TITLE: Union of Schubert cells being affine
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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.)