<p>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 <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\({{\,\textrm{cd}\,}}(G)=2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mrow> <mspace width="0.166667em" /> <mtext>cd</mtext> <mspace width="0.166667em" /> </mrow> <mo stretchy="false">(</mo> <mi>G</mi> <mo stretchy="false">)</mo> <mo>=</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(b_2^{(2)}(G)=0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>b</mi> <mn>2</mn> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </msubsup> <mrow> <mo stretchy="false">(</mo> <mi>G</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. We explore two other consequences. On the one hand, we derive that if <i>u</i> is a surface word of a finitely generated free group <i>F</i> and <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(v\in F\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>v</mi> <mo>∈</mo> <mi>F</mi> </mrow> </math></EquationSource> </InlineEquation> is measure equivalent to <i>u</i> in all finite solvable quotients of <i>F</i>, then <i>u</i> and <i>v</i> belong to the same <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\({{\,\textrm{Aut}\,}}(F)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mrow> <mspace width="0.166667em" /> <mtext>Aut</mtext> <mspace width="0.166667em" /> </mrow> <mo stretchy="false">(</mo> <mi>F</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>-orbit. Finally, we get a partial result towards Mel’nikov’s surface group conjecture. Let <i>F</i> be a free group of rank <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(n\geqslant 3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>⩾</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation> and let <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(w\in F\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>w</mi> <mo>∈</mo> <mi>F</mi> </mrow> </math></EquationSource> </InlineEquation>. Suppose that <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(G=F/\langle \!\langle w\rangle \!\rangle \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>G</mi> <mo>=</mo> <mi>F</mi> <mo stretchy="false">/</mo> <mo stretchy="false">⟨</mo> <mspace width="-0.166667em" /> <mo stretchy="false">⟨</mo> <mi>w</mi> <mo stretchy="false">⟩</mo> <mspace width="-0.166667em" /> <mo stretchy="false">⟩</mo> </mrow> </math></EquationSource> </InlineEquation> is a residually finite group all of whose finite-index subgroups are one-relator groups. Then <i>G</i> is 2-free. Moreover, we show that if <InlineEquation ID="IEq8"> <EquationSource Format="TEX">\(H^2(G; \mathbb {Z})\ne 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mi>H</mi> <mn>2</mn> </msup> <mrow> <mo stretchy="false">(</mo> <mi>G</mi> <mo>;</mo> <mi mathvariant="double-struck">Z</mi> <mo stretchy="false">)</mo> </mrow> <mo>≠</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> then <i>G</i> must be a surface group.</p>

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Prosolvable rigidity of surface groups

  • Andrei Jaikin-Zapirain,
  • Ismael Morales

摘要

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 \({{\,\textrm{cd}\,}}(G)=2\) cd ( G ) = 2 and \(b_2^{(2)}(G)=0\) b 2 ( 2 ) ( G ) = 0 . 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 \(v\in F\) v F is measure equivalent to u in all finite solvable quotients of F, then u and v belong to the same \({{\,\textrm{Aut}\,}}(F)\) Aut ( F ) -orbit. Finally, we get a partial result towards Mel’nikov’s surface group conjecture. Let F be a free group of rank \(n\geqslant 3\) n 3 and let \(w\in F\) w F . Suppose that \(G=F/\langle \!\langle w\rangle \!\rangle \) G = F / w 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 \(H^2(G; \mathbb {Z})\ne 0\) H 2 ( G ; Z ) 0 then G must be a surface group.