A question on the fundamental group of a compact orientable surface of genus >1











up vote
8
down vote

favorite
3












Let $G=pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $gge 2$. It is well known that a presentation of
$G$ is
$$G=langle x_1,y_1,dots,x_g,y_g | [x_1,y_1]cdots
[x_g,y_g]=1rangle$$
(where $[x,y]=xyx^{-1}y^{-1}$ is the
commutator).



Denote by $F$ be the free group with $2g$ generators
$x_1,y_1,dots,x_g,y_g$ and by $R$ be the normal closure of the
relation $r=[x_1,y_1]cdots [x_g,y_g]$, so $G=F/R$.



It is clear that $rin [F,F]$.




Question: Is there an elementary proof that $r=[x_1,y_1]cdots [x_g,y_g]notin [F,[F,F]]$?




This result appears when one considers the Stallings exact
sequence associated to
$$1to [G,G]to Gto G^{ab}to 1$$
to get
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
H_1(G,mathbb{Z})to H_1(G^{ab},mathbb{Z}) to 0$$



Since $H_1(G,mathbb{Z})cong H_1(G^{ab},mathbb{Z})cong G^{ab}$ we
obtain a short exact sequence
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
0$$
which should be injective at the left (see at the end some argument why).



Now, Hopf's formula gives $$H_2(F/R,mathbb{Z})=(Rcap
[F,F])/[R,F]=R/[R,F]$$
since $Rsubset [F,F]$, hence
$H_2(F/R,mathbb{Z})$ is cyclic and the generator is given by (the
class of) $r$. So the map $psi:H_2(G,mathbb{Z})to
H_2(G^{ab},mathbb{Z})$
is either injective or zero. But
$$H_2(G^{ab},mathbb{Z})cong [F,F]/[F,[F,F]]$$
since $G^{ab}cong F/[F,F]$, and so the map $psi$ is given by the
natural map
$$psi:R/[R,F]to [F,F]/[F,[F,F]]$$
coming from the inclusion $Rhookrightarrow [F,F]$, hence $psi$ is
injective if and only if $rnotin [F,[F,F]]$.



One possibility is to use another description of the map $psi$ as
$$H_2(G,mathbb{Z})to
bigwedge H_1(G^{ab},mathbb{Z})$$
that should correspond
to the dual of the cup product in cohomology via Poincaré duality
(i.e. dual universal coefficient theorem) (but I am not sure if to consider this
approach as really elementary).










share|cite|improve this question




















  • 5




    Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
    – YCor
    10 hours ago












  • (or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
    – YCor
    10 hours ago












  • @YCor I like your proof: it is really elementary!
    – Xarles
    9 hours ago















up vote
8
down vote

favorite
3












Let $G=pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $gge 2$. It is well known that a presentation of
$G$ is
$$G=langle x_1,y_1,dots,x_g,y_g | [x_1,y_1]cdots
[x_g,y_g]=1rangle$$
(where $[x,y]=xyx^{-1}y^{-1}$ is the
commutator).



Denote by $F$ be the free group with $2g$ generators
$x_1,y_1,dots,x_g,y_g$ and by $R$ be the normal closure of the
relation $r=[x_1,y_1]cdots [x_g,y_g]$, so $G=F/R$.



It is clear that $rin [F,F]$.




Question: Is there an elementary proof that $r=[x_1,y_1]cdots [x_g,y_g]notin [F,[F,F]]$?




This result appears when one considers the Stallings exact
sequence associated to
$$1to [G,G]to Gto G^{ab}to 1$$
to get
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
H_1(G,mathbb{Z})to H_1(G^{ab},mathbb{Z}) to 0$$



Since $H_1(G,mathbb{Z})cong H_1(G^{ab},mathbb{Z})cong G^{ab}$ we
obtain a short exact sequence
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
0$$
which should be injective at the left (see at the end some argument why).



Now, Hopf's formula gives $$H_2(F/R,mathbb{Z})=(Rcap
[F,F])/[R,F]=R/[R,F]$$
since $Rsubset [F,F]$, hence
$H_2(F/R,mathbb{Z})$ is cyclic and the generator is given by (the
class of) $r$. So the map $psi:H_2(G,mathbb{Z})to
H_2(G^{ab},mathbb{Z})$
is either injective or zero. But
$$H_2(G^{ab},mathbb{Z})cong [F,F]/[F,[F,F]]$$
since $G^{ab}cong F/[F,F]$, and so the map $psi$ is given by the
natural map
$$psi:R/[R,F]to [F,F]/[F,[F,F]]$$
coming from the inclusion $Rhookrightarrow [F,F]$, hence $psi$ is
injective if and only if $rnotin [F,[F,F]]$.



