0
$\begingroup$

Let $B$ denote a fixed topological space, which will be called the base space. A real vector bundle $\xi$ over $B$ consists of a total space $E=E(\xi)$, a projective map $\pi: E\to B$ and for each $b\in B$ the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$. It also satisfy the condition of local triviality.

The vector space $\pi^{-1}(b)$ is called the fiber over $b$. It may be denoted by $F_b$ or $F_b (\xi)$. The dimension $n$ of $F_b$ is allowed to be a locally constant function of $b$. However, my question is why it is also a constant function on the connected component of the topological space $B$.

ps: In most case of interest this function is constant. One then speaks of $n$-plane bundle. Any book about vector bundle are dealing with the $n$-plane bundle, particularly, the material above are from John milnor. I was wondering is there any vector bundle is not an $n$-plane bundle? vector bundle On Wikipedia, someone said it is constant on connected component. I learn from this and ask myself what could happenes if we put together two vector bundles of different (constant) dimensional of connected base space respectively? It do make sense but it is useless. Although I solved my original problem, i.e, to give a non $n$-plane bundle, I have a new one, which I ask here.

Not to this specific problem, my friends have no interest on something old stuff such as point set topology since we have to learn something new, say, characteristic classes. I have no idea about learn math but just want to solve this uselessness problem.

edit: Connected topological space is a topological space that cannot be divided into two disjoint open set which union is the original space.

$\endgroup$
4
  • 2
    $\begingroup$ It is a hindrance when the body of the Question is not self-contained and (as here) relies on the title alone to bear the burden of stating the problem to be solved/question to be answered. Please review How to Ask. $\endgroup$
    – hardmath
    Jul 16, 2016 at 11:15
  • $\begingroup$ @hardmath thx and I have edited it. $\endgroup$
    – Brooks
    Jul 17, 2016 at 8:43
  • 1
    $\begingroup$ A function $f:X\to Y$ between topological spaces is locally constant if and only if it is constant on the connected components of $X$: if $q\ne r\in Y$, then $f^{-1}\{q\}$ and $f^{-1}\{r\}$ are two disjoint open subsets of $X$. $\endgroup$
    – user228113
    Jul 17, 2016 at 10:02
  • $\begingroup$ I must make a correction to what I just wrote: the "only if" part (which is the case of interest) is always true. However, since the connected components of a topological space may not be open, the "if" part is not necessarily true even if $Y\subseteq \Bbb N$. My previous statement holds as I wrote it under the additional hypothesis that $X$ is locally connected. $\endgroup$
    – user228113
    Jul 17, 2016 at 10:18

1 Answer 1

0
$\begingroup$

Show that the following two definitions of "locally constant" for a function $f : X \to Y$ are equivalent (Edit: As Eric Wofsey says in the comments, under the hypothesis that $X$ is locally connected):

  1. $f$ is constant on the connected components of $X$.
  2. For every point $x \in X$, there is an open neighborhood $U \ni x$ on which $f$ is constant.

You can show that the dimension of the fiber of a vector bundle satisfies the second definition using local triviality.

$\endgroup$
2
  • 4
    $\begingroup$ These are not equivalent in general. (2) always implies (1), but the converse requires $X$ to be locally connected. $\endgroup$ Jul 19, 2016 at 5:13
  • $\begingroup$ @Eric: rats, I knew I should've actually checked this. In any case 2) is probably the correct definition in general. $\endgroup$ Jul 19, 2016 at 5:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .