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I understand that an algebraic group may not be a topological group because the continuity of multiplication with respect to the Zariski topology is weaker than that with respect to product topology. But is there a nice example that helps to explain this. Many thanks!

MJS
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  • For instance the additive group of the field of rational numbers. – Moishe Kohan Oct 07 '23 at 03:52
  • What do you mean by "is not a topological group"? Do you mean: is there is an example of an algebraic group $G$ over an algebraically closed field $k$ such that $G(k)$ is not a topological group when equipped with the Zariski topology? (It's dangerous to get into the habit of identifying $G$ with its set of $k$-points since this no longer makes sense when $k$ isn't algebraically closed so there's room for ambiguity here.) – Qiaochu Yuan Oct 07 '23 at 04:08

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Pick the additive group $(\mathbb{C},+)$ with the Zarisky topology. Then, the map $\mathbb{C}\times\mathbb{C}\to\mathbb{C}$ given by $(a,b)\to a+b$, is not continuous.

To see this, just consider the preimage of $0$. This is just the set of points of the form $(a,-a)$ in $\mathbb{C}\times\mathbb{C}$. This is not closed in $\mathbb{C}\times\mathbb{C}$ with the product topology, since the only closed subsets are either of the form $A\times B\subset \mathbb{C}\times \mathbb{C}$ with $A$ and $B$ finite, or of the form $A\times \mathbb{C}$ with $A$ finite.

Giulio Ricci
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