Phrased in the language of vector bundles, the Quillen-Suslin theorem states that vector bundles on $\mathbb C^n$ are algebraically trivial (for any algebraic vector bundle there exists an algebraic isomorphism to the trivial bundle).

For more general affine varieties, Grauert's Oka principle implies that the holomorphic and topological classification of vector bundles coincide. In particular, all algebraic vector bundles which are topologically trivial are also holomorphically trivial.

As far as I understand, it is not known whether topologically trivial algebraic bundles on affine varieties are algebraically isomorphic to the trivial bundle.

If this would be the case for an affine variety $X$, I would call this an analogue of the Quillen-Suslin theorem for $X$.

My question is whether such analogues of the Quillen-Suslin theorem have been proved for affine varieties other than $\mathbb C^n$.

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    $\begingroup$ That fails already for topologically trivial algebraic line bundles on the affine open complement of a singleton subset of a smooth, projective curve of positive genus. $\endgroup$ May 12, 2019 at 21:26
  • $\begingroup$ @JasonStarr Thank you, I didn't know this. (Could you indicate how to see this?) I guess a family of counter-examples does not prevent this from being true for other affine varieties, or are you suggesting it's unlikely to hold for any affine variety? $\endgroup$
    – Earthliŋ
    May 12, 2019 at 21:31
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    $\begingroup$ @Earthlin You can compute $Pic( \Sigma - p)$ as $Pic(\Sigma)/\mathbb Z$, whereas every line bundle is topologically trivial because $H^2$ vanishes. $\endgroup$ May 13, 2019 at 2:33
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    $\begingroup$ There is a lot of research about this type of question in the motivic homotopy community. In particular, on smooth affine varieties (e.g. over C) the vector bundles are classified by (motivic) homotopy classes of morphisms to a Grassmannian. If the variety is motivic contractible, all algebraic vector bundles are isomorphic to a trivial one. $\endgroup$ May 14, 2019 at 7:50
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    $\begingroup$ @Earthliŋ sorry, the right keyword to google is "A¹-contractible" or "motivic homotopy equivalent to the base scheme" and the right paper to start looking at may be arxiv.org/abs/1402.4156 of Asok and Fasel, also arxiv.org/abs/1506.07093 of Asok-Hoyois-Wendt and references therein. $\endgroup$ May 14, 2019 at 15:39

1 Answer 1


More of an aside than anything, but we will need to distinguish between holomorphic and algebraic Quillen Suslin statements. Grauert's Oka principle can show a vector bundle is holomorphically trivial but on affine varieties holomorphically trivial doesn't imply algebraically trivial.

Expanding on the comments of Starr and Tosteson to illustrate this subtlety: when $X = Y \setminus \{p\}$ is the complement of a point on a smooth projective curve $Y$ of genus $g> 0$, $\mathrm{Pic}(X) \simeq \mathrm{Pic}(Y) / \mathbb{Z} \simeq \mathrm{Pic}^{0}(Y)$ where the last isomorphism is obtained from $\mathrm{Pic}(Y) \simeq \mathrm{Pic}^{0}(Y) \times \mathbb{Z}$ (an exercise in Hartshorne II.6). On the other hand it is true that all line bundles on $X$ are holomorphically trivial: this can be seen from the exponential exact sequence $$ \cdots \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_X^\times) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}_X) \to \cdots $$ And vanishing for higher cohomology of $\mathcal{O}_X$ (Cartan AB) and $ H^2(X, \mathbb{Z})$ $X$ is homotopy equivalent to a wedge of $2g$ circles).

A neccessary conditon

If $X$ is a smooth variety and every algebraic vector bundle on $X$ is trivial, then $K_{0}(X) = \mathbb{Z}$, generated by the class of $\mathcal{O}_{X}$, a severe restriction, for instance after tensoring with $\mathbb{Q}$ we have $\mathrm{CH}^{*}(X) \otimes \mathbb{Q} \simeq K_{0}(X) \otimes \mathbb{Q} = \mathbb{Q}$.

In the example above a key point was $\mathrm{CH}^*(X) \not \simeq \mathbb{Q}$.

One answer to the question

A conjecture of Anderson/theorem of Gubeladze shows that if $X$ is an affine toric variety, then every vector bundle on $X$ is free. The primary source is

I. Gubeladze, The Anderson conjecture and projective modules over monoid algebras, Soobshch. Akad. Nauk Gruzin. SSR 125 (1987), 289–291

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    $\begingroup$ Does the existence of $\tilde{E}$ on $X\times I$ imply that $E$ is holomorphically trivial on $X$? Is there some sort of theorem that says that holomorphic v.b. on $X$ are in bijection with holomorphic-at-every-time-homotopy classes of holomorphic maps from $X$ into the Grassmannian? $\endgroup$
    – Qfwfq
    May 13, 2019 at 23:59
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    $\begingroup$ Thank you very much for your answer. Your "homotopy" point of view of Grauert's theorem seems to be at odds with the interpretation I stated in my question. In particular, making a continuous family (over $I$) between two holomorphic vector bundles into a holomorphic family doesn't say anything about the isomorphism type of the bundles at $0$ and $1$, as I guess @Qfwfq was wondering. $\endgroup$
    – Earthliŋ
    May 14, 2019 at 9:17
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    $\begingroup$ I had in mind Cor. 3.3 of J. Leiterer's chapter "Holomorphic vector bundles and the Oka-Grauert principle" in Several Complex Variables IV which states: If $E$ and $F$ are holomorphic vector bundles of a Stein analytic space $X$ which are continuously isomorphic, then $E$ and $F$ are also holomorphically isomorphic. $\endgroup$
    – Earthliŋ
    May 14, 2019 at 11:06
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    $\begingroup$ I appear to have accidentally invoked GAGA for an affine variety -- for $X$ the curve of positive genus minus a point all hlomorphic line bundles are trivial even though many algebraic ones are non-trivial! Thanks again for your comments, revisions forthcoming. $\endgroup$
    – cgodfrey
    May 14, 2019 at 16:33
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    $\begingroup$ To summarize this answer in the smooth case, topologically trivial vector bundles over smooth affine varieties are known to be trivial for $\mathbb C^n \times (\mathbb C^∗)^m$, the case $m=0$ being the classical Quillen-Suslin theorem and $m \neq 0$ being the slightly more general affine toric case of Gubeladze. $\endgroup$
    – Earthliŋ
    May 15, 2019 at 18:23

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