# Intersection of two Lie subgroup is Lie subgroup ?

This question is already asked in here but i can't find a satisfactory answer.

The question (in the title) arise from the following definition (i'm using Lee's smooth manifold p.156) : If $G$ is a Lie group and $S \subseteq G$, the $\textbf{subgroup generated by } S$ is the smallest subgroup containing $S$ (i.e., the intersection of all subgroup containing $S$).

The definition above implicitly assume that the intersection of any two Lie subgroup $S_1,S_2 \subset G$ is again Lie subgroup. I find it difficult to prove this. I can see that the intersection has the group property but i have no idea how to show that $S_1 \cap S_2$ is an immersed submanifold of $G$.

The answer by @Moishe Cohen in the given link above is using argument involving Lie algebra. But since the definition in Lee's book is given before he define Lie algebra, i assume that this problem can be solved without it (probably).

Can anyone help me with this ? Thank you.

• Lee is not claiming that the subgroup generated by $S$ is a Lie group. This is irrelevant to his discussion; we just need to know that it is a group. – Spenser Dec 17 '17 at 16:43

There are two issues here:

First, the definition of the "subgroup generated by a set $S$" has nothing to do with Lie groups -- it's purely a group-theoretic concept. So for a general set $S$, there's no assumption, implicit or explicit, that the subgroup generated by $S$ is a Lie subgroup. (And it might not be -- for example, there are dense uncountable subgroups of $\mathbb R$, which cannot be given any topology or smooth structure making them into immersed Lie subgroups, and we can take $S$ to be such a subgroup.)

Second, independently of that, it is true that the intersection of two Lie subgroups is again a Lie subgroup.

Theorem. Suppose $G$ is a Lie group, and $H_1,H_2$ are Lie subgroups of $G$. Let $H$ be the subgroup $H_1\cap H_2$. Then $H$ has a unique topology and smooth structure making it into a Lie subgroup of $G$.

EDIT: In Moishe Cohen's answer to the question you cited, he originally just stated this as a simple exercise, but in reply to my laborious argument below, he's now added a simple proof. The idea is to view the two Lie subgroups $H_1$ and $H_2$ as injective Lie homomorphisms $f_i\colon H_i\to G$, and define $H$ as the subgroup $\{(x_1,x_2)\in H_1\times H_2: f_1(x_1) = f_2(x_2)\}$ of $H_1 \times H_2$. The equivariant rank theorem shows that $H$ is an embedded Lie subgroup of $H_1\times H_2$, and for either $i=1$ or $i=2$, the following composition given an injective Lie homomorphism of $H$ into $G$: \begin{equation*} H\hookrightarrow H_1\times H_2 \overset{p_i} {\to} H_i \overset{f_i}{\to} G. \end{equation*}

I'll leave my much more laborious argument here, in case anyone's interested.

My proof:

I don't know of any proof that doesn't rely on some nontrivial facts about Lie algebras, exponential maps, and foliations. Here's a quick sketch of a proof. Can't guarantee that I haven't missed some details, but this general idea should work. (Note that I'm using the definitions from my Intro to Smooth Manifolds book -- in particular, smooth manifolds are second-countable and therefore have only countably many components, and a Lie subgroup is a subgroup endowed with a topology and smooth structure making it into a Lie group and an immersed, not necessarily embedded, submanifold.)

Proof: Let's denote the Lie alebras of $G$, $H_1$, and $H_2$ by $\mathfrak g$, $\mathfrak h_1$, and $\mathfrak h_2$, respectively. Since $\mathfrak h_1$ and $\mathfrak h_2$ are canonically identified with Lie subalgebras of $\mathfrak g,$ the set $\mathfrak h = \mathfrak h_1\cap \mathfrak h_2$ is a Lie subalgebra of $\mathfrak g$ too. Thus there is a unique connected Lie subgroup $H_0$ of $G$ whose Lie algebra is $\mathfrak h$. This means $H_0$ has a topology and smooth structure making it into an immersed smooth submanifold of $G$, and the group operations on $H_0$ are smooth with respect to this structure. If $V\subset\mathfrak g$ is a neighborhood of $0$ on which the exponential map of $G$ is a diffeomorphism, $H_0$ is generated (in the group-theoretic sense) by $\exp(V\cap\mathfrak h)$, where $\exp$ denotes the exponential map of $G$. Since $V\cap \mathfrak h\subset \mathfrak h_1\cap \mathfrak h_2$, it follows that $H_0\subset H_1\cap H_2$.

Since $H_0$ is a subgroup (in the algebraic sense) of $H$, it follows that $H$ is the disjoint union of the left cosets of $H_0$ in $H$. We need to verify that there are only countably many such cosets. I think you can prove this based on the fact that $H_1$ and $H_2$ are integral manifolds of left-invariant foliations of $G$; if we choose a flat chart for $H_1$ on some open subset $W\subseteq G$, then $H_1\cap W$ is a union of countably many disjoint slices; then we can take a connected neighborhood $Y$ of the identity in $H_2$ that is embedded in $W$, and $Y\cap H_1$ will consist of countably many connected embedded submanifolds. I haven't worked out the details.

For each $h\in H$, the map $L_h$ (left multiplication by $h$) is a diffeomorphism of $G$ that takes $H_0$ bijectively onto $hH_0$. Thus we can define a smooth manifold structure on $H$ by declaring each such bijection $H_0 \to hH_0$ to be a diffeomorphism, and viewing $H$ as the topological disjoint union of these cosets. (That is, we declare each coset to be open and closed in $H$.)

We already verified that the group operations are smooth on $H_0$. Given any two points $h_1,h_2\in H$, we can choose connected neighborhoods $U_1$ of $h_1$ and $U_2$ of $h_2$ in $H$, and then the multiplication map $m|_{U_1\times U_2}\colon U_1\times U_2\to H$ can be viewed as the following composition: \begin{equation*} m|_{U_1\times U_2} = L_{h_1}\circ R_{h_2} \circ (m|_{H_0\times H_0}) \circ ( L_{h_1^{-1}} \times R_{h_2^{-1}}). \end{equation*} It follows that the multiplication on $H$ is smooth. (Here you have to use the fact that $H$ is an integral manifold of the left-invariant foliation determined by $\mathfrak h$, and therefore it's "weakly embedded," meaning that a smooth map into $G$ that takes its values in $H$ is also smooth into $H$.) A similar argument applies to inversion. The inclusion map $i_H\colon H\hookrightarrow G$ is a smooth immersion, because on each component it can be written as a composition of the form $L_h \circ i_{H_0}\circ L_{h^{-1}}$.

Finally, uniqueness of the topology and smooth structure are left as an exercise. $\square$

• Thank you Prof. Lee. Espescially for the sketch of the proof. – kelvinn aja Dec 18 '17 at 1:42
• Dear Jack: I maintain that there is nothing difficult about this all what is needed is the constant rank theorem (and familiarity with the notion of a fiber product). Take a look at the edit of my answer. – Moishe Kohan Dec 19 '17 at 15:17
• @MoisheCohen: That's a very nice argument. Much simpler. I'll edit my answer to refer to yours. – Jack Lee Dec 19 '17 at 20:17