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](https://mathoverflow.net/questions/186832/optimal-definition-of-paving-by-affine-spaces)).


One can deduce that an affine paved variety (over $\mathbb{C}$) has no odd cohomology and its even cohomology is free abelian.


Examples:


1. Finite disjoint unions of affine space are affine paved. Let's call these examples "trivial."
2. Projective space is affine paved.
3. 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).



Affine Paved Affine Varieties; not Affine Space
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](https://sbseminar.wordpress.com/2008/12/05/the-plane-is-not-maximally-affine/). Much good discussion in the comments!


