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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:

  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).

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    $\begingroup$ what do you call "non-trivial"? $\endgroup$
    – YCor
    Commented Mar 14, 2018 at 21:22
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    $\begingroup$ 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
    Commented Mar 14, 2018 at 21:44
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    $\begingroup$ @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. $\endgroup$
    – David E Speyer
    Commented Mar 14, 2018 at 22:38
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    $\begingroup$ (Just deleted my answer: I didn't read carefully: you're looking for affine varieties that are affinely paved) $\endgroup$
    – Qfwfq
    Commented Mar 15, 2018 at 14:10

1 Answer 1

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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!

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