Let $k$ be a field of characteristic zero, $G$ be a reductive group with a Borel $B$ and $\mathcal{F}:=G/B$ the associated flag variety. Let $W$ be the Weyl-group of G.

Then let $S \subset W$ and $Z=\bigcup_{w \in S} C(w) \subset \mathcal{F}$ where $C(w)=BwB/B$ is the Schubert cell associated to $w$.

I'm interested to know when $Z$ is an affine scheme. This is for example the case if all $w \in S$ have the same length. Is this the only case?

  • $\begingroup$ It's certainly not the only case; you can take $S = W$, for example. $\endgroup$
    – LSpice
    May 25 '20 at 14:48
  • 3
    $\begingroup$ @LSpice: huh? Wouldn't $S=W$ give the whole flag variety? $\endgroup$ May 25 '20 at 14:50
  • $\begingroup$ Sorry, yes. I missed that CJS was taking the union of the cells in $\mathcal F$, not in $G$. $\endgroup$
    – LSpice
    May 25 '20 at 14:50
  • $\begingroup$ Do you have a reference for the union of cells of equal dimension being affine? (Or is it obvious and I'm just not seeing it?) $\endgroup$
    – imakhlin
    May 26 '20 at 1:37
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    $\begingroup$ @imakhlin: The union of cells of equal dimension is a disjoint union of affine varieties, hence affine. $\endgroup$
    – Sasha
    May 26 '20 at 4:53

This is essentially an extension of my comment, just to answer the actual "is this the only case?" question. It is not, $Z$ will be affine whenever $S$ is an antichain in the Bruhat order. Indeed, this condition means that no $C(w)$ with $w\in S$ intersects the closure $\overline{C(w')}$ for any other $w'\in S$ which shows that $C(w)$ is open in $Z$. Hence the $C(w)$ are the irreducible components of $Z$ and are also affine, this renders $Z$ affine itself (Hartshorne, Exercise 3.3.2).

Of course, the more interesting underlying question is whether this condition is necessary, I might update this answer if I come up with a proof. (Any algebraic geometers here? Is it at all possible for an affine space to be embedded into an affine variety as a proper open subset? If not, this would give us the answer.)

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    $\begingroup$ (To answer your parenthetical question at the end: It is indeed possible, already for surfaces. I don't remember the construction though - maybe it's something like remove a divisor from a Hirzebruch surface? In any case, these examples shouldn't work here, because if $Z$ is open in its closure, then it will contain a $\mathbb{P}^1$ and so can't be affine. On the other hand, if $Z$ is not open in its closure, then the question is a bit ambiguous because there is no natural scheme structure on $Z$.) $\endgroup$
    – dhy
    May 27 '20 at 16:59
  • $\begingroup$ @dhy Not sure if I understand the "if Z is open in its closure, then it will contain a P1" part. It would seem that a $Z$ consisting of a single cell is open in its closure but contains no $\mathbb P^1$? $\endgroup$
    – imakhlin
    May 27 '20 at 23:14
  • $\begingroup$ If it is open in its closure and is not an antichain. The point being that then it contains cells $w, w'$ with $l(w)=l(w')+1$, and the union of two such cells is a product of an affine space and a $\mathbb{P}^1.$ $\endgroup$
    – dhy
    May 27 '20 at 23:21
  • $\begingroup$ Oh, okay, I see now. Yes, if we leave out the cases where $S$ is not convex in the Bruhat order as ambiguous, the rest should be clear. $\endgroup$
    – imakhlin
    May 27 '20 at 23:33

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