If I understand correctly,

(1) the mapping class group of $S^2 \times S^1$ is $$ \mathbb{Z}_2 \times \mathbb{Z}_2, $$

  • 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.

  • $\begingroup$ 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. $\endgroup$ Jul 22, 2018 at 20:28
  • 1
    $\begingroup$ 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 $S^2 \to S^2$.) $\endgroup$
    – user98602
    Jul 22, 2018 at 20:54
  • $\begingroup$ It is a theorem of Hatcher that there is a homotopy equivalence $\text{Diff}(S^2 \times S^1) \simeq O(3) \times O(2) \times \Omega SO(3)$. $\endgroup$
    – user98602
    Jul 22, 2018 at 20:55
  • $\begingroup$ You know a lot, Mike and Abnubav. How do you define ΩSO(3) just make sure? $\endgroup$
    – wonderich
    Jul 22, 2018 at 21:27
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    $\begingroup$ 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. $\endgroup$
    – user98602
    Jul 22, 2018 at 21:29

1 Answer 1


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 $\text{Sp}_{2g}(\Bbb Z)$, 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 $\text{Out}(\pi_1 \Sigma_g)$, 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 $\text{Sp}_2(\Bbb Z) = SL_2(\Bbb Z)$, 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 $\text{MCG}(Y) \cong \text{Out}(\pi_1 Y)$ for any Haken manifold $Y$. On our example above, $\text{Out}(\pi_1 T^3) = GL_3 \Bbb Z$; 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 $(\Bbb Z/2)^2 = \text{Aut}(H_* S^1 \times S^2)$, 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 $\text{MCG}_{smooth}(X_n)$ 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 $\text{MCG}(S^2 \times D^2, \text{rel } \partial)/\text{MCG}(D^4, \text{rel } \partial) = *$ - 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 $\text{MCG}(S^2 \times S^2)$.

In dimensions $d \geq 5$, there is good, and there is bad. The good: If $M$ is simply connected, then Sullivan proved that $\text{MCG}(M)$ 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 $\text{min}(p, q) \geq 2$. 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 $\text{Aut}(H_p(S^p \times S^p)) = SL_2 \Bbb Z$ 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 $$\text{Aut}(S^p \times S^p) := \text{MCG}(S^p \times S^p)/\text{MCG}(D^{2p}, \text{rel } \partial).$$

For $p = 3, 7$ he identifies $\text{Aut}(S^p \times S^p) \cong \Bbb Z^2 \oplus SL_2 \Bbb Z$ (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 $\pi_0 \text{SDiff}(M)$, and fits into a short exact sequence $$0 \to \pi_0 \text{SDiff}(M) \to \text{MCG}^+(M) \to \Bbb Z/2.$$

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 $\Omega_6^\text{spin}(\Bbb{CP}^\infty)$. 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.


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