Let $T_pM$ be the tangent space at a point $p$ in a n-dimensional smooth manifold $M$.

In addition, if we assume $(M,g)$ as a smooth Riemannian manifold, then $T_pM$ is a n-dimensional real normed-linear space (real NLS).

Since, every finite-dimensional NLS (over same field) with same dimension are isomorphic, $T_pM$ and $\mathbb R^n$ are isomorphic.

My question is : Are the spaces $T_pM$ and $\mathbb R^n$ homeomorphic?

We know that in a finite dimensional linear space any two norms are equivalent i.e. induced topologies are same.

But, if two NLS are isomorphic, are the topological spaces induced by the norms homeomorphic?

Thanks a lot in advance. Any help will be appreciated.

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    $\begingroup$ let $f:T_p M\rightarrow \mathbf{R}^n$ be an isometric isomorphism. See where does the inverse image of the open unit ball goes. We know $||x||=||f^{-1}(x)||$ $\endgroup$ – Larsson Apr 10 at 12:54
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    $\begingroup$ @Larsson I am trying. But I can not find the inverse image of the open unit ball. It seems that the inverse image will be the open unit ball in $T_pM$. Is it? $\endgroup$ – BijanDatta Apr 10 at 18:26
  • $\begingroup$ Yeah ...and as any open ball in $\mathbf {R}^2$ is homeomorphic to it we are done. $\endgroup$ – Larsson Apr 10 at 18:31
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    $\begingroup$ Normed linear spaces of the same finite dimension need dot be isometrically isomorphic. Consider, e.g., the euclidean and the maximum norm on $\mathbb R^2$. $\endgroup$ – Jochen Apr 11 at 8:25
  • $\begingroup$ @Jochen Yes. They are not isometrically isomorphic. $\endgroup$ – BijanDatta Apr 23 at 12:03

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