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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.

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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 $f_* \mathcal O_G=\mathcal O_S$.

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