TITLE: Affine Paved Affine Varieties; not Affine Space
QUESTION [11 upvotes]: 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:

Finite disjoint unions of affine space are affine paved.  Let's call these examples "trivial."
Projective space is affine paved.  
The Bruhat cells in a flag variety show there are interesting 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).

REPLY [14 votes]: One example is $uw = v(v+1)$. The equations $u=v=0$ cut out an $\mathbb{A}^1$, and the complement is isomorphic to $\mathbb{A}^2$ by the map $(x,y) \mapsto (y, xy-1, x(xy-1))$. Note that the hypersurface $u=v=0$ is not principal; by dhy's comment, it can't be.
More generally, any of the Danielewski surfaces $u w^k = \prod_{i=1}^r (v-\alpha_i)$, with the $\alpha_i$ distinct, should be paveable with one $\mathbb{A}^2$ and $(r-1)$ copies of $\mathbb{A}^1$.
I wrote a related blogpost. Much good discussion in the comments!