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Let $G$ be a Lie group (paracompact, not necessarily compact), and $A$ an abelian Lie group. I want to write down cocycles in $\mathrm{H^n}(\mathbf{B}G,A)$, the cohomology in the cohesive $\infty$-topos of smooth $\infty$-stacks. This topos is presented by the model category ${[\mathrm{CartSp}^\mathrm{op},\mathrm{sSet}]}_{\mathrm{proj,loc}}$ of simplicial presheaves over the site of Cartesian spaces and smooth maps between them, equipped with the local projective model structure. In other words, I would like to write down morphisms $\mathbf{B}G \to \mathbf{B}^nA$ using this model.

I have particular interest in $A=U(1)$. In that case, a fibrant presentation of $\mathbf{B}^n A$ is given by $\Xi(A[n])$, the image of the chain complex $A[n]$ under the Dold-Kan correspondence (applied pointwise). This is Proposition 3.3.21 in [Schr]. The proof crucially uses the coefficient exact sequence for $\mathbb{Z} \to \mathbb{R} \to U(1)$. I would like to know an example of an abelian Lie group $A$ where the statement is wrong (and why).

Bounded hypercovers (of height k) are defined e.g. at [nLabHyp] and [ConHyp]. They are those hypercovers where the covers of matching objects in simplicial degree $\geq k$ are isomorphisms. They are relevant here because fibrant objects in the model category ${[\mathrm{CartSp}^\mathrm{op},\mathrm{sSet}]}_{\mathrm{proj,loc}}$ satisfy descent with respect to hypercovers, and because Segal's model for Lie group cohomology is computed in [BryCoh] using simplicial covers (which are essentially hypercovers). Theorem 3.3.28 in [Schr] proves that in two special cases ($A = \mathbb{R}$ or discrete), the cohomology can indeed be computed by Segal's model. I tried to trace through the proof to understand why this would go wrong for other coefficient groups, but did not succeed.

Under good assumptions, $\mathbf{B}^nA$ should be $k$-coskeletal for some finite $k$, which should allow one to use the adjunction $\mathbf{sk}_k \dashv \mathbf{cosk}_k$ to replace hypercovers by bounded hypercovers in general. This motivates my questions:


When are cocycles $\mathbf{B}G \to \mathbf{B}^nA$ represented by maps $Y \to \Xi(A[n])$ out of bounded hypercovers $Y$ of $\mathbf{B}G$ (particularly in the case $A=U(1)$)?

and

Are there good examples of Lie groups $G, A$ as above where Segal's cohomology fails to model the intrinsic cohomology of smooth $\infty$-stacks correctly?


References:
[Schr]: https://ncatlab.org/schreiber/files/cohesivedocumentv031.pdf
[nLabHyp]: https://ncatlab.org/nlab/show/hypercover
[ConHyp]: https://math.stanford.edu/~conrad/papers/hypercover.pdf
[BryCoh]: https://arxiv.org/abs/math/0011069

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I would like to know an example of an abelian Lie group A where the statement is wrong (and why).

There is no such example because the statement is true for all $A$. (This also implies trivial answers for the other two questions: always and no.)

Any abelian Lie group $A$ fits into an exact sequence $$1→π_1(A)→U→A→π_0(A)→1,$$ where $π_0(A)$ denotes the discrete abelian group of connected components of $A$, $U$ is the universal cover of the connected component of the identity in $A$, and $π_1(A)$ is the fundamental group of $A$.

Consequently, the desired claim is implied by two special cases: the case of discrete abelian groups has the same proof as for Z and the case of simply connected abelian Lie groups has the same proof as for R.

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  • $\begingroup$ Thank you very much for your answer! You have convinced me that it is enough to treat the discrete and simply-connected case. However, I had another try at tracing through the proof for $A=\mathbb{R}$ mentioned in my post. It is not obvious to me that this generalises to simply connected $A$. Do you have a reference for this claim? $\endgroup$ Sep 29, 2020 at 10:56
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    $\begingroup$ @ChristophWeis: Simply connected abelian Lie groups are finite-dimensional real vector spaces, i.e., R^n. The corresponding chain complex is the n-fold direct sum of chain complexes for R. $\endgroup$ Sep 29, 2020 at 13:09

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