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?

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    $\begingroup$ Same comment as on that one, you mean affine SPACE in your question, not affine line. $\mathbb{A}^2$ is definitely contractible. $\endgroup$ Dec 2, 2009 at 19:34

2 Answers 2


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:

Challenging problems on affine $n$-space.

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 On unipotent quotients and some A^1-contractible smooth schemes.

  • $\begingroup$ This is a very interesting example -- I wonder if it's also not birational to affine line? $\endgroup$ Dec 2, 2009 at 20:22
  • $\begingroup$ The first link's paper is really nice! $\endgroup$ Dec 3, 2009 at 6:19
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    $\begingroup$ The link to H. Kraft's paper is broken $\endgroup$
    – Qfwfq
    Feb 25, 2016 at 9:33
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    $\begingroup$ @Qfwfq, I fixed the link to Kraft's paper (I presume that I have the right one, and there is also a citation to Winkelmann's paper there). $\endgroup$ Feb 25, 2016 at 22:26

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 $\chi_\text{top}(X)=1$, obviously.

If $X$ is a curve then it must have only cusps as singularities, if any, by a simple $\chi_\text{top}$ 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:


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