i hope this is the right place for this.
In <https://math.berkeley.edu/~ogus/preprints/anonhodge.pdf> on page 15 Ogus and Vologodsky state the following: 


Let $\pi\_T: \textbf{T} \rightarrow X$ be a vector group and let $T$ be its sheaf of sections. Thus $T$ is a locally free sheaf of $\mathcal{O}\_X$-modules with dual $\Omega$ and $\textbf{T}=Spec\_X(Sym(\Omega))$. The pairing $T \times \Omega \rightarrow \mathcal{O}\_X$ extends to a pairing $T \times Sym(\Omega) \rightarrow Sym(\Omega)$, where sections of $T$ act as derivations on $Sym(\Omega)$. 
The action defines a map $\xi \mapsto D\_{\xi}: T \rightarrow {\pi\_T}\_{\star} T\_{\textbf{T}/X}$ which identifies T with the sheaf of translation invariant vector fields of $\textbf{T}$ relative to $X$. It also induces an isomorphism $ \pi\_T ^{\star}T \rightarrow T\_{\textbf{T}/X}$.


**Question:** First of all the phrase "vector group" (in this context) is unfamiliar to me, for me it looks like a vector bundle. Then i do not understand how we extend $T \times \Omega \rightarrow \mathcal{O}\_X $ to $T \times Sym(\Omega) \rightarrow Sym(\Omega)$ such that 


1. sections of $T$ act as derivations on $Sym(\Omega)$, and
2. $\xi \mapsto D\_{\xi}$ (what is $D\_{\xi}$ here?) identifies $T$ with the sheaf of translation invariant vector fields of $\textbf{T}$.


My guess at this point is that I am misinterpreting the term "vector group", which I know to be a term in Lie theory. This, and the fact that we know that the space of translation invariant vector fields of a Lie group can be identified with the tangent space (at the identity), make me believe the answer to lie in this direction. Any help would be very much appreciated.



Vector groups in Ogus-Vologodskys Nonabelian Hodge theory in characteristic p
Arthur Ogus was so kind as to explain this to me in an Email:


The use of the phrase "vector group" was motivated by the similarities to Lie theory (as stated in my question).


As to the construction: For every $\xi \in T$ we get a map $ Sym^k( \Omega) \rightarrow Sym^{k-1} ( \Omega)$ by interior multiplication by $\xi$, adding these up we get a map $D\_{\xi}: Sym( \Omega) \rightarrow Sym(\Omega)$ which is a derivation.