One possibility is to use another description of the map $psi$ as
$$H_2(G,mathbb{Z})to
bigwedge H_1(G^{ab},mathbb{Z})$$
that should correspond
to the dual of the cup product in cohomology via Poincaré duality
(i.e. dual universal coefficient theorem) (but I am not sure if to consider this
approach as really elementary).










share|cite|improve this question




















  • 5




    Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
    – YCor
    10 hours ago












  • (or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
    – YCor
    10 hours ago












  • @YCor I like your proof: it is really elementary!
    – Xarles
    9 hours ago













up vote
8
down vote

favorite
3









up vote
8
down vote

favorite
3






3





Let $G=pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $gge 2$. It is well known that a presentation of
$G$ is
$$G=langle x_1,y_1,dots,x_g,y_g | [x_1,y_1]cdots
[x_g,y_g]=1rangle$$
(where $[x,y]=xyx^{-1}y^{-1}$ is the
commutator).



Denote by $F$ be the free group with $2g$ generators
$x_1,y_1,dots,x_g,y_g$ and by $R$ be the normal closure of the
relation $r=[x_1,y_1]cdots [x_g,y_g]$, so $G=F/R$.



It is clear that $rin [F,F]$.




Question: Is there an elementary proof that $r=[x_1,y_1]cdots [x_g,y_g]notin [F,[F,F]]$?




This result appears when one considers the Stallings exact
sequence associated to
$$1to [G,G]to Gto G^{ab}to 1$$
to get
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
H_1(G,mathbb{Z})to H_1(G^{ab},mathbb{Z}) to 0$$



Since $H_1(G,mathbb{Z})cong H_1(G^{ab},mathbb{Z})cong G^{ab}$ we
obtain a short exact sequence
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
0$$
which should be injective at the left (see at the end some argument why).



Now, Hopf's formula gives $$H_2(F/R,mathbb{Z})=(Rcap
[F,F])/[R,F]=R/[R,F]$$
since $Rsubset [F,F]$, hence
$H_2(F/R,mathbb{Z})$ is cyclic and the generator is given by (the
class of) $r$. So the map $psi:H_2(G,mathbb{Z})to
H_2(G^{ab},mathbb{Z})$
is either injective or zero. But
$$H_2(G^{ab},mathbb{Z})cong [F,F]/[F,[F,F]]$$
since $G^{ab}cong F/[F,F]$, and so the map $psi$ is given by the
natural map
$$psi:R/[R,F]to [F,F]/[F,[F,F]]$$
coming from the inclusion $Rhookrightarrow [F,F]$, hence $psi$ is
injective if and only if $rnotin [F,[F,F]]$.



One possibility is to use another description of the map $psi$ as
$$H_2(G,mathbb{Z})to
bigwedge H_1(G^{ab},mathbb{Z})$$
that should correspond
to the dual of the cup product in cohomology via Poincaré duality
(i.e. dual universal coefficient theorem) (but I am not sure if to consider this
approach as really elementary).










share|cite|improve this question















Let $G=pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $gge 2$. It is well known that a presentation of
$G$ is
$$G=langle x_1,y_1,dots,x_g,y_g | [x_1,y_1]cdots
[x_g,y_g]=1rangle$$
(where $[x,y]=xyx^{-1}y^{-1}$ is the
commutator).



Denote by $F$ be the free group with $2g$ generators
$x_1,y_1,dots,x_g,y_g$ and by $R$ be the normal closure of the
relation $r=[x_1,y_1]cdots [x_g,y_g]$, so $G=F/R$.



It is clear that $rin [F,F]$.




Question: Is there an elementary proof that $r=[x_1,y_1]cdots [x_g,y_g]notin [F,[F,F]]$?




This result appears when one considers the Stallings exact
sequence associated to
$$1to [G,G]to Gto G^{ab}to 1$$
to get
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
H_1(G,mathbb{Z})to H_1(G^{ab},mathbb{Z}) to 0$$



Since $H_1(G,mathbb{Z})cong H_1(G^{ab},mathbb{Z})cong G^{ab}$ we
obtain a short exact sequence
$$ H_2(G,mathbb{Z})to H_2(G^{ab},mathbb{Z})to [G,G]/[G,[G,G]]to
0$$
which should be injective at the left (see at the end some argument why).



Now, Hopf's formula gives $$H_2(F/R,mathbb{Z})=(Rcap
[F,F])/[R,F]=R/[R,F]$$
since $Rsubset [F,F]$, hence
$H_2(F/R,mathbb{Z})$ is cyclic and the generator is given by (the
class of) $r$. So the map $psi:H_2(G,mathbb{Z})to
H_2(G^{ab},mathbb{Z})$
is either injective or zero. But
$$H_2(G^{ab},mathbb{Z})cong [F,F]/[F,[F,F]]$$
since $G^{ab}cong F/[F,F]$, and so the map $psi$ is given by the
natural map
$$psi:R/[R,F]to [F,F]/[F,[F,F]]$$
coming from the inclusion $Rhookrightarrow [F,F]$, hence $psi$ is
injective if and only if $rnotin [F,[F,F]]$.



