Let $S$ be a noetherian scheme (possibly nicer assumptions) and $G$ a smooth group scheme of finite type over $S$ with geometrically integral fibers with structure morphism $f: G\rightarrow S$ and $f_*\mathcal O_G =\mathcal O_S$ universally.

For $n\geq 0$ let $G_n$ denote the $n$-th infinitesimal neighborhood of the zero section of $S$ in $G$ and by denote $f_n: G_n \rightarrow S$ its structure morphism and denote by $g_n: S \rightarrow G_n$ the obvious closed immersion.

Let $\mathcal V$ be a coherent sheaf on $G_n$ with the following property:

$(*)$ The direct image of $\mathcal V$ along $f_n$ is locally free of finite type on $S$.

The question is if one can conclude that then the following holds:

$(**)$ The pullback $g_n^* \mathcal V$ of $\mathcal V$ along $g_n$ is locally free of finite type on $S$.

One may feel free to add additional assumptions on $G$ or $S$ or $\mathcal V$ in a discussion about when $(**)$ would hold, but the general case is of course very interesting.

Note: One could of course also ask if $(*)$ already implies that $\mathcal V$ itself is locally free of finite rank, but I guess this would be false in general.


1 Answer 1


No. Let $G=\mathbb A^2$ with the projection map to $S=\mathbb A_1$. It is a relative group scheme under addition of the $y$ coordinate. Take the module on $G_2=\textrm{Spec} k[x,y]/y^2$ with two generators, $a$ and $b$, under the relations $ya-xb=0$, $yb=0$. This gives a coherent sheaf. The direct image is the free $k[x]$-module on two generators, defining a locally free sheaf. The pullback is $k[x]\oplus k[x]/x$, which does not define a locally free sheaf.

Edit: Let $E$ be an elliptic curve and $G=E\times S$. Then $G_n$, for all $n$, is the same as in this example, and $f_* \mathcal O_G=\mathcal O_S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.