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I am having trouble to understand what topology is given to the tangent bundle of a smooth manifold that allows it to be a smooth manifold itself. In my understanding, among other things the topology must be second countable and Hausdorff. The definition of the tangent bundle $TM$ of a smooth manifold $M$ I am using is

$TM = \bigsqcup_{p\in M} T_pM$,

that is the disjoint union of all $T_pM$ where $T_pM$ is tangent space at $p$ consisting of all derivations at $p$. Since there is no further specification on what topology this space is given I assume we take the natural disjoint union topology.

However, in that case it seems that $TM$ is not second countable because then every set $(O,p)$ where $O$ is an open subset of $T_pM$ would be open and disjoint from any $(O,q)$ for $q \neq p$. So unless $M$ is countable there would be an uncountable number of disjoint open sets which contradicts second countability.

The only alternative I can think of is using the natural smooth structure of $TM$ as the topology. That is for every open subset $O$ of $M$ the open sets of $TM$ are defined as $\pi^{-1}(O)$ where $\pi$ is the natural projection $TM \rightarrow M$. But then $TM$ can not be Hausdorff, since any two elements of the same fiber of $\pi$ could not be seperated by open sets.

In conclusion, in both cases $TM$ could not be a manifold, so I must be missing something very obvious. Thus, I would really appreciate it if someone could point out my misconception.

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I'll try to find time to write an answer, but the topology definitely not the disjoint union topology. You want the different tangent spaces to vary in a nice way. –  Dylan Moreland May 10 '12 at 22:55
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For a definition of the topology of $TM$ and $T^*M$, check Warner, Foundations of Differentiable Manifolds and Lie Groups, in I think the second chapter. –  Neal May 10 '12 at 23:37
    
Thanks, he outlines the steps to construct the topology explicitly in 1.25. –  erlking May 11 '12 at 14:01
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3 Answers 3

up vote 8 down vote accepted

Take some atlas on $M$, and let $U$ be an element of that atlas. Then $TU=\pi^{-1}(U) \cong U \times \mathbb{R}^n$ as a set, so it inherits a topology. Moreover, all these topologies (for different $U$) are compatible with each other, so together they give you a topology on the total space $TM$.

Note that this is very much like your second idea, except that we don't require open sets to contain entire fibres $\pi^{-1}(x)$ -- just open subsets of them.

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You give $TM$ a topology and a manifold structure as follows. Suppose that $\varphi\colon U\subseteq M\to V\subseteq\mathbb{R}^n$ is a local chart of $M$. Let $x_1,\ldots, x_n$ be the corresponding coordinate functions, i.e., $\varphi(p) =(x_1(p),\ldots, x_n(p))$. Then you get a bijective map $\pi^{-1}(U)\to V\times \mathbb{R}^n$ given by $$\left(p, \sum_{i=1}^n\lambda_i\left.\frac{\partial}{\partial x_i}\right|_p\right)\mapsto (\varphi(p), (\lambda_1,\ldots, \lambda_n)).$$ The topology on $\pi^{-1}(U)$ is defined by pulling back the topology on $V\times \mathbb{R}^n$. Moreover, this map $\pi^{-1}(U)\to V\times\mathbb{R}^n$ is a chart map for the manifold $TM$. You must check, of course, that if you choose different charts $\varphi$, this does not change the topology or the manifold structure.

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Thank you, froggie. Isn't the topology on $\pi^{-1}(U)$ the same as the pullback topology from $U$. That is $F\subset\Pi^{-1}(U)$ is open iff $\pi(F)$ is open in $U$ with the induced topology from $M$? –  Student May 6 at 16:27
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@Student: I'm afraid with that topology you would not get a manifold structure on $TM$. Note that the topology you define is not Hausdorff, because if $p\in U$ and $x,y\in \pi^{-1}(p)$, then every open neighborhood of $x$ contains $y$ and vice versa. –  froggie May 6 at 18:38
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Maybe a broader perspective would be useful.

Let $F$ and $B$ be topological spaces, with a given open cover $\{U_\alpha\}$ of $B$ and continuous transition functions $\theta_{\alpha\beta}:U_\alpha\cap U_\beta\to \operatorname{Aut}(F)$ on nonempty intersections of the sets in the open cover, which satisfy the conditions that, always, $\theta_{\alpha\beta} = \theta_{\beta\alpha}^{-1}$, $\theta_{\alpha\alpha} = 1$, and $\theta_{\alpha\beta}\theta_{\beta\gamma} = \theta_{\alpha\gamma}$.

These data are exactly what we need to put together a fiber bundle with fiber $F$ and base $B$. We do this the following construction. One consequence of the construction is insight into the topology of the total space of the fiber bundle.

The prospective bundle atlas will be composed of prospective bundle trivializations $\{U_\alpha\times F\}$, indexed by the open cover of $B$. Now define $$E = \bigg(\coprod_\alpha U_\alpha\times F\bigg)/\sim,$$ where $(x,f)\sim (y,g)$ if and only if $x=y\in U_\alpha\cap U_\beta$ and $\theta_{\alpha\beta}(x)f = g$. (Note that the topology of each component of the disjoint union is just the product topology.)

While it is an exercise (left to you) to check that this is in fact the total space of a fiber bundle, the idea should be clear enough: we have taken the locally trivial neighborhoods and pasted them together with knowledge of how they transform into each other. We now see that the topology of the total space is just the quotient topology.

This construction encompasses all topological fiber bundles. To construct smooth fiber bundles, replace "continuous" by "smooth" and "topological space" by "smooth manifold." Nice examples include (but are certainly not limited to):

  • the Hopf fibrations $\mathbb{S}^1\to\mathbb{S}^{2n+1}\to\mathbb{C}P^n$, $\mathbb{S}^3\to\mathbb{S}^{4n+3}\to\mathbb{H}P^n$;
  • symmetric and homogeneous spaces, such as the fibration $SO(n)\to SO(n+1)\to\mathbb{S}^n$;
  • Seifert fibered spaces and surface bundles in $3$-manifold theory; and
  • all real and complex vector bundles, such as $TM$, $T^*M$, the exterior bundles $\Lambda^k(M)$, and the tensor bundles $\mathcal{T}^r_s(M)$, for a smooth manifold $M$.

In the particular case of your question, we have $F = \mathbb{R}^n$, transition functions are maps $U_\alpha\cap U_\beta \to Gl(n;\mathbb{R})$, and the topology is given locally by the product topology on $U\alpha\times\mathbb{R}^n$. The "compatibility" mentioned in the Micah's answer, and the check froggie suggests, are just verification that the topology on $TM$ is the quotient topology in my above definition.

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