Positive definiteness of degree on elliptic-curve homomorphisms ($q(\varphi )=\mathrm{deg}\varphi$)

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.

With zero degree assigned to the zero element of the homomorphism group of elliptic curves, the degree parallelogram law makes this symmetric pairing additive in both arguments. Its diagonal is the degree of an isogeny; it is positive definite on every finite-dimensional rational span after extension to the real numbers. The integral polarization $2B(\varphi ,\psi )$ is integer-valued, while $B$ can be half-integral. For instance, if an elliptic curve has an automorphism $\omega$ with ${\omega }^2+\omega +1=0$, then $1+\omega =-{\omega }^2$ has degree one, so $B(1,\omega )=-1/2$.