Why does a tangent bundle have dimension 2n instead of n?

Let $n=dim(T_pM)$ for every $p\in M$, where $M$ is a smooth manifold. I understand that specifying $p$ is not enough to determine an element of $TM$, but what if do we specify only $v\in T_pM\subset TM$? Since the tangent spaces are disjoint, specifying $v$ is all we need to determine the respective element of $TM$, which is $v$ itself. Therefore I don't understand why $dim(TM)=2n$.

PS: I know there are some related posts over here, but none of them have answered my question.

• The bundle being dimension $2n$ includes both the dimension of $M$ itself as well as the dimension of the tangent space. – Rolf Hoyer Jun 7 '16 at 16:17
• To specify a point in the bundle, you need both a point $p$ and a vector $v\in T_pM$. Just a vector, without a point to "pin" it to makes no sense. – Arthur Jun 7 '16 at 16:19
• Specifying that $v$ (i.e. specifying which of the disjoint tangent spaces we're looking at and where in the tangent space we are) requires $2n$ numbers, no matter how you do it. If you don't give enough information to say what $p$ is, then you haven't specified an element of $TM$. – Omnomnomnom Jun 7 '16 at 16:19
• Specifying $v$ requires two pieces of information: the base point $p \in M$ which has $n$ degrees of freedom; and the choice of a vector in $T_p(M)$ which has $n$ independent degrees of freedom. Add those up and you get $n+n=2n$ degrees of freedom altogether. – Lee Mosher Jun 7 '16 at 16:21

If all you specify is $v$, you haven't picked a point in $TM$, you have (simultaneously) picked a point in each $T_pM$. (Edit: Assuming that point $v$ actually is in each $T_pM$.) To pick a single point in $TM$, you need to pick a $p \in M$ and then a $v \in T_pM$ (or vice versa -- either way, two choices).

Example: $T\Bbb{R} \cong \Bbb{R}^2$ and to pick a point in the plane takes two coordinates. (This Cartesian product relation between $M$ and $TM$ is relatively uncommon, but is easy to see for Euclidean spaces.)

Edit: Further elaboration: A $T_pM$ is a vector space. It contains vectors. It doesn't contain tagged vectors. In particular, the elements of a $T_pM$ do not in any way indicate to which $p$ they are associated. $TM$ is the disjoint union $\sqcup_{p \in M} T_pM$. One standard set theoretic implementation of the disjoint union is as the set of ordered pairs $(p, v)$ (the $p$s are the index of the disjoint union). Since a given tangent vector may be in several such pairs, it is not enough to merely indicate the $v$. You must indicate both $p$ and $v$ to pick a specific element of $TM$.

• That's the thing, how can $v$ be in more than one tangent space? If we say $v\in T_pM$ we need only $n$ numbers to specify $v$ (the information that $v$ is based on $p$ is already in $v$ itself), then somehow when we embed the tangent space into the tangent bundle, saying now $v\in T_pM\subset TM$, we need $2n$. Why do we loose the information about the base point in this case? – Mr. K Jun 7 '16 at 16:36
• @Mr.K : Which $T_p\Bbb{R}$ is $\partial/\partial x$ in? Tangent vectors do not indicate their base points. – Eric Towers Jun 7 '16 at 16:45
• "the elements of a $T_p M$ do not in any way indicate to which $p$ they are associated": this depends on how you construct the tangent space, and in general they certainly do. $TM$ is a topological space, equipped with a projection $\pi : TM \to M$. How could that projection even be defined if an element of $TM$ didn't already have the information of which point it is lying over? But if say $M$ is parallelized, $TM \cong M \times \mathbb{R}^n$, then no, a bare element of $\mathbb{R}^n$ doesn't define a tangent vector, as indeed an element of $TM$ is a pair $(x, v) \in M \times \mathbb{R}^n$. – Najib Idrissi Jun 7 '16 at 17:34
• @NajibIdrissi : An element of $TM$ is an element of a disjoint union. Elements of disjoint unions are pairs, $(p,v) \in M \times T_pM$ and note that the space of the second element depends on the first element. But the elements of $T_pM$ are just vectors, not tagged vectors, so a particular $v \in T_pM$ has no way of knowing that it is based at $p$ -- it's just a vector. The pair $(p,v)$ can trivially be projected onto its first element, so there is a map from $TM \rightarrow M$, but there is no map $\pi_2(TM) \rightarrow M: v \mapsto p$ because $v$ has no information on $p$. – Eric Towers Jun 7 '16 at 23:07
• Let's start from the beginning... What's your (precise) definition of $T_pM$? – Najib Idrissi Jun 8 '16 at 6:00

