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?



Split real form of $SL(2,\mathbb{C})$ description of the two sphere?
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.


