# How to make the the tangent space TM into a fiber bundle

#### MathNugget

##### Junior Member
Can TM, the tangent space to a smooth manifold M, which is a smooth manifold itself, be seen as a fiber bundle?

The theory around this is:
given E, M smooth manifolds, I need to find [imath]\pi: E \rightarrow M[/imath] differentiable, surjective, and open sets [imath]\{U_i\}_i[/imath] covering M so that [imath]\pi^{-1}(U_i)\simeq U_i \times R^k[/imath].

I noticed wikipedia uses an arbitrary fiber F, instead of [imath]R^k[/imath], but I'll try with [imath]R^k[/imath].

Here's my thoughts so far:
let's choose [imath]\pi : TM \rightarrow M[/imath], [imath]\pi(v_p)=p[/imath], where [imath]v_p\in TM[/imath] is a tangent vector, and p is the point of M it is tangent to.
The inverse image of an open set in M would be a reunion of tangent spaces in points of M, which is open. But how would I prove [imath]\cup_{p \in I} T_pM \simeq I \times R^k[/imath]?

My vague recollection is that tangent bundle is one of the, if not the, most popular examples of fiber bundles. Homeomorphism between [imath]U_p\in M[/imath] and [imath]V \in \mathbb R^k[/imath] should lead to pretty a straightforward diffeomorphism between [imath]TU_p[/imath] and [imath]TV[/imath], with the latter pretty clearly diffeomorphic to [imath]V\times \mathbb R^k[/imath].

Thank you for your answer.

I have trouble seeing how we'd go [imath]TU_p \rightarrow U_p \rightarrow V \rightarrow TV[/imath]. Most, if not all information is lost when we go through [imath]TU_p \rightarrow U_p[/imath] ...

I suppose every vector in [imath]TU_p[/imath] is tangent to at least one curve= function in [imath]U_p[/imath], which can be parametrised as [imath]\gamma: V_\gamma \rightarrow U_p, V_\gamma \subset V[/imath] . Given there's a homeomorphism between U and V, the function [imath]V_\gamma \rightarrow U_p[/imath] is the inverse of the homeomorphism...

So [imath]\forall v \in TU_p \: \exists \gamma[/imath] such that [imath]d\gamma=v[/imath]. Then...I don't really know?

Do we just send [imath](x, v) \rightarrow (\phi(x), \gamma^{-1})[/imath]? where v is a vector in [imath]TU_p[/imath] tangent to U in x, [imath]\phi[/imath] is the homeomorphism, and [imath]\gamma[/imath] is the curve in [imath]U_p[/imath] to which v is tangent to in x?

I don't understand you last post. But I believe any differentiable map [imath]f:M\rightarrow N[/imath] induces differentiable map [imath]\bar f : TM \rightarrow TN[/imath]. Now replace [imath]M,N[/imath] with [imath]U,V[/imath].

I don't understand you last post. But I believe any differentiable map [imath]f:M\rightarrow N[/imath] induces differentiable map [imath]\bar f : TM \rightarrow TN[/imath]. Now replace [imath]M,N[/imath] with [imath]U,V[/imath].
I am not exactly certain how [imath]\bar{f}[/imath] works; I know it would send [imath]T_pM \rightarrow T_{f(p)}N[/imath], but is it possible to find / do we need to find what [imath]\bar{f}[/imath] does to a single vector? This question basically sums up my previous post ...

Can TM, the tangent space to a smooth manifold M, which is a smooth manifold itself, be seen as a fiber bundle?
Yes.

Maybe this article is of help:

The theory around this is:
given E, M smooth manifolds, I need to find [imath]\pi: E \rightarrow M[/imath] differentiable, surjective, and open sets [imath]\{U_i\}_i[/imath] covering M so that [imath]\pi^{-1}(U_i)\simeq U_i \times R^k[/imath].

I noticed wikipedia uses an arbitrary fiber F, instead of [imath]R^k[/imath], but I'll try with [imath]R^k[/imath].

