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Let $G$ and $H$ be Lie groups and $i:H\to G$ an injective group homomorphism, not necessarily continuous.

Once $i$ is continuous, it follows easily that $i$ is differentiable and even an immersion. So $H$ is a Lie subgroup of $G$ (It is not guaranteed that $H$ is a submanifold of $G$, which is only the case when $i(H)$ is closed in $G$).

However, it is even possible that $i:H\to G$ is not continous?

Klaas
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  • For these and similar questions I recommend the book of Warner (older edition also OK) http://books.google.com/books/?id=iaeUqc2yQVQC – orangeskid Jan 16 '15 at 16:11

2 Answers2

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It's possible, but the examples are weird.

Consider $H = \mathbb{R}^n$ and $G = \mathbb{R}^m$ with $n\neq m$. If we view both as vector spaces over $\mathbb{Q}$, then we see they are both vector spaces of dimension $|\mathbb{R}|$. It follows that they are isomorphic as $\mathbb{Q}$ vector spaces.

In particular, there is a bijective $\mathbb{Q}$-linear map $i:H\rightarrow G$ with $i(x+y) = i(x) + i(y)$.

Said another way, $i$ is a group homomorphism! It turns out that $i$ is not even measurable, let alone continuous. But, $i(H) = G$, so the image is a Lie subgroup. If you insist on $i(H)$ being a proper subgroup of $G$, embed $G$ into $\mathbb{R}^{m+1}$ in the usual way, then $i:H\rightarrow \mathbb{R}^{m+1}$ has image $\mathbb{R}^m$ embedded in the usual way, but is horribly discontinuous.

On the other hand, if $H$ is compact and semisimple (equivalently, if $H$ is compact and so is the universal cover of $H$) and if $G$ is compact, then van der Waerden proved that any abstract homomorphism $i:H\rightarrow G$ is automatically continuous (hence, smooth).

The reference is

B. L. van der Waerden, “Stetigkeitssätze für halbeinfache Liesche Gruppen,” Math. Z. 36, 780–786 (1933).

Jason DeVito - on hiatus
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  • This is great. Thank you! – Klaas Jan 16 '15 at 16:49
  • Glad you like it! Also, as in the second answer here: http://math.stackexchange.com/questions/565552/group-isomorphism-mathbbr-mathbbz-cong-s1 one can see that, as a group, $S^1$ is isomorphic to $S^1\times \mathbb{R}$. Then one can use the argument in my answer to create discontinuous functions with domain $S^1$. This shows that "semisimple" part of van der Waerden's theorem is necessary. – Jason DeVito - on hiatus Jan 16 '15 at 17:21
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It may seem like a silly example, but this kind is the best one can get.

Take $G=\mathbb{R}$ with the discrete topology. It is a Lie group of dimension $0$ with uncountably many connected components.

Take $H=\mathbb{R}$ with the usual structure of Lie group.

The morphism $H\to G$ is not continuous.

orangeskid
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  • I thought of this one, but $\mathbb R$ with the discrete topology is not second-countable and hence not a Lie group. Or am I missing something? – Klaas Jan 16 '15 at 16:32
  • @user1135859: Various authors still consider an uncountable discrete topological space a $0$-dimensional manifold. – orangeskid Jan 16 '15 at 16:51
  • OK, I wasn't aware of that. But Jason DeVito provided an example where the Lie groups are even second-countable: http://math.stackexchange.com/a/1106867/74153 – Klaas Jan 16 '15 at 16:54
  • @user1135859: OK, I see that. All right, I may have to take back the :"best you can" statement. – orangeskid Jan 16 '15 at 17:00
  • @user1135859: OK, I think I know what all that second countability is all about. It's about Lie semigroups. Given an abstract subgroup of $G$ there is at most one way to make it into a Lie semigroup with a countable basis. With an uncountable basis, there could be several. And the counterexample with standard and discrete $\mathbb{R}$ is just reversed. – orangeskid Jan 16 '15 at 17:04