Is there a notion of a smallest (immersed) Lie subgroup containing a given set?

Question

Let $G$ be a Lie group, $S \subseteq G$. Is there a smallest Lie subgroup of $G$ containing $S$? (i.e a Lie subgroup $H$ which is contained in every Lie subgroup which contains $S$).

We need to distinguish between two cases:

If we talk about embedded (equivalently closed) Lie subgroups, then the answer is positive: Take $\tilde H=\cap \{H \, \, | \, \, H \text{ is a closed Lie subgroup of $G$ which contains $S$}\}$.

Then $\tilde H$ is a closed subgroup, hence (by the closed subgroup theorem), it is a closed (embedded) Lie subgroup of $G$, so we are done.

What happens if we allow immersed Lie subgroups?

The definition I am using is that $H \le G$ is a Lie subgroup, if it is a subgroup endowed with some topology and smooth structure, making it a Lie group, and an immersed submanifold of $G$.

Now if we try to repeat the "intersection construction" (with immersed subgroups), the result is a group, but it's not clear that it is an immersed submanifold.

Is there in fact an example where the intersection won't be a Lie subgroup?

Answer 1

I am not very familiar with immersed submanifolds, so not entirely sure if this counts as an example or not, but it is too long for a comment.

Let $G$ be the torus $S^1 \times S^1$. If we think about it as $\mathbb{R}^2/\mathbb{Z}^2$ then I think that the subgroup $H$ which is the image of the line $y = rx$ where $r$ is a fixed irrational number, is an immerged Lie subgroup of the torus that is not a Lie subgroup in the ordinary sense: it is dense in $G$ but isomorphic to $\mathbb{R}$ when considered as a Lie group on its own. Is this correct?

Now taking the intersection of this $H$ with another, nicer, 1-dimensional subgroup $K$ (such as $S^1 \times 1$) we find that the intersection is dense and even equidistributed in $K$ (there is a lot of theory about the equidistribution of the sequence $rn \mod 1$ where $r$ is a fixed irrational and $n$ runs through the natural numbers, this should apply here too, I'd say). But as far as I can tell the intersection $H \cap K$ is not an immersed Lie subgroup of $G$ in a meaningful way.

However as I said, I am not feeling too confident about my understanding of immersions, so please share your thoughts on if and why this answer does or does not make sense!

(To be clear: I figured any two immersed subgroups whose intersection is not a submanifold would constitute an example. Taking $S$ in retrospect to be this intersection saves us from the somewhat daunting task of thinking about the intangible set of 'all subgroups containing $S$' for some pre-chosen $S$)

Answer 2

The positive answer to your question is a corollary of the following theorem (which should be somewhere in the literature).

Theorem. The intersection $\tilde{H}$ of any family $\{H_i, i\in I\}$ of immersed Lie subgroups of a Lie group $G$ is again an immersed Lie subgroup.

(My definition of a Lie group is that it is required to have a countable basis of topology. In particular, each Lie group has only countably many connected components.)

Let ${\mathfrak h}_i$ denote the Lie algebra of $H_i$, $i\in I$; each of these is a subalgebra of the Lie algebra ${\mathfrak g}$ of $G$. It is clear that there exists a finite subset $J\subset I$ such that $$ {\mathfrak h}:= \bigcap_{i\in I} {\mathfrak h}_i = \bigcap_{j\in J} {\mathfrak h}_j $$ is a Lie subalgebra of ${\mathfrak g}$. Therefore, without loss of generality, we may assume that all the subgroups $H_i$ have the same Lie algebra ${\mathfrak h}$. It follows that the identity components of all the subgroups $H_i$ are also the same, equal to some immersed Lie subgroup $H_0< G$ (equal to the subgroup generated by $\exp({\mathfrak h})$). Moreover, without loss of generality, we may assume that there exists $i_1\in I$ such that $H_i\subset H_{i_1}$ for all $i\in I$. (Otherwise, pick some $i_0\in I$ and replace each $H_i$ with $H_i\cap H_{i_1}$.) We have $H_i/H_0\cong \Gamma_i$, a countable group, for each $i\in I$. We have natural inclusions $\Gamma_i\to \Gamma_{i_1}$. Since $$ \Gamma:=\bigcap_{i\in I} \Gamma_i $$ is again countable, we conclude that $$ \tilde H/H_0 \cong \Gamma $$ is a countable group. Thus, $$ \tilde{H}:=\bigcap_{i\in I} H_i $$ contains a normal connected subgroup which an immersed Lie subgroup $H_0< G$, such that $\tilde{H}/H_0$ is countable. Hence, $\tilde{H}$ is an immersed Lie subgroup in $G$. qed