Here's my thoughts so far:
let's choose [imath]\pi : TM \rightarrow M[/imath], [imath]\pi(v_p)=p[/imath], where [imath]v_p\in TM[/imath] is a tangent vector, and p is the point of M it is tangent to.
The inverse image of an open set in M would be a reunion of tangent spaces in points of M, which is open. But how would I prove [imath]\cup_{p \in I} T_pM \simeq I \times R^k[/imath]?

I'm not sure I understand your difficulties. What is [imath] I [/imath]?

We have [imath] T_pM\cong \mathbb{R}^k [/imath] for every [imath] p\in I [/imath] hence [imath]\cup_{p \in I} T_pM \simeq I \times R^k[/imath] holds on the set level. The projection is the map onto [imath] p. [/imath]

...but is it possible to find / do we need to find what $$\bar{f}$$ does to a single vector?
How do you define vectors in [imath]T_p M[/imath]?

How do you define vectors in [imath]T_p M[/imath]?
I think for every vector v tangent in a point p to M, there is a curve such that p is on the curve and v is tangent to the curve in the point p. Doesn't it work this way?

Yes.

Maybe this article is of help:

I'm not sure I understand your difficulties. What is [imath] I [/imath]?

We have [imath] T_pM\cong \mathbb{R}^k [/imath] for every [imath] p\in I [/imath] hence [imath]\cup_{p \in I} T_pM \simeq I \times R^k[/imath] holds on the set level. The projection is the map onto [imath] p. [/imath]
[imath]I[/imath] was an arbitrary open set in M...

I'll check out the article and will return. Thank you!

I think for every vector v tangent in a point p to M, there is a curve such that p is on the curve and v is tangent to the curve in the point p. Doesn't it work this way?
It does work. And every diffeomorphism of manifolds maps curves, and thus tangent vectors. Does this help?

It does work. And every diffeomorphism of manifolds maps curves, and thus tangent vectors. Does this help?
Well then, now I can see how we'd send vectors of TM to vectors of TN, and in this case N is an [imath]\mathbb{R}^k[/imath], so [imath]TN \simeq R^k \times R^k[/imath]. I think I got it. Thank you very much.

And if the homeomorphism takes more than 1 map, we go from [imath]TU_p \rightarrow TV[/imath], where [imath](U_p, \phi_p)[/imath] is the map, and [imath]\phi(U_p)=V_p\subseteq R^k[/imath].

If I may add just this 1 last question, to make sure I understood: when we go from TM to [imath]TR^k[/imath] (or from [imath]TU_p[/imath] to [imath]TV_p[/imath]), we don't look at the [imath]\pi[/imath] function, but actually find the functions for the local trivialization, right?

If I may add just this 1 last question, to make sure I understood: when we go from TM to [imath]TR^k[/imath] (or from [imath]TU_p[/imath] to [imath]TV_p[/imath]), we don't look at the [imath]\pi[/imath] function, but actually find the functions for the local trivialization, right?
What do you mean by going from [imath] TM=\sqcup_p T_pM [/imath] to [imath] T\mathbb{R}^k [/imath]? We can go from [imath] TM [/imath] to [imath] \{p\} [/imath] or even to [imath] T_pM [/imath] but what is [imath] T\mathbb{R}^k [/imath]? [imath] T_pM \cong\mathbb{R}^k [/imath] so what does [imath] T\mathbb{R}^k \cong T(T_pM)[/imath] mean? Same with the next question: Are [imath] U,V [/imath] two different neighborhoods of [imath] \{p\} [/imath]? If so then their tangent spaces at [imath] p [/imath] should be the same.

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If I may add just this 1 last question, to make sure I understood: when we go from TM to [imath]TR^k[/imath] (or from [imath]TU_p[/imath] to [imath]TV_p[/imath]), we don't look at the [imath]\pi[/imath] function, but actually find the functions for the local trivialization, right?
Sorry, but I don't understand this question.

