Connected algebraic group need not be connected as a Lie group

Statement

It is possible to have a field $K$ admitting an analytic structure, such that there is a connected algebraic group $G$ over $K$ which, when interpreted naturally as a Lie group over $K$, is not a connected Lie group. In other words, $G$ may be connected in the Zariski topology but not in the analytic, or Lie, topology.

Related facts

• Subgroup of finite index need not be closed in algebraic group
• Subgroup of finite index need not be closed in T0 topological group

Proof

Example over the reals

Let $\mathbb {R}$ be the field of real numbers and $\mathbb {R} ^{\ast }$ be the multiplicative group.

• Since $\mathbb {R}$ is infinite, $\mathbb {R} ^{\ast }$ is connected in the Zariski topology on account of the Zariski topology being the cofinite topology. Thus, it is a connected algebraic group.
• On the other hand, as a real Lie group, it has two connected components: the positive numbers and the negative numbers. In particular, it is not connected.