For an affine algebraic group $G$ it's often convenient (and harmless) to work concretely over an algebraically
closed field of definition $k$ while identifying $G$ with its group of rational points over $k$.
Once in a while, however, results obtained over $k$ need to be compared with results over
a bigger algebraically closed field $K$.


For example, in his 1976 *Inventiones* [paper](https://gdz.sub.uni-goettingen.de/id/PPN356556735_0034) completing the proof that a semisimple algebraic group
always has a finite number of unipotent conjugacy classes, Lusztig observed that in characteristic
$p>0$ it suffices to assume that $k$ is an algebraic closure of the prime field. This in turn
allowed him to apply indirectly the Deligne-Lusztig construction of characters for subgroups of
$G$ over finite subfields of $k$. Here the number of unipotent classes is moreover the same for any algebraically closed field. To justify his reduction, he cites a "simple argument" shown to
him by Deligne (which he later told me he viewed in retrospect as "obvious").


Independently, a formal statement of the principle was written down and proved as Proposition 1.1 in a 1997 multi-author paper [here](https://doi.org/10.1006/jabr.1997.7068). The proof is fairly elementary, but requires more than a journal page to write down and involves an induction step left to the reader.



> 
> Is there a short and transparent proof (perhaps from the scheme viewpoint) that finiteness of the number of orbits of a semisimple group acting on an affine variety is the same when an algebraically closed field of definition is extended to another such field, while the number of orbits is unchanged?
> 
> 
> 


It would also be interesting to know of other situations in which such a comparison occurs.
(Historical remark: In the older version of foundations for algebraic geometry developed
by Weil and others it was standard procedure to work over a "universal domain" having infinite
transcendence degree over its prime field, to permit for instance the use of "generic points". Then it was usually troublesome to descend to a countable field.)


ADDED: To be more precise about "the number of orbits is unchanged", implicit in Lusztig's work and explicit in the 1997 paper cited is the natural requirement on such a bijection that each orbit over $K$ should contain a point over $k$. (It's hard to visualize a proof that gives a bijection without this refinement.) On the other hand, it's unclear to me whether special assumptions on $G$ such as "reductive" and "connected" are essential for the proof of a general comparison principle.



Comparing algebraic group orbits over big and small algebraically closed fields
I think this will work. There are a finite number of orbits of the action of $G$
on $X$ precisely when there is an open orbit and a finite number of orbits on
the complement of the orbit. Hence, it is enough to show that if there is an
open orbit of a point over the smaller field precisely when there is an open
orbit over the larger field. One direction is clear so it is enough to show that
there is an open orbit over the larger field, there is one over the smaller
field. However, consider the closed subscheme $S:=\{(g,x)\in G\times X | gx=x\}$
and the projection $S\to X$ on the second variable. A point of $X$ has an open
orbit precisely when the fibre has the smallest possible dimension (when $X$ is
irreducible which we may assume). However, there is an open subset (defined over
the smaller field) of $X$ with fibres of minimal dimension.


**Addendum**: As for the bijection between the orbits this is proved the same way, we have an open orbit which gives one orbit over each field and then we use Noetherian induction.


