2021-04-25_vector_bundles_ug

#homotopy/bundles #homotopy

Vector Bundles

Definition: A rank \(n\) vector bundle is a fiber bundle in which the fibers \(F\) have the structure of a vector space \(k^n\) for some field \(k\); the structure group of such a bundle is a subset of \(\operatorname{GL}(n, k)\).

Note that a vector bundle always has one global section: namely, since every fiber is a vector space, you can canonically choose the 0 element to obtain a global zero section.

Proposition: A rank \(n\) vector bundle is trivial iff it admits \(k\) linearly independent global sections.

Example: The tangent bundle of a manifold is an \({\mathbf{R}}\)-vector bundle. Let \(M^n\) be an \(n{\hbox{-}}\) dimensional manifold. For any point \(x\in M\), the tangent space \(T_xM\) exists, and so we can define \begin{align*} TM = \coprod_{x\in M} T_xM = \left\{{(x, t) \mathrel{\Big|}x\in M, t \in T_xM}\right\} \end{align*} Then \(TM\) is a manifold of dimension \(2n\) and there is a corresponding fiber bundle \begin{align*} {\mathbf{R}}^n \to TM \xrightarrow{\pi} M \end{align*} given by a natural projection \(\pi:(x, t) \mapsto x\)

Example A circle bundle is a fiber bundle in which the fiber is isomorphic to \(S^1\) as a topological group. Consider circle bundles over a circle, which are of the form \begin{align*} S^1 \to E \xrightarrow{\pi} S^1 \end{align*} There is a trivial bundle, when \(E = S^1 \times S^1 = T^2\), the torus:

There is also a nontrivial bundle, \(E = K\), the Klein bottle:

As in the earlier example involving the Mobius strip, since \(K\) is nonorientable, \(T^2 \not\cong K\) and there are thus at least two distinct bundles of this type.


Remark: A section of the tangent bundle \(TM\) is equivalent to a vector field on \(M\).

Definition: If the tangent bundle of a manifold is trivial, the manifold is said to be parallelizable.

Proposition: The circle \(S^1\) is parallelizable.

Proof Let \(M = S^1\), then there is a rank 1 vector bundle
\begin{align*}{\mathbf{R}}\to TM \to M\end{align*} and since \(TM = S^1 \times{\mathbf{R}}\) (why?), we find that \(S^1\) is parallelizable.

Proposition: The sphere \(S^2\) is not parallelizable.

Proof: Let \(M = S^2\), which is associated to the rank 2 vector bundle \begin{align*}{\mathbf{R}}^2 \to TM \to M\end{align*}

Then \(TM\) is trivial iff there are 2 independent global sections. Since there is a zero section, a second independent section must be everywhere-nonzero - however, this would be a nowhere vanishing vector field on \(S^2\), which by the Hairy Ball theorem does not exist.

Alternate proof: such a vector field would allow a homotopy between the identity and the antipodal map on \(S^2\), contradiction by basic homotopy theory.

See next: ../2021-04-25_classifying_spaces_ug

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