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.

  • $\begingroup$ To find your own questions, click at the top of the page where your reputation and your badges are displayed, and up will come a window where you can look at your own questions, your own answers, etc. $\endgroup$ – Lee Mosher Aug 22 at 13:27
  • $\begingroup$ I know that, but I still can find that specific question. $\endgroup$ – user8469759 Aug 22 at 13:58

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.

  • $\begingroup$ I've just found it : math.stackexchange.com/questions/3288659/…. At this point I think I've completely miss-understood the clue I was given. Is it possible to expound the answer given more rigorously to cover the choice of sets of the form $\mathcal{U}$? $\endgroup$ – user8469759 Aug 22 at 14:02
  • $\begingroup$ I've added a little about how to correctly formulate the basis sets $\mathcal U$. $\endgroup$ – Lee Mosher Aug 22 at 14:26
  • $\begingroup$ I still think I'm miss understanding the meaning of the word "basis" in this context. Please point out anything you think is unclear in my question. $\endgroup$ – user8469759 Aug 22 at 14:30
  • $\begingroup$ In $\mathbb{R}^2$ the corrected version of $\mathcal{U}$ would include sets of the form $(a,b)\times(u-\epsilon,u+\epsilon)$ for arbitrary $u \in \mathbb{R}$. What got lost between your previous question and this one was you switched from Lee Mosher's "neighborhood basis of the point $(p,0)$" to "basis for the topology." These are of course different concepts. $\endgroup$ – Rylee Lyman Aug 22 at 15:03
  • 1
    $\begingroup$ It might help if, in addition to a book on topological vector spaces, you keep by your side a book on topology, such as Munkres' "Topology". That way you'll be able to compare terminologies. For example, in the present situation the explanation of a "basis for a topology" found in a topology textbook is more elementary and expansive, but in the topological vector spaces textbook is much terser. $\endgroup$ – Lee Mosher Aug 22 at 16:21

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 $\mathcal{U}$.

  • $\begingroup$ I'm trying to put all the pieces of your answer together. In case of differentiable manifolds coordinate systems are also homeomorphism, right? Therefore $\phi^{-1}(\mathcal{V}_{N,\epsilon})$ is an open set in $TU$ (according to the definition of open set I've given in my question). Is this right? $\endgroup$ – user8469759 Aug 23 at 9:54
  • $\begingroup$ Also, I know I'm probably being silly but it sounds like to me that proving that $\phi^{-1}(\mathcal{V}_{N,\epsilon})$ has the form $\mathcal{U}$ is almost a straightforward calculation, I can't manage however to perform this calculation in a way that convinces me though. $\endgroup$ – user8469759 Aug 23 at 10:09

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