Relative tangent bundle of a twisted Grassmann variety

Question

Assume $S$ is a scheme over $\mathbb{C}$ (as nice as you want), $\mathcal{E}$ is a locally free $\mathcal{O}_S$-module and $\mathcal{A}$ is a coherent $\mathcal{O}_S$-algebra, locally free of finite rank, not neccesarily commutative. We assume that $\mathcal{E}$ has the structure of an $\mathcal{A}$-module.

According to [1, Exercise 8.14] one can construct the Grassmannian $\pi: Gr_k^{\mathcal{A}}(\mathcal{E})\rightarrow S$, classifying $\mathcal{A}$-submodules of $\mathcal{E}$, which are locally free $\mathcal{O}_S$-modules of rank $k$.

This is a closed subscheme of the usual Grassmannian $Gr_k(\mathcal{E})$, hence the universal subsheaf $\mathcal{S}$ and the universal quotient $\mathcal{Q}$ on $Gr_k(\mathcal{E})$ restrict to a universal $\mathcal{A}$-subsheaf $\mathcal{S}^{\mathcal{A}}$ and a universal $\mathcal{A}$-quotient $\mathcal{Q}^{\mathcal{A}}$ on $Gr_k^{\mathcal{A}}(\mathcal{E})$.

Is the relative tangent sheaf of $\pi: Gr_k^{\mathcal{A}}(\mathcal{E})\rightarrow S$ given by $\mathcal{T}_{Gr_k^{\mathcal{A}}(\mathcal{E})/S}=\mathcal{H}om_{\mathcal{A}}(\mathcal{S}^{\mathcal{A}},\mathcal{Q}^{\mathcal{A}})$?

If $\iota: Gr_k^{\mathcal{A}}(\mathcal{E}) \hookrightarrow Gr_k(\mathcal{E})$ deontes the closed $S$-immersion, we should have an injection $\mathcal{T}_{Gr_k^{\mathcal{A}}(\mathcal{E})/S} \rightarrow \iota^{*}\mathcal{T}_{Gr_k(\mathcal{E})/S}=\iota^{*}\mathcal{H}om(\mathcal{S},\mathcal{Q})=\mathcal{H}om(\mathcal{S}^{\mathcal{A}},\mathcal{Q}^{\mathcal{A}})$. So I guessed that the subsheaf should consist of those morphisms which are compatible with the $\mathcal{A}$-action. Is this true? If not: is there any description of the relative tangent sheaf in terms of the known bundles in the situation? Is there a class of algebras for which we can describe the relative tangent bundle in this way?

[1] - Görtz & Wedhorn - Algebraic Geometry I

Answer 1

I think the answer is yes. I will work out the proof when $S = \textrm{Spec}(\mathbb{C})$, but it certainly also works for $S$ being any scheme (stack?).

So let $\mathcal{A}$ be a finitely generated algebra over $\mathbb{C}$ and let $\mathcal{E}$ be a finite dimensional vector space with a structure of (left) $\mathcal{A}$-module.

I denote by $\textrm{GL}_{\mathcal{A}}(\mathcal{E})$ the group of automorphisms of $\mathcal{E}$ which respect the $\mathcal{A}$-module structure. Then $Gr^{\mathcal{A}}_k(\mathcal{E})$ is a homogeneous space for $\textrm{GL}_{\mathcal{A}}(\mathcal{E})$.

Hence, if $\mathcal{F} \subset \mathcal{E}$ be a subvector space of rank $k$ which has also a $\mathcal{A}$-module structure, then the tangent space to $Gr^{\mathcal{A}}_k(\mathcal{E})$ at $\mathcal{F}$ is the quotient of the Lie algebra of $\textrm{GL}_{\mathcal{A}}(\mathcal{E})$ by the Lie algebra of the stabilizer of $\mathcal{F}$.

The Lie algebra of $\textrm{GL}_{\mathcal{A}}(\mathcal{E})$ is $\textrm{Hom}_{\mathcal{A}}(\mathcal{E}, \mathcal{E})$. One also checks that the Lie algebra of the stabilizer of $\mathcal{F}$ is $\{ \phi \in \textrm{Hom}_{\mathcal{A}}(\mathcal{E}, \mathcal{E}), \phi(\mathcal{F}) \subset \mathcal{F} \}$.

It is now easily seen that the quotient of these two Lie algebras is $\textrm{Hom}_{\mathcal{A}}(\mathcal{F}, \mathcal{E}/\mathcal{F})$.