# Understanding the notion of a tangent bundle

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1. Feb 5, 2015

### "Don't panic!"

I've been reading up on the definition of a tangent bundle, partially with an aim of gaining a deeper understanding of the formulation of Lagrangian mechanics, and there are a few things that I'm a little unclear about.

From what I've read the tangent bundle is defined as the disjoint union of the tangent spaces to each point on a manifold, i.e. $TM = \bigcup_{p\in M} T_{p}M$. By disjoint union is it meant that the new set $TM$ is created by constructing ordered pairs $(p,\mathbf{v})$ such that each $p\in M$ indexes a family of vectors in the tangent space $T_{p}M$ to that point. In doing so, although two tangent vectors may essentially be 'the same' (in that they have the same components), they are distinct as they are 'fixed' at different points on the manifold $M$ and, as such, belong to different tangent spaces?! Is it then correct to write $TM= \lbrace (p,\mathbf{v})\;\;\vert\quad p\in M, \mathbf{v}\in T_{p}M\rbrace$?

I guess my real struggle about the concept is why we can treat the points $p\in M$ and tangent vectors $\mathbf{v}\in T_{p}M$ as independent variables? Intuitively, I understand that for each point on an $n$-dimensional manifold there is an associated $n$-dimensional tangent space, and as such, a space constructed from the union of all such tangent spaces at each point on the manifold should then be $2n$-dimensional. I also see how, at any given point $p\in M$ there exists an infinite number of vectors to choose from in the tangent space to that point, and as such given a point $p\in M$ one can essentially independently choose any vector from the tangent space at that point. However, I don't see how this would work the other way around, as surely one needs to pick a point in the manifold before one can choose a vector, as otherwise one does not know which tangent space it belongs to?! (Or is it this why the tangent bundle is constructed as a disjoint union, such that one can pick a vector first and then distinguish it from any other vector by picking a point on the underlying manifold to 'attach' it to a particular tangent space $T_{p}M$?)

2. Feb 5, 2015

### Orodruin

Staff Emeritus
Clearly the vector in $T_pM$ belongs to the tangent space at $p$, if you know which tangent space the the vector is in, you also know the projection onto the base space. By selecting an element in $TM$ you are selecting both the point $p$ and the vector at $p$. Fredrik wrote a short summary on fiber bundles (of which tangent bundles is a special case) not so long ago, I will see if I can locate it. An element of $x \in TM$ is simply any tangent vector on $M$ as it is the union of all tangent spaces, which a member of $T_pM$ for some point. You can therefore define a projection operator, which projects it to the base space as $\pi(x) = p$.

Edit: Found Fredrik's post: https://www.physicsforums.com/threads/fiber-bundles.790178/

3. Feb 5, 2015

### "Don't panic!"

So is the reason why you can treat them as independent variables , say for example in the Lagrangian, $\mathcal{L} : TM\rightarrow\mathbb{R}$, because you can choose any vector you like from any tangent space to the manifold $M$, and then associate that vector with a given point $p\in M$ on the manifold?
Sorry to harp on about this point, it's been causing me a lot of confusion and I really want to clear up the concept in my head! I really appreciate your time.

4. Feb 5, 2015

### Orodruin

Staff Emeritus
No, once you have chosen one vector, it also belongs to a tangent space and so you have chosen both the vector and the point in the base space. If the base space is $n$ dimensional, the tangent bundle is naturally still $2n$ dimensional. The more intuitive way of seeing it is first selecting a point in the base space and then selecting the tangent vector from the tangent bundle at that point. Pairs such as $(p,v)$ with $v \notin T_p M$ are not elements of $TM$.

5. Feb 5, 2015

### "Don't panic!"

Ah ok, so would it be correct to say that once I've chosen a point $p\in M$ then I'm free to choose any vector from the tangent space $T_{p}M$ at that point, so (in essence), as there will be an infinite number of vectors in the tangent space to a given point on the manifold, they are independent choices?
(I'm just trying to understand how they are treated as independent variables in the Lagrangian really?!)

One person I spoke to gave me this explanation, but I'm a little unsure about it?!

