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Practically the first property of vector bundles encountered in topology is that the pullbacks of a bundle over $Y$ by homotopic maps $X \to Y$ are isomorphic over $X$ (for reasonable spaces). Now let's consider similar situation in the algebraic setting: let $X$ be a good enough scheme over a field, and $E$ a vector bundle (locally free sheaf of finite rank) over $X \times \mathbb{A}^1$. Is it true that restrictions of $E$ to $X \times \{0\}$ and $X \times \{1\}$ are isomorphic as vector bundles over $X$? Is it even true for affine $X$?

The idea of the proof of topological statement doesn't seem to be applicable, nor does attempting to follow it immediately produce counterexamples.

Dmitry
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  • Do you assume $X$ is normal ? – Cantlog Nov 03 '13 at 17:24
  • Yeah, let it be normal. Perhaps even smooth, if that simplifies the proof. – Dmitry Nov 03 '13 at 17:38
  • Search for "Homotopy invariance" and "algebraic K-theory". It holds when $X$ is smooth. – Martin Brandenburg Nov 03 '13 at 18:07
  • @MartinBrandenburg, as far as I've understood after some superficial browsing, my question is way easier than what's usually associated with A^1 homotopy theory. Sources I've found (according to their introductions) are mostly concerned with existence of universal bundles, and are studying whether (roughly) every vector bundle over $X \times \mathbb{A}^1$ is a pullback of a bundle from $X$. I don't need anything this strong. Surely two bundles over a product space may be different but still restrict to each fiber isomorphically. – Dmitry Nov 03 '13 at 18:33
  • @Dmitry: I think this is essentially the same. As you can't really distinguish any point of $A^1$ from another one, what you want is $t \mapsto E|_{X\times { t}}$ is constant. – Cantlog Nov 03 '13 at 20:31
  • @Cantlog, does this mapping reveal a significant part of the information about $E$ itself? I've always thought it does not (as a trivial example, take $X$ to be a point: any vector bundle restricts the same to the point; does a positive dimensional $X$ really change anything?). Besides, the codomain of the map is, uhm, a set of isomorphism classes of vector bundles, so it ignores the (quite non-canonical) choice of isomorphism between restrictions to two different copies of $X$, while making coherent choices is crucial in studying vector bundles over a product, as I understand it. – Dmitry Nov 03 '13 at 20:50
  • Yes I meant $t\mapsto$ the isomorphism class of the restriction of $E$. It tells nothing on the isomorphisms. – Cantlog Nov 03 '13 at 23:23

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If $X=Spec(A)$ is smooth affine over some field $k$ then any vector bundle over $X \times \mathbb{A}^n$ is in fact a pullback of some vector bundle over $X$, in other words any family of vector bundles is trivial over smooth affine scheme. This is a non-tivial result, it was proved by Lindel (see Lindel H. "On the Bass-Quillen conjecture concerning projective modules over polynomial rings").

I don't know examples over singular affine $X$, but I think Lindel's theorem can not be true in this case.

Of course, for projective varieties we have such families in abundance, this is subject of deformation theory of sheaf. The simplest example that comes to mind: take $\mathbb{P}^1$, extensions $$ 0 \to \mathcal{O}(-1) \to E \to \mathcal{O}(1) \to 0, $$ are parametrized by $Ext^1(\mathcal{O}(1), \mathcal{O}(-1)) \cong k$, then take universal affine extension over $\mathbb{P}^1 \times \mathbb{A}^1$, that is a non-trivial family.

Alex
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