- #1
enricfemi
- 195
- 0
I don not know whether I was right or not, please give me a hint.
(R3,+) can be considered a Lie group. and its TG in 0 is still R3.
suppose X as a infinitesimal generater, it can give a left-invariant vector field and also an one-parameter subgroup.
but i think, this one-parameter subgroup is not exponential map r(t)=exp(Xt). it should be r(t)=Xt
must a 1-parameter subgroup whose tangent vector at 0 is X, have a unique exponential map?
is there any thing wrong?
(R3,+) can be considered a Lie group. and its TG in 0 is still R3.
suppose X as a infinitesimal generater, it can give a left-invariant vector field and also an one-parameter subgroup.
but i think, this one-parameter subgroup is not exponential map r(t)=exp(Xt). it should be r(t)=Xt
must a 1-parameter subgroup whose tangent vector at 0 is X, have a unique exponential map?
is there any thing wrong?