tangent space of quotient algebraic group

Question

Let $A$ be an abelian variety, $B\subseteq A$ a closed algebraic subgroup and $A/B$ the quotient abelian variety. Moreover, let $F\colon A \rightarrow A$ be an algebraic endomorphism with $F(B)=B$, so that $F$ descends to a well-defined endomorphism $\bar{F}\colon A/B \rightarrow A/B$. What is the relation between the tangent space at the identity of $A$, i.e. $\operatorname{Lie}(A)$, and that of the quotient $A/B$, i.e. $\operatorname{Lie}(A/B)$? In particular, why does $\operatorname{Lie} \bar{F}$ have the same eigenvalues on $\operatorname{Lie}(A/B)$ as $\operatorname{Lie} F$ on $\operatorname{Lie}(A)$?

Answer 1

I will write $\mathfrak a:=\operatorname{Lie}(A)$, $\mathfrak b:=\operatorname{Lie}(b)$ and also $\mathfrak f:=\operatorname{Lie}(F)$.

Note that $\mathfrak b$ is a Lie subalgebra of $\mathfrak a$, and $\operatorname{Lie}(A/B)=\mathfrak a/\mathfrak b$. A reference for this statement would be Theorem 25.4.11 in the Book by Tauvel and Yu, Lie Algebras and Algebraic Groups.

Now, since $F(B)=B$, it follows that $\mathfrak f(\mathfrak b)=\mathfrak b$. Linear Algebra tells us that the minimal polynomial of $\mathfrak f|_{\mathfrak b}$ is a factor of the minimal polynomial of $\mathfrak f$. In particular, we can choose a vector space complement $\mathfrak a=\mathfrak b\oplus\mathfrak K$ such that $\mathfrak f$ has a block decomposition $$ \mathfrak f=\begin{pmatrix} \mathfrak f|_{\mathfrak b}& 0 \\ 0 & \mathfrak g \end{pmatrix} $$ where $\mathfrak g$ represents the induced endomorphism $\mathfrak a/\mathfrak b\to \mathfrak a/\mathfrak b$.