Topological and algebraic groups

Question

I learnt that an algebraic group $G$ is not a topological group because Zariski topology on $G$ x $G$ is not the product topology. Are there any other topologies, other than Zariski, that do not generate a topological group? If so, is it for the same reason (topology on the product is not the product topology)?

Answer 1

$\mathbb R$ with the Sorgenfrey topology is not a topological group; not only with respect to the usual sum and $x\mapsto -x$, but with respect to every group operation whatsoever; to wit: there is no group operation on $\mathbb R$ making the multiplication and inversion Sorgenfrey-continuous.

Answer 2

It's unclear to me what the actual question means. That said:

A group object in a category-with-finite-products $\mathsf{C}$ is an object $G \in {\rm Ob}({\mathsf{C}})$ together with $\mathsf{C}$-morphisms $\mu : G \times G \to G$, $e : 1 \to G$, and $\sigma : G \to G$ satisfying the group axioms.

If we take $\mathsf{C}$ to be:

• the category $\mathsf{Set}$ of sets and functions, we get the usual definition of groups;
• the category $\mathsf{Top}$ of topological spaces and continuous maps, we get the notion of a topological group;
• the category $\mathsf{Var}_k$ of algebraic varieties over $k$ and regular maps, we get the notion of an algebraic group.

What do we mean when we say that "an algebraic group is not a topological group"? Well, we mean that there's a forgetful functor $F : \mathsf{Var}_k \to \mathsf{Top}$ that takes a variety to its underlying Zariski topological space, and that if you apply this functor to an algebraic group the resulting thing is not (necessarily) a topological group, because $F$ doesn't preserve products.

So I'll interpret the question to mean "are there other categories-with-finite-products $\mathsf{C}$, functors $F :\mathsf{C} \to \mathsf{Top}$, and group objects $\mathbf{G} = (G, 1, \mu, \sigma)$ in $\mathsf{C}$ such that $F(\mathbf{G})$ is not a topological group?" (Maybe we want to put some condition on $F$ to make it resemble a forgetful functor, like requiring that it have a left adjoint.)

If $F$ preserves products, then functoriality guarantees that everything else goes through, so if we interpret the question in this sense then the answer is "this is the only thing that can go wrong."

But of course we can cook up all sorts of bizarre, useless categories $\mathsf{C}$ and functors $F : \mathsf{C} \to \mathsf{Top}$ which don't preserve products, and then come up with group objects in these categories that don't get sent by $F$ to topological groups.