# Can a non-trivial algebraic variety carry a vector bundle whose total space is affine space?

Suppose $$X$$ is an algebraic variety over $$\mathbb{C}$$, and let $$Y\to X$$ be an algebraic vector bundle. Suppose $$Y$$ is algebraically isomorphic to $$\mathbb{C}^n$$ for some $$n$$. Does it follow that $$X$$ is algebraically isomorphic to $$\mathbb{C}^m$$ for some $$m$$?

• Hi Anton! This sounds related to what is known as the Cancellation Problem, asking if $X\times \mathbf{A}^n \simeq Y\times \mathbf{A}^n$ implies $X\simeq Y$. The answer to the general question is no (Google "Danielewski surfaces") and it is studied quite extensively (see e.g. papers of Dobouloz and Jelonek). For $X = \mathbf{A}^m$ as in your case, the answer is yes for $m\leq 2$, no for $m>2$ in positive characteristic arxiv.org/abs/1208.0483 , and as far as I can google, it is open for $m>2$ in characteristic zero. I'm hope someone more knowledgeable will weigh in. – Piotr Achinger May 24 '19 at 14:17
• As @PiotrAchinger says, this is unknown even for trivial bundles. In fact, your assumptions imply that the bundle must be trivial. Firstly, $X$ must be affine since a vector bundle always has a zero section. $X$ is also $\mathbb{A}^1$-contractible since $Y$, being affine space, is so Finally, any vector bundle on an $\mathbb{A}^1$-contractible (smooth) affine variety is trivial by a theorem of Morel. (There might well be a simpler proof.) – naf May 24 '19 at 22:20
• @ulrich indeed, the zero section gives a decomposition of the identity map $X \to Y \to X$ and the pull-back of the bundle to $Y$ is trivial because $Y$ is the affine space. So the pullback to $X$ must be trivial, which is the original bundle because $X\to X$ is identity. – Anton Mellit May 25 '19 at 15:26
• Nice, it is indeed much simpler! – naf May 26 '19 at 3:14
• So, to sum up: a) $E$ is a trivial bundle over $X$ of rank $n-m$. b) It is not known for $m\gt 2$ whether $X$ is isomorphic as an algebraic variety to the affine space $\mathbb C^m.$ c) For $m\leq 2$ however we do know that $X=\mathbb C^m$. – Georges Elencwajg Jun 1 '19 at 22:19

Summing up the discussion in the comments:

As user ulrich observed, the vector bundle has to be trivial. First, since $$X$$ is a closed subscheme of $$Y\simeq \mathbf{A}^n$$ via the zero section, it has to be affine. It is also smooth. Finally, it is $$\mathbf{A}^1$$-contractible since $$Y$$ is, since $$Y\to X$$ induces an equivalence in $$\mathbf{A}^1$$-homotopy. We conclude by a theorem of Morel (see Chapter 7 in $$\mathbf{A}^!$$-algebraic topology over a field, here), saying that vector bundles on a smooth affine $$\mathbf{A}^1$$-contractible variety are trivial.

A simpler argument using Quillen-Suslin was given by Anton: $$X$$ is a retract of $$Y \simeq \mathbf{A}^n$$ via the zero section, and since every vector bundle on $$Y$$ is trivial, the same is true for $$X$$.

This turns the question into an important special case of the Cancellation Problem.

Cancellation Problem. If $$X\times \mathbf{A}^m \simeq Z \times \mathbf{A}^m$$, can we conclude that $$X\simeq Z$$?

The answer to this general problem is no (there are famous counterexamples already in dimension two, known as Danielewski surfaces). However, in the special case where $$Z$$ is an affine space ($$\mathbf{A}^{r}$$, $$r+m=n$$ in our case) the answer is known to be yes for $$r\leq 2$$, and open for $$r>2$$. (In positive characteristic, the answer is no for all $$r>2$$: arxiv.org/abs/1208.0483 )

• Because it is covered by Zariski opens on which the vector bundle is trivial, and $X\times \mathbf{A}^r$ is smooth if and only if $X$ is. Alternatively, apply Stacks Project tag 02K5 stacks.math.columbia.edu/tag/02K5 (with roles of $X$ and $Y$ switched and $S= \operatorname{Spec} \mathbf{C}$). – Piotr Achinger Jun 2 '19 at 18:30