If I am correct, an endomorphism is a homomorphism from one group to itself. In the case of an elliptic curve, we'd need a map $\phi: E \rightarrow E$. Ive seen in some places that you can get such a map by defining $\phi_n(P)=[n]P=P+P...+P$ (n times). This means if we represent P as $(\wp(t),\wp'(t))$, (over some arbitrary lattice) $\phi_n((\wp(t),\wp'(t)))=(\wp(nt),\wp'(nt))$.

Using this definition of a endomorphism found that $\phi_i$ is also a valid map if $E$ is parametrized by $(\wp(t,[1,i]),\wp'(t,[1,i]))$ since $\phi_i((\wp(t,[1,i]),\wp'(t,[1,i])))=(\wp(it,[1,i]),\wp'(it,[1,i]))=(-\wp(t,[1,i]),i\wp'(t,[1,i]))$.

That case was simple because the lattice $[1,i]$ is unchanged when divided by $i$. In a more general case where $E$ is parametrized by $(\wp(t,\Lambda),\wp'(t,\Lambda))$, how would one find other endomorphisms of E? Also, how does one prove these are homomorphisms? I havent been able to find any sort of reasoning for using such a map and think it would be useful to actually see why this is a homomorphism. Thanks for any help.


You are asking for all the holomorphic (because they are algebraic) endomorphisms of $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice.

Note that they correspond to holomorphic endomorphisms $f:\mathbb{C} \rightarrow \mathbb{C}/\Lambda$.

They lift to the universal cover into $g: \mathbb{C} \rightarrow \mathbb{C}$ with $g(0)=0$. Thus $(x,y) \rightarrow g(x+y)-g(x)-g(y)$ is continuous and $\Lambda$-valued, therefore $g(z)=\alpha z$ for some $\lambda \in \Lambda$.

Since $\Lambda \subset \ker(g)$, $\alpha \Lambda \subset \Lambda$.

Assume $f$ is not trivial: thus $\alpha$ is not an integer.

$x \in \Lambda \longmapsto \alpha x \in \Lambda$ can be seen as an endomorphism of $\mathbb{Z}^2$. Let $\chi$ be its characteristic polynomial: by “Cayley-Hamilton”, $\chi(\alpha)=0$, thus $\alpha$ is a quadratic integer, and $\Lambda=e_1(\mathbb{Z}1\oplus\mathbb{Z}\alpha)$.

Thus, for most lattices $\Lambda$, (each one but countably many, up to similarity), the only endomorphisms of $\mathbb{C}/\Lambda$ are integer multiples of identity.

On the other lattices, there are some “exotic” multiplications allowed (they are the “complex multiplication” or CM elliptic curves/lattices), but all endomorphisms are multiplications by some complex number $\alpha$ such that $\alpha\Lambda \subset \Lambda$. The set of such $\alpha$ is a $\mathbb{Z}[\alpha_0]$ for some quadratic integer $\alpha_0$.

  • $\begingroup$ Is there any way to think about this in terms of the weistrass function? Why is $\wp(\alpha z,\Lambda)$ special when $\alpha \Lambda \subset \Lambda$? I believe I saw somewhere that $\wp(\alpha z,\Lambda)$ is a rational function in $\wp(z,\Lambda)$ if $\alpha \Lambda \subset \Lambda$, is this related to the fact to the endomorphisms of E or is it just a coincidence? $\endgroup$ – uhhhhidk Jul 26 at 20:44
  • $\begingroup$ I think I understand now: I forgot the condition that $\phi_a (O)=O$. The only $a$ that satisfy this property are $a$ such that $a\Lambda \subset \Lambda$ because $(\wp (a\lambda), \wp'(a\lambda))$ ($\lambda \in \Lambda$) wont be equal to the identity unless $a\Lambda \subset \Lambda$ $\endgroup$ – uhhhhidk Jul 26 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.