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I know that the MCG (isotopy-classes of orientation preserving homeomorphisms) of 3-torus $(S^1\times S^1 \times S^1)$ is $SL(3,Z)$, since it is an Eilenberg–MacLane space, giving $ MCG(T^3)=Out(\pi_1(T^3))=Out(Z^3)=SL(3,Z)$.

But $S^2\times S^1$ is not an Eilenberg–MacLane space. I want to know what $MCG(T^3\# S^2\times S^1)$ is? And, what relation does it have to $\pi_1((T^3\# S^2\times S^1)) $? Is there a general way of finding MCG of connected sum of 3-manifolds?

Thanks

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    $\begingroup$ The fact that $T^n$ is an Eilenberg-MacLane space tells you that the group of homotopy classes of homotopy equivalences is $GL_n(\mathbb{Z})$; it doesn't imply the same fact about mapping class groups, both because 1) a priori a homotopy equivalence may not be representable by a homeomorphism, and 2) a priori a homotopy between homotopy equivalences, even ones that can be represented by homeomorphisms, may not be representable by an isotopy. For the actual mapping class groups, at least when $n \ge 5$, see en.wikipedia.org/wiki/Mapping_class_group#Torus. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 29, 2016 at 4:47
  • $\begingroup$ I am not sure I understand your comment. I was just using the fact that Dehn-Nielsen-Baer theorem applies to space $K(\pi,1)$, that is, $\pi_0HomotopyEquivalences(M)$ is isomorphic to $Out(\pi)$. $\endgroup$
    – SKShukla
    Commented Jul 29, 2016 at 5:05
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    $\begingroup$ Those are two different statements; the Dehn-Nielsen-Baer theorem is a claim about the mapping class groups of surfaces, whereas your second claim is about homotopy classes of homotopy equivalences. Neither of them imply the desired result about the mapping class group of $T^3$. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 29, 2016 at 5:10
  • $\begingroup$ yes, but it generalizes to 3-manifold in this specific case. A discussion can be found here: math.stackexchange.com/questions/35702/… $\endgroup$
    – SKShukla
    Commented Jul 29, 2016 at 5:11
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    $\begingroup$ I do not see what in that post gives the result you want for $T^3$. In any case some further argument beyond what you wrote down is needed. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 29, 2016 at 5:12

1 Answer 1

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The mapping class groups of all compact orientable 3-manifolds are essentially known. A fairly detailed summary of the results, focusing on the nonprime case and with references to proofs in the literature, can be found in Section 2 of a paper of mine with Nathalie Wahl: "Stabilization for mapping class groups of 3-manifolds", Duke Math. J. 155 (2010), 205-269, arXiv:0709.2173. (Section 2 can be read independently of the rest of the paper.) The results are stated in terms of the map $\Phi:MCG(M)\to Out(\pi_1M)$. For $M$ closed and orientable the kernel of $\Phi$ is a direct sum of finitely many copies of ${\mathbb Z}_2$ generated by Dehn twists along 2-spheres. Also, $\Phi$ is surjective if this is true for each prime connected summand of $M$ and a certain other natural condition holds involving permuting different summands with isomorphic fundamental groups.

In the case of $T^3 \# (S^1\times S^2)$ the kernel of $\Phi$ is a single ${\mathbb Z}_2$ generated by a Dehn twist along the $S^2$ factor of $S^1\times S^2$. (Interestingly, the twist along the $S^2$ separating the two connected summands is isotopic to the identity.) The image of $\Phi$ is an index two subgroup of $Out({\mathbb Z}^3*{\mathbb Z})$ since one can only realize $SL(3,{\mathbb Z})$, not all of $GL(3,{\mathbb Z})$, by orientation-preserving homeomorphisms of $T^3$.

The structure of the automorphism group of a free product seems to be well understood in terms of the automorphism groups of the factors. For $M=T^3 \# (S^1\times S^2)$ this means its mapping class group is generated by the MCG's $SL(3,{\mathbb Z})$ and ${\mathbb Z}_2 \times {\mathbb Z}_2$ of the two summands, along with homeomorphisms obtained by regarding $M$ as obtained from $T^3$ by attaching a "1-handle" and dragging either end of this handle around loops in $T^3$.

$\bf Added\ later\colon$ Generators for the mapping class group of $T^3 \# (S^1\times S^2)$ can be described explicitly, using the algebraic fact that the automorphism group of a free product of groups $G_1,\cdots,G_n$ is generated by four types of automorphisms:

(1) Automorphisms of the individual factors $G_i$,

(2) Permutations of isomorphic factors.

(3) Partial conjugations, in which one $G_i$ is conjugated by an element $x_j\in G_j$ and the other factors $G_k$ are fixed. (If $i=j$ this gives an automorphism of type (1), an inner automorphism of $G_i$.)

