Let me start with an apologetic disclaimer: I am very far from an algebraic geometer, so this question might be crudely formulated.

I have a specific question about du Val singularities, but while trying to dig up information about it online I've ended up with a more general question as well. In everything below, I am assuming that I'm working over $\mathbb{C}$.

General question

A smooth variety/(reduced, separated, of finite type, etc.) scheme comes with an atlas of open affines. The variety is called uniformly rational if it admits an atlas of open affines each of which is itself a Zariski open subset of affine space $\mathbb{A}^d$.

An even stronger condition that one might ask for is an atlas where each open affine is in fact a copy of $\mathbb{A}^d$ itself. In other words, the variety should be locally isomorphic to affine space. Obvious examples are projective space itself and total spaces of line bundles over projective space, for example.

My general question is whether this class of varieties has a name, or is known to be so restrictive as to be totally uninteresting (I realize that already the requirement that the variety be rational is already quite restrictive).

Specific question

A slightly more interesting class of examples that can be seen to be locally affine in the above sense are the minimal resolutions of $A$-type du Val singularities. Blowing up the singularity provides exactly such an atlas of $\mathbb{A}^2$ patches.

It appears to me that the same is true of (minimal resolutions of) the $D$-type (and possibly $E$-type) singularities as well, though this is not provided for you automatically by doing a textbook blow-up to realize the minimal resolutions.

My specific question is whether such an atlas for resolutions of du Val singularities is a standard thing, or if this claim is somehow obvious (or, alternatively, if the claim is obviously false indicating an error on my part).

  • 2
    $\begingroup$ I think it is conjectured that every smooth projective rational variety admits an atlas by affine spaces. It sounds ridiculous, but reflects how little we know. $\endgroup$ – Piotr Achinger Jul 25 '19 at 19:46
  • 1
    $\begingroup$ @PiotrAchinger I believe this was proven false a couple years ago -- see the second answer here: mathoverflow.net/questions/99144/… $\endgroup$ – Kevin Casto Jul 25 '19 at 20:24
  • 2
    $\begingroup$ @KevinCasto Has Karzhemanov’s claim been accepted by the experts? If true this would be spectacular, but in the three years after the preprint appeared I still haven’t heard anything. $\endgroup$ – Piotr Achinger Jul 25 '19 at 20:38
  • $\begingroup$ @PiotrAchinger, I had gathered that the conjecture that gets discussed was that all smooth rational varieties are uniformly rational. Are you saying that (modulo assessment of Kakrzhemanov's claim) the stronger conjecture that all rational varieties admit atlases that are actually affine spaces, rather than open subsets of affine spaces is also viable? $\endgroup$ – Christopher Beem Jul 26 '19 at 11:31
  • $\begingroup$ @ChristopherBeem that's the conjecture I remember (for smooth and projective rational varieties), though I don't have a reference. $\endgroup$ – Piotr Achinger Jul 26 '19 at 11:37

A chapter of the unpublished PhD thesis of Rebecca Leng, a student of Miles Reid from about 2002, presents a careful study of a natural affine cover of the minimal resolution $Y={\rm GHilb}({\mathbb A^2})$ of the singularity $X={\mathbb A}^2/G$ of type D. The conclusion in Section 2.5.6 is that 􏰳􏰍she "constructs an affine open cover of ${\rm GHilb}({\mathbb A^2})$ where every open set is a smooth surface in ${\mathbb A}^4$; two of the open sets contain open subsets of ${\mathbb A}^2$".

| cite | improve this answer | |
  • $\begingroup$ This is a great reference for the details of the D-type singularities, but Leng is giving a cover by smooth affine varieties. The phenomenon I'm interested in is a cover by actual copies of $\mathbb{A}^2$ in this case. $\endgroup$ – Christopher Beem Jul 27 '19 at 11:01

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.