Isomorphic elliptic curves are projectively equivalent Let $E_1$, $E_2 \subset \mathbb{P}^d$ be two smooth elliptic curves, that are isomorphic as abstract curves. How can one prove that they are projectively equivalent? That is there is a automorphism $\phi \in \mathbb{P}GL(d+1)$ s.t. $\phi(E_1)=E_2$. 
 A: This is only correct for elliptic normal curves, i.e for genus 1 curves in $\mathbb{P}^r$ with degree $r+1$. To see this, observe by Riemann-Roch that any degree $r+1$ line bundle $L$ on a genus 1 curve $E$, is very ample (for $r\geq 2$) and has $r+1$ sections. Moreover, any line bundle of degree $r+1$ on $E$ is isomorphic to $\mathcal{O}_E((r+1)p)$ for some $p\in E$. Since any other line bundle will be of the form $\mathcal{O}_E((r+1)q)$, it will be isomorphic to $L$ under the automorphism of $E$ given by translation by $p-q$. So the map given by any other line bundle will differ from the one given by $L$ by an automorphism of $E$. Therefore all genus 1, degree $r+1$ curves in $\mathbb{P}^r$ will be given by embeddings defined by a basis of $H^0(L)$ and the map given by another basis of $H^0(L)$ will differ from the first map by the action of the base change matrix "mod scalars", which is an element of $\mathbb{P}GL(r+1)$.
Why it fails for other degrees can be seen by a parameter count. For a fixed abstract genus 1 curve $E$, the space of smooth projective embeddings of degree $d$ in $\mathbb{P}^r$ has dimension:
$\operatorname{dim Pic}^d(E)+ \operatorname{dim}G(r+1,d)+\operatorname{dim}\mathbb{P}GL(r+1)-\operatorname{dim} Aut(E)$.
This equals to $\operatorname{dim}G(r+1,d)+\operatorname{dim}\mathbb{P}GL(r+1)$ and this dimension is not equal to $\operatorname{dim}\mathbb{P}GL(r+1)$ unless $d=r+1$.
