I would like to ask for confirmation whether the following argument is correct.
We work over an algebraically closed field $k$ of characteristic $0$. For an elliptic curve $E$, the Picard variety, or dual Abelian variety, $E^\vee$ is canonically isomorphic to $E$ itself. Denote by $\rho:E \to E^\vee$ the canonical isomorphism.
I think that strictly identifying $E$ with $E^\vee$ might hide some important structure. Namely, if $\alpha \in \mathrm{End}^0(E)$, what is the dual morphism $\alpha^\vee\in \mathrm{End}^0(E^\vee) \cong \mathrm{End}^0(E)$?
Let $^\dagger$ denote the Rosati involution corresponding to the canonical isomorphism $\rho: E \to E^\vee$. The identification $\mathrm{End}^0(E^\vee) \cong \mathrm{End}^0(E)$ sends $\alpha^\vee$ to $\alpha^\dagger := \rho^{-1} \circ \alpha^\vee \circ \rho$, so the right question to ask is actually:
What is $\alpha^\dagger$, i.e., what is Rosati involution for an elliptic curve?
Since $\mathrm{End}^0(E)$ is a field, Rosati involution is a $\mathbb{Q}$-algebra automorphism. If $E$ has no complex multiplication, $\mathrm{End}^0(E)=\mathbb{Q}$, so the Rosati involution is the identity. If $E$ has complex multiplication, we can identify $\mathrm{End}^0(E) \cong \mathbb{Q}(J)$, where $J^2 =-d$ for some $d \in \mathbb{N}$. Then Rosati involution must send $J$ to $\pm J$. Therefore, $^\dagger$ should either be the identity or complex conjugation, but which of the two? Rosati positivity says that $^\dagger$ must be complex conjugation, since
$$ \mathrm{Tr}(JJ^\dagger) = \mathrm{Tr}(\mp d) = \mp 2d \overset{!}{>}0.$$
