Banach bundle

This article is about Banach bundles in differential geometry. For Banach bundles in non-commutative geometry, see Banach bundle (non-commutative geometry).

In mathematics, a Banach bundle is a vector bundle each of whose fibres is a Banach space, i.e. a complete normed vector space, possibly of infinite dimension.

Definition of a Banach bundle

Let M be a Banach manifold of class Cp with p ≥ 0, called the base space; let E be a topological space, called the total space; let π : EM be a surjective continuous map. Suppose that for each point xM, the fibre Ex = π1(x) has been given the structure of a Banach space. Let

\{ U_{i} | i \in I \}

be an open cover of M. Suppose also that for each iI, there is a Banach space Xi and a map τi

\tau_{i} : \pi^{-1} (U_{i}) \to U_{i} \times X_{i}

such that

and for each x Ui the induced map τix on the fibre Ex
\tau_{ix} : \pi^{-1} (x) \to X_{i}
is an invertible continuous linear map, i.e. an isomorphism in the category of topological vector spaces;
U_{i} \cap U_{j} \to \mathrm{Lin}(X_{i}; X_{j})
x \mapsto (\tau_{j} \circ \tau_{i}^{-1})_{x}
is a morphism (a differentiable map of class Cp), where Lin(X; Y) denotes the space of all continuous linear maps from a topological vector space X to another topological vector space Y.

The collection {(Ui, τi)|iI} is called a trivialising covering for π : EM, and the maps τi are called trivialising maps. Two trivialising coverings are said to be equivalent if their union again satisfies the two conditions above. An equivalence class of such trivialising coverings is said to determine the structure of a Banach bundle on π : EM.

If all the spaces Xi are isomorphic as topological vector spaces, then they can be assumed all to be equal to the same space X. In this case, π : EM is said to be a Banach bundle with fibre X. If M is a connected space then this is necessarily the case, since the set of points xM for which there is a trivialising map

\tau_{ix} : \pi^{-1} (x) \to X

for a given space X is both open and closed.

In the finite-dimensional case, the second condition above is implied by the first.

Examples of Banach bundles

\pi : \mathrm{T} V \to V;
(x, v) \mapsto x.
This bundle is "trivial" in the sense that TV admits a globally defined trivialising map: the identity function
\tau = \mathrm{id} : \pi^{-1} (V) = \mathrm{T} V \to V \times V;
(x, v) \mapsto (x, v).
\pi^{-1} (x) = \mathrm{T}_{x}^{*} M = (\mathrm{T}_{x} M)^{*};
also forms a Banach bundle with respect to the usual projection onto M.

Morphisms of Banach bundles

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

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

f_{0} : M \to M';
f : E \to E'.

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 Cp, 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):

commutes, and, for each x M, the induced map
f_{x} : E_{x} \to E'_{f_{0} (x)}
is a continuous linear map;
\tau : \pi^{-1} (U) \to U \times X
\tau' : \pi'^{-1} (U') \to U' \times X'
such that x0 U, f0(x0) U,
f_{0} (U) \subseteq U'
and the map
U \to \mathrm{Lin}(X; X')
x \mapsto \tau'_{f_{0} (x)} \circ f_{x} \circ \tau^{-1}
is a morphism (a differentiable map of class Cp).

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 : EN be a Banach bundle and f : MN a differentiable map (as usual, everything is Cp). Then the pull-back of π : EN is the Banach bundle f*π : f*EM satisfying the following properties:

with the top horizontal map being the identity on each fibre;
f^{*} \pi : f^{*} E = M \times X \to M
is the projection onto the first coordinate;
f^{*} (E_{V}) = (f^{*} 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.

References

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