Let $G$ be a (complex, compact, commutative) Lie group. Apparently the endomorphism ring $\textrm{End}(G)$ of $G$ (i.e., holomorphic group homomorphisms) acts on the cotangent space $T^\ast_eG$ at the identity $e \in G$.
If $\phi \in \textrm{End}(G)$ and $\omega \in T^\ast_eG$ then \begin{align} \phi \cdot \omega = \omega \circ d_e\phi \end{align} seems like a reasonable definition of this action.
I have two questions.
(i) Is this the correct action?
(ii) How can I show that the action is faithful?
I think that (ii) holds if I can show $\phi$ is a submersion. For example, if $\phi \cdot \omega = 0$ then $T^\ast_eG = \textrm{im}(d_e\phi) \subseteq \textrm{ker}(\omega)$ so $\omega = 0$. Moreover, $\phi$ will be a submersion if and only if it is an immersion, since it maps from a space to itself.
Some extra facts that may be of use:
(a) I really am interested in $G = \mathbf{C}^g/\Lambda$ where $\Lambda \subseteq \mathbf{C}^g$ is a lattice, i.e., $G$ is a $g$-dimensional torus.
(b) We can assume that any endomorphism is surjective with finite kernel. This is because my endomorphisms are isogenies of abelian varieties.
I found a solution using this question.
We have a diagram $\require{AMScd}$ \begin{CD} T^\ast_eG @>{d_e\phi}>> T^\ast_eG \\ @V{\textrm{exp}}VV @VV{\textrm{exp}}V\\ G @>{\phi}>> G. \end{CD}
Since $G$ is compact, the exponential map is surjective. Hence $\phi$ surjective implies $d_e\phi$ surjective. As remarked in the question, this shows that $\textrm{End}(G)$ acts faithfully on $T^\ast_eG$.
Please let me know if this solution is not correct.