Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Banach manifold differentials are chart independent

Statement

Let k≥1, let M be a Ck Banach manifold modelled on E (all manifolds below are Ck and smooth maps are C1), and use the tangent space, its vector space structure and the differential of Tangent space and differential on a Banach manifold. Then:

  1. ∼ is an equivalence relation on the pairs (φ,v) with p∈dom⁡φ.
  2. The differential Df(p):TpM→Tf(p)N of a C1 map f:M→N is well defined; more precisely, for any two choices of charts the resulting classes coincide, and the chart-based vector space operations on TpM are independent of the chart used.
  3. D(idM)(p)=idTpM for every p, and D(g∘f)(p)=Dg(f(p))∘Df(p) whenever f:M→N and g:N→P are C1.

Facts & Assumptions

Given: k≥1, Ck Banach manifolds M,N,P modelled on real Banach spaces E,F,G, points p∈M, and charts φ,φ′ of M at p, ψ,ψ′ of N at f(p) for the C1 map f:M→N.

[L1]

Definition of tangent vectors, their chart identifications, the vector space operations and the differential, together with the fact that transition maps of a Ck manifold are Ck and hence C1 (Tangent space and differential on a Banach manifold, Countable base Banach manifold and smooth map).

[L2]

Chain rule, sum rule and the derivative of the identity: for C1 maps of open subsets of Banach spaces, D(β∘α)(x)=Dβ(α(x))Dα(x), and D(id)(x)=I; a bounded linear map is its own derivative (Chain sum product and composition rules for Banach derivatives, Fréchet derivative between Banach spaces).

[L3]

Chart representatives of C1 maps between manifolds are C1 on their open domains, and composites of the transition maps appearing below are defined on a neighbourhood of the relevant point, because chart domains and their images are open (Countable base Banach manifold and smooth map).

Proof

technique · direct
1.1L1L2L3

(Reflexivity and symmetry.) For a chart φ at p the transition φ∘φ−1 is the identity on an open set containing φ(p), so D(φ∘φ−1)(φ(p))=I by [L2] and (φ,v)∼(φ,v). If (φ,v)∼(ψ,w), then w=D(ψ∘φ−1)(φ(p))v; the two transition maps are mutually inverse C1 maps on neighbourhoods of φ(p) and ψ(p), so differentiating the identities (φ∘ψ−1)∘(ψ∘φ−1)=id and (ψ∘φ−1)∘(φ∘ψ−1)=id with [L2] gives D(φ∘ψ−1)(ψ(p))w=v, that is (ψ,w)∼(φ,v).

2.1step 1.1L2L3

(Transitivity.) If (φ,v)∼(ψ,w) and (ψ,w)∼(χ,u), then on a neighbourhood of φ(p) the identity χ∘φ−1=(χ∘ψ−1)∘(ψ∘φ−1) holds, and [L2] gives D(χ∘φ−1)(φ(p))v=D(χ∘ψ−1)(ψ(p)) D(ψ∘φ−1)(φ(p))v=D(χ∘ψ−1)(ψ(p))w=u, that is (φ,v)∼(χ,u). Together with [step 1.1], this establishes the equivalence relation before any construction is asserted on its classes.

3.1step 1.1step 2.1L1L2L3

(The differential is well defined.) Let f:M→N be C1 and let (φ,ψ), (φ′,ψ′) be two chart pairs at p and f(p). On a neighbourhood of φ′(p) one has ψ′∘f∘φ′−1=(ψ′∘ψ−1)∘(ψ∘f∘φ−1)∘(φ∘φ′−1); if (φ′,v′)∼(φ,v), that is v=D(φ∘φ′−1)(φ′(p))v′, then [L2] gives D(ψ′∘f∘φ′−1)(φ′(p))v′=D(ψ′∘ψ−1)(ψ(f(p))) [D(ψ∘f∘φ−1)(φ(p))v], which is precisely the relation [ψ′,D(ψ′∘f∘φ′−1)(φ′(p))v′]=[ψ,D(ψ∘f∘φ−1)(φ(p))v] defining ∼ on the target manifold. Because [step 1.1] and [step 2.1] have already proved that ∼ is an equivalence relation, this comparison proves independence of both the representative and the chart pair.

3.2step 1.1step 2.1L1L2algebra

(The vector space operations are chart independent.) If (φ,v)∼(φ′,v′) and (φ,w)∼(φ′,w′), then the shared transition derivative T:=D(φ′∘φ−1)(φ(p)) is linear with v′=Tv, w′=Tw; hence v′+w′=T(v+w) and λv′=T(λv), that is (φ,v+w)∼(φ′,v′+w′) and (φ,λv)∼(φ′,λv′). Since ∼ is an equivalence relation by [step 1.1] and [step 2.1], these representative calculations define operations on the quotient classes.

4.1step 3.1L1L2L3

(Functoriality.) The maps in this step are well defined on tangent classes by [step 3.1]. For the identity, D(idM)(p)[φ,v]=[φ,D(φ∘idM∘φ−1)(φ(p))v]=[φ,D(id)(φ(p))v]=[φ,v] by [L2]. For a composite, fix charts φ at p, ψ at f(p) and ρ at g(f(p)); then ρ∘(g∘f)∘φ−1=(ρ∘g∘ψ−1)∘(ψ∘f∘φ−1) near φ(p), and applying [L2] to this identity of open-subset maps gives equality of the two well-defined maps D(g∘f)(p) and Dg(f(p))∘Df(p) on every class represented in the chart φ.

5.1step 1.1step 2.1step 3.1step 3.2step 4.1∎

Assertion 1 is [step 1.1] with [step 2.1]; assertion 2 is [step 3.1] and [step 3.2]; assertion 3 is [step 4.1].

Depends on

Used by

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources