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.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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. $\square$

  • $\begingroup$ Thank you Prof. Lee. Espescially for the sketch of the proof. $\endgroup$ – kelvinn aja Dec 18 '17 at 1:42
  • $\begingroup$ 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. $\endgroup$ – Moishe Kohan Dec 19 '17 at 15:17
  • $\begingroup$ @MoisheCohen: That's a very nice argument. Much simpler. I'll edit my answer to refer to yours. $\endgroup$ – Jack Lee Dec 19 '17 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.