I'm a little confused about fiber bundles. I have a specific example and would appreciate someone clarifying this for me. Let $f:M\times U\rightarrow TM$ be a map from a product space $M\times U\rightarrow$ to the tangent bundle TM of M. Is ($M\times U$,f,TM) a fiber bundle with $M\times U\equiv E$ the total space and TM the base space? If so, which is the fiber space? Or should the base be just M? For example, let $f(x,u)=\dot x=g(x)h(u), x \in M, u \in U$, for some g,h. Is f the projection map of the a fiber bundle? Should it be TM or $T_{x} M$, the tangent space of M at x? Thanks in advance!
$\begingroup$
$\endgroup$
11
-
2$\begingroup$ You seem more than a little confused. $M \times U$ is a trivial fibre bundle over $M$, with fibre $U$. The projection map is the map $M \times U \to M$ such that $(m, u) \mapsto m$. $TM$ is another fibre bundle over $M$, with projection map $TM \to M$ such that $v \mapsto x$ for all $v \in T_x M$. $TM$ is generally not a trivial bundle over $M$. $\endgroup$– Zhen LinDec 16, 2011 at 3:27
-
$\begingroup$ What's $\dot x$ here? $\endgroup$– Dylan MorelandDec 16, 2011 at 3:30
-
$\begingroup$ Thanks Zhen but i've already understood what you're saying. My main problem is the map f:M×U→TM. Can this be a bundle? $\endgroup$– JorgeDec 16, 2011 at 3:34
-
1$\begingroup$ @Jorge But $x \in M$, isn't it? I don't really see what $f, g, h$ are supposed to be. It would be good to write down what it means to be a fibre bundle over $TM$, and see whether you've satisfied the requirements. $\endgroup$– Dylan MorelandDec 16, 2011 at 3:36
-
2$\begingroup$ Please provide more information. From the definition you need at the very least $f:M\times U\to TM$ to be surjective. For a general map this is obviously false (for example, let $f = \iota\circ\mathrm{pr}_1$ where $\mathrm{pr}_1$ is the projection from $M\times U$ to $M$ and $\iota$ is the inclusion of $M$ into $TM$ as the 0 section). So please be more precise and edit into the question what you know about the map $f$. $\endgroup$– Willie WongDec 16, 2011 at 15:48
|
Show 6 more comments
1 Answer
$\begingroup$
$\endgroup$
In the definition of a fiber bundle, the map from the total space $E$ to the base space $M$ is the canonical projection, meaning locally $f:M\times F\to M$ takes $f(x,p)=x$ for fiber $F$. In your example, I guess $f:M\times U\to TM$ could be the projection of a fiber bundle if $TM$ $was$ a trivial bundle, say $M=\mathbb{R}^4$. But also note that most people would write $TM$ as the the entire bundle, meaning $TM=M\times \mathbb{R}^n$ locally, not just $TM=\mathbb{R}^n$.
EDIT: Note that I keep saying "locally". For any fiber bundle $f:E\to M$ with fiber $F$, you can write $f_i:U_i\times F\to M$ for a local covering $\{U_i\}$ of $M$. I won't write down all the details of that, but you should be familiar with it as well.
