Let $G$ be a linear algebraic group, i.e. a Zariski-closed subgroup of $GL(n,\mathbb{C})$ as embedded in $\mathbb{C}^{n^2+1}$ via $A \mapsto (A,\det(A)^{-1})$. Then $G$ is also a complex Lie group with the Euclidean topology, considered as a subset of $\mathbb{C}^{n^2}$.
- A Borel subgroup of $G$ as algebraic group is a maximal (Zariski-)connected solvable subgroup.
- A Borel subgroup of $G$ as Lie group is a maximal (Euclidean-)connected solvable subgroup.
I'm trying to show that both notions are equal. Is this proof correct? Why does it feel way too complicated to me?
$\implies$: Let $U$ be a maximal Zariski-connected subgroup. Then $U$ is Zariski-closed by maximaility, for otherwise its Zariski-closure is a larger such group, and hence $U$ is an algebraic group. As a connected algebraic group, $U$ is also irreducible. For irreducible varieties, it is known that they are connected in the Euclidean topology, so this is the case for $U$. Now assume $U \subseteq V$ for a larger Euclidean-connected solvable subgroup $V$. Since $V$ is also Zariski-connected, it must already be $U$ by the maximality of $U$. So $U$ is a maximal Euclidean-connected solvable subgroup.
$\impliedby$: Let $U$ be a maximal Euclidean-connected solvable subgroup. Then $U$ is also Zariski-connected. Assume $U \subseteq V$ for a larger Zariski-connected solvable subgroup $V$. Then the Zariski-closure $\overline{V}$ of $V$ is a Zariski-closed and Zariski-connected solvable group, hence an irreducible variety and hence connected in the Euclidean topology. So by maximality of $U$, $\overline{V}$ must be equal to $U$ and hence $U=V$. This shows that $U$ is a maximal Euclidean-connected solvable subgroup.