Does there exist a simple characterization of epimorphisms between affine group schemes over a field ? Are they faithfully flat morphisms ?

2$\begingroup$ The closed immersion of $\textbf{SO}_n$ in $\textbf{SL}_n$ is an epimorphism in the category of affine group schemes over a fixed field. Are you asking about group scheme morphisms that are also epimorphisms in the category of schemes? $\endgroup$ – Jason Starr May 11 '16 at 9:52

$\begingroup$ The answer is yes for commutative affine group schemes (DemazureGabriel, III, §3, 7.4). $\endgroup$ – Thomas Poguntke May 12 '16 at 13:03
There are papers by Bien and Borel on epimorphisms in the category of algebraic groups over a closed field. According to these papers, the answer to your question is no: there is no simple characterization. A necessary criterion for an immersion $H\hookrightarrow G$ to be an epimorphism is that all regular functions on $G/H$ are constant. So $SO(n)$ in $SL(n)$ is not epimorphic but a Borel subgroup in any connected group is.
Edit: I just looked it up. The criterion above is also sufficient.

$\begingroup$ My mistake, but the point is still valid with the Borel subgroup. $\endgroup$ – Jason Starr May 11 '16 at 11:32

$\begingroup$ Wait a minute . . . are you considering nonfinite type group schemes? Every finite dimensional algebraic representation of $\textbf{SL}_n$ whose restriction to a maximal torus is trivial is itself a direct sum of copies of the trivial representation. So the inclusion of a maximal torus seems to be an epimorphism in the category of (finite type) affine group schemes, yet the quotient has many nonconstant regular functions. $\endgroup$ – Jason Starr May 11 '16 at 11:58

1$\begingroup$ @Jason: Take a repesentation $\phi:G\to GL(V)$ which admits an invertible $H$endomorphism $\rho$ which is not $G$linear. Then $\phi$ and $\rho\phi\rho^{1}$ are equal on $H$ but not on $G$. For $SO(n)\subset SL(n)$ one can take $V=S^2(k^n)$. In general, one can construct $\phi$, $\rho$ from any nonconstant function on $G/H$. See BienBorel. $\endgroup$ – Friedrich Knop May 11 '16 at 14:08

$\begingroup$ Great! Now I see it. I was assuming that we could choose one of the two representations to be constant on $H$, but that is wrong. $\endgroup$ – Jason Starr May 11 '16 at 14:28
Today Brion posted a paper titled Epimorphic subgroups of algebraic groups on the arXiv.
The theorems in this paper directly answer your question for all fields (not just algebraically closed fields), generalizing the aforementioned results of Bien and Borel.
In particular, quoting from the abstract:
"We show that the epimorphic subgroups of an algebraic group are exactly the pullbacks of the epimorphic subgroups of its affinization. We also obtain epimorphicity criteria for subgroups of affine algebraic groups, which generalize a result of Bien and Borel."