# Are there non-embedded tori in Lie groups?

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.

• $S^1$ 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. – Wojowu Apr 22 '19 at 20:30
• Lie groups are groups + (manifold) topology. If you are willing to ignore the topology aspect, you might as well ignore the algebraic aspect ... – Hagen von Eitzen Apr 22 '19 at 20:32
• @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. – Conifold Apr 22 '19 at 21:15

## 1 Answer

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.