Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

From a post to The Jouanolou trick:

Are all topologically trivial (contractible) complex algebraic varieties necessarily affine? Are there examples of those not birationally equivalent to an affine space?

The examples that come to my mind are similar to a singular $\mathbb P^1$ without a point given by equation $x^2 = y^3$. This particular curve is clearly birationally equivalent to affine line.

Perhaps the "affine" part follows from a comparison between Zariski cohomology and complex cohomology?

share|improve this question
Same comment as on that one, you mean affine SPACE in your question, not affine line. $\mathbb{A}^2$ is definitely contractible. –  Charles Siegel Dec 2 '09 at 19:34
add comment

2 Answers

up vote 17 down vote accepted

No. Counterexamples were first constructed by Winkelmann, as quotients of $\mathbb A^5$ by algebraic actions of $\mathbb G_{\text{a}}$. I learned this from Hanspeter Kraft's very nice article available here:


Recently Aravind Asok and Brent Doran have been studying these kinds of examples in the setting of $\mathbb A^1$-homotopy theory, on the arxiv as math/0703137.

share|improve this answer
This is a very interesting example -- I wonder if it's also not birational to affine line? –  Ilya Nikokoshev Dec 2 '09 at 20:22
The first link's paper is really nice! –  Csar Lozano Huerta Dec 3 '09 at 6:19
add comment

About the rationality of contractible varieties: Yes for curves and surfaces and is an open question for higher dimensions.

Any such contractible variety $X$ has $\chi_{top}(X)=1$, obviously.

If $X$ is a curve then it must have only cusps as singularities, if any, by a simple $\chi_{top}$ calculation. Now let $Y$ be a projective model of $X$ such that it is smooth at the points in $Y-X$. Topologically, $Y$ is a real surface without boundary such that a few punctures make it contractible. The only real surface with this property is $S^2$, obviously. Hence $Y$ better be rational and so is $X$.

If $X$ is an algebraic surface then it was a conjecture of Van de Ven that such a surface must be rational (actually his conjecture is for any homologically trivial $X$). This was proved by Gurjar & Shastri in:

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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