# Algebraic group not implies topological group

## Statement

An algebraic group comes equipped with a Zariski topology on its underlying set. With this topology, it need not be a topological group.

## Related facts

### Failure of Hausdorffness

For topological groups, the T0 condition is equivalent to the T1 condition, which is equivalent to the Hausdorff condition, which is equivalent to regularity and to complete regularity. See T0 topological group and equivalence of definitions of T0 topological group.

However, for quasitopological groups, being a T0 quasitopological group is equivalent to being $T_1$ but it is not equivalent to the Hausdorff condition. In fact, if the underlying variety is irreducible (which is always the case for a connected algebraic group), and the group is nontrivial, then it is not Hausdorff.

## Proof

### Where the naive reasoning fails

The naive reasoning may be that for an algebraic group $G$, the group multiplication is an algebraic map from $G \times G$ to $G$, hence it is a continuous map from $G \times G$ to $G$.

The problem is that the Zariski topology on $G \times G$ is not the product topology on $G \times G$ arising from Zariski topologies on $G$. In fact, the Zariski topology on $G \times G$ is a lot finer, i.e., has a lot more open sets. Thus, continuity of multiplication with respect to the Zariski topology is a lot weaker than continuity with respect to the product topology.

### Example

Consider any example of a one-dimensional algebraic group over an infinite field $K$. For concreteness, we take the additive group of $K$. The group operation $K \times K \to K$ is given by $(g,h) \mapsto g + h$. The Zariski topology is the cofinite topology in this case. However, we know that infinite group with cofinite topology is not a topological group, so $K$ is not a topological group with the Zariski topology.

### Proof based on how topological groups behave

See the discussion #Failure of Hausdorffness. This can be converted to a proof.