"For a tangent bundle $TM$, the key fact is that given any particular tangent vector $(p,v)\in TM$, you can choose smooth local coordinates $(x^i)$ on some neighbourhood $U$ of $p$, and then points in $TU\subseteq TM$ correspond to elements of $\mathbb R^{2n}$ via the correspondence
$$(x^1(q),\dots,x^n(q),v^1,\dots,v^n) \leftrightarrow v^1 \left.\frac{\partial}{\partial x^1}\right|_q+\dots+ \left.v^n \frac{\partial}{\partial x^n}\right|_q.$$
As long as you stay in $TU$, you can freely choose the coordinates $(x^i,v^i)$, and you can choose them in any order."

Last edited: Feb 5, 2015
6. Feb 5, 2015

### Orodruin

Staff Emeritus
Yes, that is fine. If you think about the most boring case possible, i.e., a particle in one dimension, the variables would be the particle position and the particle momentum (or velocity). If you take a particle on the surface of a sphere, you have the particle position on the sphere and the particle velocity (which has to in the tangent space of the point on the sphere if the particle is constrained to be on the sphere). Regardless of the particle position, you will have a choice of velocity vector in the tangent space and that is an additional $n$ degrees of freedom as compared to only having the base manifold.

7. Feb 5, 2015

### lavinia

Locally the tangent bundle is a product $U x R^n$. So over the domain,$U$, the tangent vectors and coordinates in $U$ are independent variables.

Globally though there may not be a parameteriation of the tangent bundle. For most manifolds the tangent bundle is not "trivial". For example the tangent spaces to the 2 sphere have no global parameterization.

So locally yes, the tangent vectors are independent parameters but globally they can not be parameterized.

Last edited: Feb 5, 2015
8. Feb 5, 2015

### "Don't panic!"

Why is this the case? What makes them completely independent?

9. Feb 5, 2015

### lavinia

Think of ordinary Euclidean space. Tangent vector coordinates and space coordinates are independent. On a manifold one can always find a domain around any point that is diffeomorphic to an open domain in Euclidean space.

Last edited: Feb 5, 2015
10. Feb 5, 2015

### "Don't panic!"

Is this to do with the fact that one can choose the components of a vector (with respect to a basis) and those components don't depend on any particular point in Euclidean space (i.e. One can translate such a vector to any other point and its components won't change - it will still be 'the same' vector regardless of the coordinates of the point that it is tangent to)?!

I guess my issue is that I really struggle to get past the notion of a tangent vector being the rate of change of a coordinate curve at some point (I think it's somewhat blinkering me in the concept)!

11. Feb 5, 2015

### lavinia

Yes.

This is again the same idea as the velocity vector of a curve in Euclidean space. On a manifold one needs to know when two velocity vectors in different coordinates are really the same vector. This is done by the usual change of coordinates formulas.

12. Feb 5, 2015

### "Don't panic!"

So, is it correct to say that, for an open subset $U_{i} \subset M$, then we can choose a local coordinate system in the neighbourhood of some point $p\in U_{i}$ through a mapping $\varphi : U_{i} \rightarrow \mathbb{R}^{n}$ which assigns a set of $n$ coordinates to each point $p\in U_{i}$, i.e. $p\mapsto (x^{1}, \ldots , x^{n})$. Given this, we have open subset of the tangent bundle $TU_{i} \subset TM$ whose points are ordered pairs $(p, \mathbf{v})$ corresponding to elements in $\mathbb{R}^{2n}$, i.e. $(p, \mathbf{v}) \mapsto (x^{q}, \ldots , x^{n}, v^{1}, \ldots , v^{n})$. Now, as we are in $\mathbb{R}^{2n}$ we are free to choose the coordinates $x^{i}$ and $v^{i}$ in any order (i.e. they are truly independent), as the components of any vector we choose in this neighbourhood do not depend on the coordinates in this neighbourhood (if we choose the vector components first and then the space coordinates, then we a choosing a particular vector and subsequently placing it in a particular tangent space). Is this (roughly) right?
Sorry to go on a bit, just want to check whether I'm understanding this correctly or not.

13. Feb 5, 2015

### Orodruin

Staff Emeritus
That is fine. If you want to select vector components first without reference to a local coordinate chart, you can try to find a set of vector fields that span the tangent space at every point and quote the components with respect to this basis first. However, it is not always possible to find such vector fields.

