Skip to main content
2 of 2
added tags
Konrad Waldorf
  • 4.5k
  • 4
  • 29
  • 37

3-cocycles on outer automorphism groups

Given a group $G$, the outer automorphism group $Out(G)$ acts on the center by $Z(G)$ by lifting an outer automorphism to an actual automorphism and evaluating this on elements of $Z(G)$. What is classified by the degree three group cohomology $H^3(Out(G),Z(G))$ ?

Given an algebra $A$, say finite-dimensional over a field, the outer automorphism group $Out(A)$ acts on the group of central units $Z(A)^\times$ by lifting an outer automorphism to an actual automorphism and then evaluating. The action is trivial if $A$ is a central algebra. What is classified by the degree three group cohomology $H^3(Out(A),Z(A)^\times)$ ?

My motivation for this question is the following. If $A$ is my algebra, then we have a crossed module $$ A^\times \to Aut(A) $$ given by inner automorphisms and the evaluation action of $Aut(A)$ on $A^\times$. The homotopy groups of this crossed module are $\pi_0=Out(A)$ and $\pi_1=Z(A)^\times$, and the usual action of $\pi_0$ on $\pi_1$ is the one described above. Crossed modules have a so-called k-invariant, which is precisely a class $\xi\in H^3(\pi_0,\pi_1)$.

Baez and Lauda have shown that crossed modules are classified in a sense by triples $(\pi_0,\pi_1,\xi)$. The classification is saying that the crossed module - viewed as a monoidal category - is equivalent to the usual monoidal category constructed from the 3-cocycle $\xi$.

Baez, John C.; Lauda, Aaron D., Higher-dimensional algebra. V: 2-Groups, Theory Appl. Categ. 12, 423-491 (2004). ZBL1056.18002.

It is also known that if $A$ and $B$ are Picard-surjective and Morita equivalent, then their crossed modules are equivalent and so their classes coincide. Moreover, if $A$ is Picard-surjective and central-simple, then its class vanishes.

Summarizing, associated to every algebra is a class in $H^3(Out(A),Z(A)^\times)$. What is the intrinsic meaning of this class, apart from classifying some crossed module?

For groups instead of algebras it is basically the same story, and the question is analogous.

Konrad Waldorf
  • 4.5k
  • 4
  • 29
  • 37