Let $f:M_n(\mathbb{R})\rightarrow \mathbb{R}^{n^{2}}$ be the isomorphism which maps matrices to tuples of their components: $$f(A)=(a_{11},a_{12},...,a_{21},a_{22},...a_{nn})$$ I wonder if this function is smooth. My guess is that the topology and the differential structure in $M_n(\mathbb{R})$ are defined precisely as those who make this function smooth. I can't find any sources to confirm this guess. Is that what is happening here?
Isomorphism mapping matrices to components is smooth
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0It is however a "smooth" function, in that the components of $f(A)$ are infinitely differentiable functions of the components of $A$... namely monomials of degree one. – 2017-02-25
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0@user251257 Of course: I should have written $M_n(\mathbb{R}$. I'll correct that. – 2017-02-25
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0@hardmath Ok, but I was thinking in terms if differentiability of functions between manifolds. So $f$ is smooth iff for all matrices $A$ one has that for some chart $(\phi,U)$ of $M_n(\mathbb{R})$ about $A$, $f\circ \phi^{-1}$ is smooth. – 2017-02-25
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0You've changed the domain to $M_n(\mathbb{R})$, so perhaps you want to make a specification of the "charts" for $M_n(\mathbb{R})$. The dimension is $n^2$ and the obvious choice of charts to set up the diffeomorphism will make both $f$ and $f^{-1}$ "smooth" as in my first Comment. – 2017-02-25
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0So you ask if the differential structure induced by the norm topology is the same as usual frechet differentiability? – 2017-02-25