Prosolvable rigidity of surface groups

Abstract Surface groups are known to be the Poincaré duality groups of dimension two since the work of Eckmann, Linnell and Müller. 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 $${{\,\textrm{cd}\,}}(G)=2$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mrow> <mml:mspace/> <mml:mtext>cd</mml:mtext> <mml:mspace/> </mml:mrow> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mn>2</mml:mn> </mml:mrow> </mml:math> and $$b_2^{(2)}(G)=0$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:msubsup> <mml:mi>b</mml:mi> <mml:mn>2</mml:mn> <mml:mrow> <mml:mo>(</mml:mo> <mml:mn>2</mml:mn> <mml:mo>)</mml:mo> </mml:mrow> </mml:msubsup> <mml:mrow> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>)</mml:mo> </mml:mrow> <mml:mo>=</mml:mo> <mml:mn>0</mml:mn> </mml:mrow> </mml:math> . 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 $$v\in F$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>v</mml:mi> <mml:mo>∈</mml:mo> <mml:mi>F</mml:mi> </mml:mrow> </mml:math> is measure equivalent to u in all finite solvable quotients of F , then u and v belong to the same $${{\,\textrm{Aut}\,}}(F)$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mrow> <mml:mspace/> <mml:mtext>Aut</mml:mtext> <mml:mspace/> </mml:mrow> <mml:mo>(</mml:mo> <mml:mi>F</mml:mi> <mml:mo>)</mml:mo> </mml:mrow> </mml:math> -orbit. Finally, we get a partial result towards Mel’nikov’s surface group conjecture. Let F be a free group of rank $$n\geqslant 3$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>n</mml:mi> <mml:mo>⩾</mml:mo> <mml:mn>3</mml:mn> </mml:mrow> </mml:math> and let $$w\in F$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>w</mml:mi> <mml:mo>∈</mml:mo> <mml:mi>F</mml:mi> </mml:mrow> </mml:math> . Suppose that $$G=F/\langle \!\langle w\rangle \!\rangle $$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>G</mml:mi> <mml:mo>=</mml:mo> <mml:mi>F</mml:mi> <mml:mo>/</mml:mo> <mml:mo>⟨</mml:mo> <mml:mspace/> <mml:mo>⟨</mml:mo> <mml:mi>w</mml:mi> <mml:mo>⟩</mml:mo> <mml:mspace/> <mml:mo>⟩</mml:mo> </mml:mrow> </mml:math> 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 $$H^2(G; \mathbb {Z})\ne 0$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:msup> <mml:mi>H</mml:mi> <mml:mn>2</mml:mn> </mml:msup> <mml:mrow> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>;</mml:mo> <mml:mi>Z</mml:mi> <mml:mo>)</mml:mo> </mml:mrow> <mml:mo>≠</mml:mo> <mml:mn>0</mml:mn> </mml:mrow> </mml:math> then G must be a surface group.