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

Example over the reals
Let $$\R$$ be the field of real numbers and $$\R^\ast$$ be the multiplicative group.


 * Since $$\R$$ is infinite, $$\R^\ast$$ 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.