The list of connected subgroups of $SL(2,\mathbb{R})$

Question

So, the question is to find all connected subgroups of $SL(2,\mathbb{R})$. I understand, how to find all closed connected subgroups: they are in one-to-one correspondence with Lie subalgebras of $\mathfrak{sl}(2,\mathbb{R})$. Up to conjugation, they are: orthogonal, diagonal, upper triangular and upper unitriangular matrices. I also know, how to find all path-connected subgroups: the list is the same as in the case of closed subgroups. But I don't know, whether or not it exhausts all connected subgroups.

EDIT: To clarify, the question can be formulated, as whether it is true, that for subgroups of $SL(2,\mathbb{R})$ connectedness and path-connectedness is the same property.

Have you already checked this ? https://math.stackexchange.com/questions/646183/list-of-connected-lie-subgroups-of-mathrmsl2-mathbbc?rq=1 It can give some insight — julio_es_sui_glace, Jun 23 '23 at 19:28
Yes, but I think in that question they only consider closed connected subgroups. And I'm interested in all connected subgroups — NicStr, Jun 23 '23 at 19:34
Sometimes people say "connected subgroups" and mean "connected Lie subgroups". See for example this post. Why are you interested in the non-closed subgroups? — Dietrich Burde, Jun 23 '23 at 19:51
@DietrichBurde, No, I mean precisely connected subgroups, not connected Lie subgroups. That's the problem I'm trying to solve. — NicStr, Jun 23 '23 at 20:08

Dietrich Burde · Accepted Answer · 2023-06-24T11:55:46.307

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Your question is related to the existence of connected but not path-connected subgroups of $SL_2(\Bbb R)$. I am not sure that there exists a list of all connected subgroups of $SL_2(\Bbb R)$. But it seems that one example showing that connectedness and path-connectedness is not the same property for subgroups of Lie groups is enough (see your edit). There is already a long discussion about the construction of connected, not path-connected subgroups of Lie groups here, with links to MO-posts. The construction is given for the $2$-torus. A similar construction could work for $SL_2(\Bbb R)$.

The answer given here mentiones a Theorem by Thomas:

Theorem (Thomas): Every Lie group of dimension greater than one always admits a connected subgroup which is not a Lie subgroup (and hence isn't path connected, since connected Lie subgroups are path-connected).

However, it is not clear if the proof is entirely correct. Moreover, his definition of a Lie subgroup is different from other definitions (Thomas calls it an "analytic" subgroup).

edited Jun 24 '23 at 11:55

answered Jun 24 '23 at 09:42

Dietrich Burde

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It is not true, that path-connected subgroups of Lie group are closed, as shown by the example of irrational winding of a 2-torus. It is not closed because the closure of this irrational winding is the whole torus. – NicStr Jun 24 '23 at 10:31
I see. But at least the result of Yamabe (1950) states that every arcwise connected subgroup of a real Lie group is Lie subgroup. For $G=SL_2(\Bbb R)$ this should be the same as path-connected. – Dietrich Burde Jun 24 '23 at 11:28
The title here is confusing. In fact, arcwise connected is assumed. In the first answer a link to a book by Hilgert and Neeb is given. There is more about connected and path-connected subgroups of Lie groups (which might be useful, I haven't checked yet). – Dietrich Burde Jun 24 '23 at 11:36

The list of connected subgroups of $SL(2,\mathbb{R})$

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