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Mapping class group of $S^p times S^q$
Clash Royale CLAN TAG#URR8PPP
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If I understand correctly,
(1) the mapping class group of $S^2 times S^1$ is
$$
mathbbZ_2 times mathbbZ_2,
$$
- how do I understand these two generators?
(2) What is the mapping class group of $S^p times S^q$ in general?
Here $S^d$ is a d-sphere.
general-topology differential-topology homology-cohomology geometric-topology mapping-class-group
asked Jul 22 at 19:49
wonderich
1,64821226
 |Â
show 1 more comment
up vote
3
down vote
favorite
If I understand correctly,
(1) the mapping class group of $S^2 times S^1$ is
$$
mathbbZ_2 times mathbbZ_2,
$$
- how do I understand these two generators?
(2) What is the mapping class group of $S^p times S^q$ in general?
Here $S^d$ is a d-sphere.
general-topology differential-topology homology-cohomology geometric-topology mapping-class-group
asked Jul 22 at 19:49
wonderich
1,64821226
For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
1
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
1
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If I understand correctly,
(1) the mapping class group of $S^2 times S^1$ is
$$
mathbbZ_2 times mathbbZ_2,
$$
- how do I understand these two generators?
(2) What is the mapping class group of $S^p times S^q$ in general?
Here $S^d$ is a d-sphere.
general-topology differential-topology homology-cohomology geometric-topology mapping-class-group
asked Jul 22 at 19:49
wonderich
1,64821226
If I understand correctly,
(1) the mapping class group of $S^2 times S^1$ is
$$
mathbbZ_2 times mathbbZ_2,
$$
- how do I understand these two generators?
(2) What is the mapping class group of $S^p times S^q$ in general?
Here $S^d$ is a d-sphere.
general-topology differential-topology homology-cohomology geometric-topology mapping-class-group
asked Jul 22 at 19:49
wonderich
1,64821226
asked Jul 22 at 19:49
wonderich
1,64821226
asked Jul 22 at 19:49
wonderich
1,64821226
asked Jul 22 at 19:49
wonderich
1,64821226
1,64821226
For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
1
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
1
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29
 |Â
show 1 more comment
For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
1
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
1
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29
For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
1
1
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
1
1
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29
 |Â
show 1 more comment
1 Answer
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I think any answer deserves to start with a landscape of the field of mapping class groups, if for no reason but a word of warning.
Allow me to restrict to orientation-preserving transformations as a mild convenience.
For 2-dimensional manifolds, mapping class groups are very well-studied objects. For surfaces of genus $g geq 2$, the mapping class group is a complicated (but finitely presented) group; it always surjects onto the group $textSp_2g(Bbb Z)$, the group of automorphisms of $H_1(Sigma_g;Bbb Z) cong Bbb Z^2g$ preserving the intersection form $H_1(Sigma_g) otimes H_1(Sigma_g) to Bbb Z$.
In general, the mapping class group of a closed oriented surface is isomorphic to $textOut(pi_1 Sigma_g)$, the space of automorphisms $pi_1 Sigma_g to pi_1 Sigma_g$, modulo those automorphisms induced by conjugation $varphi_g: gamma mapsto g gamma g^-1$.
For the 2-torus $S^1 times S^1$, the mapping class group surjects onto $textSp_2(Bbb Z) = SL_2(Bbb Z)$, and in fact this surjection is an isomorphism.
In 3 dimensions, mapping class groups are still well-studied and well-understood. For any "Haken manifold", which is an irreducible 3-manifold (that is, every $S^2 hookrightarrow M$ bounds a 3-ball) so that there is some surface of positive genus $Sigma hookrightarrow M$ which is "incompressible", meaning that no embedded loop in $Sigma$ bounds an embedded disc in $M$ unless it already bounds a disc in $Sigma$. $T^3$ is a simple example of a Haken manifold, with incompressible surface $T^2 hookrightarrow T^3$.
Waldhausen proved that $textMCG(Y) cong textOut(pi_1 Y)$ for any Haken manifold $Y$. On our example above, $textOut(pi_1 T^3) = GL_3 Bbb Z$; the orientation-preserving mapping class group is thus $SL_3 Bbb Z$.
Now you cared about $S^1 times S^2$; its mapping class group can be calculated as $(Bbb Z/2)^2 = textAut(H_* S^1 times S^2)$, where the first $Bbb Z/2$ factor corresponds to a map that induces $-1$ on $H_i(S^1 times S^2)$ (for $i = 1,2$), and the third factor corresponds to the nontrivial loop in $SO(3)$.
If you want to get a good feel for mapping class groups of 3-manifolds, you can't do much better than starting with Hatcher's note A 50-year view of diffeomorphism groups.
Up through here, you could use homeomorphisms or diffeomorphisms more or less interchangeably.
In dimension 4 essentially nothing is known. Danny Ruberman has a paper constructing an infinite sequence of simply connected 4-manifolds $X_n$ such that the map from the smooth mapping class group (of diffeomorphisms mod smooth isotopy) to the continuous mapping class group (of homeomorphisms mod continuous isotopy) has infinitely generated kernel. That's wild! It is worth noting that this does not prove that $textMCG_smooth(X_n)$ is infinitely generated, but it seems plausible.
In dimensions up to 3, these groups were actually the same. I will stick with smooth mapping class groups from now on.
The best you can say is a recent theorem of Gabai (called the 4D lightbulb theorem), which has as a corollary that $textMCG(S^2 times D^2, textrel partial)/textMCG(D^4, textrel partial) = *$ - that is, every diffeomorphism of $S^2 times D^2$ rel the boundary is given by doing some diffeomorphism of the 4-ball somewhere in the interior. It seems very difficult to extend this to say anything about $textMCG(S^2 times S^2)$.
In dimensions $d geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $textMCG(M)$ is finitely presented. (This can go as badly as possible when $M$ has infinite fundamental group, eg the example of $T^n$ given there.)
The bad: These groups are somewhere between very hard and impossible to calculate.
We restrict to $S^p times S^q$ where $p + q geq 5$ and $textmin(p, q) geq 2$. There is a dichotomy between the cases where $p = q$ and where $p neq q$. This is analagous to the difference between $S^1 times S^1$ and $S^1 times S^2$: the point is that $textAut(H_p(S^p times S^p)) = SL_2 Bbb Z$ when $p$ is odd (and $D_8$ when $p$ is even; the difference is whether the intersection form is symmetric or skew-symmetric). However, there are many fewer automorphisms of $H_p(S^p times S^q) = Bbb Z$ where $p neq q$ (there is just multiplication by -1).
Now to quote some results.
First, for $S^p times S^p$, our starting point is this paper by Diarmuid Crowley. Just like in the statement of Gabai's theorem above, we should start by quotienting by the action of diffeomorphisms of the $2p$-ball (so that we don't really have to think about them). Call $$textAut(S^p times S^p) := textMCG(S^p times S^p)/textMCG(D^2p, textrel partial).$$
For $p = 3, 7$ he identifies $textAut(S^p times S^p) cong Bbb Z^2 oplus SL_2 Bbb Z$ (the $Bbb Z^2$ factor acts trivially on homology). When $p = 4j-1 > 7$, the $SL_2 Bbb Z$ is replaced by automorphisms of a more complicated quadratic form.
It seems you should be able to get more results about these manifolds from the article of Kreck here, but I did not have the patience to work out the answer.
When $p neq q$, here are some results.
First, Turner in 1969 calculated, when $k < l < 2k - 3$, that the mapping class group of diffeomorphisms that act trivially on homology is given as a semidirect product $$(pi_l SO_k+1 oplus Gamma_k+l+1) rtimes pi_k SO,$$ where $Gamma_j$ is the group of oriented exotic $j$-spheres modulo oriented diffeomorphism. (This mapping class group should be called $pi_0 textSDiff(M)$, and fits into a short exact sequence $$0 to pi_0 textSDiff(M) to textMCG^+(M) to Bbb Z/2.$$
I will provide one example not in that case: the mapping class group acting trivially on the homology $S^2 times S^3$ (which fits into a sequence as above). Diarmuid Crowley here says that this group is calculated by work of Fang as being isomorphic to $Omega_6^textspin(BbbCP^infty)$. With some help from homotopy theorists on this site, a difficult calculation shows that this is isomorphic to $Bbb Z^2$.
I hope this shows how wild, interesting, and very difficult to study are mapping class groups (and diffeomorphism groups) in high dimensions.
answered Jul 26 at 9:33
Mike Miller
33.9k364128
add a comment |Â
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
I think any answer deserves to start with a landscape of the field of mapping class groups, if for no reason but a word of warning.
Allow me to restrict to orientation-preserving transformations as a mild convenience.
For 2-dimensional manifolds, mapping class groups are very well-studied objects. For surfaces of genus $g geq 2$, the mapping class group is a complicated (but finitely presented) group; it always surjects onto the group $textSp_2g(Bbb Z)$, the group of automorphisms of $H_1(Sigma_g;Bbb Z) cong Bbb Z^2g$ preserving the intersection form $H_1(Sigma_g) otimes H_1(Sigma_g) to Bbb Z$.
In general, the mapping class group of a closed oriented surface is isomorphic to $textOut(pi_1 Sigma_g)$, the space of automorphisms $pi_1 Sigma_g to pi_1 Sigma_g$, modulo those automorphisms induced by conjugation $varphi_g: gamma mapsto g gamma g^-1$.
For the 2-torus $S^1 times S^1$, the mapping class group surjects onto $textSp_2(Bbb Z) = SL_2(Bbb Z)$, and in fact this surjection is an isomorphism.
In 3 dimensions, mapping class groups are still well-studied and well-understood. For any "Haken manifold", which is an irreducible 3-manifold (that is, every $S^2 hookrightarrow M$ bounds a 3-ball) so that there is some surface of positive genus $Sigma hookrightarrow M$ which is "incompressible", meaning that no embedded loop in $Sigma$ bounds an embedded disc in $M$ unless it already bounds a disc in $Sigma$. $T^3$ is a simple example of a Haken manifold, with incompressible surface $T^2 hookrightarrow T^3$.
Waldhausen proved that $textMCG(Y) cong textOut(pi_1 Y)$ for any Haken manifold $Y$. On our example above, $textOut(pi_1 T^3) = GL_3 Bbb Z$; the orientation-preserving mapping class group is thus $SL_3 Bbb Z$.
Now you cared about $S^1 times S^2$; its mapping class group can be calculated as $(Bbb Z/2)^2 = textAut(H_* S^1 times S^2)$, where the first $Bbb Z/2$ factor corresponds to a map that induces $-1$ on $H_i(S^1 times S^2)$ (for $i = 1,2$), and the third factor corresponds to the nontrivial loop in $SO(3)$.
If you want to get a good feel for mapping class groups of 3-manifolds, you can't do much better than starting with Hatcher's note A 50-year view of diffeomorphism groups.
Up through here, you could use homeomorphisms or diffeomorphisms more or less interchangeably.
In dimension 4 essentially nothing is known. Danny Ruberman has a paper constructing an infinite sequence of simply connected 4-manifolds $X_n$ such that the map from the smooth mapping class group (of diffeomorphisms mod smooth isotopy) to the continuous mapping class group (of homeomorphisms mod continuous isotopy) has infinitely generated kernel. That's wild! It is worth noting that this does not prove that $textMCG_smooth(X_n)$ is infinitely generated, but it seems plausible.
In dimensions up to 3, these groups were actually the same. I will stick with smooth mapping class groups from now on.
The best you can say is a recent theorem of Gabai (called the 4D lightbulb theorem), which has as a corollary that $textMCG(S^2 times D^2, textrel partial)/textMCG(D^4, textrel partial) = *$ - that is, every diffeomorphism of $S^2 times D^2$ rel the boundary is given by doing some diffeomorphism of the 4-ball somewhere in the interior. It seems very difficult to extend this to say anything about $textMCG(S^2 times S^2)$.
In dimensions $d geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $textMCG(M)$ is finitely presented. (This can go as badly as possible when $M$ has infinite fundamental group, eg the example of $T^n$ given there.)
The bad: These groups are somewhere between very hard and impossible to calculate.
We restrict to $S^p times S^q$ where $p + q geq 5$ and $textmin(p, q) geq 2$. There is a dichotomy between the cases where $p = q$ and where $p neq q$. This is analagous to the difference between $S^1 times S^1$ and $S^1 times S^2$: the point is that $textAut(H_p(S^p times S^p)) = SL_2 Bbb Z$ when $p$ is odd (and $D_8$ when $p$ is even; the difference is whether the intersection form is symmetric or skew-symmetric). However, there are many fewer automorphisms of $H_p(S^p times S^q) = Bbb Z$ where $p neq q$ (there is just multiplication by -1).
Now to quote some results.
First, for $S^p times S^p$, our starting point is this paper by Diarmuid Crowley. Just like in the statement of Gabai's theorem above, we should start by quotienting by the action of diffeomorphisms of the $2p$-ball (so that we don't really have to think about them). Call $$textAut(S^p times S^p) := textMCG(S^p times S^p)/textMCG(D^2p, textrel partial).$$
For $p = 3, 7$ he identifies $textAut(S^p times S^p) cong Bbb Z^2 oplus SL_2 Bbb Z$ (the $Bbb Z^2$ factor acts trivially on homology). When $p = 4j-1 > 7$, the $SL_2 Bbb Z$ is replaced by automorphisms of a more complicated quadratic form.
It seems you should be able to get more results about these manifolds from the article of Kreck here, but I did not have the patience to work out the answer.
When $p neq q$, here are some results.
First, Turner in 1969 calculated, when $k < l < 2k - 3$, that the mapping class group of diffeomorphisms that act trivially on homology is given as a semidirect product $$(pi_l SO_k+1 oplus Gamma_k+l+1) rtimes pi_k SO,$$ where $Gamma_j$ is the group of oriented exotic $j$-spheres modulo oriented diffeomorphism. (This mapping class group should be called $pi_0 textSDiff(M)$, and fits into a short exact sequence $$0 to pi_0 textSDiff(M) to textMCG^+(M) to Bbb Z/2.$$
I will provide one example not in that case: the mapping class group acting trivially on the homology $S^2 times S^3$ (which fits into a sequence as above). Diarmuid Crowley here says that this group is calculated by work of Fang as being isomorphic to $Omega_6^textspin(BbbCP^infty)$. With some help from homotopy theorists on this site, a difficult calculation shows that this is isomorphic to $Bbb Z^2$.
I hope this shows how wild, interesting, and very difficult to study are mapping class groups (and diffeomorphism groups) in high dimensions.
answered Jul 26 at 9:33
Mike Miller
33.9k364128
add a comment |Â
up vote
4
down vote
I think any answer deserves to start with a landscape of the field of mapping class groups, if for no reason but a word of warning.
Allow me to restrict to orientation-preserving transformations as a mild convenience.
For 2-dimensional manifolds, mapping class groups are very well-studied objects. For surfaces of genus $g geq 2$, the mapping class group is a complicated (but finitely presented) group; it always surjects onto the group $textSp_2g(Bbb Z)$, the group of automorphisms of $H_1(Sigma_g;Bbb Z) cong Bbb Z^2g$ preserving the intersection form $H_1(Sigma_g) otimes H_1(Sigma_g) to Bbb Z$.
In general, the mapping class group of a closed oriented surface is isomorphic to $textOut(pi_1 Sigma_g)$, the space of automorphisms $pi_1 Sigma_g to pi_1 Sigma_g$, modulo those automorphisms induced by conjugation $varphi_g: gamma mapsto g gamma g^-1$.
For the 2-torus $S^1 times S^1$, the mapping class group surjects onto $textSp_2(Bbb Z) = SL_2(Bbb Z)$, and in fact this surjection is an isomorphism.
In 3 dimensions, mapping class groups are still well-studied and well-understood. For any "Haken manifold", which is an irreducible 3-manifold (that is, every $S^2 hookrightarrow M$ bounds a 3-ball) so that there is some surface of positive genus $Sigma hookrightarrow M$ which is "incompressible", meaning that no embedded loop in $Sigma$ bounds an embedded disc in $M$ unless it already bounds a disc in $Sigma$. $T^3$ is a simple example of a Haken manifold, with incompressible surface $T^2 hookrightarrow T^3$.
Waldhausen proved that $textMCG(Y) cong textOut(pi_1 Y)$ for any Haken manifold $Y$. On our example above, $textOut(pi_1 T^3) = GL_3 Bbb Z$; the orientation-preserving mapping class group is thus $SL_3 Bbb Z$.
Now you cared about $S^1 times S^2$; its mapping class group can be calculated as $(Bbb Z/2)^2 = textAut(H_* S^1 times S^2)$, where the first $Bbb Z/2$ factor corresponds to a map that induces $-1$ on $H_i(S^1 times S^2)$ (for $i = 1,2$), and the third factor corresponds to the nontrivial loop in $SO(3)$.
If you want to get a good feel for mapping class groups of 3-manifolds, you can't do much better than starting with Hatcher's note A 50-year view of diffeomorphism groups.
Up through here, you could use homeomorphisms or diffeomorphisms more or less interchangeably.
In dimension 4 essentially nothing is known. Danny Ruberman has a paper constructing an infinite sequence of simply connected 4-manifolds $X_n$ such that the map from the smooth mapping class group (of diffeomorphisms mod smooth isotopy) to the continuous mapping class group (of homeomorphisms mod continuous isotopy) has infinitely generated kernel. That's wild! It is worth noting that this does not prove that $textMCG_smooth(X_n)$ is infinitely generated, but it seems plausible.
In dimensions up to 3, these groups were actually the same. I will stick with smooth mapping class groups from now on.
The best you can say is a recent theorem of Gabai (called the 4D lightbulb theorem), which has as a corollary that $textMCG(S^2 times D^2, textrel partial)/textMCG(D^4, textrel partial) = *$ - that is, every diffeomorphism of $S^2 times D^2$ rel the boundary is given by doing some diffeomorphism of the 4-ball somewhere in the interior. It seems very difficult to extend this to say anything about $textMCG(S^2 times S^2)$.
In dimensions $d geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $textMCG(M)$ is finitely presented. (This can go as badly as possible when $M$ has infinite fundamental group, eg the example of $T^n$ given there.)
The bad: These groups are somewhere between very hard and impossible to calculate.
We restrict to $S^p times S^q$ where $p + q geq 5$ and $textmin(p, q) geq 2$. There is a dichotomy between the cases where $p = q$ and where $p neq q$. This is analagous to the difference between $S^1 times S^1$ and $S^1 times S^2$: the point is that $textAut(H_p(S^p times S^p)) = SL_2 Bbb Z$ when $p$ is odd (and $D_8$ when $p$ is even; the difference is whether the intersection form is symmetric or skew-symmetric). However, there are many fewer automorphisms of $H_p(S^p times S^q) = Bbb Z$ where $p neq q$ (there is just multiplication by -1).
Now to quote some results.
First, for $S^p times S^p$, our starting point is this paper by Diarmuid Crowley. Just like in the statement of Gabai's theorem above, we should start by quotienting by the action of diffeomorphisms of the $2p$-ball (so that we don't really have to think about them). Call $$textAut(S^p times S^p) := textMCG(S^p times S^p)/textMCG(D^2p, textrel partial).$$
For $p = 3, 7$ he identifies $textAut(S^p times S^p) cong Bbb Z^2 oplus SL_2 Bbb Z$ (the $Bbb Z^2$ factor acts trivially on homology). When $p = 4j-1 > 7$, the $SL_2 Bbb Z$ is replaced by automorphisms of a more complicated quadratic form.
It seems you should be able to get more results about these manifolds from the article of Kreck here, but I did not have the patience to work out the answer.
When $p neq q$, here are some results.
First, Turner in 1969 calculated, when $k < l < 2k - 3$, that the mapping class group of diffeomorphisms that act trivially on homology is given as a semidirect product $$(pi_l SO_k+1 oplus Gamma_k+l+1) rtimes pi_k SO,$$ where $Gamma_j$ is the group of oriented exotic $j$-spheres modulo oriented diffeomorphism. (This mapping class group should be called $pi_0 textSDiff(M)$, and fits into a short exact sequence $$0 to pi_0 textSDiff(M) to textMCG^+(M) to Bbb Z/2.$$
I will provide one example not in that case: the mapping class group acting trivially on the homology $S^2 times S^3$ (which fits into a sequence as above). Diarmuid Crowley here says that this group is calculated by work of Fang as being isomorphic to $Omega_6^textspin(BbbCP^infty)$. With some help from homotopy theorists on this site, a difficult calculation shows that this is isomorphic to $Bbb Z^2$.
I hope this shows how wild, interesting, and very difficult to study are mapping class groups (and diffeomorphism groups) in high dimensions.
answered Jul 26 at 9:33
Mike Miller
33.9k364128
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I think any answer deserves to start with a landscape of the field of mapping class groups, if for no reason but a word of warning.
Allow me to restrict to orientation-preserving transformations as a mild convenience.
For 2-dimensional manifolds, mapping class groups are very well-studied objects. For surfaces of genus $g geq 2$, the mapping class group is a complicated (but finitely presented) group; it always surjects onto the group $textSp_2g(Bbb Z)$, the group of automorphisms of $H_1(Sigma_g;Bbb Z) cong Bbb Z^2g$ preserving the intersection form $H_1(Sigma_g) otimes H_1(Sigma_g) to Bbb Z$.
In general, the mapping class group of a closed oriented surface is isomorphic to $textOut(pi_1 Sigma_g)$, the space of automorphisms $pi_1 Sigma_g to pi_1 Sigma_g$, modulo those automorphisms induced by conjugation $varphi_g: gamma mapsto g gamma g^-1$.
For the 2-torus $S^1 times S^1$, the mapping class group surjects onto $textSp_2(Bbb Z) = SL_2(Bbb Z)$, and in fact this surjection is an isomorphism.
In 3 dimensions, mapping class groups are still well-studied and well-understood. For any "Haken manifold", which is an irreducible 3-manifold (that is, every $S^2 hookrightarrow M$ bounds a 3-ball) so that there is some surface of positive genus $Sigma hookrightarrow M$ which is "incompressible", meaning that no embedded loop in $Sigma$ bounds an embedded disc in $M$ unless it already bounds a disc in $Sigma$. $T^3$ is a simple example of a Haken manifold, with incompressible surface $T^2 hookrightarrow T^3$.
Waldhausen proved that $textMCG(Y) cong textOut(pi_1 Y)$ for any Haken manifold $Y$. On our example above, $textOut(pi_1 T^3) = GL_3 Bbb Z$; the orientation-preserving mapping class group is thus $SL_3 Bbb Z$.
Now you cared about $S^1 times S^2$; its mapping class group can be calculated as $(Bbb Z/2)^2 = textAut(H_* S^1 times S^2)$, where the first $Bbb Z/2$ factor corresponds to a map that induces $-1$ on $H_i(S^1 times S^2)$ (for $i = 1,2$), and the third factor corresponds to the nontrivial loop in $SO(3)$.
If you want to get a good feel for mapping class groups of 3-manifolds, you can't do much better than starting with Hatcher's note A 50-year view of diffeomorphism groups.
Up through here, you could use homeomorphisms or diffeomorphisms more or less interchangeably.
In dimension 4 essentially nothing is known. Danny Ruberman has a paper constructing an infinite sequence of simply connected 4-manifolds $X_n$ such that the map from the smooth mapping class group (of diffeomorphisms mod smooth isotopy) to the continuous mapping class group (of homeomorphisms mod continuous isotopy) has infinitely generated kernel. That's wild! It is worth noting that this does not prove that $textMCG_smooth(X_n)$ is infinitely generated, but it seems plausible.
In dimensions up to 3, these groups were actually the same. I will stick with smooth mapping class groups from now on.
The best you can say is a recent theorem of Gabai (called the 4D lightbulb theorem), which has as a corollary that $textMCG(S^2 times D^2, textrel partial)/textMCG(D^4, textrel partial) = *$ - that is, every diffeomorphism of $S^2 times D^2$ rel the boundary is given by doing some diffeomorphism of the 4-ball somewhere in the interior. It seems very difficult to extend this to say anything about $textMCG(S^2 times S^2)$.
In dimensions $d geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $textMCG(M)$ is finitely presented. (This can go as badly as possible when $M$ has infinite fundamental group, eg the example of $T^n$ given there.)
The bad: These groups are somewhere between very hard and impossible to calculate.
We restrict to $S^p times S^q$ where $p + q geq 5$ and $textmin(p, q) geq 2$. There is a dichotomy between the cases where $p = q$ and where $p neq q$. This is analagous to the difference between $S^1 times S^1$ and $S^1 times S^2$: the point is that $textAut(H_p(S^p times S^p)) = SL_2 Bbb Z$ when $p$ is odd (and $D_8$ when $p$ is even; the difference is whether the intersection form is symmetric or skew-symmetric). However, there are many fewer automorphisms of $H_p(S^p times S^q) = Bbb Z$ where $p neq q$ (there is just multiplication by -1).
Now to quote some results.
First, for $S^p times S^p$, our starting point is this paper by Diarmuid Crowley. Just like in the statement of Gabai's theorem above, we should start by quotienting by the action of diffeomorphisms of the $2p$-ball (so that we don't really have to think about them). Call $$textAut(S^p times S^p) := textMCG(S^p times S^p)/textMCG(D^2p, textrel partial).$$
For $p = 3, 7$ he identifies $textAut(S^p times S^p) cong Bbb Z^2 oplus SL_2 Bbb Z$ (the $Bbb Z^2$ factor acts trivially on homology). When $p = 4j-1 > 7$, the $SL_2 Bbb Z$ is replaced by automorphisms of a more complicated quadratic form.
It seems you should be able to get more results about these manifolds from the article of Kreck here, but I did not have the patience to work out the answer.
When $p neq q$, here are some results.
First, Turner in 1969 calculated, when $k < l < 2k - 3$, that the mapping class group of diffeomorphisms that act trivially on homology is given as a semidirect product $$(pi_l SO_k+1 oplus Gamma_k+l+1) rtimes pi_k SO,$$ where $Gamma_j$ is the group of oriented exotic $j$-spheres modulo oriented diffeomorphism. (This mapping class group should be called $pi_0 textSDiff(M)$, and fits into a short exact sequence $$0 to pi_0 textSDiff(M) to textMCG^+(M) to Bbb Z/2.$$
I will provide one example not in that case: the mapping class group acting trivially on the homology $S^2 times S^3$ (which fits into a sequence as above). Diarmuid Crowley here says that this group is calculated by work of Fang as being isomorphic to $Omega_6^textspin(BbbCP^infty)$. With some help from homotopy theorists on this site, a difficult calculation shows that this is isomorphic to $Bbb Z^2$.
I hope this shows how wild, interesting, and very difficult to study are mapping class groups (and diffeomorphism groups) in high dimensions.
answered Jul 26 at 9:33
Mike Miller
33.9k364128
I think any answer deserves to start with a landscape of the field of mapping class groups, if for no reason but a word of warning.
Allow me to restrict to orientation-preserving transformations as a mild convenience.
For 2-dimensional manifolds, mapping class groups are very well-studied objects. For surfaces of genus $g geq 2$, the mapping class group is a complicated (but finitely presented) group; it always surjects onto the group $textSp_2g(Bbb Z)$, the group of automorphisms of $H_1(Sigma_g;Bbb Z) cong Bbb Z^2g$ preserving the intersection form $H_1(Sigma_g) otimes H_1(Sigma_g) to Bbb Z$.
In general, the mapping class group of a closed oriented surface is isomorphic to $textOut(pi_1 Sigma_g)$, the space of automorphisms $pi_1 Sigma_g to pi_1 Sigma_g$, modulo those automorphisms induced by conjugation $varphi_g: gamma mapsto g gamma g^-1$.
For the 2-torus $S^1 times S^1$, the mapping class group surjects onto $textSp_2(Bbb Z) = SL_2(Bbb Z)$, and in fact this surjection is an isomorphism.
In 3 dimensions, mapping class groups are still well-studied and well-understood. For any "Haken manifold", which is an irreducible 3-manifold (that is, every $S^2 hookrightarrow M$ bounds a 3-ball) so that there is some surface of positive genus $Sigma hookrightarrow M$ which is "incompressible", meaning that no embedded loop in $Sigma$ bounds an embedded disc in $M$ unless it already bounds a disc in $Sigma$. $T^3$ is a simple example of a Haken manifold, with incompressible surface $T^2 hookrightarrow T^3$.
Waldhausen proved that $textMCG(Y) cong textOut(pi_1 Y)$ for any Haken manifold $Y$. On our example above, $textOut(pi_1 T^3) = GL_3 Bbb Z$; the orientation-preserving mapping class group is thus $SL_3 Bbb Z$.
Now you cared about $S^1 times S^2$; its mapping class group can be calculated as $(Bbb Z/2)^2 = textAut(H_* S^1 times S^2)$, where the first $Bbb Z/2$ factor corresponds to a map that induces $-1$ on $H_i(S^1 times S^2)$ (for $i = 1,2$), and the third factor corresponds to the nontrivial loop in $SO(3)$.
If you want to get a good feel for mapping class groups of 3-manifolds, you can't do much better than starting with Hatcher's note A 50-year view of diffeomorphism groups.
Up through here, you could use homeomorphisms or diffeomorphisms more or less interchangeably.
In dimension 4 essentially nothing is known. Danny Ruberman has a paper constructing an infinite sequence of simply connected 4-manifolds $X_n$ such that the map from the smooth mapping class group (of diffeomorphisms mod smooth isotopy) to the continuous mapping class group (of homeomorphisms mod continuous isotopy) has infinitely generated kernel. That's wild! It is worth noting that this does not prove that $textMCG_smooth(X_n)$ is infinitely generated, but it seems plausible.
In dimensions up to 3, these groups were actually the same. I will stick with smooth mapping class groups from now on.
The best you can say is a recent theorem of Gabai (called the 4D lightbulb theorem), which has as a corollary that $textMCG(S^2 times D^2, textrel partial)/textMCG(D^4, textrel partial) = *$ - that is, every diffeomorphism of $S^2 times D^2$ rel the boundary is given by doing some diffeomorphism of the 4-ball somewhere in the interior. It seems very difficult to extend this to say anything about $textMCG(S^2 times S^2)$.
In dimensions $d geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $textMCG(M)$ is finitely presented. (This can go as badly as possible when $M$ has infinite fundamental group, eg the example of $T^n$ given there.)
The bad: These groups are somewhere between very hard and impossible to calculate.
We restrict to $S^p times S^q$ where $p + q geq 5$ and $textmin(p, q) geq 2$. There is a dichotomy between the cases where $p = q$ and where $p neq q$. This is analagous to the difference between $S^1 times S^1$ and $S^1 times S^2$: the point is that $textAut(H_p(S^p times S^p)) = SL_2 Bbb Z$ when $p$ is odd (and $D_8$ when $p$ is even; the difference is whether the intersection form is symmetric or skew-symmetric). However, there are many fewer automorphisms of $H_p(S^p times S^q) = Bbb Z$ where $p neq q$ (there is just multiplication by -1).
Now to quote some results.
First, for $S^p times S^p$, our starting point is this paper by Diarmuid Crowley. Just like in the statement of Gabai's theorem above, we should start by quotienting by the action of diffeomorphisms of the $2p$-ball (so that we don't really have to think about them). Call $$textAut(S^p times S^p) := textMCG(S^p times S^p)/textMCG(D^2p, textrel partial).$$
For $p = 3, 7$ he identifies $textAut(S^p times S^p) cong Bbb Z^2 oplus SL_2 Bbb Z$ (the $Bbb Z^2$ factor acts trivially on homology). When $p = 4j-1 > 7$, the $SL_2 Bbb Z$ is replaced by automorphisms of a more complicated quadratic form.
It seems you should be able to get more results about these manifolds from the article of Kreck here, but I did not have the patience to work out the answer.
When $p neq q$, here are some results.
First, Turner in 1969 calculated, when $k < l < 2k - 3$, that the mapping class group of diffeomorphisms that act trivially on homology is given as a semidirect product $$(pi_l SO_k+1 oplus Gamma_k+l+1) rtimes pi_k SO,$$ where $Gamma_j$ is the group of oriented exotic $j$-spheres modulo oriented diffeomorphism. (This mapping class group should be called $pi_0 textSDiff(M)$, and fits into a short exact sequence $$0 to pi_0 textSDiff(M) to textMCG^+(M) to Bbb Z/2.$$
I will provide one example not in that case: the mapping class group acting trivially on the homology $S^2 times S^3$ (which fits into a sequence as above). Diarmuid Crowley here says that this group is calculated by work of Fang as being isomorphic to $Omega_6^textspin(BbbCP^infty)$. With some help from homotopy theorists on this site, a difficult calculation shows that this is isomorphic to $Bbb Z^2$.
I hope this shows how wild, interesting, and very difficult to study are mapping class groups (and diffeomorphism groups) in high dimensions.
answered Jul 26 at 9:33
Mike Miller
33.9k364128
answered Jul 26 at 9:33
Mike Miller
33.9k364128
answered Jul 26 at 9:33
Mike Miller
33.9k364128
answered Jul 26 at 9:33
Mike Miller
33.9k364128
33.9k364128
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For the first question, if you consider a Heegard decomposition by two solid tori, then one of the generator interchange two solid torus, where the other generator will do a self map of degree 2 in each solid torus.
– Anubhav Mukherjee
Jul 22 at 20:28
1
I guess by mapping class group you mean the oriented mapping class group. Then one generator is given by the antipodal map on $S^2$ cross a reflection on $S^1$ (the point being that they are orientation reversing on each fiber). The more interesting generator is given by writing $gamma: S^1 to SO(3)$ for a nontrivial loop, and writing $varphi: S^2 times S^1 to S^2 times S^1$ as $varphi(x,theta) = (gamma_theta(x), theta)$. (Remember that $gamma_theta$ is a diffeomorphism $S^2 to S^2$.)
– Mike Miller
Jul 22 at 20:54
It is a theorem of Hatcher that there is a homotopy equivalence $textDiff(S^2 times S^1) simeq O(3) times O(2) times Omega SO(3)$.
– Mike Miller
Jul 22 at 20:55
You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure?
– wonderich
Jul 22 at 21:27
1
This is the based loop space, the space of maps $S^1 to SO(3)$ sending $1$ to the identity. You could write just as well $O(2) times LO(3)$, where $L$ is the free loop space, where we make no demands on $gamma(1)$; the appearance of $O(3)$ in the above formula accounts for where $gamma(1)$ goes.
– Mike Miller
Jul 22 at 21:29