Difference between a function and a section of a fibre bundle

Question

Suppose $E \rightarrow B$ with projection map $\pi$ and fibre F is a fibre bundle, with a section $\sigma$. How is $\sigma$ different from a function $f:B \rightarrow F$? The standard answer I find everywhere is that $\sigma$ is only a function if E is the trivial bundle $B \times F$, or that in general no such function exists with 'global structure', but I do not see why the former is true, and I do not understand what the latter means. A section seems like it is associating an element of the fibre F to every point in B. How is this not a function, even if E has some sort of non trivial topology? Why should 'twists' or other aberrations from a product in the space E matter, since the fibre over any point in B is the same?

Answer 1

Locally, the bundle looks like a direct product, so a section can indeed be locally identified with a map to the standard fibre $F$ via a non-canonic isomorphism. The point is that when you are trying to contruct a global section, you are using different trivial neighbouhoods covering $B$, and different trivializing charts. You have to take into account how your trivialisations glue together.

As an example, take the möbius strip as a line bundle over the circle. If you remove one point of the circle denote it $\infty$), you get a trivial bundle of the form $]0:1[\times \mathbb{R}$. But if you are trying to find a global section, you must take care that both sides of the section "glue at $\infty$". And you get that any global section has to vanish at least at one point.

Actually, it is a characterisation for line bundles, that such a bundle is trivial if and only if there is a nowhere vanishing section.

With this example, we also get a hint that, for non trivial bundles to exist, the base $B$ needs to "have some topology" (fiber bundles over contractible spaces are trivial)

Answer 2

A function $f : B \to E$ can be for example a constant function $f(p) = (0,X)$ where $X$ is an element of the fiber over a point $0 \in B$.

On the contrary, a section $\sigma: B \to E$ is forced to preserve the base point: the value of a point $p$ has to be in the fiber over $p$, which means that $\sigma(p) = (p,X_p)$ where $X_p \in \pi^{-1}(p)$ with $\pi: E \to B$ the projection.

In addition, for trivial fiber bundle $E = B \times F$, any function $f:B \to F$ gives birth to a section $\sigma_f(p) = (p,f(p))$, but in general, there is no reason for a bundle to have any global section. But by the very definition of a fiber bundle, you always have local section, that is, there exist open subsets $U \subset B$ on which you can find sections.

Edit As stated in the comments, I misunderstood your question. If you have a function $f:B\to F$, you can try to create a section $\sigma_f : B \to E$ by the locally trivialization property. But in fct, you can have topological obstruction to it. For example, say the mobius strip is a fiber bundle $\pi: M \to \mathbb{S}^1$ whose fiber is $F = [-1,1]$. Every point $p$ in $\mathbb{S}^1$ has an open neighboorhood $U$ such that $\pi^{-1}(U) \simeq U\times [-1,1]$. Implicitly, there exists such an isomorphism $i_U$. Take $f : \mathbb{S}^1 \to [-1,1]$. You can try to create $\sigma_f : \mathbb{S}^1 \to M$ by saying $\sigma_f(p) = {i_U}^{-1}(p,f(p))$ if $p \in U$. But there is a problem: $p$ can be in two diffents open neighboorhood $U$ and $V$, and maybe ${i_U}^{-1}(p,f(p)) \neq {i_V}^{-1}(p,f(p))$!

In case $f$ is the constant function $f(p)=1$, you cannot do this because the gluing of $U$ and $V$ will invert $1$ and $-1$ sometimes, and $\sigma_f$ would not be well defined.

As said in different comments, what is hidden behind is that there is no canonical way to identify the fibers. In the very definition of fiber bundle, close fibers are identified to each other thanks to local trivializations. But there is no global identification.

Answer 3

Consider the map $\mathbb{C}^\times \to \mathbb{C}^\times$ given by $z \mapsto z^2$.

This is a fiber bundle with fibers isomorphic to the two point discrete space. It doesn't have a (continuous) section. But there are two maps from $\mathbb{C}^\times$ to the two point discrete space.

Answer 4

On a fiber bundle $\pi: E\to B$ with fiber $F$, $\sigma: B\to E$ is similar to $f:B \to F$ in that $\sigma(b) \in \pi^{-1}(B) \simeq F$ just like $f(b) \in F$.

The difference is that $\sigma$ is mapping into $E$ which is a space consisting of many copies of $F$ stitched together in possibly non-trivial ways whereas $f$ only maps into a single copy of $F$. Now, if $\sigma$ and $f$ are arbitrary functions this is actually no issue. We can map elements of $B$ to $F$ and also map those elements to the corresponding elements in the corresponding fibers within $E$. The issue comes up when we require $f$ and $\sigma$ to be continuous. Now these maps must, in some sense, be sensitive to the global topology of $F$ and $E$ respectively.

If $E$ is a trivial fiber bundle such that $E \simeq B\times F$ globally then mapping into $E$ globally and continuously is pretty much the same as mapping into $F$ continuously. But if $E$ has a more sophisticated topology then now mapping into $E$ globally and continuously is different than mapping into $F continuously.

Answer 5

Long story short and summarizing the answers, since I think the OP had my same doubt, a function $f : B \rightarrow F$ takes any base point $m$ and maps it to an element of the same, fixed, topological space $F$. The intuition behind a section $\sigma : B \rightarrow E, \pi(\sigma(m))=m$, is that it maps any point $m$ to a different space, namely the fiber $\pi^{-1}(m)$, which are different copies of F, one for each base point $m$.

Only in particular cases ( e.g. when the bundle is trivial ) the two operations are equivalent in a well defined way.