Let $\mathbb G$ be an abelian vatiety over an $\mathbb E_\infty$-ring $A$. That is to say, it consists of an abelian group object in the $\infty$-category of relative schemes $\mathbb G\to \operatorname{Spec} A$ which are flat (+ maybe more conditions). Denote its underlying classical abelian variety over $\pi_0A$ by $\mathbb G_0$.

Recall from Lurie's "Survey of Elliptic Cohomology":

Definition: A preorientation on $\mathbb G$ is a map of abelian topological groups $ \mathbf{CP}^\infty\to \mathbb G(A) $ or equivalently a map of formal spectral group schemes $ \operatorname{Spf}A^{\mathbf {CP}^\infty}\to\mathbb G, $ or equivalently yet an element of $\pi_2\mathbb G(A)$.

Viewed as a map $S^2\to\mathbb G(A)$, a preorientation induces through some adjunctions and restriction to $\pi_0$ a map $\beta :\omega\to \pi_2$, where $\omega$ denotes the invariant differentials of $\mathbb G_0$ over $\pi_0A$.

Question: In what way are the following two conditions related?

A) The preorientation map exhibits an equivalence $\operatorname{Spf}A^{\mathbf {CP}^\infty}\simeq \widehat{\mathbb G}$, where the RHS is the formal completion of $\mathbb G$ at the identity.

B) The preorientation is an orientation, in the sense that

1. The map of underlying ordinary schemes $\mathbb G_0\to \operatorname{Spec} \pi_0A$ is smooth of relative dimnension $1$.
2. For all $n,$ the composition $$\pi_nA\otimes_{\pi_0A}\omega\xrightarrow{\operatorname{id}\otimes\beta}\pi_nA\otimes_{\pi_0A}\pi_2A\to\pi_{n+2}A,$$ where the unlabeled arrow is the multiplication in the graded ring $\pi_*A$, is an isomorphism.

Lurie seems to assert in "Survey" that they are equivalent, or at the very least that B) implies A). I would be very happy if somebody could explain why that is.

Remark: A surely relevant fact is that for the classical formal group $\mathbb G_0 = \operatorname{Spf}\pi_0\left(A^{\mathbf {CP}^\infty}\right)=\operatorname{Spf}A^0(\mathbf {CP}^\infty)$, the $n$-th tensor power of its module of invariant differentials $\omega^n$ is isomorphic to $\pi_{2n}A$, as proved e.g. in Rezk's notes. But I don't see how this shows that B) $\Rightarrow$ A).

Namely, I feel like this should be saying something about the $\operatorname{Spf}A^{\mathbf{CP}^\infty}$ and its module (spectrum) of invariant differentials, but all I see is a statement about (tensor powers of) the module of invariant differentials of its underlying classical counterpart.

Any help will be warmly appreciated!

Let $\mathbb G$ be an abelian vatiety over an $\mathbb E_\infty$-ring $A$. That is to say, it consists of an abelian group object in the $\infty$-category of relative schemes $\mathbb G\to \operatorname{Spét} A$ which are flat (+ maybe more conditions). Denote its underlying classical abelian variety over $\pi_0A$ by $\mathbb G_0$.

Recall from Lurie's "Survey of Elliptic Cohomology":

Definition: A preorientation on $\mathbb G$ is a map of abelian topological groups $ \mathbf{CP}^\infty\to \mathbb G(A) $ or equivalently a map of formal spectral group schemes $ \operatorname{Spf}A^{\mathbf {CP}^\infty}\to\mathbb G, $ or equivalently yet an element of $\pi_2\mathbb G(A)$.

Viewed as a map $S^2\to\mathbb G(A)$, a preorientation induces through some adjunctions and restriction to $\pi_0$ a map $\beta :\omega\to \pi_2$, where $\omega$ denotes the invariant differentials of $\mathbb G_0$ over $\pi_0A$.

Question: In what way are the following two conditions related?

A) The preorientation map exhibits an equivalence $\operatorname{Spf}A^{\mathbf {CP}^\infty}\simeq \widehat{\mathbb G}$, where the RHS is the formal completion of $\mathbb G$ at the identity.

B) The preorientation is an orientation, in the sense that

1. The map of underlying ordinary schemes $\mathbb G_0\to \operatorname{Spec} \pi_0A$ is smooth of relative dimnension $1$.
2. For all $n,$ the composition $$\pi_nA\otimes_{\pi_0A}\omega\xrightarrow{\operatorname{id}\otimes\beta}\pi_nA\otimes_{\pi_0A}\pi_2A\to\pi_{n+2}A,$$ where the unlabeled arrow is the multiplication in the graded ring $\pi_*A$, is an isomorphism.

Lurie seems to assert in "Survey" that they are equivalent, or at the very least that B) implies A). I would be very happy if somebody could explain why that is.

Remark: A surely relevant fact is that for the classical formal group $\mathbb G_0 = \operatorname{Spf}\pi_0\left(A^{\mathbf {CP}^\infty}\right)=\operatorname{Spf}A^0(\mathbf {CP}^\infty)$, the $n$-th tensor power of its module of invariant differentials $\omega^n$ is isomorphic to $\pi_{2n}A$, as proved e.g. in Rezk's notes. But I don't see how this shows that B) $\Rightarrow$ A).

Namely, I feel like this should be saying something about the $\operatorname{Spf}A^{\mathbf{CP}^\infty}$ and its module (spectrum) of invariant differentials, but all I see is a statement about (tensor powers of) the module of invariant differentials of its underlying classical counterpart.

Any help will be warmly appreciated!