What do you mean by going from [imath] TM=\sqcup_p T_pM [/imath] to [imath] T\mathbb{R}^k [/imath]? We can go from [imath] TM [/imath] to [imath] \{p\} [/imath] or even to [imath] T_pM [/imath] but what is [imath] T\mathbb{R}^k [/imath]? [imath] T_pM \cong\mathbb{R}^k [/imath] so what does [imath] T\mathbb{R}^k \cong T(T_pM)[/imath] mean? Same with the next question: Are [imath] U,V [/imath] two different neighborhoods of [imath] \{p\} [/imath]? If so then their tangent spaces at [imath] p [/imath] should be the same.
I am sorry, I'll try to write it more clearly. I skipped too many steps, and my writing became rather ambiguous (and I was probably mistaken, so I'll try a different idea).

Say, M is the smooth manifold, and suppose for simplicity there's a single global map [imath](M, \phi)[/imath], [imath]M \rightarrow^\phi V \subseteq R^k[/imath].

We identify [imath]TM[/imath] with [imath]M \times R^k[/imath] (I think now this is actually sort of a definition of TM, given that it exists because we bring the basis [imath]x_1,...x_k[/imath] though [imath]\phi^{-1}[/imath], so each [imath]T_pM[/imath] has to be like [imath]R^k[/imath]) and then we consider the projection on first argument:
[imath]TM \rightarrow^\pi M[/imath]. Then [imath]\forall x \in M[/imath], [imath]\pi^{-1}(x)=T_xM[/imath].

1) The tangent spaces [imath]T_xM[/imath] in each point are supposed to be the fibers?

Although now that I look again at the formal definitions, I still think what I am looking for is [imath]\pi: TM \rightarrow TV[/imath], going through this path:
[imath]TM \rightarrow M \rightarrow V \rightarrow TV[/imath]

1) The tangent spaces TxMT_xMTxM in each point are supposed to be the fibers?
It does seem to feet the definition in Wikipedia.

So now if we look for the trivialisation maps, we have to find "?" and [imath]\theta[/imath] such that
1) [imath]TM= \pi^{-1}(M)\rightarrow^\theta M \times ? \rightarrow^{proj_1} M[/imath]
2) [imath]\pi(v_p)=proj_1(\theta(v_p))[/imath], where [imath]v_p \in TM[/imath]

What's the strategy here? "?" I suppose is always [imath]R^k[/imath], with k being the dimension of M... would I identify a vector [imath]v_p \in TM[/imath] with the pair [imath](p, \overline{v_p})[/imath], where[imath]\overline{v_p}[/imath] is just [imath]v_p[/imath] seen in [imath]R^k[/imath]? (We could just consider the same components but seen in the basis of [imath]R^k[/imath])...

I am sorry everyone. I am really slow when it comes to differential geometry...

Now I am confused even more. Do you think it is not enough to show that [imath]TU_p \equiv TV[/imath]?
And a more general question: do you have a textbook on differential geometry? Does not it discuss tangent bundles?

Now I am confused even more. Do you think it is not enough to show that [imath]TU_p \equiv TV[/imath]?
I think it is enough... I understand roughly, the layman's explanation of it available on the internet: we want to locally see the manifold as a product space. I can grasp that the cylinder is the product of a line and a circle, while a mobius strip is just locally the product of a line and an arc of a circle, because of the twist...
The mathematical part though, about finding [imath]\pi, B\times F[/imath], the local trivialization... it's not so obvious for me .

And a more general question: do you have a textbook on differential geometry? Does not it discuss tangent bundles?
Yes, I have textbook in my own language. It gives [imath]TM[/imath], [imath]T^*M[/imath] (cotangent space) as examples of trivial fiber bundles, without any explanation. There's also an example of finding the (line fibers?) of the circle [imath]S^1[/imath], and it's explained using the complex plane, and seeing the points of the circle as complex numbers...couldn't quite follow-up any of that, but it ends up having 2 very nice trivialisation maps. Weird enough, the short ppt presentations I could find online, and wikipedia articles, mostly go through similar explanation paths, so I can only assume these things are supposed to be obvious .

