Compactness of Lie groups

Question

Let $G$ be a Zariski-closed subgroup of $GL(V)$, where $V$ is an $n$-dimensional complex vector space.

Question. Does $G$ have the structure of a compact Lie group?

Such $G$ certainly is a Lie group. Being an algebraic variety, it is compact (quasi-compact is the common terminology), but it is not compact in the analytic topology, as it is affine. So, maybe, my actual question is:

Question $\star$. When an algebraic variety has the structure of a Lie group, is this structure understood via the Zariski topology, or the analytic topology?

Thanks!

Answer 1

1. No, e.g. $G$ could be all of $\text{GL}(V)$. (But $\text{GL}_n(\mathbb{C})$, despite not being compact, turns out to be controlled in a precise sense by its maximal compact subgroup $\text{U}(n)$.)

2. The analytic topology; a variety with the Zariski topology isn't even Hausdorff, let alone a manifold.