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!
Clarification on Fiber bundles
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fiber-bundles
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2You 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 $M$. – 2011-12-16
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0What's $\dot x$ here? – 2011-12-16
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0Thanks 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? – 2011-12-16
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0@Dylan, $\dot x$ is the derivative of x. – 2011-12-16
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1@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. – 2011-12-16
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0@Dylan, yes x∈M, and $\dot x$ is the time derivative of x. – 2011-12-16
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0Where does this time parameter come from? – 2011-12-16
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0@Dylan, $\dot x=g(x)h(u)$ is the equation of a control-linear system (a dynamical system). Apparently x=x(t), and u=u(t). – 2011-12-16
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2Please 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 $f$. – 2011-12-16
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0In the question statement, you assumed that $f(x,u) = g(x)h(u)$. We know that $f(x,u)\in TM$. But you didn't specify where $g(x)$ and $h(u)$ lives. Seeing that $TM$ is a vector bundle, possibly you are letting one of $g$ and $h$ take real values and the other take value in $TM$? – 2011-12-16
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0@Willie, please see the system in matrix notation: $\dot x=[\dot x_{i}]^T, i=1...n, h(u)=[h_{j}(u)]^T, j=1...m, g(x)=n\times m matrix. So, \dot x=g(x)h(u)$ – 2011-12-16
1 Answers
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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.