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Given a Riemannian manifold $M$, What is the topology of its tangent space?

There are two interpretations:

1). View each $T_pM$ as a subspace of the tangent bundle, then its topology should be the subspace topology.

2). Since $M$ is a Riemannian manifold, $T_pM$ is a normed vector space. Then it has a canonical smooth structure as a normed vector space.

I think the second interpretation is more reasonable. Since it is well know that the exponential map $exp$ when restricted to $T_pM$ is a local diffeomorphism around zero. To talk about local diffeomorphism, apparently $T_pM$ should have a smooth structure.

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  • $\begingroup$ Being normed vector space tells nothing about smooth structures. But it tells everything about its topology; and it coincides with the subspace topology induced from the total tangent bundle. (To see this, recall that the tangent bundle over trivializing cover is isomorphic to the product bundle over where the claim is trivial.) $\endgroup$
    – cjackal
    May 18, 2017 at 19:44
  • $\begingroup$ Can you expand that to an answer? $\endgroup$
    – z.z
    May 18, 2017 at 19:51
  • $\begingroup$ Maybe could clarify @cjackal's comment a bit: recall that any two norms on a finite-dimensional real or complex vector space are equivalent; so any norm induces the usual Euclidean topology on $\mathbb{R}^n$. $\endgroup$
    – Daniel Schepler
    May 18, 2017 at 19:54
  • $\begingroup$ Look at the construction of atlasses on $TM$, which induces the topology on $TM$. You will see that your two interpretations are both correct. $\endgroup$
    – Frieder Jäckel
    May 18, 2017 at 21:12
  • $\begingroup$ I have updated the question $\endgroup$
    – z.z
    May 19, 2017 at 0:00

1 Answer 1

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$T_pM$ is a finite-dimensional normed vector space, so the topology is that of $\mathbb{R}^d$, where $d = \mathrm{dim}\; M$.

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