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Let $C^{\infty,\star}(\mathbb{R}^d)$ denote the non-commutative topological group of smooth diffeomorphisms from $\mathbb{R}^d$ to itself with $\circ$ as multiplication and let $\emptyset\subset X\subseteq C^{\infty,\star}(\mathbb{R}^d)$. Are there (non-trivial) conditions one can use to verify if the closure of the group generated by $X$, $\overline{\langle X\rangle}$, equals to all of $C^{\infty,\star}(\mathbb{R}^d)$?

I'm thinking of some type of a group-theoretic version of the Weierstrass's result (since obviously these objects need not be algebras).

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    $\begingroup$ The fine topology? $\endgroup$
    – Ben McKay
    Apr 19, 2020 at 15:56
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    $\begingroup$ Yes exactly. But if no reasonable cirteria are availalbe in that setting, I'd settle with soemtihng weaker (though prefarably not). $\endgroup$
    – ABIM
    Apr 19, 2020 at 16:06
  • $\begingroup$ The question is fuzzy; consider the case $d=1$. Do you mean uniform convergence of functions and all their derivates on compact subsets? In that case, the real-analytic diffeomorphisms are dense by Weierstrass. $\endgroup$ Apr 20, 2020 at 17:08
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    $\begingroup$ There is this result: STONE-WEIERSTRASS THEOREMS FOR GROUP-VALUED FUNCTIONS: link.springer.com/content/pdf/10.1007/BF02772227.pdf $\endgroup$
    – ABIM
    Apr 21, 2020 at 8:51

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