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Banach manifold differentials are chart independent

Statement

Let k1, 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 pdomφ.
  2. The differential Df(p):TpMTf(p)N of a C1 map f:MN 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(gf)(p)=Dg(f(p))Df(p) whenever f:MN and g:NP are C1.

Facts & Assumptions

Given: k1, Ck Banach manifolds M,N,P modelled on real Banach spaces E,F,G, points pM, and charts φ,φ of M at p, ψ,ψ of N at f(p) for the C1 map f:MN.

[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(βα)(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
1.1

(Reflexivity and symmetry.) For a chart φ at p the transition φφ1 is the identity on an open set containing φ(p), so D(φφ1)(φ(p))=I by [L2] and (φ,v)(φ,v). If (φ,v)(ψ,w), then w=D(ψφ1)(φ(p))v; the two transition maps are mutually inverse C1 maps on neighbourhoods of φ(p) and ψ(p), so differentiating the identities (φψ1)(ψφ1)=id and (ψφ1)(φψ1)=id with [L2] gives D(φψ1)(ψ(p))w=v, that is (ψ,w)(φ,v).

L1L2L3
2.1

(Transitivity.) If (φ,v)(ψ,w) and (ψ,w)(χ,u), then on a neighbourhood of φ(p) the identity χφ1=(χψ1)(ψφ1) holds, and [L2] gives D(χφ1)(φ(p))v=D(χψ1)(ψ(p))D(ψφ1)(φ(p))v=D(χψ1)(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.

step 1.1L2L3
3.1

(The differential is well defined.) Let f:MN 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 ψfφ1=(ψψ1)(ψfφ1)(φφ1); if (φ,v)(φ,v), that is v=D(φφ1)(φ(p))v, then [L2] gives D(ψfφ1)(φ(p))v=D(ψψ1)(ψ(f(p)))[D(ψfφ1)(φ(p))v], which is precisely the relation [ψ,D(ψfφ1)(φ(p))v]=[ψ,D(ψfφ1)(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.

step 1.1step 2.1L1L2L3
3.2

(The vector space operations are chart independent.) If (φ,v)(φ,v) and (φ,w)(φ,w), then the shared transition derivative T:=D(φφ1)(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.

step 1.1step 2.1L1L2algebra
4.1

(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(φidMφ1)(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(gf)φ1=(ρgψ1)(ψfφ1) near φ(p), and applying [L2] to this identity of open-subset maps gives equality of the two well-defined maps D(gf)(p) and Dg(f(p))Df(p) on every class represented in the chart φ.

step 3.1L1L2L3
5.1

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].

step 1.1step 2.1step 3.1step 3.2step 4.1

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