Why defining tangent bundle?

Question

I'm learning a bit about smooth manifolds, and currently I'm learning about tangent bundles (just definitions mainly) and vector field.

This is my reference : Tu's Introduction to Manifolds. I was also watching a video about tangent bundles (because I was struggling with the concept).

Once I understood the definition however I realized I have an issue understanding why we need the notion of tangent bundle. I can't remember where I read this but am I right when I say that tangent bundles are necessary if we want to generalize the notion of function on a manifold?

Consider a smooth manifold $M$, if we wanted to define what a vector field is to me the definition should reflect the fact that for each $p \in M$ we have a $v \in T_p M$, therefore it should be a map.

This is probably the key why such association isn't good as definition because a map needs both domain (in this case $M$) and an image space, however my naive definition involves for each $p$ a different space $T_p M$ and this is why we need the notion of tangent bundle.

Is this observation correct?

Answer 1

I think the simplest motivation is that of vector fields. We want to be able to assign a tangent vector to each point in the manifold, giving us a "field" of vectors on the manifold. That is, $F$ should be some map such that

$$ F(p)\in T_pM $$

for $p\in M$. So, what's wrong with just saying that? Well, nothing really, if you're only interested in the value of vector fields at a point. If you ever want to look beyond singular points, you need some structure connecting your different tangent spaces. For example, to look at continuity of $F$, we need the space of outputs of $F$ to have a topological structure; to look at differentiability, we need a differentiable structure.

So, we need some space which

1. contains all the tangent vectors of $M$
2. has the same "level" of structure that $M$ has

To satisfy (1), we simply glue together all the tangent spaces by taking a union. Since the tangent spaces are completely disconnected from one another, we can exemplify this fact by using a disjoint union

$$ TM = \coprod\limits_{p\in M} T_pM $$

The rest of the bundle structure is just there to "lift" the structure of $M$ onto $TM$. With that, we can now define vector fields as functions in the usual way

$$ F: M\to TM $$

and we are able to discuss continuity and differentiability to the extent that $M$ admits such properties. However, this definition isn't "complete" because it allows for e.g. attaching a tangent vector from $T_qM$ to $p$, which doesn't fit our idea of a vector field. This gives us another requirement,

3. we need a way to determine which point a vector is tangent to

This is the bundle projection map, $\pi: TM\to M$, and so we can add the requirement on $F$ that $\pi(F(p)) = p$ everywhere.

---

Here, we created a bundle with a base manifold and a vector space at each point, but we could imagine a more general concept of bundle which just has some kind of space $B$ with some other kind of space $F_p B$ at each point $p\in B$. Even in this general setting, we can see the utility of attaching to each point of the base space some element of its attached space. We define a function

$$ \sigma: B\to FB = E $$

with $\pi(\sigma(p)) = p$ as a cross-section (or just section) of the total space $E$.

Applying this terminology to our original example, we can then reconstruct the more terse definition:

$$ \text{a vector field on } M \text{ is a section }\sigma\text{ of the tangent bundle } TM$$

Answer 2

Remember that the tangent bundle is the disjoint union of the tangent spaces: $$TM = \coprod_{P \in M} T_P M.$$ It has the topology of a smooth manifold in the following manner. Let $(U_\alpha, \phi_\alpha)$ be an atlas for $M$, and let $\pi: TM \longrightarrow M$ be the natural projection, i.e. if $(P, v) \in T_P M \subset TM$, then $\pi(P, v) = P$.

Edit: On the Why? bit

The point of this is just to establish a mathematical framework in which we can talk about points on a base manifold, and all the possible curves through any point on the manifold. We can talk about the base points $x$, together with possible directions at $x$.

Answer 3

You are concerned that for each $p$ you need a different space $T_p(M)$ when building a vector field. By putting all the tangent spaces together (disjoint union, with a suitable topology and even a smooth manifold structure) to form the tangent bundle $TM$ you get one target space for all vector fields on $M$. For a vector field on $M$ you want the tangent vector attached to $p$ to be living in $T_p(M)$ and not $T_q(M)$ for some $q \not= p$. A precise way of specifying this condition is to say a vector field on $M$ is a map $X \colon M \rightarrow TM$ where $X(p) \in T_p(M)$ for all $p \in M$. Or, in terms of the natural surjective map $\pi \colon TM \rightarrow M$ that sends a point $(p,v)$ in $T(M)$ to the point $p$ at which it is based, a vector field on $M$ is a map $X \colon M \rightarrow TM$ such that $\pi \circ X \colon M \rightarrow M$ is the identity. We call $X$ a "section" of $\pi$ (or a section of the tangent bundle). Quite generally, when $f \colon A \rightarrow B$ is a surjective map, a section of $f$ is mapping in the other direction $g \colon B \rightarrow A$ where every $g(b)$ is in the fiber $f^{-1}(b)$, which is another way of saying $f(g(b)) = b$ for all $b \in B$, or equivalently $f \circ g \colon B \rightarrow B$ is the identity.

The map $\pi\colon TM \rightarrow M$ is smooth, and we call a vector field $X \colon M \rightarrow T(M)$ continuous or smooth when $X$ is continuous or smooth as a mapping.

Answer 4

Actually we know how to do calculus (differentiation, integration...) for functions on $R^n$. And if, $f:M\rightarrow N$ is smooth map then its differential $df:TM\rightarrow TN$ Will be map between tangent spaces so overall we will need knowledge of tangent spaces if we want to study smooth maps between manifolds.