Algebraic group implies quasitopological group
- The multiplication map is separately continuous in the two inputs, i.e., it is continuous in each input holding the other input constant.
- The inverse map is continuous.
Proof for the multiplication map
Given: An algebraic group .
To prove: The multiplication map is continuous in each variable. More explicitly, for any , the multiplication maps and are continuous maps from to itself with the Zariski topology.
Proof: We will fix for the proof.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The inclusion maps and are continuous maps from to where both and are given the respective Zariski topologies. Note that the Zariski topology on arises from the product variety structure but is not the same as the product topology arising from the Zariski topology on .||(cite relevant facts from algebraic geometry)|
|2||The multiplication map is continuous as a map from equipped with the Zariski topology arising from the product variety structure to .||definition of algebraic group||By the definition of algebraic map, the multiplication map is algebraic (i.e., it is a regular map). Any algebraic map is continuous with respect to the Zariski topology.|
|3||The multiplication maps and are both continuous as maps from to itself with the Zariski topology.||a composite of continuous functions is continuous||Steps (1) and (2)||step-fact-direct|
Alternative formulation of the proof for multiplication
Another way of thinking of the proof is as follows. Consider the following three possible topologies on :
- The product topology on arising from the Zariski topology on .
- The Zariski topology arising from the product variety structure on that in turn arises from the variety structure on .
- The coarsest topology on such that any separately continuous map (for the Zariski topology on ) is continuous from that topology on to the Zariski topology on . Explicitly, this is a topology where a subset of is open if its intersection with each fiber of the form is open in that copy of and its intersection with each fiber of the form is open in that copy of .
Proof for the inverse map
The inverse map is an algebraic map by the definition of algebraic group, hence is continuous in the Zariski topology by the definition of Zariski topology.