# Does the section and the bottom manifold equivalent in the bundle?

Here the equivalent means:in the bundle ,B is bottom manifold(also called base space),S is section,we can also treat S as a bottom and B as a section in the same bundle. Is it true in the vector bundle? and also in the fiber bundle?

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Maybe you could explain a bit more what the context is, and what you want? –  Mariano Suárez-Alvarez Mar 6 '11 at 5:30
I think vector bundle and fiber bundle have stand definition.Oh,a mistake of spelling,let me change it. –  Strongart Mar 6 '11 at 10:37
they do have standard definitions, I simply do not understand what you are asking... While you are correcting spelling, it's manifold, not mainfold, and I imagine that when you say bottom manifold you really mean base manifold. –  Mariano Suárez-Alvarez Mar 6 '11 at 17:41
Yes,my English is poor and the question is not copied from some book,so maybe there are some small mistake in it.But I think the question is important,if that is true, it can be uncentre with the bottom manifold(also called base space).Here is the definition of fibre bundle:en.wikipedia.org/wiki/Fiber_bundle –  Strongart Mar 7 '11 at 5:03
the typos are mostly irrelevant: my problem with what you wrote, as I said above, is that I simply cannot understand what you are asking. Hopefully someone does... But there is one way you can make it easier for people to help you: explain more what you want to know. –  Mariano Suárez-Alvarez Mar 7 '11 at 5:17

First, a minor point: As Mariano notes in a comment, in English one says base manifold, not bottom manifold. In any case, I will let $E$ denote the total space of the fibre bundle and $B$ denote the base.

Let $\pi: E \to B$ be a fibre bundle, and let $\sigma:B \to E$ be a section, with image $S = \sigma(B) \subset E$.

Your question is, as far as I understand, the following:

can one find a projection $\pi': E \to S$ which makes $E$ a bundle over $S$, so that furthermore $B$ is now realized as a section.

In some sense the answer is "Yes", but for trivial reasons:

One can always find such a projection $\pi'$, because $\pi$ induces a diffeomorphism $S \to B$, so that abstractly $B$ and $S$ are the same space. More concretely, define $\pi': E \to S$ via $\pi' = \sigma\pi.$ This realizes $E$ as a fibre bundle over $S$. However, nothing very exciting has happened; we have just replaced the base $B$ by the diffeomorphic manifold $S$.

We may now define the section $\sigma': S \to E$ via $\sigma'(s) = s.$ The image of $\sigma'$ is just $S$ again, thought of as a submanifold of $E$, but if you want we can identify this with $B$ via the projection $\pi$. Again, this is not very exciting.

My sense is that you are thinking of something a little deeper than this, but I'm not sure what. One problem in general is that the base $B$ and a section $S$ have a different nature: $S$ maps into $E$, while $E$ maps onto $B$ (so $B$ is not really "in" the fibre bundle at all).

Perhaps the vector bundle case is the one you want to focus on: then $B$ can be made into a section via the zero section. But I still don't see what you are going to do with this. (One thing you can do is that you can intersect $B$ and $S$, and this is how one produces characteristic classes geometrically; is that of any interest to you at all?)

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I have not learned characteristic classes,then I know that is true. So we kill the central position of the bottom manifold because we can use any global section to take place of it.But some fibre bundles can have no global section,what a pity! –  Strongart Mar 7 '11 at 10:37

I think what you meant is some duality between $\Gamma(E)$ and $B$. I have never seen bottom manifold, etc defined in literature. I think there is no such relationship. Maybe you can ask some Chinese for help in English so others can understand you.

The best way to treat sections is to think them as an equivalence between the category of vector bundles over $X$ and the categrory of finitely generated projective modules over $\mathbb{C}(X)$ (complex value functions on $X$).

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