Edit. Much of this question stems from the confusing definition of an "immersed Lie subgroup". Let me rewrite your definition in the language that I prefer. First of all, the concept of an "immersed submanifold in $M$" is just a slang for: "an immersion $N\to M$, where $N$ is a (smooth) manifold". (Here I am adopting the viewpoint that an $n$-dimensional manifold is a 2nd countable Hausdorff topological space equipped with a maximal differentiable atlas with values in $R^n$; sometimes, it is convenient to be more flexible.) Now, with this in mind, your definition of "an immersed Lie subgroup $H$ in a Lie group $G$" simply means "a monomorphism $f: H\to G$ in the category of Lie groups," i.e. a monomorphism from a Lie group $H$ to a Lie group $G$. We will see (soon) that one can uniquely recover (up to a natural isomorphism) such a monomorphism from its image.

The following simple lemma relates embedded and immersed subgroups:

Lemma 1. A monomorphism of Lie groups $f: H\to G$ is an embedding if and only if its image is closed.

Proof. This lemma of course, follows from the general (and nontrivial) theorem (due to Elie Cartan) that a subgroup of a Lie group is a Lie subgroup if and only if it is closed, but my goal is to trivialize everything, so I will give a direct and easy proof.

One direction of this lemma is a special case of Arens Lemma in the theory of topological transformation groups and the proof is exactly the same. Suppose $f(H)$ is closed. Since $G$ (being a manifold) is completely metrizable, it follows that $f(H)$ is a Baire space. It suffices to show that $f$ is an open map to its image $E:=f(H)$ (equipped with the subspace topology). Suppose that there exists an an open subset whose image is not open. Since $f$ is a homomorphism, it follows that for every nonempty open subset $U\subset H$ the image is not open in $E$; it then also follows that this image has empty interior in $E$. No, I will use the fact that Lie groups are second countable: I can find a countable open cover $\{U_{\alpha}: \alpha\in A\}$ of $H$ by (nonempty) open subsets. The image of each has empty interior in $E$, hence, since $E$ is a Baire space, their union also has empty interior in $E$. But this is absurd since their union equals $E$.

Now, let's prove the converse: Let $f: H\to G$ be an embedding. If $E=f(H)$ is not closed in $G$, there exists a sequence $h_i\to \infty$ in $H$ (in the one-point compactification $H\cup \{\infty\}$ of $H$) such that $f(h_i)\to g\notin E$. Pick a relatively compact neighborhood of identity $K\subset H$. Then, after extraction, we can assume that $h_{i+1}\notin h_iK$ for every $i$. On the other hand, $f(h_{i+1}h_i^{-1})$ converges to $1$ in $G$. Since $f$ is an embedding, it follows that $h_{i+1}h_i^{-1}\to 1\in H$, which is a contradiction. qed

Next, I will do an exercise that you asked me about in a comment and that Jack Lee finds nontrivial: "The intersection of two immersed Lie subgroups is again an immersed Lie subgroup''. (Same for embedded Lie subgroups, see below.) As I said in the comment, the only tool (besides undergraduate-level general topology) that we will need is the Constant Rank Theorem.

