Take differential operator as polynomial of letter $d$ with coefficients in some function field, where $d$ act by derivation in this function field. Call it a differential field. For simplicity let work on $P^1$ with field of rational functions in $t$ variable and $d:[d,t\*]=1, d(t)=1$. We can go to differential extensions of $P^1$ attaching solutions of differential equations and ask how to describe all possible factorizations of original differential operator $D=d^n+f\_1(t)d^{n-1}+\ldots$ into composition of irreducible components defined over fixed differential field extension (require that top coefficient equal to 1).   
  

 The few obvious things about this. Over closure of initial field with $n$-dimensional vector space $V$ of solutions $Dv=0$ we have nice variety parametrizing all possible factorizations - it is complete flag variety. If we looking for factorizations defined not over differential closure, but smallest field, than some irreducible components have degrees $>1$ and reading factorization from right to left $D=\ldots D\_{k-1}D\_k$ we can attach to this non-complete flag of solutions $D\_kv=0, D\_{k-1}D\_kv=0,\ldots$ and, because solutions of differential equation restore it's own equation, all possible factorizations naturaly lies in some flag varieties. You can see that full flag variety in first case obtained bacause in differential field closure every solution, every subspace of solutions, defines differential equation over closure and this can be realised as component of $D$, generally not every subspace gives factorization component defined over our fixed field extension. 
  
  
So, this is proper subvariety. All this talk is a (known?) topic of differential Galois theory, but i'm not familar with that. Only thing that i know is Galois group, which naturally acts on this varieties. So, my first question is this varieties are single orbits of Galois group action or not? Unfortunately i does not have examples, but have speculation why it is not. And if they actually not, who they are? I mean subvarieties naturally obtained geometricaly as sections of bundles, as single orbits of group action, but what construction here? Thanks and sorry for my bad english.



Variety of factorizations of differential operator
A factorization would correspond to any flag such that the subspaces are invariant to the action of the differential Galois group and that each graded piece is an irreducible representation of the differential Galois group. It is easy to compute the space of such flags from the representation.


If the representation is semisimple, then the variety of factorizations will be a finite union of components, each component being a product of complete flag varieties. More precisely, if the irreducible representations $V\_1,\dots, V\_k$ occur with multiplicity $m\_1 , \dots , m\_k$ then the number of components in the multinomial:


$$\left( \begin{array}{c} m\_1 + \dots + m\_k \\ m\_1, \dots, m\_k \end{array}\right)$$


because this just counts the number of ways to order the irreducible factors corresponding to the different representations, and the flag variety is the product of the complete flag varieties of vector spaces of dimension $m\_1,\dots,m\_k$, because once you choose the order, the flags of the subspaces corresponding to different irreducible representations are totally independent.


In the non-semisimple case things get more complicated, but should still basically look like some form of flag variety. I think it might always be a disjoint union of flag varieties for the centralizer of the differential Galois group.


