Why is the tangent bundle defined using a disjoint union?

Question

In textbooks about differential geometry, one finds often the disjoint union in the definition of the tangent bundle (e.g. in "Lee: Introduction to smooth manifolds", or "Amann, Escher: Analysis III"):

$$T\mathcal{M}:=\coprod_{p\in\mathcal{M}}T_{p}\mathcal{M}:=\bigcup_{p\in\mathcal{M}}\{p\}\times T_{p}\mathcal{M}$$

I have the following questions:

(1) Why is this necessary? Is it necessary for the definition of the smooth structure on $T\mathcal{M}$?

(2) When we define the tangent space as the set of all derivations (like in Lee's book), aren't $T_{p}\mathcal{M}$ and $T_{q}\mathcal{M}$ for $p\neq q$ allready disjoint? Why is it then necessary to use the disjoint union? Is it just a notation in this cases? How about other definitions of tangent spaces, like the geometric defined tangent space, via equivalence classes of curves?

(3) How about other bundles? Also in the definition of tensor fields and differential forms, these authors often use the disjoint union...

Thank you all!

Answer 1

In my book Introduction to Smooth Manifolds, I define $T_pM$ as the set of all linear maps $v\colon C^\infty(M)\to\mathbb R$ that satisfy $$ v(fg) = f(p)vg + g(p)vf $$ for all $f,g\in C^\infty(M)$. Note that the zero map is an element of all these spaces, so $T_pM$ and $T_qM$ are not disjoint for $p\ne q$.

For other definitions of tangent vectors, disjointness might follow automatically. Using the disjoint union in the definition of the bundle is just a handy way to ensure that the vector spaces associated with different points are disjoint regardless of the definition being used.

Answer 2

You are right, the "usual" constructions of $T_p M$ as a set of derivations or a set of equivalence classes of curves produces disjoint tangent spaces at distinct points. Thus you could define $T'M = \bigcup_{p \in M} T_pM$ which is the union of pairwise disjoint vector spaces. I used $T'M$ to distinguish it from $TM = \bigcup_{p \in M} \{p\} \times T_pM$.

Has $T'M$ any benefit in comparison to $TM$? There exists a canonical bijection between both sets, and if we introduce the usual smooth structure on $TM$ and (mutatis mutandis) on $T'M$, then this bijection turns out to be a diffeomorphism.

You might argue that the bundle projection $\pi' : T'M \to M$ has the property that the fiber over $p$ is the "true" tangent space $T_pM$, whereas $\pi : TM \to M$ has the fiber $\{p\} \times T_pM$ which is only a copy of $T_pM$. But as you know, there are various constructions for $T_pM$ which yield formally different vector spaces. Hence it is a rather philosophical question what the true variant should be, and $\{p\} \times T_pM$ has the same right as $T_pM$.

In my opinion it is just a matter of taste if you want to work with $TM$ or $T'M$. The same applies of course to other bundles like tensor bundles, cotangent bundles, etc.

Here is an example where the additional $\{p\}$ is useful. For smooth submanifolds $M \subset \mathbb R^N$ there is a nice geometric construction of the tangent space at $p$: Take all smooth curves $c$ in $\mathbb R^N$ whose image is contained in $M$ and which go through $p$ at some $t_0$. Then the set of all derivatives $c'(t_0)$ forms an $\dim M$-dimensional linear subspace $\tilde T_pM$ of $\mathbb R^N$. These $\tilde T_pM$ are not pairwise disjoint. We need the point $p$ to make them disjoint and then define $$\tilde T M = \bigcup_{p \in M} \{p\} \times \tilde T_pM \subset \mathbb R^{2N}.$$ It turns out that this a smooth submanifold of $\mathbb R^{2N}$ which is diffeomorphic to the abstract $TM$.