2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So is it that because the isomorphisms from each fibre to F are not the same, we can't glue these pieces together? What does it mean to glue a section at such a point? $\endgroup$ – Varun May 26 at 13:39
  • 1
    $\begingroup$ That is the idea. To define a continous function, we only need to prescribe its behavior on an open covering, if we make sure that the definitions coincide with each other on the intersections(that is the glueing). Now, if we have a fiber bundle, we are given an open cover of $B$, say $B = \cup_i U_i$, and isomorphisms $\phi_i : \pi^{-1}(U_i) \rightarrow U_i\times F$. So when you are checking the compatibility condition on $U_i\cap U_j$, you have to remember that you used the $\phi_i$ to define your section. $\endgroup$ – Maxime Cazaux May 26 at 13:49
  • $\begingroup$ Makes perfect sense, thank you. $\endgroup$ – Varun May 26 at 13:56
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But isn’t the fibre over any point in B just F? If F is a topological space in its own right, what is the difference between (0, X) and say (b, X) for any other b in B? $\endgroup$ – Varun May 26 at 13:21
  • 2
    $\begingroup$ @Varun the fiber over any point in $B$ is isomorphic to $F$ but not canonically $\endgroup$ – hunter May 26 at 13:21
  • $\begingroup$ That helps a tonne. Just to make sure I understand the point, the fibre over any point in B can be identified with F but every fibre cannot be identified with F in the ‘same way’ so they are in fact distinct? $\endgroup$ – Varun May 26 at 13:24
  • 1
    $\begingroup$ Imagine the mobius strip. You draw a circle on it, and say it is the altitude $0$. Can you draw a line on the mobius strip that is always at altitude $1$? If so, you can continue drawing, and after a few moments, you will find that you are drawing at altitude $-1$! This is because, even if locally, above every point of altitude $0$, you can say "this point is at altitude $1$" that this notion is globally defined. In this example, $B = \mathbb{S}^1$ the circle and the fiber is isomorphic to $[-1,1]$. But there is no continuous function $B \to \mathrm{Mobius}$ that takes value $1$ in fibers $\endgroup$ – DIdier_ May 26 at 13:25
  • 1
    $\begingroup$ Well I think the OP is asking the difference between a sectuon and a function from B to the fiber F.... $\endgroup$ – Arctic Char May 26 at 13:36
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why are there no sections? $\endgroup$ – Varun May 26 at 13:31
  • 1
    $\begingroup$ @Varun the question is why there is no continuous way to choose a square root of every complex number. Intuitively, put your right hand at $1$ and with your left hand, make a choice of a square root of 1, e.g. choose 1 or -1, and imagine your right hand going around a circle counterclockwise till you get back to 1. Your "choice of square root" in the left hand is forced to move at half speed and when the right hand gets back to 1 there's no way to have made the same choice. Formally, you could use the fundamental group or the winding number to prove this. $\endgroup$ – hunter May 26 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.