I am very confused about the base affine space. Let $G$ be a complex reductive algebraic group and $H=B/U$ the abstract torus.


If I understand it correctly (Edit: which turns out not to be the case! See Sams answer), the base affine space is an $H$ bundle on the flag variety, such that every fiber is canonically isomorphic to $H$? Now common sense tells that a bundle where all fibers are canonically the same should be trivial, right !?


On the other hand, I think I calculated that if base affine space was trivial, all the interesting TDOs $\mathcal D\_\chi$ for $\chi \in \mathfrak h^\* $ on the flag variety would be isomorphic...


So my questions are:


Is base affine space trivial? If not, why does common sense fail here? 


$\textbf{Bonus question:}$ Can you give (other) examples of nontrivial bundles with all fibers canonically isomorphic?



Is base affine space a trivial fibration?
I am having a hard time believing your intrinsic construction of the basic affine space $G/U$ (given in the comments). 


If I understand your description correctly, you want to build it by constructing the tautological bundle $\widetilde{G} \to G/B$, and quotienting out by the bundle of unipotent radicals $\widetilde{\mathcal U} \to G/B$. Here $\widetilde{G}$ is the space consisting of a Borel and an element of the Borel, and $\widetilde{\mathcal U}$ is the same but with the element unipotent.


As you point out all the fibres of the quotient bundle are canonically isomorphic to $H=B/U$, and this is because the bundle is trivial! (It seems to me that the act of writing down this identification of each fibre gives this trivialization).


However this bundle is not the same as the basic affine space $G/U \to G/B$. The basic affine space is constructed by starting with the principal $B$-bundle $G \to G/B$ (for a particular choice of $B$), then taking the associated $H$-bundle $G \wedge \_B H$. This is non-trivial (e.g. for $SL\_2$ it is essentially the Hopf fibration).


This leaves us without an intrinsic construction of $G/U$ I guess... maybe an expert could comment on this?


