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 $m$?

10$\begingroup$ 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. $\endgroup$ – Piotr Achinger May 24 '19 at 14:17

7$\begingroup$ 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.) $\endgroup$ – naf May 24 '19 at 22:20

2$\begingroup$ @ulrich indeed, the zero section gives a decomposition of the identity map $X \to Y \to X$ and the pullback 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. $\endgroup$ – Anton Mellit May 25 '19 at 15:26

$\begingroup$ Nice, it is indeed much simpler! $\endgroup$ – naf May 26 '19 at 3:14

$\begingroup$ So, to sum up: a) $E$ is a trivial bundle over $X$ of rank $nm$. 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 $X=\mathbb C^m$. $\endgroup$ – 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 QuillenSuslin 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 )
This answer is "Community Wiki".


1$\begingroup$ 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 $S= \operatorname{Spec} \mathbf{C}$). $\endgroup$ – Piotr Achinger Jun 2 '19 at 18:30