Kourovka Notebook Problem 21.70
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Definitions and mathematical statement
For $R=\mathbb Z$ or a field, write $RG$ for the group ring and regard $R$ as a trivial $RG$-module through the augmentation $\sum r_g g\mapsto\sum r_g$. A group is of type $FP$ over $R$ here if this module has a finite-length resolution by finitely generated projective $RG$-modules. Projective means a direct summand of a free module; the terms need not themselves be free.Group cohomology with coefficients in the group ring is $H^i(G;RG)=\operatorname{Ext}_{RG}^i(R,RG)$, computed by applying $\operatorname{Hom}_{RG}(-,RG)$ to a projective resolution of $R$. Right multiplication on the coefficient ring gives these cohomology groups a right $RG$-module structure. An orientable Poincaré duality group of dimension $n$ over $R$ is a group of type $FP$ over $R$ satisfying \[ H^i(G;RG)=0\quad(i\ne n),\qquad H^n(G;RG)\cong R \] as right $RG$-modules, with trivial $G$-action on the right-hand $R$. The module condition is the orientability requirement, not merely an isomorphism of underlying abelian groups.
Fix $n\ge0$. If a group $G$ has this property in dimension $n$ over every field, must it have the same property over $\mathbb Z$? The dimension is common to all the field hypotheses; the resolutions may depend on the field. The conclusion includes integral type $FP$ as well as integral cohomology vanishing and the trivial integral dualizing module. Finite presentability of $G$ is not assumed.
Short English statement
Does orientable Poincaré duality in the same dimension over every field imply orientable integral Poincaré duality?Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2579 (KOU-21.70), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.70. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.
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