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 $m$?



Can a non-trivial algebraic variety carry a vector bundle whose total space is affine space?
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](http://www.mathematik.uni-muenchen.de/~morel/Prepublications/A1TopologyLNM.pdf)), 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.



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


