8
$\begingroup$

Suppose $(C,\odot,\Bbb I)$ is an additive category with a compatible symmetric monoidal structure and $Pic(C)$ is the group of isomorphism classes of objects which have an inverse under $\odot$. For $\alpha \in Pic(C)$, we would like define the $\alpha$'th coefficient functor by choosing a representative $P^\alpha \in C$ for the isomorphism class, and letting $\pi_\alpha X = Hom(P^\alpha,X)$.

In Section 14 of Morava K-theories and localisation, Hovey and Strickland discuss an obstruction to extending this to something compatible with the monoidal structure (possessing appropriate maps $\pi_\alpha X \otimes \pi_\beta Y \to \pi_{\alpha \odot \beta}(X \odot Y)$ satisfying all the desired naturality properties). They comment that there is an obstruction cocycle in $H^3(Pic(C), Aut(\Bbb I))$, and that if this obstruction vanishes then the possible choices of this natural isomorphism are parametrized by $H^2(Pic(C), Aut(\Bbb I))$. They did not know of an explicit reference for this fact.

  1. Are now there references where this obstruction is worked out in detail?

  2. This cohomology group is a group of Postnikov invariants, parametrizing spaces whose only two homotopy groups are $\pi_1$ and $\pi_2$. This is equivalent to the theory of (nonsymmetric) Picard groupoids: monoidal categories where all morphisms are invertible and where every object has an inverse under the monoidal structure. There is a natural candidate monoidal category: the category of tensor-invertible objects in $C$ and isomorphisms between them. What is the relationship between these two classes in $H^3$?

  3. If the answer to (2) is roughly "these are the same", then because $C$ is symmetric monoidal this Postnikov invariant deloops, at least as far as an element of $H^5(K(Pic(C),3),Aut(\Bbb I))$. A back-of-the-envelope calculation seems to indicate that the obstruction must vanish because Postnikov invariants won't deloop at this low stage. Is this the case?

  4. If we suspect that this is true, then: Is there an explicit proof using the symmetry isomorphism that Hovey-Strickland's obstruction cocycle must vanish? (If this does vanish, then the $H^5$-cocycle is still an obstruction to making this compatible with the symmetry isomorphism.)

$\endgroup$
1
3
$\begingroup$

Let $\mathbf{Pic}(C)$ be the monoidal subgroupoid of tensor invertible objects and isomorphisms between them, so $\pi_0\mathbf{Pic}(C)=Pic(C)$. For the definition of those appropriate maps, it is clearly necessary and sufficient to choose representatives in such a way that we get a monodical splitting $Pic(C)\to \mathbf{Pic}(C)$ of the quotient map $\mathbf{Pic}(C)\to Pic(C)$. The obstruction to the definition of such a splitting is precisely the cohomology class $H^3(Pic(C),Aut(\mathbb I))$ you mention in 2, compare Joyal-Street. In the braided case, this class comes from $H^4$, and in the symmetric case from $H^5$, which is stable, as you say in 3. Therefore the class in $H^3$ vanishes in the symmetric case, compare this paper for a proof. You can in principle obtain the splitting from the symmetry using proofs therein, but the paper is written in terms of stricter but equivalent structures, so you'd have to strictify and go back and forth. It's also remarkable that the trivialisation seems to be unnatural in $C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.