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Logarithmic derivative of the local Fredholm determinant

Statement

Assume the Axiom of Countable Choice. Let $H$ be a separable complex Hilbert space and let $T : H \to H$ be trace class. Write $D_{T}$ for the locally constructed determinant of Local separable trace-class determinant construction. For each $z \in C$ for which $I + z T$ has a bounded inverse $R_{z} \in B (H)$ , $D_{T}^{'} (z) = D_{T} (z) tr (T R_{z}) .$ Equivalently, $R_{z} = (I + z T)^{- 1}$ in the displayed formula.

Facts & Assumptions

Given: Countable Choice, a separable complex Hilbert space $H$ , a trace-class operator $T : H \to H$ , and, at the point $z$ under consideration, a bounded two-sided inverse $R_{z}$ of $I + z T$ .

[A1]

Countable Choice, written ${A C}_{ω}$ , is the declared choice assumption (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A2]

For a trace-class operator $U$ on a separable complex Hilbert space, the local determinant is $D_{U} (w) = \sum_{n \geq 0} w^{n} tr (Λ^{n} U),$ is entire, and satisfies $D_{U} (0) = 1$ and $D_{U}^{'} (0) = tr (U)$ (Local separable trace-class determinant construction).

[A3]

Trace-class operators form a linear space; if $U$ is trace class and $V$ is a bounded operator, then $U V$ and $V U$ are trace class (Trace class operator, Trace class is a two sided Banach operator ideal).

[A4]

For $n \geq 1$ , the tensor definition of exterior powers gives $Λ^{n} (c U) = c^{n} Λ^{n} U$ for $c \in C$ , while $Λ^{0} (c U) = I_{C}$ (Hilbert exterior powers and induced operators).

[A5]

For trace-class $A, B$ on the same separable complex Hilbert space, $D_{A + B + A B} (1) = D_{A} (1) D_{B} (1)$ (Trace-norm continuity, growth and multiplicativity of the local determinant).

[A6]

Membership $R_{z} \in B (H)$ means that $R_{z}$ is a bounded linear operator, as required in the trace-class ideal estimate (A bounded linear operator between normed spaces).

Source audit: Dyatlov–Zworski, Mathematical Theory of Scattering Resonances, Appendix B §B.5.2, Lemma B.26 (PDF p. 509) proves the finite-rank formula for a $C^{1}$ path $A_{t}$ whose ranges lie in one fixed finite-dimensional subspace, assuming $I - A_{t}$ remains invertible along the path. Its proof reduces to Jacobi's finite-dimensional determinant formula. The local proof below does not extend that finite-rank statement by assertion: it derives the trace-class identity from the already proved determinant multiplicativity and the derivative at zero. Van Neerven, Functional Analysis, §14.5.a, Definition 14.34 (printed p. 585; PDF p. 597) gives the entire exterior-trace series and its first-order expansion, and Lemma 14.39 (printed p. 587; PDF p. 599) gives multiplicativity. Kostenko, Trace Ideals with Applications, §3.4.3, Corollary 3.4.2 (printed p. 40; PDF p. 49) also gives the product identity. Those passages were read in full; none states the general infinite-dimensional logarithmic-derivative formula as used here. They inform the audit, while the argument below proves the claim locally. No source uncertainty remains.

Proof

technique · direct

1.1A3A6algebra

Fix such a $z$ , write $R = R_{z}$ , put $M : = I + z T$ , and set $S : = T R$ . Since $M T = T M$ and $R M = M R = I$ , the equality $R (M T) R = R (T M) R$ gives $T R = R T$ , hence $M S = M T R = T M R = T$ . The operator $S$ is trace class by [A3], because $T$ is trace class and $R$ is bounded; therefore the scalar multiples $z T$ and $h S$ are trace class for every $h \in C$ .

1.2A2A4

For every trace-class $U$ and $c \in C$ , [A2] and [A4] give $D_{c U} (1) = \sum_{n \geq 0} tr (Λ^{n} (c U)) = D_{U} (c)$ : the degree-zero term is $1$ on both sides and for $n \geq 1$ the degree- $n$ term is $c^{n} tr (Λ^{n} U)$ , including when $c = 0$ .

2.1step 1.1algebra

For each $h \in C$ , step 1.1 gives $(I + z T) S = T$ , so $I + (z + h) T = (I + z T) (I + h S),$ because the coefficient of $h$ on the right is $S + z T S = (I + z T) S = T$ .

3.1A5step 1.2step 2.1

Apply [A5] to trace-class $z T$ and $h S$ ; by step 2.1, $(z + h) T = z T + h S + z h T S$ , so $D_{(z + h) T} (1) = D_{z T} (1) D_{h S} (1)$ and step 1.2 turns this into $D_{T} (z + h) = D_{T} (z) D_{S} (h) .$

4.1A2step 1.1step 3.1

Subtract $D_{T} (z)$ from step 3.1 and divide by $h \neq 0$ ; as $h \to 0$ , [A2] gives $D_{T}^{'} (z) = D_{T} (z) D_{S}^{'} (0) = D_{T} (z) tr (S)$ , and $S = T R = T (I + z T)^{- 1}$ gives the claimed formula. Taking $h = - z$ in step 3.1 and using $D_{T} (0) = 1$ from [A2] gives $1 = D_{T} (z) D_{S} (- z)$ , so $D_{T} (z) \neq 0$ and the identity is its ordinary logarithmic-derivative formula as well.

5.1

If $T = 0$ , then $S = 0$ , $D_{T} \equiv 1$ , and both sides are zero; this also covers $H = {0}$ . On $H = C$ with $T = t I$ and $1 + z t \neq 0$ , $D_{T} (z) = 1 + z t$ and $R = (1 + z t)^{- 1} I$ , so both sides equal $t$ . At $z = 0$ , $R = I$ and step 4.1 gives $D_{T}^{'} (0) = tr (T)$ . The complex derivative exists on the whole plane by [A2], and the difference quotient uses complex $h$ , so there is no one-sided endpoint case. Countable Choice is exactly [A1], inherited through the trace-class, determinant, ideal, and multiplicativity suppliers; this proof makes no additional choices. The claim is not an equivalence, so both iff directions are inapplicable. [A1, A2, A3, A5, A6, step 1.1, step 3.1, step 4.1] \qed

Depends on

Used by

Spectral product from traces of powers Lemma
Fredholm determinant properties for trace-class operators Proposition

Dependency tree · two levels

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Sources

Kostenko, Trace Ideals with Applications, §3.4.3, Corollary 3.4.2, printed p. 40 (PDF p. 49) (standard reference, not scraped)
van Neerven, Functional Analysis, §14.5.a, Definition 14.34 and Lemma 14.39, printed pp. 585 and 587 (PDF pp. 597 and 599) (standard reference, not scraped)
Dyatlov–Zworski, Mathematical Theory of Scattering Resonances, Appendix B §B.5.2, Lemma B.26, PDF p. 509 (standard reference, not scraped)