One possibility is to use another description of the map $psi$ as
$$H_2(G,mathbb{Z})to
bigwedge H_1(G^{ab},mathbb{Z})$$
that should correspond
to the dual of the cup product in cohomology via Poincaré duality
(i.e. dual universal coefficient theorem) (but I am not sure if to consider this
approach as really elementary).







at.algebraic-topology gr.group-theory free-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 11 hours ago

























asked 11 hours ago









Xarles

702713




702713








  • 5




    Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
    – YCor
    10 hours ago












  • (or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
    – YCor
    10 hours ago












  • @YCor I like your proof: it is really elementary!
    – Xarles
    9 hours ago














  • 5




    Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
    – YCor
    10 hours ago












  • (or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
    – YCor
    10 hours ago












  • @YCor I like your proof: it is really elementary!
    – Xarles
    9 hours ago








5




5




Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
– YCor
10 hours ago






Yes it's elementary: consider the group $H$ of upper triangular integral matrices with 1 on the diagonal (integral Heisenberg group): it's 2-nilpotent. Write $e_{ij}(t)=1+tE_{ij}$. Consider the homomorphism $Fto H$ mapping $x_i$ to $e_{12}(1)$ and $y_i$ to $e_{23}(1)$. Hence it maps $[x_i,y_i]$ to $e_{13}(1)$, and hence $r=prod [x_i,y_i]$ to $e_{13}(g)$. So $rnotin [F,[F,F]]$, since otherwise it would be in the kernel of every homomorphism to every 2-nilpotent group.
– YCor
10 hours ago














(or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
– YCor
10 hours ago






(or do the same with killing $x_i,y_i$ for $ige 2$, again with $x_1mapsto e_{12}(1)$, $y_1mapsto e_{23}(1)$)
– YCor
10 hours ago














@YCor I like your proof: it is really elementary!
– Xarles
9 hours ago




@YCor I like your proof: it is really elementary!
– Xarles
9 hours ago










2 Answers
2






active

oldest

votes

















up vote
6
down vote













Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $Sigma_g$ itself is an Eilenberg-MacLane space. Let ${a_1,b_1,ldots,a_g,b_g}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ be the map whose kernel is spanned by ${[a_2],[b_2],ldots,[a_g],[b_g]}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $mathbb{Z}^2$. To prove that $psi$ is injective, it is enough to prove that the composition
$$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
is injective. But $phi$ is easy to understand geometrically: the surface $Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $mathbb{Z}^2$, and $phi$ is induced by the map $Phicolon Sigma_g rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,ldots,a_g,b_g$. The point here is that it is obvious that $Phi_{ast}$ takes the fundamental class of $Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.



You can soup up this argument to show that $psi$ takes the fundamental class of $Sigma_g$ to the element $a_1 wedge b_1 + cdots + a_g wedge b_g$ of $H_2(mathbb{Z}^{2g}) cong wedge^2 mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.






share|cite|improve this answer






























    up vote
    2
    down vote













    The dual to the map
    $psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$. Clearly, the latter map is surjective; hence, the former map must be injective.






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "504"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














       

      draft saved


      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315971%2fa-question-on-the-fundamental-group-of-a-compact-orientable-surface-of-genus-1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      6
      down vote













      Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $Sigma_g$ itself is an Eilenberg-MacLane space. Let ${a_1,b_1,ldots,a_g,b_g}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ be the map whose kernel is spanned by ${[a_2],[b_2],ldots,[a_g],[b_g]}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $mathbb{Z}^2$. To prove that $psi$ is injective, it is enough to prove that the composition
      $$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
      is injective. But $phi$ is easy to understand geometrically: the surface $Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $mathbb{Z}^2$, and $phi$ is induced by the map $Phicolon Sigma_g rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,ldots,a_g,b_g$. The point here is that it is obvious that $Phi_{ast}$ takes the fundamental class of $Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.



      You can soup up this argument to show that $psi$ takes the fundamental class of $Sigma_g$ to the element $a_1 wedge b_1 + cdots + a_g wedge b_g$ of $H_2(mathbb{Z}^{2g}) cong wedge^2 mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.






      share|cite|improve this answer



























        up vote
        6
        down vote













        Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $Sigma_g$ itself is an Eilenberg-MacLane space. Let ${a_1,b_1,ldots,a_g,b_g}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ be the map whose kernel is spanned by ${[a_2],[b_2],ldots,[a_g],[b_g]}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $mathbb{Z}^2$. To prove that $psi$ is injective, it is enough to prove that the composition
        $$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
        is injective. But $phi$ is easy to understand geometrically: the surface $Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $mathbb{Z}^2$, and $phi$ is induced by the map $Phicolon Sigma_g rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,ldots,a_g,b_g$. The point here is that it is obvious that $Phi_{ast}$ takes the fundamental class of $Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.



        You can soup up this argument to show that $psi$ takes the fundamental class of $Sigma_g$ to the element $a_1 wedge b_1 + cdots + a_g wedge b_g$ of $H_2(mathbb{Z}^{2g}) cong wedge^2 mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.






        share|cite|improve this answer

























          up vote
          6
          down vote










          up vote
          6
          down vote









          Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $Sigma_g$ itself is an Eilenberg-MacLane space. Let ${a_1,b_1,ldots,a_g,b_g}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ be the map whose kernel is spanned by ${[a_2],[b_2],ldots,[a_g],[b_g]}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $mathbb{Z}^2$. To prove that $psi$ is injective, it is enough to prove that the composition
          $$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
          is injective. But $phi$ is easy to understand geometrically: the surface $Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $mathbb{Z}^2$, and $phi$ is induced by the map $Phicolon Sigma_g rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,ldots,a_g,b_g$. The point here is that it is obvious that $Phi_{ast}$ takes the fundamental class of $Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.



          You can soup up this argument to show that $psi$ takes the fundamental class of $Sigma_g$ to the element $a_1 wedge b_1 + cdots + a_g wedge b_g$ of $H_2(mathbb{Z}^{2g}) cong wedge^2 mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.






          share|cite|improve this answer














          Probably the easiest way to see that the map $psicolon H_2(G) rightarrow H_2(G^{text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $Sigma_g$ itself is an Eilenberg-MacLane space. Let ${a_1,b_1,ldots,a_g,b_g}$ be the usual collection of oriented simple closed curves that one draws whose homology classes form a basis for $H_1(Sigma_g)$. Thus $a_i$ intersects $b_i$ once, and otherwise the curves are disjoint. Let $fcolon G^{text{ab}} rightarrow mathbb{Z}^2$ be the map whose kernel is spanned by ${[a_2],[b_2],ldots,[a_g],[b_g]}$ and which takes $[a_1]$ and $[b_1]$ to the usual basis for $mathbb{Z}^2$. To prove that $psi$ is injective, it is enough to prove that the composition
          $$phicolon H_2(G) stackrel{psi}{longrightarrow} H_2(G^{text{ab}}) stackrel{f_{ast}}{longrightarrow} H_2(mathbb{Z}^2)$$
          is injective. But $phi$ is easy to understand geometrically: the surface $Sigma_g$ is an Eilenberg-MacLane space for $G$, a torus $T$ is an Eilenberg-MacLane space for $mathbb{Z}^2$, and $phi$ is induced by the map $Phicolon Sigma_g rightarrow T$ that collapses a genus $(g-1)$-subsurface with one boundary component to a point. This subsurface contains $a_2,b_2,ldots,a_g,b_g$. The point here is that it is obvious that $Phi_{ast}$ takes the fundamental class of $Sigma_g$ to the fundamental class of $T$, and thus induces an isomorphism on $H_2$.



          You can soup up this argument to show that $psi$ takes the fundamental class of $Sigma_g$ to the element $a_1 wedge b_1 + cdots + a_g wedge b_g$ of $H_2(mathbb{Z}^{2g}) cong wedge^2 mathbb{Z}^{2g}$. For more details, see Theorem 2.7 of the lecture notes from my Park City course on the Torelli group, which are available here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 10 hours ago

























          answered 10 hours ago









          Andy Putman

          30.7k5132210




          30.7k5132210






















              up vote
              2
              down vote













              The dual to the map
              $psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$. Clearly, the latter map is surjective; hence, the former map must be injective.






              share|cite|improve this answer

























                up vote
                2
                down vote













                The dual to the map
                $psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$. Clearly, the latter map is surjective; hence, the former map must be injective.






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  The dual to the map
                  $psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$. Clearly, the latter map is surjective; hence, the former map must be injective.






                  share|cite|improve this answer












                  The dual to the map
                  $psicolon H_2(G,mathbb{Z}) to H_2(G^{operatorname{ab}},mathbb{Z})$ is the cup-product map $cupcolon H^1(G,mathbb{Z})wedge H^1(G,mathbb{Z}) to H^2(G,mathbb{Z})$. Clearly, the latter map is surjective; hence, the former map must be injective.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  Alex Suciu

                  1,8331415




                  1,8331415






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315971%2fa-question-on-the-fundamental-group-of-a-compact-orientable-surface-of-genus-1%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Costa Masnaga

                      Fotorealismo

                      Sidney Franklin