- #1
Arkuski
- 38
- 0
I want to show that if G is a smooth manifold and the multiplication map m:G×G\rightarrow G defined by m(g,h)=gh is smooth, then G is a Lie group.
All there is to show is that the inverse map i(g)=g^{-1} is also a smooth map. We can consider a map F:G×G\rightarrow G×G where F(g,h)=(g,gh) and its inverse is F^{-1}(x,y)=(x,x^{-1}y). If I show that F is a diffeomorphism, then it should be the case that i(x), which is the second component of F^{-1}(x,e), is smooth.
I have found an answer online that describes F as a very long chain of translations, but can't we simply say that F=Id_G×L_g? And since both Id_G and L_g are diffeomorphisms, isn't their product be a diffeomorphism as well?
All there is to show is that the inverse map i(g)=g^{-1} is also a smooth map. We can consider a map F:G×G\rightarrow G×G where F(g,h)=(g,gh) and its inverse is F^{-1}(x,y)=(x,x^{-1}y). If I show that F is a diffeomorphism, then it should be the case that i(x), which is the second component of F^{-1}(x,e), is smooth.
I have found an answer online that describes F as a very long chain of translations, but can't we simply say that F=Id_G×L_g? And since both Id_G and L_g are diffeomorphisms, isn't their product be a diffeomorphism as well?