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.


  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).

  • 1
    $\begingroup$ what do you call "non-trivial"? $\endgroup$
    – YCor
    Mar 14, 2018 at 21:22
  • 4
    $\begingroup$ Any affine variety containing $\mathbb{A}^n$ as an open subset must be $\mathbb{A}^n$ itself. An affine open embedding corresponds to a localization at the level of rings, but there are no non-constant invertible functions in $k[x_1,\cdots,x_n]$. $\endgroup$
    – dhy
    Mar 14, 2018 at 21:44
  • 8
    $\begingroup$ @dhy This is wrong. That shows that $\mathbb{A}^n$ can not be a principal open, but it can be the complement of a nonprincipal hypersurface. I'll put up an example shortly. $\endgroup$ Mar 14, 2018 at 22:38
  • 1
    $\begingroup$ (Just deleted my answer: I didn't read carefully: you're looking for affine varieties that are affinely paved) $\endgroup$
    – Qfwfq
    Mar 15, 2018 at 14:10

1 Answer 1


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.