Do Carmo's Riemannian Geometry - Basis element of a topology in tangent bundle

Question

In definition 2.4. Chapter 2 "The Geodesic Flow" of Riemannian Geometry the following open set $\mathcal{U}$ in $TU$ (subset of a tangent bundle) is defined

$$ \mathcal{U} = \left\{(q,v) \in TU ; q \in V, v \in T_q, |v| < \epsilon \right\} $$

What I'm trying to understand is why sets of such form define a basis for the topology in the tangent bundle.

The tangent bundle of a differentiable manifold $\mathcal{M}$ is defined as

$$ T\mathcal{M} = \left\{(p,v); p \in \mathcal{M}, v \in T_q\mathcal{M} \right\} $$

It is possible to prove that the tangent bundle is a differentiable manifold of dimension $2n$ more specifically the differentiable structure is given by $\left\{U_{\alpha} \times \mathbb{R}^n, y_{\alpha} \right\}$ where

$$ y_\alpha = (x_1^{\alpha},\ldots,x^n_{\alpha}, u_1, \ldots u_n) $$

(the first $n$ coordinates are a system of coordinates in $\mathcal{M}$ and the last $n$ coordinates are coordinates in the basis $\partial_{x_1},\ldots, \partial_{x_n}$ in $T_p \mathcal{M}$.

In general given a differentiable manifold $\mathcal{M}$ we say that $A \subset \mathcal{M}$ is open if $x^{-1}_{\alpha}(A \cap x_{\alpha}(U_\alpha))$ is an open set.

Now first of all how does this definition of open set specialize for the tangent bundle? Second is it possible from this special case of definition of open set prove that an open set is union of elements of the form $\mathcal{U}$ defined at the beginning of my question?

Thank you

PS. I'm pretty sure I've asked a related question but not the same question, I simply cannot retrieve it and link to this one.

Answer 1

You are trying to prove a false statement: the sets of the form $\mathcal U$ do not form a basis for the topology.

You can see this most easily with the example $M=\mathbb R$ where the tangent bundle projection map $TM \mapsto M$ is isomorphic to the projection $\mathbb R^2 \mapsto \mathbb R$ onto the first factor. Your "basis" for this example would consist of sets of the form $\mathcal U = (a,b) \times (-\epsilon,+\epsilon)$, where $V=(a,b)$, but these sets clearly do not form a basis for the standard topology on $\mathbb R^2$.

You asked for another description of basis elements. I am not able to do this matching notation that you might know from Do Carmo, so I'll just try to do it with as simple notation as possible.

What you could do is something like this. Letting $p : TM \to M$ be the projection, $M$ is covered by open sets $W$ equipped with diffeomorphisms $$\phi : p^{-1}(W) \to W \times \mathbb R^m $$ subject to overlap conditions which I am sure you understand. (This is something like that $y_\alpha$ in your question, but I did not entirely understand what you wrote about that.)

For each such $W$, you can choose an open set $V \subset W$, a vector $\vec u \in \mathbb R^m$, and an $\epsilon > 0$, and you could then define a basis element of the form $$\mathcal U(V,W,\vec u,\epsilon) = \{\phi^{-1}(q,\vec v) \mid q \in W, \vec v \in \mathbb R^n, |\vec u - \vec v| = \epsilon \} $$ Picking the $\vec u \in \mathbb R^m$ is the major thing missing in your description of $\mathcal U$ in your question.

Answer 2

This is more than a comment, but in view of the comments and @LeeMosher's answer in a previous question of yours, it's really an answer about how to show that the set $\mathcal{U}$ you gave is a neighborhood basis for the point $(p,0)$, where $U$ is an open set containing $p \in M$.

For simplicity, let's assume that there is a chart/coordinate system/homeomorphism $\phi\colon TU \to O \times\mathbb{R}^n \subset \mathbb{R}^{2n}$, where $O$ is an open subset of $\mathbb{R}^n$. It's clear we can always do this (perhaps after shrinking $U$), as you described in the question. Furthermore, we may assume $\phi(p,0) = (v,0)$. The exercise now becomes showing that sets of the following type form a neighborhood basis of $(v,0)$ in $\mathbb{R}^{2n}$. Let $N$ be an open set in $\mathbb{R}^n$ containing $v$. $$\mathcal{V}_{N,\epsilon} = \{(q,x)\ \mid q \in N, |x| < \epsilon \}. $$ Once you've done this, you can check that $\phi^{-1}(\mathcal{V_{N,\epsilon}})$ has the form of your set $\mathcal{U}$.