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If we denote the parabolic subgroup of $SL(2,\mathbb{C})$ by $P$, then we have the well known isomorphism $SL(2,\mathbb{C})/P \simeq S^2$, where $S^2$ is the two sphere. Now the compact real form of $SL(2,\mathbb{C})$ is given by $SU(2,\mathbb{C})$, and it gives us the alternative presentation $SU(2,\mathbb{C})/S^1 \simeq S^2$, where $S^1$ is the circle group. What I would like to know is whether one has a similar description of $S^2$ using $SL(2,\mathbb{R})$ the split real form of $SL(2,\mathbb{C})$, i.e does there exist a subgroup $L \subset SL(2,\mathbb{R})$, such that $SL(2,\mathbb{R})/L \simeq S^2$, and if so, what is $L$? Moreover, is this description of $S^2$ (if it exists) an affine variety?

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    $\begingroup$ Jason gave the proof of impossibility. Let me suggest the correct question, which is, what do the orbits of a real form of $G$ (say $SL(2,{\mathbb C})$) look like on $G/B$? If the real form is the maximal compact, then there's just one orbit, but I feel that misled you. In the case you ask about, $SL(2,{\mathbb R})$ acts with three orbits: the equator and the open hemispheres. In general a real form will act with finitely many orbits and the poset thereof was studied by Richardson and Springer. Also one should look up "Matsuki correspondence". $\endgroup$ Oct 21, 2013 at 16:51
  • $\begingroup$ Small comment on fuzzy terminology: "the parabolic subgroup" doesn't make sense, since there are infinitely many such subgroups. Moreover, they form two conjugacy classes because the whole group is by definition "parabolic", along with all Borel subgroups. $\endgroup$ Oct 21, 2013 at 17:35
  • $\begingroup$ P.S. It's not clear to me how you view homogeneous spaces here: just as topological spaces, or as manifolds, or as algebraic varieties? Usually the first quotient you describe is viewed algebraically as a projective lthe quotient of a semisimple algebraic group by any parabolic subgroup being a projective variety). $\endgroup$ Oct 21, 2013 at 17:42
  • $\begingroup$ I view them as algebraic varieties. $\endgroup$ Oct 23, 2013 at 10:46

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That seems to be impossible. First of all, I assume that you meant to write $\textbf{SL}(2,\mathbb{R})/L \cong S^2$, since otherwise the dimensions don't work out. Denote by $L_0$ the connected component of $L$ containing the identity element. Then $L_0$ is a connected subgroup of $\textbf{SL}(2,\mathbb{R})$. Thus the quotient $\textbf{SL}(2,\mathbb{R})/L_0$ is an unbranched cover of $\textbf{SL}(2,\mathbb{R})/L$, i.e., of $S^2$. Since $S^2$ is simply connected, it follows that $L$ is connected. Thus, either $L$ is isomorphic to $\mathbb{R}$ or $L$ is isomorphic to $S^1$. Either way, using the long exact sequence of homotopy groups associated to a fibration, it follows that $\pi_3(\textbf{SL}(2,\mathbb{R}))$ equals $\pi_3(S^2)$, which is $\mathbb{Z}$ generated by the Hopf fibration (since $\pi_2$ and $\pi_3$ of $L$ are both trivial).

This is impossible. Consider the standard Borel subgroup $B$ of $\textbf{SL}(2,\mathbb{R})$, i.e., upper triangular matrices whose first entry is positive. Then $\textbf{SL}(2,\mathbb{R})/B$ is isomorphic to the circle $S^1$. Also $B$ is contractible: as a group it is a semidirect product of $\mathbb{R}_{>0}$ and $\mathbb{R}$, thus as a topological space it is $\mathbb{R}^2$. Hence, again by the long exact sequence, $\pi_3(\textbf{SL}(2,\mathbb{R}))$ equals $\pi_3(S^1)$, which is $\{*\}$. This is a contradiction that proves that $\textbf{SL}(2,\mathbb{R})$ contains no subgroup $L$ such that $\textbf{SL}(2,\mathbb{R})/L$ is isomorphic to $S^2$.

Probably there are more elementary proofs.

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