ScalingStacks

Proof. We follow similar arguments to [Sau16, Lemma 1].

By the explicit description of nilpotent representations, $M$ is a direct sum of modules of the form $E(i,l).$ We can hence choose a basis of $\operatorname{soc}(M)$ by a choosing non-zero vectors in each $\operatorname{soc}(E(i,l)).$ With respect to this basis, $P$ is generated by elementary matrices and the proof can be reduced to the case $M=E(i,l_{1})\oplus E(i,l_{2}).$ We denote the standard basis vectors of $E(i,l_{1})$ and $E(i,l_{2})$ by $e_{j}$ and $f_{j},$ respectively.

The socle of $\operatorname{soc}(M)=\operatorname{soc}(E(i,l_{1}))\oplus\operatorname{soc}(E(i,l_{2}))=ke_{i}\oplus kf_{i}$ is two-dimensional in degree $i.$ Assume that $l_{1}>l_{2}.$ Then the radical filtration of the socle is

$$I^{\prime}=(0\subset ke_{i}\subset ke_{i}\oplus kf_{i}).$$

Let $g\in P.$ For $j<l_{2}$ we define the action $\theta(g)$ on $ke_{i-j}\oplus kf_{i-j}$ via the natural isomorphism $ke_{i-j}\oplus kf_{i-j}\cong ke_{i}\oplus kf_{i}.$ For $j\geq l_{2},$ we define the action $\theta(g)$ on $ke_{i-j}$ via the natural isomorphism $ke_{i-j}\cong ke_{i}.$ It can easily be checked that $\theta(g)\in\operatorname{Aut}(M)$ and $\operatorname{Res}\theta(g)=g.$ The case $l_{1}=l_{2}$ is proven similarly. ∎