22 votes

Fixed-point free diffeomorphisms of surfaces fixing no homology classes

Here's an example with $g=2$. Let $T$ be the torus $\mathbb C/L$, where $L$ is the lattice spanned by $1$ and $\zeta=e^{2\pi i/6}$. Let $f:T\to T$ be induced by multiplication by $\zeta$. This is a ...
Tom Goodwillie's user avatar
18 votes
Accepted

On trivial mapping class group of 3-manifolds

Dave Gabai proved that the mapping class group of a closed hyperbolic 3-manifold is isomorphic to its isometry group. For a hyperbolic knot $K$ without any symmetries, for large enough $n$, $S^3_{1/n}(...
Ian Agol's user avatar
  • 66.5k
16 votes

Mapping Class Group (MCG) of connected sum of 3-torus and $S^2\times S^1$

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 ...
Allen Hatcher's user avatar
16 votes
Accepted

Conjugacy classes of the mapping class group

An exponential-time solution to the conjugacy problem in the mapping class group was given by Jing Tao, in: Tao, Jing(1-OK) Linearly bounded conjugator property for mapping class groups. (English ...
HJRW's user avatar
  • 23.8k
15 votes

Product of conjugate matrices in $\mathrm{SL}(2, \mathbb{Z})$

To expand my comment, and combine it with some points from Zach Teitler's answer, but more in a generators and relations framework: ${\rm SL}(2,\mathbb{Z})$ is well-known to be isomorphic to the group ...
Geoff Robinson's user avatar
15 votes
Accepted

Product of conjugate matrices in $\mathrm{SL}(2, \mathbb{Z})$

See Keith Conrad's notes http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf, particularly Example 2.5. Let us write (as Conrad does) $S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{...
Zach Teitler's user avatar
  • 6,187
15 votes
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Pseudo-Anosov maps with same dilatation.

Yes. If two pseudo-Anosov mapping classes are conjugate then they must have the same dilatation. So take any pseudo-Anosov $f$ and any mapping class $h$ not in the centraliser of $f$ and let $g = h f ...
Mark Bell's user avatar
  • 3,125
15 votes
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Fixed-point free diffeomorphisms of surfaces fixing no homology classes

Goodwillie's construction (in genus two) generalises to all higher genus as follows. Let $P_n$ be the regular $n$-gon in the plane with vertices at roots of unity. When $n$ is even, we can glue ...
Sam Nead's user avatar
  • 25.4k
13 votes

Compact manifolds with big mapping class group

Take $M^d$ to be a connected sum of $n$ copies of $S^1\times S^{d-1}$, where $d\ge 3$. Then $M$ is a closed, orientable manifold of dimension $d$ with $\pi_1(M)=F_n$, the free group of rank $n$. If $...
Alex Suciu's user avatar
  • 2,153
13 votes
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Mapping class group of certain 3-manifolds

Since you write ${\rm Diff}_+(M)$ you are probably assuming $M$ is orientable and diffeomorphisms of $M$ are orientation-preserving. Every diffeomorphism of $M$ can be isotoped to take fibers to ...
Allen Hatcher's user avatar
13 votes

Mapping class group and pure mapping class group

Just to give an explicit description of the difference: if one takes a loop "around a boundary component," the Dehn twist around this loop is not isotopic to the identity. On the other hand, ...
Daniel Litt's user avatar
12 votes

Compact manifolds with big mapping class group

In Infinitesimal computations in topology Sullivan shows in Theorem 13.3 that if $M$ is a simply-connected manifold of dimension $>5$, then $\pi_0(\mathrm{Diff\,} M)$ is commensurable to an ...
Igor Belegradek's user avatar
12 votes
Accepted

Mapping class group of torus, why is $(ST)^3=S^2$?

Flip the direction of rotation for $S$, or choose the other meridian for $T$. We can see this at the level of matrices. Define $$S_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad ...
dvitek's user avatar
  • 1,691
11 votes

Homeomorphic but Non-Conjugate Mapping Tori

Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$ and with pseudo-Anosov monodromy, and ...
Lee Mosher's user avatar
  • 15.3k
11 votes
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Mapping class group and pure mapping class group

I do not recall the conventions adopted in the Primer, but there is a wide difference between boundary components (which will be embedded circles or lines) and punctures (which are “missing points” ...
Sam Nead's user avatar
  • 25.4k
10 votes
Accepted

