Why a submanifold is an open subset of its closure?

Question

When I am reading Lie Groups and Lie Algebras I by Onishchik, I come across the claim that "As any submanifold, a Lie subgroup is an open subset of its closure." From this the author deduces that all cosets of the subgroup is open. Why is the quoted claim true? Also, why does a Lie subgroup being open in its closure imply a Lie subgroup is open in the entire Lie group?

Answer 1

I may misunderstand something but from my perspective it looks like actually the opposite is true. Even in more general, topological case. If $H$ is a subgroup of $G$ and $H$ is not closed then it is also not open in $\overline{H}$. In particular the statement is only true when $H$ is closed to begin with, which makes it trivially true.

Lemma 1. If $H$ is a subgroup of a topological group $G$ with a nonempty interior then $H$ is clopen.

Proof. Let $U\subseteq H$ be a nonempty open subset. Given $h\in U$ we get that $e\in h^{-1}U$ and still $h^{-1}U\subseteq H$. And so WLOG we may assume that $e\in U$. But then $H=\bigcup_{h\in H}hU$ making $H$ open. On the other hand $G-H=\bigcup_{h\in G-H}hU$ (because cosets partition $G$) and so $G-H$ is open as well making $H$ closed. $\Box$

Lemma 2. If $H$ is a subgroup of a topological group $G$ then $\overline{H}$ is a subgroup as well.

Proof. Let $a,b\in\overline{H}$. Let $W$ be a neighbourhood of $ab^{-1}$. Consider $f:G\times G\to G$, $f(x,y)=xy^{-1}$. By the continuity of $f$ and the definition of product topology there is open neighbourhood $U$ of $a$ and open neighbourhood $V$ of $b$ such that $U\times V\subseteq f^{-1}(W)$. But then $U\cap H\neq\emptyset$ and $V\cap H\neq \emptyset$ since $a,b\in\overline{H}$. Let $x\in U\cap H$ and $y\in V\cap H$. It follows that $xy^{-1}\in W\cap H$, and so $W\cap H\neq\emptyset$. By the arbitrary choice of $W$ we get that $ab^{-1}\in\overline{H}$. $\Box$

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Both lemmas applied imply that if a subgroup $H$ is open in $\overline{H}$ then $H$ is also closed and thus $H=\overline{H}$. In particular if $H$ is not closed then it is never open in $\overline{H}$.

Answer 2

Let $X$ be a topological space. A set $A\subset X$ is open in its closure if and only if for every $a\in A$ there is a neighbourhood $U$ of $a$ in $X$ such that the intersection $A\cap U$ is closed in $U$. I know this property by the name "locally closed".

We will need the reverse implication, which for completeness we can prove like this. For $a\in A$, pick a neighbourhood $U$ of $a$ in $X$ as in the definition. Then $A\cap U$ is actually open in $\bar A$ - indeed, $A\cap U$ is equal to its closure in $U$, which is $\bar A \cap U$ (since any closed set in $U$ is of the form $F\cap U$ for $F$ closed in $X$). This proves that a neighbourhood of $a$ in $\bar A$ is completely contained in $A$, as needed.

Consider now a smooth manifold $M$ and a(n embedded) submanifold $S$. Then every $s\in S$ has a coordinate neighbourhood $U$ with the property that $S\cap U$ is given by the vanishing of a subset of the coordinates - in particular, $S\cap U$ is closed in $U$. This means that $S$ is locally closed, and so it's open in its closure.

Finally, let's consider the case of a Lie group $G$ and a Lie subgroup $H$. The upshot of the paragraph cited in the question is that $H$ must be closed in $G$. We verify this by showing $H = \bar H$. As $H$ is a submanifold of $G$, it is open in $\bar H$. On the other hand, $\bar H$ is itself a subgroup of $G$. Indeed, if $(x_i)$ and $(y_i)$ are sequences in $H$ convergent to $x,y$ respectively, then continuity of multiplication implies that $(x_i y_i)$ converges to $xy$, which is therefore in $\bar H$. Similar considerations hold for the inverse.

This means we have that $H$ is an open subgroup of $\bar H$. Since translation by an element is a homeomorphism, each coset of $H$ is also open in $\bar H$. Since $H$ is the complement in $\bar H$ of its nontrivial cosets, it is closed. But this means that $H = \bar H$, as claimed.