(4) If some $G_i$ is infinite cyclic generated by $g_i$, say, and $x_j$ is an arbitrary element of some $G_j$ with $j\neq i$, there is an automorphism sending $g_i$ to $g_ix_j$ or $x_jg_i$ and fixing the other factors $G_k$, $k\neq i$.

For $T^3 \# (S^1\times S^2)$ the type (1) automorphisms for the ${\mathbb Z}^3$ factor form a copy of $SL(3,{\mathbb Z})$ which is the mapping class group of $T^3$ (a result originally due to Waldhausen I think). Since one can isotope an orientation-preserving homeomorphism of $T^3$ to fix a ball pointwise, this homeomorphism of $T^3$ gives rise to a homeomorphism of the connected sum which is the identity on the other summand. For the $S^1\times S^2$ summand the automorphism group of $\pi_1$ is ${\mathbb Z}_2$ with a generator realized by a homeomorphism of $S^1\times S^2$ that reflects both factors simultaneously. This too can be isotoped to extend to the connected sum via the identity on the complement.

Note that $SL(3,{\mathbb Z})$ is generated by elementary matrices that differ from the identity matrix in a single off-diagonal entry, which is 1, and these generating automorphisms of ${\mathbb Z}^3$ are realized by Dehn twists along tori in $T^3$.

There are no type (2) automorphisms of ${\mathbb Z}^3 * {\mathbb Z}$. For type (3) automorphisms, conjugation of the ${\mathbb Z}^3$ factor by an element of the ${\mathbb Z}$ factor is actually an inner automorphism of the whole group since the ${\mathbb Z}$ factor is abelian. We are interested in $Out({\mathbb Z}^3 * {\mathbb Z})$ so inner automorphisms are factored out. On the other hand, conjugating the ${\mathbb Z}$ factor by an element of ${\mathbb Z}^3$ is realizable by composing two type (4) automorphisms.

Type (4) automorphisms are realized by the homeomorphisms that I described earlier as dragging one end of the "1-handle" $S^1\times S^2$ around a loop $\gamma$ in the $T^3$ summand. Concretely, this homeomorphism is a Dehn twist along a certain torus $T_\gamma\subset T^3 \# (S^1\times S^2)$ defined as follows. First take a solid torus neighborhood of the loop $\gamma$. We can view connected sum with $S^1\times S^2$ as removing two disjoint open balls in $T^3$ and identifying the resulting two boundary spheres. Choose one of these two balls in the interior of the solid torus neighborhood of $\gamma$ and choose the other ball in the exterior of this solid torus. The boundary torus of the solid torus then gives the $T_\gamma$ we want. The homeomorphism $h_\gamma$ we are looking for is a twist in the longitudinal direction along $T_\gamma$. The automorphism of $\pi_1={\mathbb Z}^3*{\mathbb Z}$ induced by $h_\gamma$ is the identity on the factor $\pi_1(T^3)$ since generators for this ${\mathbb Z}^3$ can be chosen disjoint from the solid torus. If $x\in \pi_1(T^3)$ is the element represented by $\gamma$ and $g$ is a generator of $\pi_1(S^1\times S^2)$, then it's not hard to check that the automorphism induced by $h_\gamma$ sends $g$ to $xg$ or $gx$. (The two automorphisms given by $xg$ and $gx$ differ by an inner automorphism in the special case we are considering.) Thus a type (4) generator is realized by a twist along a torus.

In summary, the mapping class group of $T^3\#(S^1\times S^2)$ is generated by twists along tori and 2-spheres, and also the homeomorphism reflecting the two factors of $S^1\times S^2$.

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  • $\begingroup$ Thank you very much for your answer! I am reading your paper and indeed it's clarifying a lot of things. However, I am new to the subject (and I am a physicist) so I still need to learn a few things. I think I understand the points regarding $ker(\Phi)$ and $imag(\Phi)$ but I am having trouble picturing geometrically what $\Phi$ itself is for elements outside $ker(\Phi)$. In 2D, MPG is generated by Dehn twist along different loops in the fundamental group. What kind of twisting is typically involved in 3D? thanks! $\endgroup$
    – SKShukla
    Commented Jul 30, 2016 at 2:55
  • $\begingroup$ Thank you very much for such a detailed reply. It has been very helpful! Being in physics, I work with the triangulated manifolds (lets say a $\Delta$-complex). For example, $MCG(T^n)$ action on triangulated n-torus is widely used in physics. But when I tried to represent Dehn Twisting sphere as action on a complex, I run into problems. Does the essence of MCG described above remain the same going to triangulated manifolds, or does it change nontrivially? I see a paper by you on MCG action on 2D- complex. I cannot find one on 3D- complex though. Thank you. $\endgroup$
    – SKShukla
    Commented Aug 7, 2016 at 9:34

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