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

Topological and semitopological groups

• Algebraic group implies semitopological group: This basically says that the left and right multiplication maps and the inverse map are continuous.
• Topological group implies semitopological group

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 semitopological groups, being a T0 semitopological 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 $(g,h)\mapsto g+h$.

If we take the product topology on $K\times K$ with respect to the Zariski topology on $K$, we get a topology with a basis given by sets of the form $U\times V$ where both $U,V$ are cofinite. In particular, this means that for any proper closed subset $C$ of $K\times K$, there are finite subsets $A,B$ of $K$ such that for each element of $C$, either the first coordinate lies in $A$ or the second coordinate lies in $B$.

Consider the inverse image of 0 under the addition map $K\times K\to K$. Since the topology on $K$ is T1, the point $\{0\}$ is closed. However, its inverse image under the addition map is not closed with respect to the topology described above. The upshot is that the addition map is not continuous with respect to the topology.

Proof based on how topological groups behave

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