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LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-04 (gpt-6.1-sol)
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Nilpotent parts on a joint cyclic subspace give finite logarithmic dilation factors

Statement

Let V be a complex vector space, let A,B:V→V be commuting linear maps, and let v∈V satisfy Arv=0 and Bsv=0 for positive integers r,s. Let h,hˉ∈C. Let Uz,Uζ⊂C× be simply connected open sets, with chosen holomorphic logarithms ℓz and ℓζ.

The span

Wv=span⁡C{AiBjv:0≤i<r, 0≤j<s}

is finite dimensional and invariant under A and B. On Wv, define

Dv(z,ζ)=exp⁡ ⁣(ℓz(z)(hI+A))exp⁡ ⁣(ℓζ(ζ)(hˉI+B)).

Then

Dv(z,ζ)v=ehℓz(z)+hˉℓζ(ζ)∑i=0r−1∑j=0s−1ℓz(z)iℓζ(ζ)ji! j!AiBjv.

For every linear functional λ:Wv→C, the normalized matrix coefficient e−hℓz−hˉℓζλ(Dvv) is the finite polynomial

∑i=0r−1∑j=0s−1λ(AiBjv)i! j!ℓziℓζj.

Thus the nilpotent generator parts A∣Wv,B∣Wv supply possible logarithmic terms; a chosen pairing may annihilate their positive-degree coefficients. Under a Euclidean loop about the origin, ℓz changes by 2πi and ℓζ by −2πi, so the continued dilation is

e2πi(hI+A)e−2πi(hˉI+B)Dv.

Single-valuedness therefore requires a separate condition on the paired fields and matrix coefficients.

Facts & Assumptions

Given: The vector space V, commuting endomorphisms A,B, a vector v with the stated nilpotence orders, scalars h,hˉ, and the two chosen logarithm branches.

[A1]

The positive integers r,s satisfy Arv=0 and Bsv=0.

[A2]

The branches ℓz,ℓζ are holomorphic logarithms on the specified simply connected subsets of C×; on the Euclidean locus ζ=zˉ, continuation around zero changes the branches by 2πi and −2πi, respectively.

Proof

1.1givenA1algebra

Since A and B commute, every AiBjv with i≥r equals BjArv=0, and every one with j≥s equals AiBsv=0. Thus Wv is spanned by at most rs vectors. Applying A or B to its spanning vectors either gives another listed vector or zero, so Wv is invariant.

2.1givenA1step 1.1algebra

On Wv, the restrictions of A and B are nilpotent and commute. Expanding the two matrix exponentials and using the binomial identity for commuting operators gives Dvv=ehℓz+hˉℓζexp⁡(ℓzA)exp⁡(ℓζB)v. Every term with i≥r or j≥s vanishes by [A1], so the product of exponential series is exactly the displayed finite double sum. Applying λ gives the normalized matrix coefficient.

3.1givenA2step 2.1algebra∎

Continuing the two logarithms by the changes in [A2] multiplies the first exponential by e2πi(hI+A) and the second by e−2πi(hˉI+B). The factors commute with each other and with Dv, yielding the stated continuation formula.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources