Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1?A finite index subgroup of the Mapping Class GroupOrigin of the name “Torelli group”Distorsion of subgroups of the mapping class groupTorsion elements in the mapping class groupVeech group and mapping class groupCategorical mapping class group actionA question about the abelianization of the Johnson kernel
Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1?
A finite index subgroup of the Mapping Class GroupOrigin of the name “Torelli group”Distorsion of subgroups of the mapping class groupTorsion elements in the mapping class groupVeech group and mapping class groupCategorical mapping class group actionA question about the abelianization of the Johnson kernel
$begingroup$
Write Mod(g,n) for the mapping class group of a genus-$g$ surface $Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(Sigma,mathbfZ)$.
The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $mathrmHom(H,wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.
Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in
$H^1(mathrmMod(g,1), mathrmHom(H,wedge^2 H))$
where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $pi_1(Sigma)$ by the third term of its lower central series.
All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?
Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?
at.algebraic-topology gt.geometric-topology mapping-class-groups
$endgroup$
add a comment |
$begingroup$
Write Mod(g,n) for the mapping class group of a genus-$g$ surface $Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(Sigma,mathbfZ)$.
The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $mathrmHom(H,wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.
Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in
$H^1(mathrmMod(g,1), mathrmHom(H,wedge^2 H))$
where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $pi_1(Sigma)$ by the third term of its lower central series.
All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?
Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?
at.algebraic-topology gt.geometric-topology mapping-class-groups
$endgroup$
add a comment |
$begingroup$
Write Mod(g,n) for the mapping class group of a genus-$g$ surface $Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(Sigma,mathbfZ)$.
The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $mathrmHom(H,wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.
Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in
$H^1(mathrmMod(g,1), mathrmHom(H,wedge^2 H))$
where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $pi_1(Sigma)$ by the third term of its lower central series.
All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?
Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?
at.algebraic-topology gt.geometric-topology mapping-class-groups
$endgroup$
Write Mod(g,n) for the mapping class group of a genus-$g$ surface $Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(Sigma,mathbfZ)$.
The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $mathrmHom(H,wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.
Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in
$H^1(mathrmMod(g,1), mathrmHom(H,wedge^2 H))$
where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $pi_1(Sigma)$ by the third term of its lower central series.
All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?
Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?
at.algebraic-topology gt.geometric-topology mapping-class-groups
at.algebraic-topology gt.geometric-topology mapping-class-groups
edited 2 hours ago
JSE
asked 8 hours ago
JSEJSE
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15.4k5 gold badges56 silver badges119 bronze badges
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(textAut(F_m), textHom(H, wedge^2 H)),$$
where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $textMod_g, nto textAut(pi_1(Sigma_g, n-1))simeq F_2g+n-2$ for $ngeq 2$, given by the conjugation action of $textMod_g,n$ on the point-pushing subgroup, namely $pi_1(Sigma_g, n-1)$.
A construction goes as follows. Let $mathbbZ[F_m]$ be the group ring of $F_m$, and let $mathscrI$ be the augmentation ideal. Then $H simeq mathscrI/mathscrI^2$ canonically (via the map sending $g$ to $g-1$) and $mathscrI^2/mathscrI^3simeq H^otimes 2$ canonically (via the multiplication map). There is a short exact sequence of $textAut(F_m)$ modules $$0to mathscrI^2/mathscrI^3to mathscrI/mathscrI^3to mathscrI/mathscrI^2to 0,$$ which we can think of as an extension of $H$ by $H^otimes 2$, and hence gives a class in
$$H^1(textAut(F_m), textHom(H, H^otimes 2)).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $textHom(H, textAlt^2(H)).$
$endgroup$
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
|
show 1 more comment
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$begingroup$
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(textAut(F_m), textHom(H, wedge^2 H)),$$
where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $textMod_g, nto textAut(pi_1(Sigma_g, n-1))simeq F_2g+n-2$ for $ngeq 2$, given by the conjugation action of $textMod_g,n$ on the point-pushing subgroup, namely $pi_1(Sigma_g, n-1)$.
A construction goes as follows. Let $mathbbZ[F_m]$ be the group ring of $F_m$, and let $mathscrI$ be the augmentation ideal. Then $H simeq mathscrI/mathscrI^2$ canonically (via the map sending $g$ to $g-1$) and $mathscrI^2/mathscrI^3simeq H^otimes 2$ canonically (via the multiplication map). There is a short exact sequence of $textAut(F_m)$ modules $$0to mathscrI^2/mathscrI^3to mathscrI/mathscrI^3to mathscrI/mathscrI^2to 0,$$ which we can think of as an extension of $H$ by $H^otimes 2$, and hence gives a class in
$$H^1(textAut(F_m), textHom(H, H^otimes 2)).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $textHom(H, textAlt^2(H)).$
$endgroup$
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
|
show 1 more comment
$begingroup$
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(textAut(F_m), textHom(H, wedge^2 H)),$$
where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $textMod_g, nto textAut(pi_1(Sigma_g, n-1))simeq F_2g+n-2$ for $ngeq 2$, given by the conjugation action of $textMod_g,n$ on the point-pushing subgroup, namely $pi_1(Sigma_g, n-1)$.
A construction goes as follows. Let $mathbbZ[F_m]$ be the group ring of $F_m$, and let $mathscrI$ be the augmentation ideal. Then $H simeq mathscrI/mathscrI^2$ canonically (via the map sending $g$ to $g-1$) and $mathscrI^2/mathscrI^3simeq H^otimes 2$ canonically (via the multiplication map). There is a short exact sequence of $textAut(F_m)$ modules $$0to mathscrI^2/mathscrI^3to mathscrI/mathscrI^3to mathscrI/mathscrI^2to 0,$$ which we can think of as an extension of $H$ by $H^otimes 2$, and hence gives a class in
$$H^1(textAut(F_m), textHom(H, H^otimes 2)).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $textHom(H, textAlt^2(H)).$
$endgroup$
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
|
show 1 more comment
$begingroup$
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(textAut(F_m), textHom(H, wedge^2 H)),$$
where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $textMod_g, nto textAut(pi_1(Sigma_g, n-1))simeq F_2g+n-2$ for $ngeq 2$, given by the conjugation action of $textMod_g,n$ on the point-pushing subgroup, namely $pi_1(Sigma_g, n-1)$.
A construction goes as follows. Let $mathbbZ[F_m]$ be the group ring of $F_m$, and let $mathscrI$ be the augmentation ideal. Then $H simeq mathscrI/mathscrI^2$ canonically (via the map sending $g$ to $g-1$) and $mathscrI^2/mathscrI^3simeq H^otimes 2$ canonically (via the multiplication map). There is a short exact sequence of $textAut(F_m)$ modules $$0to mathscrI^2/mathscrI^3to mathscrI/mathscrI^3to mathscrI/mathscrI^2to 0,$$ which we can think of as an extension of $H$ by $H^otimes 2$, and hence gives a class in
$$H^1(textAut(F_m), textHom(H, H^otimes 2)).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $textHom(H, textAlt^2(H)).$
$endgroup$
The answer is "yes" -- in fact one can do better and get a class in
$$H^1(textAut(F_m), textHom(H, wedge^2 H)),$$
where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $textMod_g, nto textAut(pi_1(Sigma_g, n-1))simeq F_2g+n-2$ for $ngeq 2$, given by the conjugation action of $textMod_g,n$ on the point-pushing subgroup, namely $pi_1(Sigma_g, n-1)$.
A construction goes as follows. Let $mathbbZ[F_m]$ be the group ring of $F_m$, and let $mathscrI$ be the augmentation ideal. Then $H simeq mathscrI/mathscrI^2$ canonically (via the map sending $g$ to $g-1$) and $mathscrI^2/mathscrI^3simeq H^otimes 2$ canonically (via the multiplication map). There is a short exact sequence of $textAut(F_m)$ modules $$0to mathscrI^2/mathscrI^3to mathscrI/mathscrI^3to mathscrI/mathscrI^2to 0,$$ which we can think of as an extension of $H$ by $H^otimes 2$, and hence gives a class in
$$H^1(textAut(F_m), textHom(H, H^otimes 2)).$$
But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $textHom(H, textAlt^2(H)).$
answered 4 hours ago
Daniel LittDaniel Litt
15.2k3 gold badges62 silver badges113 bronze badges
15.2k3 gold badges62 silver badges113 bronze badges
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
|
show 1 more comment
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
Actually this will all be in a paper I'm writing with one of your students, among others!
$endgroup$
– Daniel Litt
4 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
To fix the conjugation action, I am worried you need to fix a base point which is $operatornameMod(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something.
$endgroup$
– Will Sawin
3 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
I'm disturbed by the fact that I can't figure out which of my students you're referring to!
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
Do you think when you restrict from Aut(F_m) to Mod(g,n) you still get image in something that looks like wedge^3 H?
$endgroup$
– JSE
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
$begingroup$
@JSE: I don't think so--that identification relies on Poincare duality. And I think in the un-punctured case, one only gets that after taking Inn-coinvariants to get a class that comes from H^1(Out(F_m), ...)... And the paper is the thing with Wanlin et al from the MRC in Rhode Island, as well as (I hope) multiple sequels I think will come out of it.
$endgroup$
– Daniel Litt
2 hours ago
|
show 1 more comment
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StackExchange.helpers.onClickDraftSave('#login-link');
);
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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