Banach bundle

Examples of Banach bundles

• If V is any Banach space, the tangent space T_xV to V at any point x ∈ V is isomorphic in an obvious way to V itself. The tangent bundle TV of V is then a Banach bundle with the usual projection

${\displaystyle \pi :\mathrm{T}V\to V;}$

${\displaystyle (x,v)\mapsto x.}$

This bundle is "trivial" in the sense that TV admits a globally defined trivialising map: the identity function

${\displaystyle \tau =\mathrm{i}\mathrm{d}:{\pi }^{-1}(V)=\mathrm{T}V\to V\times V;}$

${\displaystyle (x,v)\mapsto (x,v).}$

• If M is any Banach manifold, the tangent bundle TM of M forms a Banach bundle with respect to the usual projection, but it may not be trivial.
• Similarly, the cotangent bundle T*M, whose fibre over a point x ∈ M is the topological dual space to the tangent space at x:

${\displaystyle {\pi }^{-1}(x)={\mathrm{T}}_x^{\ast }M=({\mathrm{T}}_{x}M{)}^{\ast };}$

also forms a Banach bundle with respect to the usual projection onto M.

• There is a connection between Bochner spaces and Banach bundles. Consider, for example, the Bochner space X = L²([0, T]; H¹(Ω)), which might arise as a useful object when studying the heat equation on a domain Ω. One might seek solutions σ ∈ X to the heat equation; for each time t, σ(t) is a function in the Sobolev space H¹(Ω). One could also think of Y = [0, T] × H¹(Ω), which as a Cartesian product also has the structure of a Banach bundle over the manifold [0, T] with fibre H¹(Ω), in which case elements/solutions σ ∈ X are cross sections of the bundle Y of some specified regularity (L², in fact). If the differential geometry of the problem in question is particularly relevant, the Banach bundle point of view might be advantageous.

Morphisms of Banach bundles

The collection of all Banach bundles can be made into a category by defining appropriate morphisms.

Let π : E → M and π′ : E′ → M′ be two Banach bundles. A Banach bundle morphism from the first bundle to the second consists of a pair of morphisms

${\displaystyle f_0:M\to M^{\prime };}$

${\displaystyle f:E\to E^{\prime }.}$

For f to be a morphism means simply that f is a continuous map of topological spaces. If the manifolds M and M′ are both of class C^p, then the requirement that f₀ be a morphism is the requirement that it be a p-times continuously differentiable function. These two morphisms are required to satisfy two conditions (again, the second one is redundant in the finite-dimensional case):

• the diagram

commutes, and, for each x ∈ M, the induced map

${\displaystyle f_x:E_x\to E_{f_0(x)}^{\prime }}$

is a continuous linear map;

• for each x₀ ∈ M there exist trivialising maps

${\displaystyle \tau :{\pi }^{-1}(U)\to U\times X}$

${\displaystyle {\tau }^{\prime }:{\pi }^{\prime -1}(U^{\prime })\to U^{\prime }\times X^{\prime }}$

such that x₀ ∈ U, f₀(x₀) ∈ U′,

${\displaystyle f_0(U)\subseteq U^{\prime }}$

and the map

${\displaystyle U\to \mathrm{L}\mathrm{i}\mathrm{n}(X;X^{\prime })}$

${\displaystyle x\mapsto {\tau }_{f_0(x)}^{\prime }\circ f_x\circ {\tau }^{-1}}$

is a morphism (a differentiable map of class C^p).

Pull-back of a Banach bundle

One can take a Banach bundle over one manifold and use the pull-back construction to define a new Banach bundle on a second manifold.

Specifically, let π : E → N be a Banach bundle and f : M → N a differentiable map (as usual, everything is C^p). Then the pull-back of π : E → N is the Banach bundle f*π : f*E → M satisfying the following properties:

• for each x ∈ M, (f*E)_x = E_f(x);
• there is a commutative diagram

with the top horizontal map being the identity on each fibre;

• if E is trivial, i.e. equal to N × X for some Banach space X, then f*E is also trivial and equal to M × X, and

${\displaystyle f^{\ast }\pi :f^{\ast }E=M\times X\to M}$

is the projection onto the first coordinate;

• if V is an open subset of N and U = f⁻¹(V), then

${\displaystyle f^{\ast }(E_V)=(f^{\ast }E{)}_U}$

and there is a commutative diagram

where the maps at the "front" and "back" are the same as those in the previous diagram, and the maps from "back" to "front" are (induced by) the inclusions.