0
$\begingroup$

Let $H_1, H_2, G$ be Lie groups, and suppose that $i_j:H_j\to G$ are two injective Lie group homomorphisms such that $i_1(H_1)=i_2(H_2)$. The exercise that I am trying to solve aims to show that a subgroup $H$ of $G$ carries at most one structure of a Lie subgroup, in a couple of steps.

The first thing is to show that an isomorphism $f:H_1\to H_2$ of groups is smooth iff it is smooth on a neighbourhood of $e$, and that it's a diffeomorphism iff it's a local diffeomorphism at $e$. This is straightforward. The second step is to show that there exists a unique map $\phi:H_1\to H_2$ such that $i_2\circ\phi=i_1$, and that this map is an isomorphism of groups. This is also straightforward.

The third step is to show that there exists a unique map $\tau:\mathfrak{h}_1\to\mathfrak{h}_2$ such that $i_{2\ast}\circ\tau=i_{1\ast}$ (the stars denote derivatives at the identity elements), and that this map is an isomorphism of Lie algebras. Uniqueness, injectivity, and the fact that it is a morphism of Lie algebras are all trivial. However, I have not been able to establish the existence, which wil probably straightforwardly show that it's surjective as well. Surjectivity is probably also implied by combining a dimension argument with the fact that $i_1(H_1)=i_2(H_2)$, but I'm not sure one can say anything yet since it's not clear to me that $i_1(H_1)$ and $i_2(H_2)$ are Lie subgroups of $G$.

I have tried to construct the map $\tau:\mathfrak{h}_1\to\mathfrak{h}_2$ using the flow of the left-invariant vector field determined by some $X_1\in\mathfrak{h}_1$, but to no avail. Any help, hints, clarifications are much appreciated. (Doing this exercise as prep for an exam in Lie groups/algebras.)

$\endgroup$
  • $\begingroup$ A Lie group homomorphism is just a smooth morphism ? Or do you require that it be an immersion or something like that ? $\endgroup$ – Max Jun 19 '18 at 8:12
  • $\begingroup$ Lie group morphism = smooth group morphism $\endgroup$ – B. Pasternak Jun 19 '18 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.