 Positive definiteness of degree on elliptic-curve homomorphisms

Let $q(\varphi )=\mathrm{deg}\varphi$ 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 $E\times E$,where the two diagonals are $P=Q$ and $P=-Q$. The resulting line bundle isomorphism is $a^{\ast }{\mathcal{O}}_E(O)\otimes s^{\ast }{\mathcal{O}}_E(O)\cong {\mathrm{pr}}_1^{\ast }{\mathcal{O}}_E(O{)}^{\otimes 2}\otimes {\mathrm{pr}}_2^{\ast }{\mathcal{O}}_E(O{)}^{\otimes 2}$, for the sum and difference maps $a,s$. Pull it back along $(\varphi ,\psi )$ 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 $q(n\varphi )=n^{2}q(\varphi )$ by the second-difference recursion, and polarization gives the symmetric bilinear form $B(\varphi ,\psi )=(q(\varphi +\psi )-q(\varphi -\psi ))/4$. 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.