One way to prove this statement about the intersection of two immersed subgroups is to use the notion of the "fiber product'' (also known as the "equalizer'') of two immersed subgroups $f_i: H_i\to G, i=1,2$. This construction is actually quite familiar to every differential geometry/topologist since it appears when one constructs the pull-back of a fiber bundle. The fiber product of two immersed subgroups is defined as $$ H:=H_1\times_G H_2= Eq(f_1,f_2)=\{(x_1,x_2)\in H_1\times H_2: f_1(x_1)=f_2(x_2)\}= F^{-1}(Diag(G\times G))= Ker( \phi: H_1\times H_2\to G), $$ where $F=(f_1,f_2)$ and $\phi:=f_1f_2^{-1}$ is the map (in general, not a homomorphism) sending $(x_1,x_2)$ to $f_1(x_1)f_2(x_2^{-1})$.

The next exercise is to check that the smooth map $\phi: H_1\times H_2\to G$ has constant rank. It follows from transitivity of the action of $H_1\times H_2$ on itself via left multiplication: $$ (h_1,h_2)\cdot (x_1,x_2) = (h_1x_1, h_2 x_2). $$ Since $$ \phi( (h_1,h_2)\cdot (x_1,x_2) )= f_1(h_1) \phi(x_1,x_2) f_2(h_2^{-1}) $$ and the left/right multiplications via $f_1(h_1)$, $ f_2(h_2^{-1})$ on $G$ are diffeomorphisms, it follows that $\phi$ can constant rank. In particular the subgroup $H=\phi^{-1}(1)$ is a (smooth) submanifold of $H_1\times H_2$, hence, a Lie subgroup.

By restricting the the homomorphism $F$ to $H= H_1\times_G H_2$, we obtain a homomorphism $f: H\to G$. I will leave it to you to check that the natural projections $$ p_i: H\to H_i, i=1, 2. $$ are monomorphisms and, hence, the composition $f=f_i\circ p_i$ is also monomorphism (both are general properties of fiber products of monic morphisms). It is also equally easy to check that $$ f(H)= f_1(H_1)\cap f_2(H_2). $$

Hence, the pull-back of two "immersed Lie subgroups" is an "immersed Lie subgroup" and, in your language, equals the intersection of the above subgroups.

As a bonus, let's use this to show that one can uniquely recover an immersed subgroup from its image. Let $f_i: H_i\to G, i=1, 2$ be two monomorphisms with the same image $E$. Then their fiber product $f: H:=H_1\times_G H_2\to G$ also has the image $E$. Then the compositions $$ f_i^{-1}\circ f: H\to H_i $$ is an isomorphism of abstract groups, hence, each morphism $p_i: H\to H_i$ is an isomorphism of abstract groups. Now, we use the general fact (another application of the constant rank theorem) that a bijective morphism of Lie groups is an isomorphism of Lie groups. Hence, both $p_i: H\to H_i$ are isomorphism which concludes the proof.

Now, let's deal with the intersection of Lie subgroups.

Lemma 2. The intersection of two (embedded) Lie subgroups is again a Lie subgroup.

Proof. Let $f_i: H_i\to G$ be two Lie embeddings. Then the fiber product $$ F: H:=H_1\times_G H_2\to G $$ is a monomorphism whose image equals $f_1(H_1)\cap f_2(H_2)$. Since both $f_i(H_i)$ are closed, their intersection is also closed, hence, by Lemma 1, $F: H\to G$ is an embedding. This, its image, is a Lie subgroup. qed

One can use the fiber product construction (called in general a "colimit") to handle intersections of arbitrary collections of "immersed Lie subgroups". The trouble is that an arbitrary product of Lie groups is not a Lie group. One way to handle this is to enlarge the category by allowing infinite dimensional Lie groups modelled on certain locally convex topological vector spaces (including direct products of finite dimensional vector spaces). I might add an argument along these lines later on. What I wrote initially was a "more pedestrian approach" to the intersection problem which avoids dealing with infinite dimensional Lie groups.