There are various definitions of the tangent space. A possible one is, for $p \in M$: $$T_p M = \{ \gamma : (-\epsilon, \epsilon) \to M \mid \gamma(0) = p \} / \sim$$ where $\sim$ is, informally, the equivalence relation $\gamma_1 \sim \gamma_2 \iff \gamma_1'(0) = \gamma_2'(0)$ ($\gamma'(0)$ is not actually well defined, it is only defined if you choose a chart around $p$; but given two charts, the derivatives are equal in one iff they are equal in the other, so the equivalence relation is well-defined).

As a set, the tangent space $TM$ is then defined as the union of all the $T_pM$, i.e. $$TM = \bigcup_{p \in M} T_p M.$$

Using a chart, it's easy to see that if $M$ is an $n$-manifold, then $T_p M$ is a vector space of dimension $n$. And a tangent vector $v = [\gamma]$ does carry the information of which point $p$ it lies over, because $p = \gamma(0)$ (and this is independent of the representative of the equivalence class). There is no such thing as a "bare" tangent vector that would somehow be tangent to the manifold without specifying at which point it is tangent.*

But then why isn't $TM$ of dimension $n$ too? The thing is that there is a ton of vector spaces glued together to make up $TM$; in fact, one for each point of $M$. Just like $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ is made up of a ton of vector spaces of dimension one glued together (e.g. the $\{x\} \times \mathbb{R}$ for $x \in \mathbb{R}$) and has dimension two, we cannot expect $TM$ to only have dimension $n$.

Heuristically, as Lee Mosher explains in the comments, we have $n$ degrees of freedom to choose the point $p \in M$, then another $n$ degrees of freedom to choose a tangent vector $v \in T_pM$. So in total $TM$ should have dimension $2n$.

To make this precise, you first need to give a topology to $TM$; informally, $[\gamma_1]$ and $[\gamma_2]$ will be "close" if 1/ $\gamma_1(0)$ and $\gamma_2(0)$ are "close" in $M$ and if in a chart, $\gamma'_1(0)$ and $\gamma'_2(0)$ are close in $\mathbb{R}^n$. Your differential topology textbook probably makes that precise. Then you need to cook up charts for $TM$, and in fact if $U \subset M$ is a chart $U \cong \mathbb{R}^n$, then $TU \subset TM$ will be homeomorphic to $U \times \mathbb{R}^{n} \cong \mathbb{R}^{2n}$ $\gets$ this is where the dimension $2n$ appears! Then you need to check that the change of chart maps are smooth.

* I think a possible source of confusion on this comes from framed manifolds. If your manifold $M$ is framed, then you have a bundle isomorphism $\phi : TM \to M \times \mathbb{R}^n$. It's then tempting to think of $x \in \mathbb{R}^n$ as a tangent vector to $M$. This is perhaps what some people in this thread are calling a "bare" vector, or an "untagged" vector. But really, this $x$ is a (tangent) vector field, given by $$\xi_x(p) = \phi(p,x) \in TM.$$ So it is no surprise that this "tangent vector" seems to lie over no point in particular, because it's not a tangent vector: it's something that associates a tangent vector to each point (and in a smooth way). This is for example the case for $\partial/\partial x$, which is really a tangent vector field over $\mathbb{R}$ and not just a tangent vector.

In general manifolds aren't framed though, so this kind of confusion cannot arise. If you define carefully the tangent space, then you will see that a tangent vector does have the information of both a point of $M$ and some kind of tangent vector in a chart or something like that. That's $2n$ coordinates worth of information, hence $\dim TM = 2n$.