]

On a separable complex Hilbert space the local determinant satisfies the singular-value product and exponential bounds, minimal exponential type, the displayed trace-norm continuity estimate, multiplicativity at $z = 1$ , and locally uniform convergence of finite-rank determinants in trace norm (Trace-norm continuity, growth and multiplicativity of the local determinant).

[F5]

On a separable complex Hilbert space, the local determinant vanishes exactly when $I + z S$ is not boundedly invertible, and its zero at $- 1 / λ$ has order $m_{a l g} (λ; S)$ (Zeros of the local Fredholm determinant).

[F6]

On a separable complex Hilbert space, at every invertibility point, $D_{S}^{'} (z) = D_{S} (z) tr (S (I + z S)^{- 1})$ (Logarithmic derivative of the local Fredholm determinant).

[F7]

Trace-class operators form a linear two-sided ideal, their trace norm is a norm, and singular values are the positive eigenvalues of $∣ T ∣$ (Trace class is a two sided Banach operator ideal, Absolute value and singular values of a compact operator).

[F8]

The trace is basis-independent and agrees with every nuclear trace sum (Trace is absolutely convergent and basis independent).

Proof

technique · direct

1.1F1F7F8given

Choose a nuclear support $M$ from [F1] and write $H = M \oplus M^{⊥}$ and $T = S \oplus 0$ . The nuclear trace formula gives ${tr}_{H} (T) = {tr}_{M} (S)$ : the same nuclear vectors lie in $M$ , so their scalar inner products are unchanged. The block identity $T^{*} T = S^{*} S \oplus 0$ and uniqueness of the compact positive square root, as established in Fredholm determinant of a trace-class operator, give $∣ T ∣ = ∣ S ∣ \oplus 0$ . Thus $T$ and $S$ have the same nonzero singular values, including multiplicity, and $∥ T ∥_{1} = ∥ S ∥_{1}$ . For each $λ \neq 0$ , $(T - λ I)^{r} = (S - λ I_{M})^{r} \oplus (- λ)^{r} I_{M^{⊥}}$ ; hence their nonzero eigenvalues and algebraic multiplicities agree.

2.1F1F2F3F4step 1.1

By [F1], $D_{T} = D_{S}$ is entire, normalized, and has the stated locally uniform spectral product. Applying [F3] to $S$ and step 1.1 gives $\sum_{j} ∣ λ_{j} (T) ∣ \leq ∥ T ∥_{1}$ . Applying [F2] gives $D_{T}^{'} (0) = D_{S}^{'} (0) = {tr}_{M} (S) = {tr}_{H} (T)$ . The singular-value product, exponential bound and minimal exponential type in [F4] transfer from $S$ using the same singular-value list and trace norm. This also covers finite and empty lists.

2.2F1F5F6F8step 1.1

For each $z$ , $I + z T = (I_{M} + z S) \oplus I_{M^{⊥}}$ . It has a bounded inverse exactly when $I_{M} + z S$ does, because the inverse of a block diagonal operator is the block inverse and restriction of a bounded inverse to the reducing summand is bounded. The generalized kernels in step 1.1 preserve algebraic multiplicity. Hence [F5] gives both directions of the stated zero criterion and the exact zero order at $- 1 / λ$ . If $I + z T$ is invertible, its inverse is $(I_{M} + z S)^{- 1} \oplus I_{M^{⊥}}$ ; consequently $T (I + z T)^{- 1} = S (I_{M} + z S)^{- 1} \oplus 0$ . The same nuclear trace formula as step 1.1 equates these traces, so [F6] gives the displayed logarithmic derivative.

2.3F1F4F7step 1.1

For trace-class $A, B$ on $H$ , take nuclear representations for both and let $N$ be the closed span of all their input and output vectors. The finite union of the two countable vector lists has a countable dense set of finite Gaussian-rational combinations; the support proof in [F1] shows $N$ is separable and reduces both operators, with $A = A_{N} \oplus 0$ and $B = B_{N} \oplus 0$ . By [F7], $C : = A + B + A B$ is trace class and $C = (A_{N} + B_{N} + A_{N} B_{N}) \oplus 0$ ; the same $N$ supports $A - B$ . The block singular-value argument of step 1.1 gives $∥ A - B ∥_{1} = ∥ A_{N} - B_{N} ∥_{1}$ and $∥ A ∥_{1} = ∥ A_{N} ∥_{1}$ , $∥ B ∥_{1} = ∥ B_{N} ∥_{1}$ . Apply the separable continuity estimate and multiplicativity in [F4] on $N$ , and use support independence in [F1] for all three determinants. These are exactly the displayed arbitrary-space formulas, including the same numerical exponential constant.

2.4F1F4step 1.1

Suppose finite-rank $T_{n} \to T$ in trace norm. AC chooses nuclear representations for $T$ and the countable family $(T_{n})$ ; taking the closed span of every input and output vector gives one separable reducing support $N$ for all of them. Their restrictions $S_{n}, S$ obey $∥ S_{n} - S ∥_{1} = ∥ T_{n} - T ∥_{1} \to 0$ by the block singular-value argument of step 1.1. Apply the locally uniform finite-rank approximation in [F4] on $N$ . By [F1], $D_{S_{n}} = D_{T_{n}}$ and each is the ordinary determinant on any finite-dimensional subspace containing $ran T_{n}$ ; the same support lemma shows independence of that subspace. Therefore those ordinary determinants converge locally uniformly to $D_{S} = D_{T}$ .

3.1F1F2F4F5F6step 2.1step 2.2step 2.4∎

If $H = {0}$ or $T = 0$ , [F1] gives $D_{T} \equiv 1$ , its derivative and trace are zero, and the product and singular-value lists are empty. For $z = 0$ , normalization holds, and the logarithmic-derivative formula follows from step 2.1. A finite-rank operator is covered by [F1] and step 2.4. Full AC supplies the separable-support and AC-qualified spectral suppliers and permits the countable family of nuclear representations in step 2.4; its Countable Choice consequence supplies [F2], [F4] and [F6]. Both directions of the zero criterion were established in step 2.2.

Depends on

Used by

Lidskii trace formula for trace-class operators Theorem

Dependency tree · two levels

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Sources

Aleksey Kostenko, Trace Ideals with Applications — Sections 3.4–3.5, printed pp. 34–45 (standard reference, not scraped)