Let $A$ be an abelian variety defined over $\overline{\mathbb{Q}}$ and with complex multiplication by a CM field $K$. Looking at the action of $K$ on $H^0(A, \Omega^1_A)$ one gets a CM type of $K$, that is, a subset $\Phi$ of $\mathrm{Hom}(K, \overline{\mathbb{Q}})$ such that $\Phi \amalg \bar{\Phi}=\mathrm{Hom}(K, \overline{\mathbb{Q}}).$

If now $B$ is another CM abelian variety with the same CM type, one can show that $A$ and $B$ are isogenous over $\overline{\mathbb{Q}}$. In the reference I have found this seems to be an "if and only if". However, I'm confused by the example of an elliptic curve, where you have an imaginary quadratic field, hence two CM types, but only one isogeny class of elliptic curves (if I understood correctly, taking the other CM type you get the dual elliptic curve).

So what is the precise relation between CM types and isogeny classes?

nottrue that CM elliptic curves $E$ and $E'$ over $\overline{\mathbf{Q}}$ with specified CM by $K$ are$K$-linearly isogenous. But after precomposing the "action" of $K$ on $E'$ with complex conjugation on $K$ if necessary, we can ensure $E$ and $E'$ have the same CM type and so are $K$-linarly isogenous (so $E$ and $E'$ are isogenous always!). In higher dimension the CM types on $K$ are generallynota single ${\rm{Aut}}(K)$-orbit, so the gimmick in dimension 1 does not carry over. $\endgroup$ – grghxy Jul 1 '15 at 1:11