Lie subgroups are certainly not always embedded (there is the example of the $\mathbb{R} \to S^1 \times S^1$ given by a line of irrational slope).

Can you have a torus that is a subgroup of a Lie group, but not embedded?

To me it seems like the image of a compact set is compact and hence closed (since manifolds are Hausdorff) if the inclusion is continuous. So we want a torus subgroup included in a Lie group in a non-continuous way.

I really have no idea how to come up with such an example.

  • $\begingroup$ $S^1$ is isomorphic to the product of $\mathbb Q/\mathbb Z$ and a direct sum of continuum many copies of $\mathbb Q$. It's easy to embed this in $S^1\times S^1$ (for instance) discontinuously. $\endgroup$ – Wojowu Apr 22 '19 at 20:30
  • 1
    $\begingroup$ Lie groups are groups + (manifold) topology. If you are willing to ignore the topology aspect, you might as well ignore the algebraic aspect ... $\endgroup$ – Hagen von Eitzen Apr 22 '19 at 20:32
  • $\begingroup$ @HagenvonEitzen I do not see why. Certainly not every group is algebraically isomorphic to a torus, even ignoring topology. And it makes sense to ask if such groups can be (algebraically) subgroups of a Lie group. Non-closed subgroups of Lie groups are, in fact, studied, see Virgos's paper and related thread Non-closed subgroups of Lie groups. $\endgroup$ – Conifold Apr 22 '19 at 21:15

Let's consider the torus $T=S^1$. As an abstract group, it is isomorphic to $$ {\mathbb Q}/{\mathbb Z} \times \bigoplus_{t\in {\mathbb R}} {\mathbb Q}. $$ Therefore, it contains a proper subgroup
$$ T'= {\mathbb Q}/{\mathbb Z} \times \bigoplus_{t\in {\mathbb R} -\{0\} } {\mathbb Q}. $$ Clearly, $T'$ is isomorphic to $T$ as an abstract group. However, if we equip $T$ with the standard topology of $S^1$ (making it a Lie group) then $T'$ cannot be a Lie subgroup of $T$. Thus, you get an example of a subgroup of a Lie group which is isomorphic to a torus (as an abstract group) but is not a Lie subgroup.


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.