0
$\begingroup$

I am just a new student in this field. My question is from the following statement:

"Let $G$ be a Lie group and $H$ be its subgroup, if $H$ is a smooth manifold, then $H$ is a Lie group."

My question is under such condition, is it possible that $H$ is not a smooth manifold? (I mean if there is no condition "if $H$ is a smooth manifold")

My question lies in since $G$ is a Lie group, $G$ is a smooth manifold. So all transition functions are differentiable. So for $H$.

I am not quite sure about this. Could someone point something wrong out?

$\endgroup$
  • 2
    $\begingroup$ It depends your definition. For example, consider $G = T^2$ and $H \cong S^1$ a torus with irrationnal slope. It will be dense in $G$ and I am not sure you want to call it a submanifold since it is not closed. $\endgroup$ – user171326 Mar 30 '17 at 6:16
  • $\begingroup$ I am not sure if I can say for example if $H$ is closed, then its all charts cannot cover $H$, then $H$ may not be locally Euclidean so $H$ is not a topological manifold. $\endgroup$ – sleeve chen Mar 30 '17 at 6:32
  • $\begingroup$ $H$ is abstractly a manifold since it is isomorphic to $\mathbb R$ (and NOT $\mathbb S^1$ as I wrote), but it is not a submanifold since the intersection of $H$ with any open in $G$ will have infinitely many component. $\endgroup$ – user171326 Mar 30 '17 at 6:36
  • 1
    $\begingroup$ Is your question whether all subgroups of a Lie groups are manifolds (the answer is no) or whether a subgroup that is a manifold is automatically a smooth manifold? In the latter case you probably need some relation between the manifold structures of $G$ and $H$ or else there is not much reason to suspect this to be true. $\endgroup$ – Marc van Leeuwen Mar 30 '17 at 6:41
1
$\begingroup$

Suppose $G=\Bbb{R}$ (under addition) and $H=\Bbb{Q}$. Then $H$ is a subgroup of $G$ which is definitely not a Lie group. In general just being a subgroup doesn't tell you anything about the topology of $H$ (e.g., it doesn't guarantee that it's even a topological manifold).

$\endgroup$
  • $\begingroup$ Actually, $\mathbb Q$ is an immersed Lie subgroup with the discrete topology. See my answer below. $\endgroup$ – Jack Lee Mar 30 '17 at 18:20
2
$\begingroup$

There are several important points to be made here.

First, the usual definition of a Lie subgroup of a Lie group $G$ is a subgroup $H\subseteq G$ that has some manifold topology and smooth structure under which it is a Lie group and an immersed submanifold of $G$. Thus a Lie subgroup need not be an embedded submanifold, and need not have the subspace topology. One important reason to allow non-embedded Lie subgroups is so that every Lie subalgebra of $\operatorname{Lie}(G)$ will correspond to a connected Lie subgroup of $G$.

Two simple examples of nonembedded Lie subgroups were mentioned in previous comments and answers: (1) any irrational $1$-dimensional subgroup of the torus, which is a dense Lie subgroup, and (2) $\mathbb Q\subseteq \mathbb R$, which is a zero-dimensional Lie subgroup with the discrete topology. In fact, any countable subgroup of a Lie group is automatically a Lie subgroup with the discrete topology.

Second, an extremely important theorem in Lie theory is the closed subgroup theorem, which says that any subgroup (in the algebraic sense) of a Lie group that is also a closed subset is automatically an embedded Lie subgroup. You can find a proof, for example, in my Introduction to Smooth Manifolds (2nd ed.), Theorem 20.12.

Third, there are examples of (necessarily non-closed and uncountable) subgroups that are not Lie groups with any smooth manifold structure. One class of examples is provided by uncountable proper additive subgroups of $\mathbb R$. (This MSE answer lists a number of ways to construct such a subgroup.) If $H\subseteq\mathbb R$ is such a subgroup, it cannot be a zero-dimensional Lie group because it's uncountable; it cannot be a $1$-dimensional Lie group because $\mathbb R$ itself is the only such subgroup; and it cannot be a Lie group of dimension greater than $1$ because in that case the inclusion map could not be an immersion.

$\endgroup$

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.