Let $A$ be an abelian variety.(i.e. It has a non-degenerate positive divisor $D$ which has positive definite Hermitian form $H$.)
Polarization on $A$ is a set $P$ of positive non-degenerate divisors s.t.
- $\forall a,b\in P,\exists m,n\in Z$ $na$ and $mb$ have the same Hermitian form $T$.
- $P$ is maximal subject to (1)
$\textbf{Q.1:}$Do I need $P$'s divisor's Hermitian form all equal to $H$ here?
$\textbf{Q.2:}$It is clear that if $na,mb$ have the same hermitian form then $na-mb$ is linear equivalent to $D-D_x$ where $D_x$ is translation by $x$. Then $na,mb$ induces projective embedding differs by a translation by $x$. The books says they also differ by a veronese embedding. Where is veronese embedding coming from? If necessary, I could take $3na$ and $3mb$ and both of which sections are globally generated.
$\textbf{Q.3:}$ The book also says "any projective embedding determines a polarization. " Any projective embedding guarantees existence of a non-degenerate positive divisor $D$. Where does polarization coming into the play?
$\textbf{Q.4:}$ Suppose given $\phi:A\to\hat{A}$ where $\hat{A}$ is the dual of $A$ as abelian manifold and $\phi$ depends only on Hermitian form $H$ as before. Then Rosati involution for $End_0(A)=End(A)\otimes_ZQ$ is defined as $\alpha\in End_0(A)\to\phi^{-1}\alpha^T\phi$ where $\alpha$ is the formal adjoint of $\alpha$. Book claim that Rosati involution depends on polarization.(How does polarization determines Hermitian form $H$ here?) Why so?
Ref: Analytic Theory of Abelian Varieties by Swinnerton-Dyer.