8
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • $\begingroup$ It's certainly not the only case; you can take $S = W$, for example. $\endgroup$
    – LSpice
    May 25, 2020 at 14:48
  • 3
    $\begingroup$ @LSpice: huh? Wouldn't $S=W$ give the whole flag variety? $\endgroup$
    – Sam Hopkins
    May 25, 2020 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, 2020 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?) $\endgroup$
    – Igor Makhlin
    May 26, 2020 at 1:37
  • 1
    $\begingroup$ @imakhlin: The union of cells of equal dimension is a disjoint union of affine varieties, hence affine. $\endgroup$
    – Sasha
    May 26, 2020 at 4:53

1 Answer 1

Reset to default
3
$\begingroup$

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.)

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ (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, 2020 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 $\mathbb P^1$? $\endgroup$
    – Igor Makhlin
    May 27, 2020 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, 2020 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. $\endgroup$
    – Igor Makhlin
    May 27, 2020 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.