14. Feb 5, 2015

### lavinia

You got it
The tangent bundle looks exactly like $U x R^n$ if $U$ is a coordinate chart.

Last edited: Feb 5, 2015
15. Feb 6, 2015

### "Don't panic!"

Thanks for all your help!

With respect to a Lagrangian then, would it be correct to say that $\mathcal{L} : TM\rightarrow\mathbb{R}$ such that locally, one can independently choose coordinates $\lbrace q^{i}\rbrace$ and tangent vector components $\lbrace\dot{q}^{i}\rbrace$, which return a value $\mathcal{L}= \mathcal{L}(q^{1},\ldots ,q^{n},\dot{q}^{1},\ldots ,\dot{q}^{n})$. Upon choosing a curve $q: M \rightarrow\mathbb{R}^{n}$ (parametrised by $t$ such that , for a given value of $t$, $q(t)=(q^{1},\ldots ,q^{n})$, with tangent vector components $\dot{q}(t)=(\dot{q}^{1},\ldots ,\dot{q}^{n})$, requiring that $\dot{q}(t)=\frac{dq(t)}{dt}$ so that $\dot{q}(t)$ is the vector in the tangent space to that point that is actually tangent to the curve one can evaluate the Lagrangian on this curve to give values $\mathcal{L}= \mathcal{L}(q(t),\dot{q}(t))$ for each value of $t$ (corresponding to each point $q(t)$ along the curve), that are characteristic to that particular curve?!

Could one the go on further to say that a variation in the curve $q(t)$ produced a 'new' curve $\delta q(t)$. Such a variation will induce a variation in the tangent vector to that curve $\dot{q}(t)$, again producing a new tangent vector $\delta \dot{q}(t)$ (which is tangent to this new varied curve). The variations in these two quantities are not independent, however, as the tangent vector to the 'new' varied curve $\delta q(t)$ is given by $\frac{d }{dt}\left(\delta q(t)\right)$ such that $\delta \dot{q}(t)= \frac{d }{dt}\left(\delta q(t)\right)$?!

Last edited: Feb 6, 2015
16. Feb 6, 2015

### Orodruin

Staff Emeritus
If you fix the time parameter $t$, the curve is just one out of a full equivalence class of curves that define a tangent vector in $T_{q(t)} M$. If you have the full curve, you actually have a map $q: I \to M$ (not $q: M \to \mathbb R^n$) where $I$ is an interval in $\mathbb R$. The quantity associated with the full curve is the action, which is a map on the set of all possible curves to $\mathbb R$, i.e., the functional
$$S[\gamma] = \int_{t_0}^{t_1} \mathcal L(\dot \gamma(t)) dt$$

Out of curiosity: What source are you using for reading up on this? I would recommend looking up Arnold's classical mechanics textbook.

17. Feb 6, 2015

### "Don't panic!"

Sorry, yes I had meant to write that (was copy and pasting from a previous part and forgot to change it).

Is this part correct?

Would it be right to say that one chooses a curve and then evaluates the Lagrangian on this curve, which returns a function of the time parameter $t$ such that one can then integrate between two points in time to determine a value for the action. My confusion arises slightly from reading these notes http://www.physics.usu.edu/torre/6010_Fall_2010/Lectures/02.pdf (page 6 onwards); they seemed to clear a lot of things up for me conceptually, but still left me feeling a little unsure.

I've been reading from various sources (none of which I felt gave a completely satisfactory description), and also trying to go through the logical process myself to try and understand it all at a deeper level instead of just accepting it and using the maths without really understanding it!
Thanks for the book recommendation by the way :-)

18. Feb 10, 2015

### Fredrik

Staff Emeritus
Yes, but it's also OK to just define $TM$ as the union of all the $T_pM$. The advantage of your definition is that it makes it easy to define the projection map $\pi:TM\to M$. You just define $\pi(p,v)=p$ for all $p,v$ such that $p\in M$ and $v\in T_pM$. If you define $TM=\bigcup_{p\in M}T_pM$, then the definition of the projection is slightly more complicated. For example, if you use the equivalence-classes-of-curves definition of $T_pM$, then you can define $\pi(p,v)=C(0)$, where $C$ is any curve in the equivalence classs $v$.