Let $V$ be an algebraic variety. If there is a finite ascending chain of Zariski closed sets $\emptyset=V_0\subset V_1\subset \cdots \subset V_n=V$ such that $V_i-V_{i-1}$ is a fintie disjoint union of copies of affine space $\mathbb{A}^i$ we say $V$ is affine paved (so $V$ is "algebraically cellular").
Note: there are non-equivalent variations of this definition (see here).
One can deduce that an affine paved variety (over $\mathbb{C}$) has no odd cohomology and its even cohomology is free abelian.
Examples:
- Affine space (or finiteFinite disjoint unions thereof) is (are)of affine space are affine paved: $\emptyset\subset \mathbb{A}^n$. Let's call these examples "trivial."
- Projective space is affine paved since removing a point gives an affine open.
- The Bruhat cells in a flag variety show there are non-trivialinteresting projective examples.
Question: Are there non-trivial affine paved affine varieties?
This very well might be a silly question. Perhaps the affine cone over an affine paved projective variety always works? It doesn't seem clear to me, and I figured someone out there might have thought about such examples before (Google doesn't seem to know any).