4
$\begingroup$

I'm watching at this lecture by F.P. Schuller. At minute 20.45 he defines the topology assigned to the tangent bundle $TM$ as the initial topology from the topology on the base manifold $M$ under the projection map $\pi$. In other words he assigns $TM$ the coarsest topology such that the projection map $\pi$ turns out to be continuous.

My concern is that, in this way, $TM$ as topological space is not even Hausdorff nor locally homeomorphic to open sets of $\mathbb R^{2d}$ in which $d$ is the dimension of $M$.

$\endgroup$
1
  • 1
    $\begingroup$ The topology on $TM$ is such that the topology on any local trivialization is the product topology. $\endgroup$ Commented Apr 14 at 18:40

2 Answers 2

Reset to default
7
$\begingroup$

This is plainly false. $TM$ must not be given the initial topology induced by the projection map $\pi$.

Later in the video an atlas for $TM$ is defined. Indeed Schuller defines bijections $\xi_x$ from $TU = \bigcup_{p \in U} \{p\}\times T_pM$ to $U' \times \mathbb R^d$ (which is of course endowed with the product topology) for all charts $x : U \to U' \subset \mathbb R^d$ on $M$, but in order that they form an atlas on $TM$ the $TU$ have to be open in $TM$ and the $\xi_x$ have to be homeomorphisms. With the initial topology on $TM$ induced by $\pi$ the $TU$ are open, but the the initial topology is is too coarse to make the $\xi_x$ homeomorphisms.

So what is the right approach to define a topology on $TM$?

Changing perspective, we have a family of injections $$\xi'_x : U' \times \mathbb R^d \xrightarrow{\xi^{-1}_x} TU \hookrightarrow TM .$$

Now give $TM$ the final topology induced by the $\xi'_x$, i.e. the finest topology such that all $\xi'_x$ become continuous.

It is easy to see that all $TU$ are open in $TM$. Showing that all $\xi'_x$ become homeomorphisms and $TM$ is Hausdorff and second countable requires some work .

$\endgroup$
3
  • $\begingroup$ Just a terminology check: the finest topology on $TM$ means that picking on it a topology finer than that does not result in all $\xi'_x$ being continuous, right ? $\endgroup$ Commented Apr 14 at 12:02
  • $\begingroup$ Another point: in the above claim the topology on $U' \times \mathbb R^d$ is supposed to be the product topology, right ? $\endgroup$ Commented Apr 14 at 12:06
  • $\begingroup$ @CarloC Yes to both questions! $\endgroup$ Commented Apr 14 at 12:43
3
$\begingroup$

You are right, that shouldn't be the initial topology, it should be the topology such that all the inverses of the chart maps are local homeomorphisms.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.