Fiber Bundles

A piece of paper is a collection of vertical strips. A piece of paper glued together at its ends remains as such, as does a piece of paper twisted and then glued together at its ends into a Mobius strip also remains as such.

Fiber bundles: an innocent re-representation of topological structures that has its own distinct generalizations.

Typically bundle is over a vector space. A real vector bundle $\xi$ over $B$ consists of

i) a topological space $E = E(\xi)$ called the total space

ii) a continuous map $\pi: E \to B$ called the projection map

iii) for each $b \in B$, the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$.

iv) local triviality condition: for each $b$ of $B$ a neighborhood $U \subset B$, an integer $n \geq 0$, and a homeomorphism $h : U \times \mathbb{R}^n \to \pi^{-1}(U)$ such that, for each $b \in U$, the map $x \mapsto h(b, x)$ defines an isomorphism between the vector space $\mathbb{R}^n$ and the vector space $\pi^{-1}(b)$.

A trivial bundle is one homeomorphic to $B \times \mathbb{R}^n$: the piece of paper remains in its original form.

As with the corresponding trivial bundle, a fiber bundle has no natural association between fibers. A ‘twisted’ cylinder remains topologically a cylinder: a twisting of coordinates $(x, \phi) \mapsto (x, (\phi + \epsilon x)\mod 2 \pi)$ gives an equally valid homeomorphism. The logic applies to every bundle region $\pi^{-1}(U)$.

The vector bundle is a fiber bundle with a vector space structure for the fibers. Other structures are possible; replace the vector space with the desired structure and required isomorphism in iv).

A principal bundle is a fiber bundle with groups as fibers, groups acting freely and transitively on fibers, the base space then the quotient of the total space by the action of the group.

A section $\sigma: B \to E$ is a smooth right inverse of $\pi$

visualized as a plot in the Cartesian plane with $E$ the plane, $B$ the x-axis, and each fiber a vertical strip of the plane.

It can be demonstrated that a vector bundle is trivial if and only if there exist a collection of sections that form a basis for the corresponding fiber at every point.

Tangent bundle: in a sense the ‘original’ fiber bundle, motivated by the problem of defining smooth vector fields on manifolds. The problem: no relation between tangent spaces at distinct points on a manifold, so how to define smooth vector fields? The classical definition for smooth vector fields is in terms of the local coordinate chart about any given point. This is well-defined, but unsatisfactory from the point of view of being defined in terms of a chart (into $\mathbb{R}^N$) rather than as a map into each (distinct) tangent space. The tangent bundle solution: let $TM = \bigcup_i T_{p_i}M$, the union of tangent spaces, each considered as a set. Define a vector field as a map $v: M \to TM$, with each point mapping into its corresponding tangent space. $TM$ can be turned into a manifold by defining the chart mapping a point and vector in the corresponding tangent space to its coordinate and vector coordinate representation: $(p, v) \mapsto (x_i, v_i)$. Then $v$ is a smooth vector field on $M$ if it is simply a smooth function on $TM$.

Examples

• Can demonstrate tangent space $T\mathbb{S_1}$ of the 1-sphere $\mathbb{S_1}$ is trivial, $T\mathbb{S_2}$ is not.

• The cylinder $\mathbb{S_1} \times \mathbb{R}$ (with $\mathbb{S_1}$ considered the base space and $\mathbb{R}$ a fiber) is a trivial bundle, while the Mobius strip given as the quotient space $(\mathbb{S_1} \times \mathbb{R})/ \sim$ with the equivalence relation $(p, x) \sim (p + 2\pi, -x)$ is a non-trivial bundle. It can be demonstrated non-trivial by looking for sections that form a basis. Sections are of form $\sigma(x) = (x, f(x))$ for $f: \mathbb{S_1} \to \mathbb{R}$. For the section to be a basis f must be non-zero everywhere, and for the section to be smooth (continuous) $f$ must invert at the boundaries, $f(2\pi) = -f(0)$. But the intermediate value theorem then dictates that f is zero somewhere, so no such section exists and the Mobius band is non-trivial.