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To deal with fiber bundles is to a large extent linguistical. It is the one before the last stop on the path of generalization from a directional derivative to differential geometry. The last stop is connections. The treatment depends a bit on the authors so consulting different sources before you understand your textbook can be confusing.

I would use the nomenklatura of your textbook only and do everything on an example in parallel. Since we eventually want to draw something, it is better to use real examples. Flat Euclidean spaces will add confusion since they as a manifold will be equal to themselves as a tangent space. So it's better to use something curved. My preferred examples would be the unit sphere [imath] \mathbb{S}^2\subseteq \mathbb{R}^3 [/imath] with local polar coordinates, and the two-dimensional non-abelian Lie group generated by $G= \bigl\langle \begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}\, , \,\begin{pmatrix}1&c\\0&1\end{pmatrix} \bigr\rangle .$These are nice and smooth manifolds, and a possible function, if necessary, could be a three-dimensional rotation of the sphere, or an application of elements from [imath] G [/imath] on [imath] (x,y)\in \mathbb{S}^2. [/imath] I would avoid the stereographic projection because it would involve projective spaces that would make the situation unnecessarily more complicated. For a function on [imath] G, [/imath] again if necessary, one can take the determinant or the group multiplication with a fixed element [imath] g\in G\, , \, h\mapsto g\cdot h \,. [/imath]

Now you can deal in parallel with the theoretical questions like the triviality of the tangent bundles, and an specific example. The group [imath] G [/imath] is easy and low-dimensional enough to produce nice curves on it, and complicated enough to serve as a meaningful example.

I think the reason your author didn't deal with the fact that the tangent bundle is trivial is the following:

a) A vector bundle, here the tangent bundle, is trivial if it has a global trivialization.

b) A trivialization is if there is a homeomorphism [imath] \varphi : U\times \mathbb{R}^n\longmapsto \pi^{-1}(U) \subseteq E \text{ total space }[/imath] for every [imath] p\in U\subseteq B \text{ base space } [/imath] that behaves nicely with the projections: [imath] \pi \circ \varphi =\operatorname{proj}_1 [/imath] and such that [imath] \{p\}\times \mathbb{R}^n \cong_\varphi \pi^{-1}(p). [/imath] (Up to the usual quantifiers: for every [imath] p\in M [/imath] there is a open neighborhood [imath] p\in U\subseteq B [/imath] such that ...)

In the case of a tangent bundle, we have
$\displaystyle{E=TM= \bigsqcup_{p\in M} \{p\}\times T_pM \, , \,B=M\, , \,\pi:E\longrightarrow B\, , \,\pi(p,T_pM)=p}$Let's check the trivialization now. Global means we require [imath] U=B=M. [/imath] For the trivialization, we set
$\varphi : M \times \mathbb{R}^n \longmapsto \pi^{-1}(M).$ This is clearly a homeomorphism since [imath] \varphi (p,\mathbb{R}^n) =\pi^{-1}(p)=(p,T_pM)[/imath] and [imath] T_pM\cong \mathbb{R}^n [/imath] is a simple identification. Remains to show whether the projections behave nicely. This is a formality since we only operate with simple identifications and natural projections.
$(\pi \circ \varphi )(p,\mathrm{v})=\pi( \varphi (p,\mathrm{v}) )= \pi ( \pi^{-1} (p) )=p=\operatorname{proj}_1 (p,\mathrm{v}) \;\text{ for all }\mathrm{v}\in \mathbb{R}^n$and of course
$\{p\}\times \mathbb{R}^n \cong \{p\}\times T_pM = \pi^{-1}(p)\;.$

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