Connected subgroups of $SL(2,C)$ Where can I find a list of all connected real Lie groups inside the 6-dimensional real Lie group $SL(2,C)$, up to conjugacy? How can one verify that a partial list is complete?
I found on wikipedia a list of Lie subalgebras of so(1,3), but it is not clear to me how to see which of these exponentiate to the double cover SL(2,C) of SO^+(1,3).
 A: Let me describe the Lie subalgebras. 
Up to conjugation by $\mathrm{PGL}_2(\mathbf{C})$, the complex subalgebras are $\{0\}$, the diagonal subalgebra, the upper unipotent subalgebra, the upper triangular subalgebra, and the whole $\mathfrak{sl}_2(\mathbf{C})$. Their real dimension are 0, 2, 2, 4, 6 respectively.
The purely real subalgebras ($\mathfrak{h}$ such that $\mathfrak{h}\cap i\mathfrak{h}=\{0\}$) are real forms of these. There is $\{0\}$; in dimension 1 we have the upper unipotent real subalgebra, the subalgebra generated by the diagonal matrix $(e^{it},-e^{it})$, say for $0\le t<\pi$. In dimension 2 we have the upper triangular real Lie algebra. In dimension 3 we have both $\mathfrak{sl}_2(\mathbf{R})$ and $\mathfrak{su}(2)$.
There is no subalgebra of real codimension 1. The remaining $\mathfrak{h}$ have dimension 3 or 4, and $\mathfrak{h}\cap i\mathfrak{h}$ has dimension 2; then $\mathfrak{h}+i\mathfrak{h}=\{0\}$ has dimension 4 or 6. the second case is not possible because $\mathfrak{h}$ would be non-solvable and only possible simple subalgebras are 3-dimensional, and are maximal sulalgebras. 
The remaining ones have codimension 1 in the upper triangular subalgebra: thus we get, up to conjugation, the (3-dimensional) subalgebra generated by upper unipotent matrices and the diagonal matrix $(e^{it},-e^{it})$, for $0\le t<2\pi$.
I guess this is comprehensive, I haven't 100% checked that there is no redundancy; anyway details are of exercise level (only using as a black box the classification of real simple Lie algebras up to dimension 6). In addition, they all exponentiate to closed Lie subgroups.
