I was reading this first look at differential algebra, when I came across to the following result:
any proper connected complex subgroup of ${\rm SL}_2(\mathbb{C})$ is conjugate to one of the following subgroups:
$\{ I_2\}, \pmatrix{a & 0 \cr 0 & a^{-1}}, a\in\mathbb{C}^\times, \pmatrix{1 & b \cr 0 & 1}, b\in\mathbb{C},\pmatrix{a & b \cr 0 & a^{-1}}, a\in\mathbb{C}^\times,b\in\mathbb{C}. $
There is no proof or reference given in the pdf, so I guess it is well-known from specialists and/or easy.
I can see why these guys are connected. I can also see why a connected subgroup, after conjugation if needed, would contain a matrix of one the previous four shapes, but that's it...
If I had to prove it myself, my first try would be to use the exponential map to reduce to finding connected complex subgroups of the additive group of matrices with trace zero, which is canonically isomorphic to $\mathbb{C}^3$, but I don't know how to do it.
Can someone give a reference, or even better, arguments to get this result ? Elementary arguments are particularly appreciated (if one could avoid the theory of Lie groups, it would be great).
Edit. Even if this is not mentionned in the paper, the connected subgroups are supposed to be complex subgroups. Otherwise,the list above is incomplete (see comments).