be the quotient abelian surface $A/G$ and let $\varphi:A\to B$ be the natural map. Then there is a unique principal polarization $\mu$ of $B$ that makes the following diagram commute: $$\begin{matrix}
      A      & {\mathop{{\relbar\joinrel\longrightarrow}}\limits^{n\lambda}} &      \hat{A}        \\
{\vbox{\vbox to 4pt{}\vbox{\hbox{
     \Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle\varphi$}}$}}\vfill}}} &                & {\vbox{\vbox to 4pt{}\vbox{\hbox{
     \Big\uparrow\rlap{$\vcenter{\hbox{$\scriptstyle\hat{\varphi}$}}$}}\vfill}}} \\
      B      &   {\mathop{{\relbar\joinrel\longrightarrow}}\limits^{\mu}}    &      \hat{B}
\end{matrix}$$

Now suppose that $A$ has an automorphism $\alpha$ such that $G':=\alpha(G)$ is also an isotropic subgroup of $A[n]$, and let $(B',\mu')$ be the principally polarized abelian surface obtained from $G'$ as above.

\[P-basic\] The automorphism $\alpha$ of $A$ provides an isomorphism $B\to B'$. If we identify $B'$ with $B$ via this automorphism, then $\mu' = \hat{\beta} \mu \beta$, where $\beta$ is the image of $\alpha^{-1}$ in $({\operatorname{End}}B)\otimes {{\mathbb Q}}$.

We have a commutative diagram with exact rows: $$\begin{matrix}
0 &{\longrightarrow}&      G     &{\longrightarrow}&      A     &{\mathop{{\relbar\joinrel\longrightarrow}}\limits^{\varphi}} &B  &{\longrightarrow}& 0\\
  &     &{\vbox{\vbox to 4pt{}\vbox{\hbox{
     \Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle\alpha$}}$}}\vfill}}} &     &{\vbox{\vbox to 4pt{}\vbox{\hbox{
     \Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle\alpha$}}$}}\vfill}}} &              &   &     &  \\
0 &{\longrightarrow}&      G'    &{\longrightarrow}&      A     &{\mathop{{\relbar\joinrel\longrightarrow}}\limits^{\varphi'}}&B' &{\longrightarrow}& 0,
\end{matrix}$$ where $\varphi$ and $\varphi'$ are the natural maps from $A$ to $B$ and $B'$, respectively. Completing the diagram, we find an isomorphism $B\to B'$. This proves the first statement of the theorem. The second statement 