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Redundancy of Lie Group Conditions

  1. Oct 29, 2013 #1
    I want to show that if [itex]G[/itex] is a smooth manifold and the multiplication map [itex]m:G×G\rightarrow G[/itex] defined by [itex]m(g,h)=gh[/itex] is smooth, then [itex]G[/itex] is a Lie group.

    All there is to show is that the inverse map [itex]i(g)=g^{-1}[/itex] is also a smooth map. We can consider a map [itex]F:G×G\rightarrow G×G[/itex] where [itex]F(g,h)=(g,gh)[/itex] and its inverse is [itex]F^{-1}(x,y)=(x,x^{-1}y)[/itex]. If I show that [itex]F[/itex] is a diffeomorphism, then it should be the case that [itex]i(x)[/itex], which is the second component of [itex]F^{-1}(x,e)[/itex], is smooth.

    I have found an answer online that describes [itex]F[/itex] as a very long chain of translations, but can't we simply say that [itex]F=Id_G×L_g[/itex]? And since both [itex]Id_G[/itex] and [itex]L_g[/itex] are diffeomorphisms, isn't their product be a diffeomorphism as well?
     
  2. jcsd
  3. Oct 30, 2013 #2

    quasar987

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    But why is L_g smooth in a Lie group if not for the fact that m is smooth?
     
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