If i take $V$ a finte dimensional vector space on the real number (or complex number). Setting $n=dim_{\mathbb{R}}(V)$, i know that there is a isomorphism of vector spaces so $V \simeq \mathbb{R}^n$. So can i define a topology on $V$ via this idenification? is $V$ with this topology a differential manifold as $\mathbb{R}^n$?


Fix a topology $(\mathbb R^n, \mathcal T)$ and let $f$ be an isomorphism $V\to\mathbb R^n$. Then you can define a topology on $V$ via $$\mathcal T_V = \{ f^{-1}(O): O \in \mathcal T\}$$

Take $\{(V,f)\}$ as an atlas (i.e. $(V,f)$ as the only chart). The only transition map, you have to consider, is $\operatorname{id}:\mathbb R^n\to \mathbb R^n$. $\operatorname{id}$ is differentiable under each topology (due to $\operatorname{id}(x+\epsilon)=\operatorname{id}(x)+\operatorname{id}(\epsilon)=\operatorname{id}(x)+\epsilon\cdot I$). So $V$ is a differential manifold.

  • $\begingroup$ The problem is that if i take $V=T_{m_0}M$, that is the tangent space to a differential variety in $m_0$, this is a vector space but can i say that the tangent space is a differential manifold? this because i want to define an application between an open subset of the tangent space and an open subset of $M$ and i would like to say when this application is differentiable $\endgroup$ – dario May 19 '15 at 15:19
  • $\begingroup$ ok now is clear thanks $\endgroup$ – dario May 19 '15 at 15:28

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