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!

  • 2
    $\begingroup$ This is now written up in Lurie's "Elliptic II" paper (see the whole of section 4, but particularly Remark 4.2.18 and Proposition 4.3.23). I also wrote up a (much smaller) document at mit.edu/~sanathd/orientations.pdf as notes for a seminar talk on this topic, in case it's helpful. $\endgroup$
    – skd
    Jun 19, 2018 at 17:51

1 Answer 1


Okay, I think I may have figured out how B) $\Rightarrow$ A). Because nobody else has given an answer yet, I'm spelling it out (I hope that's not considered bad form). And if what I wrote doesn't make sense, please point it out.

Condition B) is equivalent to saying that:

  1. $A$ is even weakly periodic and
  2. $\operatorname{Spf}A^0(\mathbf {CP}^\infty)\cong \widehat{\mathbb G}_0$, i.e. statement A) on the level of underlying classical formal groups.

For both we use the fact that for formal group $\mathbb G_A:=\operatorname{Spf} A^0(\mathbf{CP}^\infty)$ invariant differentials are $$ \omega^n_{\mathbb G_A}\cong \pi_{2n}A. $$ The preorientation supplies, upon passage to underlying classical schemes and formal completion, a map of formal groups $\operatorname{Spf}A^0(\mathbf{CP}^\infty)\to \widehat {\mathbb G}_0$. This map is, thanks to assumption 1. of B), an isomorphism iff it induces an isomorphism on invariant differentials, which 2. of B), used for $n=1$, now implies. Then we get weak periodicity of $A$ directly from 2. of B) by using it for arbitrary $n$.

So the claim we're after, to show that B) implies A), is that a map of formal spectral schemes $$\sigma:\operatorname{Spf}A^{\mathbf{CP}^\infty}\to \widehat{\mathbb G}$$ is an equivalence iff it induces an isomorphism on the underlying formal groups.

This should I think follow from the flatness assumption on $\mathbb G$, since $A^{\mathbf {CP}^\infty}$ is also flat over $A$ by virtue of $A$ being even weakly periodic.

We already have a map of formal groups $\sigma$, so we just need to show that it is an equivalence. But since we are trying to prove this assertion for formal completions, it suffices to restrict on both sides, i.e. on $\operatorname{Spét}A^{\mathbf{ CP}^\infty}$ and on $\mathbb G$, to étale open affine neighbourhoods of the identity. Affine spectral schemes can always be exchanged for connective $\mathbb E_\infty$-rings by reversing the arrows, and since flatness is an étale-local property, we have reduced to showing that a certain map of flat $\mathbb E_\infty$-$A$-algebras $B\to B'$ is an equivalence.

Here we can use Lemma from "Higher Algebra" which says that, in the presence of flatness, this will happen iff the induced map $\pi_0B\to \pi_0B'$ is an isomorphism. But repeating the arguments of the previous paragraph in reverse, we find this is equivalent to the condition that $\sigma$ induces an isomorphism on the underlying formal schemes $\mathbb G_A\to \widehat{\mathbb G}_0$, which we already know to be the case. Thus $\sigma$ is indeed an equivalence, proving A).


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