Trouble understanding the tangent bundle

Question

First of all, have I understood the preliminary notion of a tangent space to a point on a manifold correctly?

To each point $p\in\mathcal{M}$ on an $n$-dimensional manifold $\mathcal{M}$ there exists a tangent space $T_{p}\mathcal{M}$ whose elements are the set of vectors $\lbrace\mathbf{v}\rbrace$ whose 'base' is fixed at $p\in\mathcal{M}$. Given a local coordinate chart, these vectors can be identified with the set of derivative operators which encode information about all possible speeds and directions in which one could pass through said point (at which they are bound to). The tangent space to a particular point will be an $n$-dimensional vector space, and thus, in general, contains an infinite number of vectors (as one can construct an infinite number of vectors from a chosen $n$-dimensional basis). As such, once a point is chosen, one can (essentially) independently choose a vector in the tangent space to that point [This part I'm very unsure of?!]

So, as I understand it, given this one can construct a new manifold by taking the disjoint union of the tangent spaces of all the points on the manifold. This new manifold is called the tangent bundle, $$\mathcal{TM}= \bigcup_{p\in\mathcal{M}} T_{p}\mathcal{M}$$ Intuitively I kind of see how this is a $2n$-dimensional manifold, as for each point $p\in\mathcal{M}$ on the $n$-dimensional manifold $\mathcal{M}$ there is an $n$-dimensional tangent space $T_{p}\mathcal{M}$, and as such a point in the manifold $\mathcal{TM}$ is an ordered pair, $(p,\mathbf{v})\in \mathcal{TM}$, uniquely determined by specifying a point $p\in\mathcal{M}$ and a vector $\mathbf{v}\in T_{p}\mathcal{M}$ in the tangent space to that point. However, my confusion arises in why we can treat $p$ and $\mathbf{v}$ as independent variables (surely we need to specify a particular value of $p$ be we can choose a tangent vector $\mathbf{v}$. If they were truly independent shouldn't it be possible to choose them in any order, e.g. choose a $\mathbf{v}$ be choosing a value $p$)?

Answer 1

The variable $v$ is dependent on the point $p$ in general. The notation as a pair $(p,v)$ arises from the definition of a fiber bundle. A fiber bundle $E$ consists of a manifold $M$ and a fiber $F$ that is attached on the manifold with a projection map $\pi:E \rightarrow M$. In this case, $p \in M$ and $v \in F$ and then $(p,v)$ is a section of a cartesian product of $M$ and $F$.

Answer 2

I think you totally have the right idea.

I would work out some concrete instances, like the tangent bundle to the circle, or to the 2-sphere.

Every point in the tangent bundle has an associated point in the original manifold. This is what makes it a tangent "bundle" instead of just a tangent "space". Nevertheless, you are right that $p$ and $v$ are not truly "independent". The tangent bundle is not the same as the trivial bundle $M \times \mathbb{R}^n$. For example, the tangent bundle to the two sphere is not equivalent (as a bundle) to the trivial bundle.

One step to further solidify and formalize your knowledge is to work out the charts for the tangent bundle, and why the projection to the original manifold is smooth.

Answer 3

The intuition behind saying that something is a $2n$-dimensional manifold is that locally it can be described by $2n$ independent parameters. This means that each point has an open neighborhood homeomorphic to an open subset in $\mathbb R^{2n}$. The "homeomorphic" part roughly means that nearby points correspond to nearby choices of parameters, and the "open" part means you can freely and independently vary the parameters as long as you don't go too far from your starting point.

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 neighborhood $U$ 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.