Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?

Question

It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of \begin{align} TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\ (v,p)&\mapsto p \end{align} (I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.

Answer 1

There are two major issues here.

First of all, the only way for the union to not be disjoint and make any sense at all is if the tangent spaces are assumed to live in a common universal set, which isn't generally the case.

Secondly, even if you embed the tangent spaces in a common universe, e.g. by embedding the manifold inside some Euclidean space, then you lose crucial information by taking a regular union, since the line between tangent vectors and points becomes blurred and points living in distinct tangent spaces may get identified. For example, consider the tangent bundle of the circle embedded in $\Bbb{R}^2$, as in the following picture:

By taking a regular union, you get the subset of $\Bbb{R}^2$ consisting of all the points in red as your tangent bundle. But then all the points which lie on multiple lines "forget" which tangent space they belong to. For example, the point $(1,1)$ lies on the tangent space at $(1,0)$ as well as the tangent space at $(0,1)$. Therefore you lose the canonical surjection $\bigcup_p T_pM \to M$ which is used, among many other things, to endow the tangent bundle with a manifold structure. Disaster!

Finally, to adress the part where you mention vector fields: it is important to note that such objects are not mere set-theoretic sections of the map $\bigcup_p T_pM \to M$; they are continuous or smooth sections. And in order for this to make sense, we need a topology/smooth structure on the tangent bundle.

Answer 2

There are lots of good answers here, all of which elucidate certain parts of the situation. But there's one important point that hasn't been mentioned -- in the definition of tangent spaces that I use in my Smooth Manifolds book, the zero derivation is an element of $T_pM$ for every $p\in M$, so if you don't use disjoint union in the definition of the tangent bundle, the tangent spaces will all intersect. See also this answer.

Answer 3

Suppose $M \subset \mathbb{R}^n$ is a submanifold. Then one may define its tangent bundle to be the union of all the tangent vector spaces $T_pM$ to $M$ at points $p$. The problem is what is to be understood as "union" here.

Suppose you define it to be the union as subsets of $\mathbb{R}^n$. For example, if $M = \mathbb{R} \subset \mathbb{R}$, then at each point the tangent space is $\mathbb{R}$, so all the tangent subsets are equal to the same subspace of $\mathbb{R}$, namely $\mathbb{R}$, and so is their union.

Now suppose $M = \mathbb{S}^1 \subset \mathbb{R}^2$. Then any vector line $D$ can be seen as the tangent space of a point of the circle. So here, the usual union as subsets of $\mathbb{R}^2$ will be the union of all vector lines of $\mathbb{R}^2$, which is $\mathbb{R}^2$.

These constructions are "extrinsic" as they depend not only on $M$ but also on the ambient space $\mathbb{R}^n$. For example, if one says $M \subset \mathbb{R}^n \simeq \mathbb{R}^n\times\{0\} \subset \mathbb{R}^m$, then one can define a "different $TM$", depending on the viewpoint (even if they will be isomorphic).

But in these two examples, if you take a random element of the union, you don't know at which point it is tangent. You lose a lot of geometric sense.

The idea to avoid this is to take a disjoint union, namely $$TM = \bigcup_{p\in M} \{p\}\times T_pM.$$ An element of this union is of the form $(p,v)$ with $v \in T_pM$, so each element has in its construction more data than in the previous example.

For the first example, this construction gives $T\mathbb{R} = \bigcup_{p\in\mathbb{R}} \mathbb{R} = \mathbb{R}\times \mathbb{R}$, and any tangent vector is of the form $(x,t)$ where $t$ is tangent to $x$.

For the circle, it gives $T\mathbb{S}^1 = \bigcup_{\theta \in \mathbb{S}^1} \{\theta\} \times T_{\theta}\mathbb{S}^1$, etc.

For an abstract manifold, there is no "ambient space", so the usual union of the tangent spaces cannot be defined as an union of subspaces of the same fixed set. Therefore, it would have been a bad construction as we would not have been able to extend it. But the disjoint-union allows us to define, for a general manifold $M$ that is not embedded in a Euclidean space, $$T_pM = \bigcup_{p\in M}\{p\}\times T_pM,$$ where $T_pM$ is an intrinsic notion in $M$, depending only on the differential structure.

In addition, this construction shows that there is a natural structure of fiber-bundle on the tangent space $TM$ (this is a more general concept), and this construction automatically gives a smooth function $\pi : TM \mapsto M$ that is just the projection $(p,v) \mapsto p$.

If one wants to define a vector field by the idea that "at each point $p$ it has a tangent vector to $p$", then it can be rigorously defined using this construction as a continuous map $X : M \to TM$ such that $X(p) = (p,v_p)$. This is equivalent to saying that $X$ is a (continuous) section of $\pi$, that is $\pi\circ X = \mathrm{id}_M$. Usually we require the vector field to be smooth, which means $X$ is smooth (a smooth section).

Edit: It is a common problem for geometers, when they have to give a talk to non-specialists, to give a clear presentation while the audience has no idea of the principal objects we use, such as manifolds, vector bundles, etc. My experience is this: do not lose time giving overly complicated definitions if what is really important is the geometric sense. Just say that a manifold is a geometric notion that can extend the definition of surfaces, etc. Define tangent vectors visually. Say that a tangent field is a field of tangent vectors without talking about bundles. Same for covectors. If you have to talk about operators in bundles, just talk about how they act on vectors. You will gain a lot of time and the audience will probably understand many more things than if you had given some overly complicated rigorous statements.

Answer 4

This is just rewriting what has been written a few times, but when $M\subset \Bbb R^N$, then $$TM = \{(x,v): x\in M, v\in T_xM\}\subset M\times\Bbb R^N.$$ There's your universe. For an abstract manifold, of course, this makes no sense, because there is no sensible thing to substitute for $\Bbb R^N$.

Answer 5

Here's an explanation that will fit into the time allotted for your crash course:

Physicists will sometime say that two vectors are the same if they point in the same direction and have the same base point.

As the abstraction that mathematicians have adopted for vectors doesn't include the base point, "$\times \{p\}$" is how they label each vector with its base point. By the way, this is also probably how computer programmers would go about it too.