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Prosolvable rigidity of surface groups
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DOI：10.1007/s00029-025-01082-1.png)
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Abstract
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Surface groups are known to be the Poincar & eacute; duality groups of dimension two since the work of Eckmann, Linnell and M & uuml;ller. We prove a prosolvable analogue of this result that allows us to show that surface groups are profinitely (and prosolvably) rigid among finitely generated groups that satisfy cd(G)=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\,\textrm{cd}\,}}(G)=2$$\end{document} and b2(2)(G)=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b_2<^>{(2)}(G)=0$$\end{document}. We explore two other consequences. On the one hand, we derive that if u is a surface word of a finitely generated free group F and v is an element of F\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v\in F$$\end{document} is measure equivalent to u in all finite solvable quotients of F, then u and v belong to the same Aut(F)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\,\textrm{Aut}\,}}(F)$$\end{document}-orbit. Finally, we get a partial result towards Mel'nikov's surface group conjecture. Let F be a free group of rank n >= 3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\geqslant 3$$\end{document} and let w is an element of F\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$w\in F$$\end{document}. Suppose that G=F/<< w >>\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$G=F/\langle \!\langle w\rangle \!\rangle $$\end{document} is a residually finite group all of whose finite-index subgroups are one-relator groups. Then G is 2-free. Moreover, we show that if H2(G;Z)not equal 0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H<^>2(G; \mathbb {Z})\ne 0$$\end{document} then G must be a surface group.
Keywords:
POINCARE-DUALITY GROUPS
PROFINITE
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