# Can two Lie groups be "almost" homeomorphic?

I have two different Lie groups, one is an arcwise connected Lie subgroup

$$G_1 \subseteq \mathbb{R}^m \times (S^1)^n$$

and the other is

$$G_2 = \mathbb{R}^k \times (S^1)^l$$

While $$G_1$$, $$m$$ and $$n$$ are fixed, I can change $$G_2$$ by varying $$k$$ and $$l$$ as I please. I'm interested in continuous

$$f : G_1 \to G_2$$

$$g : G_2 \to G_1$$

and was wondering if anything at all can be said about $$f$$ or $$g$$? Ideally, I would like to find any condition for $$l$$ and $$k$$, so that $$f$$ or $$g$$ become "almost" homeomorphic or that both are surjective. By "almost" I'm trying to express that there exists an $$f$$ and $$g$$ for which

$$(g \circ f)(p) = p$$

is true except for a tiny finite subset. Is that possible?

• Conditions on $f,g$ --- any function? smooth map? Abstract group homomorphisms? Lie group homomorphisms? ... Jun 3, 2019 at 10:53
• I've added the condition that $f,g$ are continuous.
– Mike
Jun 3, 2019 at 10:58
• There is a continuous surjection $G_1\to G_2$ regardless of what $m, n, k,l$ are as long as at least $m\geq 1$. It is a variation of a space filling curve. Now for the stronger condition $g\circ f=id$ it means that $G_1$ can be embedded as a retract of $G_2$. And I think that this cannot be done when $n>l$ or $m>k$ by careful homology analysis and/or invariance of domain. BTW how is that question related to the Lie theory? How is the group/differential structure of $\mathbb{R}^n$, $S^1$ relevant here? Are you assuming that $f,g$ are Lie homomorphisms (which makes that really simple)? Jun 3, 2019 at 13:18
• Thank you for pointing out space filling curves, hadn't thought of those yet! I was hoping that restricting $G_1$ to be a Lie subgroup would make things easier. In general I am interested in all connected differentiable submanifolds of $\mathbb{R}^m \times (S^1)^n$.
– Mike
Jun 3, 2019 at 16:23