= Positive definiteness of degree on elliptic-curve homomorphisms
{title2=$q(\phi)=\deg\phi$}
Let $q(\phi)=\deg\phi$ for nonzero elements of the <homomorphism group of elliptic curves>, and $q(0)=0$. For an <elliptic curve> $E$, the degree-two coordinate $x$ on a <Weierstrass model> has a double pole at its identity and identifies a generic point only with its inverse. Thus on $E\times E$,
$$
\operatorname{div}(x(P)-x(Q))=\Delta+\Delta_- -2(\{O\}\times E)-2(E\times\{O\}),
$$
where the two diagonals are $P=Q$ and $P=-Q$. The resulting <line bundle> isomorphism is $a^*\mathcal O_E(O)\otimes s^*\mathcal O_E(O)\cong\operatorname{pr}_1^*\mathcal O_E(O)^{\otimes2}\otimes\operatorname{pr}_2^*\mathcal O_E(O)^{\otimes2}$, for the sum and difference maps $a,s$. Pull it back along $(\phi,\psi)$ and take degrees. This gives the <degree parallelogram law>, even when one homomorphism equals plus or minus the other: bundle restriction remains valid in these cases.
It implies $q(n\phi)=n^2q(\phi)$ by the second-difference recursion, and polarization gives the symmetric bilinear form $B(\phi,\psi)=(q(\phi+\psi)-q(\phi-\psi))/4$. To check additivity, the parallelogram law gives $B(x+z,y)+B(x-z,y)=2B(x,y)$; interchange $x,z$ and use oddness to get $B(x+z,y)-B(x-z,y)=2B(z,y)$. Adding proves bilinearity. Since a nonzero isogeny has positive degree, $q$ is positive on every nonzero element. On every finite-dimensional rational span the matrix of $B$ is rational. Positivity on rational vectors implies nonnegativity on real vectors by density; a nontrivial kernel of a rational matrix would contain a nonzero rational vector. Hence the real extension is positive definite. This proves the <positive-definite quadratic form> property, rather than only a parallelogram identity.
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