# Attempted proof of an open mapping theorem for Lie groups

The Classical open mapping theorem for Banach spaces tells that if $T:X \to Y$ is a continuous surjective linear map, then it is open.

I have attempted to essentially "adapt" the proof for Lie groups:

Let $G,H$ be connected Lie groups (embedded in $\mathbb R^n$, I've used second-countability and completeness so far). If $\phi: G \to H$ is a surjective Lie group homomorphism, then the image of an open neighborhood $U$ of the identity in $G$ is again a neighborhood with nonempty interior about $1 \in H$.

Pick some neighborhood $U$ of $1 \in G$. We want to pick some open ball $V$in $U$ so that the closure of $V$ is compact and contained in $U$. Hence,

$$G=\{x V \mid x \in G\}$$

but we can choose countably many such $x \in G$. We can call this collection $\{x_n\}$, and consider the image of $G$ under $\phi$, which is all of $H$. In particular, $H:=\{\phi(x_n \overline{V}) \mid n \in \mathbb N\}$. But since $\phi$ is a homomorphism, this is the same thing as considering a whole bunch of $\phi(x_n)\phi(\overline{V})$, whose images are compact and hence closed. By Baire category, one of these guys has empty interior, say $\phi(x_n)\phi(\overline{V})$. But then $\phi(\overline{V})$ has nonempty interior, since multiplication by an element is a homeomorphism. But then $\phi(U)$ has nonempty interior.

From here, one can finish by first showing that this implies that there is a basis $\mathcal U$ about the origin so that the image of each element contains the identity in $H$, which is sufficient to show the main theorem, since we can then use the homomorphism property to show that for every open set $W$ about $w \in G$, the image $f(w) \in \mathrm{Int}(W)$, proving the theorem.

1. Is My proof correct?

2. I've seen one reference "An Open Mapping Theorem for Topological Groups". but this ultimately just redirects to a source that I cannot find. Is there a better way to show this Theorem? I'm not looking for maximal generality (weakest hypotheses) just yet, only to understand why this theorem might be true.

A sufficient answer in my eyes is either a convincing argument for why my proof fails (including something in the way of "is this idea recoverable?") or some affirmation that it is indeed correct.

• Maybe it's just too late at night for me, but what's left to finish? It appears you've shown that $\phi(\overline{V})$ has nonempty interior, hence so does $\phi(U)$, as desired. – Nate Eldredge Apr 11 '17 at 6:04
• @NateEldredge you're right. I'm just nervous that this proof is shady somewhere, I should have phrased it differently-- I assumed it was just incomplete. – Andres Mejia Apr 11 '17 at 6:11

When you have an action of a topological group $K$ over a space $X$, the quotient $X \to X/K$ is an open map. This is very easy to prove.

Now let $K$ be the kernel of your surjective map $\phi \colon G \to H$. The group $K$ acts on $G$ by multiplication on the right and the quotient is $G/K$. The map $\phi$ factors through a group homomorphism $\bar{\phi} \colon G/K \to H$ which is bijective and continuous. If this map is open (hence an isomorphism of topological groups), then $\phi$ is open. So another way of looking at your question is, does the first isomorphism theorem hold for connected Lie groups?

The answer is yes and one proof goes along the following lines:

1) The quotient $G \to G/K$ is a submersion.

2) If $\phi$ is smooth, so is $\bar{\phi}$.

3) An injective homomorphism of Lie groups must be an immersion.

4) The map $\bar{\phi}$ is a submersion and an immersion, so it is a local diffeomorphism, hence a diffeomorphism and so an isomorphism of topological groups.

This is essentially the proof of Theorem 17.4 in

https://www.staff.science.uu.nl/~ban00101/lie2016/lie2010.pdf

I could not see any flaw in your proof. I am not sure if you find that this way is better, but I found more intuitive to think about it in terms of the first isomorphism theorem (which does not hold in general for general topological groups).

• Oh, this is really nice. I like the direction of this proof as well. The reason I did my proof in this convoluted way is that it essentially mimics the proof from functional analysis. I also think that it works for a large class of topological groups (complete metric, second-countable.) It is refreshing to see a different perspective, however. – Andres Mejia Apr 14 '17 at 6:39
• Interesting to see that this relies on the Baire Category theorem as well (for locally compact Hausdorff spaces), via proposition 16.6. – Bart Michels Apr 27 '17 at 11:04
• Yes, although for smooth manifolds you could also prove it using Sard's theorem instead of Baire category. – Goa'uld Apr 28 '17 at 13:39