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?

  • 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
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    $\begingroup$ 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

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).

  • $\begingroup$ 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

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.


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