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This is an excerpt of Lee's book on smooth manifolds.

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As far as I understand, if there is a (vector/linear) isomorphism between two vector spaces, then they are indistinguishable in the eyes of "vector mathematics", meaning that any proposition whose hypotheses contain nothing else but concepts defined in terms of the axioms for vector spaces vector space operations is either true for both vector spaces or false for both vector spaces.

Now: two isomorphic vector spaces which are also manifolds may not be indistinguishable from one another in the eyes of "differential mathematics", because their differential structure may be different. That said, I do not understand why Lee writes that, since $GL(V)$ and $GL(n,\mathbb{R})$ and the latter is a Lie group, then $GL(V)$ is also a Lie group. After all, the definition of Lie group contains concepts which are defined outside of the so called "vector mathematics".

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  • $\begingroup$ With an atlas of $GL(n)$ and that isomorphism you have an atlas of $GL(V)$. Now you just need to check that multiplication is smooth. But this is the case, because of the above atlas. $\endgroup$ – T'x May 8 '17 at 19:52
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Any vector space isomorphism from $V$ to $\mathbb{R}^n$ is also smooth. This is because linear maps are smooth.

I think the reason for your confusion is that your vague definition of vector space isomorphism is not correct. I suggest reviewing the definition of a linear transformation and isomorphism, and then rethinking your explanation of isomorphism. Any proprosition which can be proved using the vector space axioms is true for all vector spaces, but not all vector spaces are isomorphic. Any two vector spaces of the same dimension are isomorphic, and you can use this to refine your statement to something true ... however, an isomorphism is an explicit transformation witnessing the equivalence (which yes, allows you to transfer structure), and it is important to understand this. You should know why a basis for a vector space $V$ (finite dimensional) determines a linear isomorphism with $\mathbb{R}^n$.

Understanding linear algebra on this level is definately a pre-requisite for learning manifold theory.

I hope I have answered your question at the appropriate level. Otherwise, please follow up.

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  • $\begingroup$ I think I was not clear: I did not mention "propositions which can be proved using the vector axioms". These are, of course, true for all vector spaces. I meant the following: anything you say about a vector space using only concepts of linear algebra also holds for spaces isomorphic to our space. Just like anything you say about a topological space that depends only on the topology of that space is true for any other topological space homeomorphic the former. $\endgroup$ – Soap May 8 '17 at 21:15
  • $\begingroup$ I edited my question. If it is still confusing or it seems that I am wrong nonetheless, please tell me. $\endgroup$ – Soap May 8 '17 at 21:34
  • $\begingroup$ @soap I think the feeling of isomorphism you are alluding to is useful (it is often called, transfer of structure), and while I guess it can be made precise (perhaps using the Yoneda lemma?), it is not the really a definition $\endgroup$ – Lorenzo Najt May 8 '17 at 21:38
  • $\begingroup$ And if you study carefully the definition of isomorphism of vector space (especially regarding basis ...) I think your confusion will be resolved. $\endgroup$ – Lorenzo Najt May 8 '17 at 21:39
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    $\begingroup$ @soap maybe you will feel convinced if you write a diffeomorphism between the two groups that respects the group structure? You will see that it uses the isomorphism of vector spaces in a linear algebraic kind of way, in particular the basis ... $\endgroup$ – Lorenzo Najt May 8 '17 at 22:16

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