Mapping class group and representation of fundamental group of Riemann surfaces

There are counterexamples as soon as $g > 1$. Let $n$ be the number of surjective homomorphisms $\pi_1(S) \to A_5$, up to $S_5$-conjugacy. (We can see that $n \geq 1$ using the fact that $A_5$ ...
Will Sawin's user avatar
  • 133k
10 votes
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All non-compact simply connected $2$-manifolds with boundary

Here is one proof, using the Uniformization Theorem. This proof will be easier in the setting of the "Primer" since the authors are considering universal covering spaces of complete ...
Moishe Kohan's user avatar
  • 9,624
9 votes

Homeomorphic but Non-Conjugate Mapping Tori

McMullen and Taubes 4-manifolds with inequivalent symplectic forms and 3-manifolds with inequivalent fibrations constructs 3-manifolds $N$ with different fibrations, whose Euler classes do not lie in ...
ThiKu's user avatar
  • 10.2k
9 votes

Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1?

The answer is "yes" -- in fact one can do better and get a class in $$H^1(\text{Aut}(F_m), \text{Hom}(H, \wedge^2 H)),$$ where $F_m$ is the free group on $m$ generators and $H$ is the ...
Daniel Litt's user avatar
9 votes
Accepted

The largest group acting on a non-orientable surface of genus 5

The group $F$ is isomorphic to the symmetric group $S_5$. In fact, since $N_5$ is non-orientable of genus $5$, both $F$ and the extended group $F^*$ (of order twice the order of $F$) act on its ...
Francesco Polizzi's user avatar
9 votes
Accepted

Minimal number of (Dehn twists) generators of the mapping class group of a marked sphere

The minimum number of Dehn twist generators (and in fact the minimum number of generators of any kind) for $\Gamma_{0,n}$ is ${n-1 \choose 2} - 1$. Here's why. A presentation for $\Gamma_{0,n}$ is ...
Ty Ghaswala's user avatar
9 votes

Mapping Out(F_n) to the mapping class group

As pointed out by YCor, $\mathrm{Out}(F_g)$ is not linear (for $g \geq 3$). Also, the linearity of $\mathrm{Mod}(S_g)$ is unknown. So the existence of such an embedding would solve a long-standing ...
Sam Nead's user avatar
  • 25.4k
8 votes

Compact manifolds with big mapping class group

There are situations in which surfaces are the "unique" examples with big mapping class groups. One such is closed manifolds of negative sectional curvature. Theorem (Paulin): If $M$ is a closed $n$-...
HJRW's user avatar
  • 23.8k
8 votes
Accepted

Categorical mapping class group action

[This is an elaboration of parts of Mark Penney's answer] A natural source of categorical actions of the mapping class group is the category assigned by any 4d TFT to a surface. Such categories are ...
David Ben-Zvi's user avatar
8 votes
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Well definedness of square roots of separating Dehn Twists

They are different. In fact, they act differently on $H_1(\Sigma_2;\mathbb{Z})$. Let $V \subset H_1(\Sigma_2;\mathbb{Z})$ be the span of the homology classes of $c_1$ and $c_2$, and let $W \subset ...
Andy Putman's user avatar
  • 43.2k
8 votes
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Abelianization of mapping class groups $\Gamma_{g,n}$

The following statement can be found in Section 5 of Low-dimensional homology groups of mapping class groups: a survey: Theorem: Let $g \geq 1$. Then $$H_1(\Gamma_{g,r}^n,\mathbb{Z}) \simeq \left\{ ...
AGenevois's user avatar
  • 7,471
8 votes
Accepted

Why is the mapping class group of a surface with nonempty boundary torsion-free?

I think the reason that Dehn twists enter is that we can take the differential of a (orientation-preserving) diffeomorphism $f$ of $S_g$ that fixes a chosen basepoint $\ast \in S_g$ at this point, ...
Jens Reinhold's user avatar
8 votes

$ \mathbb{R}P^n $ bundles over the circle

No. Every smooth bundle over $S^1$ with fiber $M$ is the mapping torus of some diffeomorphism $f:M\to M$. Isomorphism classes of bundles correspond to conjugacy classes in the group of isotopy classes ...
Tom Goodwillie's user avatar
8 votes
Accepted

$ \mathbb{R}P^n $ bundles over the circle

Your answer is correct if appropriately understood, but it's a little subtle. Here I should note that I'm interpreting your question as a purely homotopy theoretic one (in particular ignoring smooth ...
Dmitry Vaintrob's user avatar

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