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Question: What are the connected subgroups of $SL(2,\mathbb{R})$?

Connected Lie subgroups of $SL(2,\mathbb{R})$ are in 1:1-correspondence with Lie subalgebras of $\mathfrak{sl}(2,\mathbb{R})$. If we denote the usual root decomposition of $\mathfrak{sl}(2,\mathbb{R})$ by

$$ \mathfrak{sl}(2,\mathbb{R}) = \mathfrak{g}_0 \oplus \mathfrak{g}_\alpha \oplus \mathfrak{g}_{-\alpha}, $$

then we clearly have the following Lie subalgebras: $\{0\}, \mathfrak{g}_0, \mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha}, \mathfrak{so}(2), \mathfrak{g}_0 \oplus \mathfrak{g}_\alpha, \mathfrak{sl}(2,\mathbb{R})$ and all conjugates of those, e.g. $Ad(g)(\mathfrak{g}_\alpha)$ for $g \in SL(2, \mathbb{R})$.

But I have a hard time imagining what other groups are there.

Background: I want to know all subgroups of the isometry group of $Isom(\mathbb{RH}^2 \times \mathbb{RH}^2)=SL(2,\mathbb{R}) \times SL(2,\mathbb{R})$. That classification will follow with Goursat's lemma from the answer to my question.

user538044
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  • See the references here. – Dietrich Burde Jun 17 '18 at 14:47
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    Using the root decomposition is not the right approach. The way to do (by oneself, which is better that reading a reference): classify 1-dimensional ones, then 2-dimensional ones. 2-dimensional Lie algebras are either either abelian, or their derived subgroup is 1-dimensional. To classify the nonabelian ones in $sl_2$ one has to compute the normalizer of the 1-dimensional subalgebras. – YCor Jun 18 '18 at 09:10
  • @YCor Thanks. Any suggestion how to check if two subalgebras are conjugate? That's already difficult for me to do for the 1-dimensional subalgebras, let alone the 2-dimensional ones. – user538044 Jul 01 '18 at 13:14
  • Hint: for a element, the unordered pair of eigenvalues is a conjugacy invariant. Hence, for a 1-dimensional subalgebra, the unordered pair of eigenvalues, up to scalar multiplication, is a conjugacy invariant. – YCor Jul 01 '18 at 14:14
  • @YCor Okay, this tells me which subalgebras are conjugate by an element in $GL(2,\mathbb{R})$. The other ingredient that I needed was "If two invertible matrices are conjugate by a matrix with negative determinant, then they are not conjugate by a matrix with positive determinant". I am left to check by hand which singular matrices in $\mathfrak{sl}(2,\mathbb{R})$ are conjugate. The only subalgebras of singular matrices are $\mathfrak{g}\alpha$ and $\mathfrak{g}{-\alpha}$ and they are not conjugate by a matrix in $SL(2,\mathbb{R})$, as direct calculation shows. Thanks. – user538044 Jul 02 '18 at 11:16
  • They're conjugate by the matrix $\begin{pmatrix}0 & -1\ 1 & 0\end{pmatrix}$ which is in $\mathrm{SL}_2$. – YCor Jul 02 '18 at 11:35
  • In yoccoz article about SL(2,R),there is a theorem about conjugacy of bounded sequence of these matrices which got from cocycles.you can check that maybe works for you.but i think first you should know the classification of SL(2,R) and then each class has its own classes which is in "BMS Particles in Three Dimensions from Oblak". – pershina olad Apr 20 '20 at 21:32

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