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I do not really think it fits MO, but I posted it in MathStackExchange with little success, so...

Assume that we have a short exact sequence of, say, Lie groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$, where $A$ is abelian and ( probably it is necessary but I am not sure) $A$ maps to the center of $B$, and a topological space $X$. Then there is an associated (semi)long exact sequence of Cech cohomology $sets$: $*\rightarrow H^0(X,A)\rightarrow...\rightarrow H^1(X,C)\rightarrow H^2(X,A).$

My question is how to prove that naturally defined last differential $H^1(X,C)\rightarrow H^2(X,A)$ actually produces a cocycle, and so an element in $H^2(X,A)$.

When I tried to verify the cocycle condition, I certainly used that $A$ is commutative, to some extent used that $A$ maps to the center of $B$, but still was not able to finish this quite tricky caclulation due to non-commutativity of $B$.

I would appreciate the actual calculation or any reference where it is done.

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  • $\begingroup$ Reference: A. Grothendieck, A General Theory of Fibre Spaces with Structure Sheaf, 5.7, and Sur quelques points d'algèbre homologique, Proposition 3.4.2. In the first paper he uses Cech cohomology and needs $X$ to be paracompact; this hypothesis is removed in the second paper by using "true" cohomology. $\endgroup$
    – abx
    Feb 27, 2017 at 17:47
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    $\begingroup$ Funny enough I just looked at the first reference and in this reference ( page 94), instead of verification of cocycle condition he writes " the verification is some lines longer", and did not actually verify. But that " verification " is exactly the issue of my question: is there a reference where it is actually verified, and not just written " verification is easy enough". $\endgroup$
    – berndt
    Feb 27, 2017 at 19:05
  • $\begingroup$ Every principal C-bundle gives a 'lifting' A-bundle gerbe. See Murray's paper on bundle gerbes for definitions using the case A is the nonzero complex numbers. Note that you need to assume B is locally trivial as an A-bundle, and X is such that every A-bundle on it is locally trivial. I can write a proper answer later. $\endgroup$
    – David Roberts ♦
    Feb 27, 2017 at 20:20
  • $\begingroup$ Thanks David, actually you can just assume that $X$ is a compact manifold, and $A, B, C$ are compact Lie groups. I do not think it would somehow affect the calculation. I just want to see how some terms cancel out. This seems to be some standard calculation. $\endgroup$
    – berndt
    Feb 27, 2017 at 20:35
  • $\begingroup$ Yes, you should assume that $A$ is central. If you assume that $X=Y/G$, where $Y$ is a contractible manifold and $G$ is a discrete group freely acting on $Y$, then $H^i(X,\cdot)=H^i(G,\cdot))$, so you get an exact sequence of nonabelian group cohomology. $\endgroup$
    – Mikhail Borovoi
    Mar 14, 2017 at 21:45

1 Answer 1

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Alright, let's do this exercise. Let $U_i$ be an open cover of $X$ such that every intersection $U_{i_0} \cap U_{i_1} \cap \cdots \cap U_{i_q}$ is either contractible or empty. For example, choose a triangulation of $X$, and, for each vertex $i$ of the traingulation, let $U_i$ be the union of the relative interiors of the faces containing $i$. Then we can use the $U_i$ as a Cech cover to compute sheaf cohomology. We abbreviate $U_{i_0} \cap U_{i_1} \cap \cdots \cap U_{i_q}$ to $U_{i_0 i_1 \cdots i_q}$.

A $1$-cocycle is, for each $U_{ij}$, a section $c_{ij}$ of $C$ over $U_{ij}$, obeying $c_{ij} = c_{ji}^{-1}$ and $c_{ij} c_{jk} c_{ki}=1$ whenever $U_{ijk}$ is nonempty. Lift each $c_{ij}$ to a section $b_{ij}$ of $B$, with our lifts obeying $b_{ji} = b_{ij}^{-1}$, and put $a_{ijk} = b_{ij} b_{jk} b_{ki}$. Note that $a_{ijk}$ lies in $A$ by the cocycle condition on $c_{ij}$.

We have $$a_{jki} = b_{jk} \left( b_{ki} b_{ij} b_{jk} \right) b_{jk}^{-1} = b_{jk} a_{ijk} b_{jk}^{-1}$$ which, since $A$ is central in $B$, collapses to $a_{ijk} = a_{jki} = a_{kij}$. Also, $$a_{ikj} = b_{ik} b_{kj} b_{ji} = b_{ki}^{-1} b_{jk}^{-1} b_{ij}^{-1} = \left( b_{ij} b_{jk} b_{ki} \right)^{-1} = a_{ijk}^{-1},$$ so $a_{ikj} = a_{kji} = a_{jik} = a_{ijk}^{-1}$. In short, $a$ has all the antisymmetry properties of a $2$-cochain.

We now want to verify that $a$ is a $2$-cocycle. This means we want to check that $$a_{ijk} a_{ikl} = a_{lij} a_{ljk}.$$ The LHS is $$b_{ij} b_{jk} b_{ki} b_{ik} b_{kl} b_{li} = b_{ij} b_{jk} b_{kl} b_{li}$$ and the RHS is $$b_{li} b_{ij} b_{jl} b_{lj} b_{jk} b_{kl} = b_{li} b_{ij} b_{jk} b_{kl} = b_{li} \left( b_{ij} b_{jk} b_{kl} b_{li} \right) b_{li}^{-1}.$$ Since $b_{ij} b_{jk} b_{kl} b_{li}$ is in $A$ and $A$ is central, this simplifies to $b_{ij} b_{jk} b_{kl} b_{li}$. Now the LHS and RHS match and we are done. $\square$

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  • $\begingroup$ I found it helpful to draw a tetrahedron with vertices $(i,j,k,l)$, and interpret each product as the monodromy of some path along the $1$-skeleton of the tetrahedron. $\endgroup$
    – David E Speyer
    Mar 17, 2021 at 13:30

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