# Inclusion map between Lie groups $H \hookrightarrow G$ is smooth

I am trying to understand some basic stuff about Manifolds and Lie Groups.

Suppose $$H$$ is a Lie group which is also a subgroup of a Lie Group $$G$$. Suppose the inclusion map is smooth. Then $$H$$ is a Lie Subgroup of $$G$$.

I am not really sure if this result is true. I am trying to give the following argument.

Since the inclusion map is smooth, we can take the differential. The differential map must be an inclusion map, hence injective. Thus the original map is an immersion. Thus $$H$$ is an immersed submanifold of $$G$$. Hence its a Lie subgroup.

Is it correct?

• What is your definition of the inclusion map being "smooth"? It's not at all obvious what this should mean, given that $H$ is not assumed to be a manifold... – Eric Wofsey Sep 22 '18 at 17:53
• Or do you mean to assume $H$ is a Lie group too, so your question is really whether a smooth injective homomorphism between Lie groups must be an immersion? – Eric Wofsey Sep 22 '18 at 17:56
• @EricWofsey Yes $H$ is Lie group too. Actually I was trying to show $H/N$ is a Lie subgroup of $G/N$ given that $N\subset H\subset G$ where $N$ is normal and $N,G,H$ are Lie subgroups. I know $H/N$ and $G/N$ are Lie groups. Using quotient manifold theorem, I was able to show the inclusion map $H/N \to G/N$ is smooth. I am following Lee's book on Intro to Smooth Manifolds. – Sayan Sep 22 '18 at 18:15

## 1 Answer

Your argument is essentially correct, but the step where you conclude that the differential is injective requires justification. It is not true that a the differential of a smooth injection between manifolds must be injective everywhere, so you will need to use something special about Lie groups.

Here's one way to do it. Let $$i:H\to G$$ be the inclusion map. The claim is that $$di_e:T_eH\to T_eG$$ is injective. To prove this, observe that since $$i$$ is a homomorphism, $$di$$ has the same rank at every point ($$di$$ at any point can be obtained from $$di_e$$ by composing with the differentials of translations, and translations are diffeomorphisms). So, by the constant rank theorem, if $$di_e$$ were not injective, then $$i$$ would not be injective.

(Or, you can avoid invoking the constant rank theorem by observing that the exponential map gives explicit local coordinates at the identity in which $$i$$ looks linear.)