Union of Schubert cells being affine

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?

• It's certainly not the only case; you can take $S = W$, for example. May 25 '20 at 14:48
• @LSpice: huh? Wouldn't $S=W$ give the whole flag variety? May 25 '20 at 14:50
• Sorry, yes. I missed that CJS was taking the union of the cells in $\mathcal F$, not in $G$. May 25 '20 at 14:50
• 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?) May 26 '20 at 1:37
• @imakhlin: The union of cells of equal dimension is a disjoint union of affine varieties, hence affine. 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.)

• (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 $Z$.)
– dhy
May 27 '20 at 16:59
• @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 $\mathbb P^1$? May 27 '20 at 23:14
• 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 $\mathbb{P}^1.$
– dhy
May 27 '20 at 23:21
• 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. May 27 '20 at 23:33