Banach bundle (non-commutative geometry)

In mathematics, a Banach bundle is a fiber bundle over a topological Hausdorff space, such that each fiber has the structure of a Banach space.

Let ${\displaystyle X}$ be a topological Hausdorff space, a (continuous) Banach bundle over ${\displaystyle X}$ is a tuple ${\displaystyle {\mathfrak {B}}=(B,\pi )}$, where ${\displaystyle B}$ is a topological Hausdorff space, and ${\displaystyle \pi \colon B\to X}$ is a continuous, open surjection, such that each fiber ${\displaystyle B_{x}:=\pi ^{-1}(x)}$ is a Banach space. Which satisfies the following conditions:

1. The map ${\displaystyle b\mapsto \|b\|}$ is continuous for all ${\displaystyle b\in B}$
2. The operation ${\displaystyle +\colon \{(b_{1},b_{2})\in B\times B:\pi (b_{1})=\pi (b_{2})\}\to B}$ is continuous
3. For every ${\displaystyle \lambda \in \mathbb {C} }$, the map ${\displaystyle b\mapsto \lambda \cdot b}$ is continuous
4. If ${\displaystyle x\in X}$, and ${\displaystyle \{b_{i}\}}$ is a net in ${\displaystyle B}$, such that ${\displaystyle \|b_{i}\|\to 0}$ and ${\displaystyle \pi (b_{i})\to x}$, then ${\displaystyle b_{i}\to 0_{x}\in B}$, where ${\displaystyle 0_{x}}$ denotes the zero of the fiber ${\displaystyle B_{x}}$.^[1]

If the map ${\displaystyle b\mapsto \|b\|}$ is only upper semi-continuous, ${\displaystyle {\mathfrak {B}}}$ is called upper semi-continuous bundle.

Let A be a Banach space, X be a topological Hausdorff space. Define ${\displaystyle B:=A\times X}$ and ${\displaystyle \pi \colon B\to X}$ by ${\displaystyle \pi (a,x):=x}$. Then ${\displaystyle (B,\pi )}$ is a Banach bundle, called the trivial bundle

• Banach bundles in differential geometry

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