34
$\begingroup$

In Une suite exacte de Mayer-Vietoris en K-théorie algébrique (1972) Jouanolou proves that for any quasi-projective variety $X$ there is an affine variety $Y$ which maps surjectively to $X$ with fibers being affine spaces. This was used e.g. by D. Arapura to (re)prove that the Leray spectral sequence of any morphism of quasi-projective varieties is equipped from the second term on with a natural mixed Hodge structure.

Here is a proof when $X$ is $\mathbf{P}^n$ over a field $k$: take $Y$ to be the affine variety formed by all $n+1 \times n+1$ matrices which are idempotent and have rank 1. This is indeed affine since it is given by the equations $A^2=A$, the characteristic polynomial of $A$ is $x^n(x-1)$. Moreover, $Y$ is mapped to $\mathbf{P}^n(k)$ by taking a matrix to its image. The preimage of a point of $\mathbf{P}^n(k)$ is "the set of all hyperplanes not containing a given line", which is isomorphic to an affine space.

The general (quasi-projective) case follows easily from the above. However, it is not clear how to generalize Jouanolou's trick for arbitrary varieties. Nor is it clear (to me) that this is impossible.

  1. Is there an analogue of the Jouanolou lemma for arbitrary (not necessarily quasi-projective) varieties (i.e. reduced separated schemes of finite type over say an algebraically closed field)?

  2. (weaker version of 1 over complex numbers) Is there, given a complex algebraic variety $X$, an affine variety $Y$ that maps surjectively to $X$ and such that all fibers are contractible in the complex topology? A negative answer would be especially interesting.

  3. (the following question is a bit vague, but if it has a reasonable answer, then it would probably imply a positive answer to 2.) Is there a quasi-projective analog of the topological join of two projective spaces? I.e., if $P_1$ and $P_2$ are two complex projective spaces, is there a quasi-projective variety $X$ which "contains the disjoint union of $P_1$ and $P_2$ and is formed by all affine lines joining a point in $P_1$ with a point in $P_2$"?

Edit 1: in 1. and 2. the varieties are required to be connected (meaning that the set of closed points is connected in the Zariski topology; in 2 one could use the complex topology instead).

Edit 2: as Vanya Cheltsov explained to me, the answer to question 3 is most likely no.

Improve this question
$\endgroup$

4 Answers 4

Reset to default
14
$\begingroup$

Jouanolou's trick has been extended to schemes with an "ample family of line bundles" by Thomason; see Weibel: Homotopy Algebraic K-theory, Proposition 4.4. This includes all smooth varieties and more generally all varieties with torsion local class groups. However, there exist (positive dimensional) proper varieties with no non-trivial line bundles on them; it seems possible that on such varieties there are no affine bundles with affine total space.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks, unknown, this is indeed helpful! One remark though: requiring the bundle to be affine may be a bit too much. I'd be more than happy with all fibers being affine spaces. $\endgroup$
    – algori
    Commented Apr 16, 2010 at 18:06
  • $\begingroup$ @naf Could you give a reference where I can find a (positive dimensional) proper varieties with no non-trivial line bundles on them? (OK, Picard is trivial!) $\endgroup$
    – Lao-tzu
    Commented Apr 8, 2022 at 15:08
  • $\begingroup$ For an explicit example, see the paragraph after the third exercise on p. 65 of Fulton's "Toric varieties". (The actual example is in Section 1.5.) $\endgroup$
    – naf
    Commented Apr 10, 2022 at 10:57
8
$\begingroup$

Jounalou's trick is great isn't it? Off the top of my head, I don't know anything similar for non quasiprojective varieties or schemes. One can certainly use Chow's lemma to reduce to the quasiprojective case, but it's a lot messier ...

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thanks, Donu! Re: Jounalou's trick is great isn't it? Indeed it is! And it is true that replacing e.g. a complete variety by a cubical projective one results in a mess. So my question can be rephrased: is this mess avoidable or not? For example, does there exist a complete variety so bad that its cohomology (seen as an algebra equipped with a mixed Hodge structure) is different from the cohomology of any affine variety? $\endgroup$
    – algori
    Commented Apr 14, 2010 at 19:59
  • $\begingroup$ To be clear, I think your question(s) is(are) interesting but difficult. $\endgroup$
    – Donu Arapura
    Commented Apr 15, 2010 at 11:50
  • 1
    $\begingroup$ Donu -- the moment I can handle all interesting and difficult questions, I'll just start posting easy and boring ones;) $\endgroup$
    – algori
    Commented Apr 16, 2010 at 17:44
4
$\begingroup$

Well, as no one else has said anything, I'll say the only thing that's come to mind, which is fairly elementary. It should be enough to do this for complete varieties, so restrict to that case. The only other thing I've got is that by Chow's lemma, there's $\bar{X}\to X$ a projective variety over $X$ and a birational map. So this tells us that over an open set, everything should work, so we can reduce to the exceptional locus. It will itself be an open subset of a complete variety, so we can at least get something like this for a stratification of an arbitrary variety, so if we're willing to cheat horribly, we can use the disjoint union of these varieties, to do it, though my thought is that the dimension of the affine space over a point will be semicontinuous rather than constant, so it's of much less use. I don't see immediately how to get an irreducible one.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks, Charles! You are right, I should have added the connectedness condition in my posting: the point of all this is to find an affine variety which has the same homotopy type as a given variety (in the complex case). So the dimension of the fiber may vary, but the affine variety should be connected. $\endgroup$
    – algori
    Commented Dec 2, 2009 at 1:34
  • $\begingroup$ Is connected good enough? Or do you want irreducible? It seems to be that irreducible is the natural thing to hope for...is this known not to be true? $\endgroup$
    – Charles Siegel
    Commented Dec 2, 2009 at 12:59
  • $\begingroup$ Yes, connected is good enough. But for all I know, there are no reasons which would prevent one from finding an irreducible one, once $X$ is irreducible. $\endgroup$
    – algori
    Commented Dec 2, 2009 at 15:06
1
$\begingroup$

I might be missing something about question 3. Here's a simple construction:

Consider a projective space $P$ of dimension $\text{dim}\\, P_1 + \text{dim}\\, P_2$ that contains both $P_1$ and $P_2$ in general position. Then each point of $P-P_1-P_2\ $ lies on exactly one line connecting $x$ from $P_1$ with $y\in P_2$. Is this the kind of join you're looking for?

About question 2, I have a simpler thing that isn't clear to me (now posted as a question):

is a complex algebraic variety which is topologically contractible necessarily affine?

Improve this answer
$\endgroup$
3
  • $\begingroup$ You most definitely mean affine space there. $\endgroup$
    – Charles Siegel
    Commented Dec 2, 2009 at 19:34
  • $\begingroup$ Yes, thanks! Something got into my mind. $\endgroup$
    – Ilya Nikokoshev
    Commented Dec 2, 2009 at 19:39
  • $\begingroup$ Thanks, Ilya! In the topological join points are joined with segments, which are contractible. So if we want to mimic this, the points should be joined with $\mathbf{A}^1$'s, not $mathbf{P}^1$'s. Re your second question: there are many contractible affine varieties: take any affine cone i.e. a projective cone minus a hyperplane section not passing through the vertex. I don't know off hand whether there are any non-affine examples. $\endgroup$
    – algori
    Commented Dec 2, 2009 at 20:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .