A question on the fundamental group of a compact orientable surface of genus >1
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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
add a comment |
up vote
8
down vote
favorite
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
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
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
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
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
at.algebraic-topology gr.group-theory free-groups
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
add a comment |
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
add a comment |
2 Answers
2
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up vote
6
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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.
add a comment |
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.
add a comment |
2 Answers
2
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2 Answers
2
active
oldest
votes
active
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active
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up vote
6
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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.
add a comment |
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.
add a comment |
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.
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.
edited 10 hours ago
answered 10 hours ago
Andy Putman
30.7k5132210
30.7k5132210
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add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered 4 hours ago
Alex Suciu
1,8331415
1,8331415
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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