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?

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    $\begingroup$ Presumably the base affine space is $G/U$ (I haven't heard the terminology 'base affine space' before)? I think it's helpful to consider the case where $G=SL_2$, so that the flag variety is $\mathbb{P}^1$ and $G/U$ is the complement of the origin in $\mathbb{A}^2$, which isn't a trivial $\mathbb{G}_m$-bundle over $\mathbb{P}^1$. $\endgroup$ – Mike Skirvin Nov 8 '11 at 19:45
  • $\begingroup$ Thank you! Base affine space is isomorphic to $G/U$. An intrinsic way to define it is as follows: Consider the tautological bundle $b$ on the flag variety (=space of Borels). i.e. $b$ attaches to each Borel itself. Now we have a subbundle $u$ which attaches to each Borel its unipotent radical. The base affine space is the quotient bundle $b/u$. Using this definition one can easily check that its fibers are all canonically isomorphic. $\endgroup$ – Jan Weidner Nov 8 '11 at 20:34
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    $\begingroup$ The terminology "base affine space" goes back at least to the paper "Differential Operators on the Base Affine Space and a Study of g-Modules, Lie Groups and Their Representations" by Bernstein I.N., Gelfand I.M., Gelfand S.I. $\endgroup$ – Dan Fox Nov 9 '11 at 10:54

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?

  • $\begingroup$ Thank you very much Sam! So my confusion came from this wrong construction I had in mind. As you say, it would be interesting to see a correct intrinsic description. $\endgroup$ – Jan Weidner Nov 9 '11 at 8:55
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    $\begingroup$ I'm late to this party, but in case someone still cares, here's an intrinsic description lifted directly from "A Proof of Jantzen Conjectures" by Beilinson and Bernstein. A point of the base affine space is a Borel subalgebra $\mathfrak{b} \subset \mathfrak{g}$ together with, for each simple root $\alpha$, a basis vector for the $\alpha$-root space in $\mathfrak{g}/\mathfrak{b}$. $\endgroup$ – Justin Campbell Oct 21 '13 at 17:27

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