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I can't proof that these definitions of lie groups are equivalent. Can anyone prove it directly?.

  1. A Lie group is a set which carries the algebraic structure of a group and the differentiable structure of a smooth manifold such that the mapping $$ G\times G \rightarrow G \hspace{1cm} (a,b) \mapsto ab^{-1}$$ is smooth

  2. A Lie group is a set which carries the algebraic structure of a group and the differentiable structure of a smooth manifold such that the mappings $$ G\times G \rightarrow G \hspace{1cm} (a,b) \mapsto ab$$ and $$ G \rightarrow G \hspace{1cm} g \mapsto g^{-1}$$ are smooth

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  • $\begingroup$ Presumably the restriction $1\times G\to G$ (i.e. inversion) must be smooth and hence the composition of inversion with $ab^{-1}$ (i.e. multiplication $ab$) is also smooth. $\endgroup$
    – anon
    Apr 4, 2020 at 0:37
  • $\begingroup$ In fact, it suffices to require smoothness of the multiplication, smoothness of the inversion then follows from the inverse function theorem. $\endgroup$ Apr 4, 2020 at 8:55

1 Answer 1

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1 implies 2.

Since $m:(a,b)\rightarrow ab^{-1}$ is smooth, and $i(a)=(e,a)$ is smooth, $i_v(a)=m\circ i(a)=a^{-1}$ is smooth.

$ f:(a,b)\rightarrow (a,i_v(b))=(a,b^{-1})$ is smooth and $m\circ f(a,b)=ab$ is smooth.

  1. implies $a$,

let $g(a,b)=ab, m=g\circ (Id_G,i_v)$.

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  • $\begingroup$ Excellent proof. Thanks $\endgroup$
    – Strauca
    Apr 4, 2020 at 2:25

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