Open Mapping Theorem for Sufficiently Nice Lie Groups

Today, I was in class, and was given a question of the form “show that h:G \to H cannot be surjective, and I asked my professor, “can we assume some basic functional analysis? The open mapping theorem implies that this is is open, and hence a homeomorphism, which it is not.” A fellow student remarked that manifolds aren’t Linear, which I accidentally misheard as “please quit mathematics immediately.” Luckily, I am very bull-headed and I’ve spent the day reformulating the proof for the open mapping theorem in Banach spaces to Lie groups. I think it works–  I was unable to find any such result in my books on Lie theory, and I didn’t want to google it initially (we live with our contradictions), but after proving the lemma, I looked it up just in case and posted this question  on SE. I guess we will just have to see.

Here, let’s just assume that G is a Lie group embedded in \mathbb R^n with all the good stuff (connected, smooth, without boundary, etc.)

I will make use of Baire Category and this nice idea from Fulton and Harris:

Nice Idea: Let U \subseteq \mathbb G \subseteq \mathbb R^n be any neighborhood of the identity. Then U “generates” G.

This essentially follows by showing that A:=\{g_1 \cdot\dots\cdotg_n \mid g_1, \dots,g_n \in U\} is clopen. First, we note that for any x \in A, xU \in A, so it is open, and it is closed since its complement is open as well: x \notin A \implies xU \notin A.

 

This is really the result that gives me hope for such a theorem to exist, because it relates the topological structure of a small ball around the origin to some algebraic notion of “generates.” From this, I hope that a theorem very much analogous to the proof for the open mapping theorem in functional analysis will work.

We will also make use of the following facts about \mathbb R^n:

Quick Fact: G is second-countable, and a Baire space. Without jargon: the topology on G has a countable basis (so every cover admits countable subcover), and G is not the union of countably many closed sets with empty interior.

In the case that this method works, and you care about such things, one could probably generalize the following theorem mildly, but this blogger already has a mental cramp, so we give our main lemma:

 

Lemma 1: Let G,H be Lie groups of arbitrary dimension. 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 a neighborhood with nonempty interior about 1 \in H.

 

As expected, pick some neighborhood U of 1 \in G.  We want to pick some open ball Vin U so that the closure of V is compact and contained in U. Remember, we want to make use of Baire category, which is no good for open sets. By our nice idea,

    \[G=\{x V \mid x \in G\}\]

but we can choose countably many such x \in G, by our quick fact.  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. Hence, \phi(U) has nonempty interior.

 

 

This implies the theorem, since once we have this, we may pick some symmetric subset V of any open neighborhood so that V^2 \subsetneq U. Since \phi(V) has nonempty interior, we determine that for any f(w) \in \mathrm{Int}( \phi(V)). Noting that w^{-1} V contains the identity of G so that 1_{G} \in w^{-1}V \subset V^2, we can do the following finagling:

    \[1_{H}=f(1_G)=f(w w^{-1})=f(w_)f(w^{-1}) \in \mathrm{Int}(f(w^{-1}V))\]

meaning that we can find a suitable basis \mathcal U so that for each U \in \mathcal U, f(U) contains the identity for H. From this, we use the homomorphism property to conclude that for each open W \subset G, we have that f(w) \in \mathrm{Int} f(W).

 

see for this, for example.