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$\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\PMod{PMod}\DeclareMathOperator\Homeo{Homeo}$I am very confused about the definition of mapping class group and pure mapping class group (and their generating sets).

In Primer on Mapping class groups:

$\Mod(S)$ is the group of isotopy classes of elements of $\Homeo^+(S, \partial S)$, where isotopies are required to fix the boundary pointwise

and

Let $\PMod(S_{g,n})$ denote the pure mapping class group of $S_{g,n}$, which is defined to be the subgroup of $\Mod(S_{g,n})$ consisting of elements that fix each puncture individually.

Question 1: From these two definitions, is it true that for compact surfaces with boundary (no puncture), the mapping class group is the same as the pure mapping class group?

Generating sets of mapping class group:

THEOREM 4.1

For $g\ge 0$, the mapping class group $\Mod(S_g)$ is generated by finitely many Dehn twists about nonseparating simple closed curves.

Corollary 4.15

For any $g$, $n\ge 0$, the group $\Mod(S_{g,n})$ is generated by a finite number of Dehn twists and half-twists.

Corollary 4.16

Let $S$ be any surface of genus $g\ge 2$. The group $\PMod(S)$ is generated by finitely many Dehn twists about nonseparating simple closed curves in $S$.

Assume that Question 1 is has a positive answer, meaning that for compact surfaces with boundary, the mapping class group is the same as the pure mapping class group. Then we ask

Question 2 For compact surfaces with boundary, is the mapping class group generated by a finite number of Dehn twists?

Question 3: If Question 2 has a positive answer, what makes the difference between punctures and boundary, why don't we define

$\Mod(S)$ is the group of isotopy classes of elements of $\Homeo^+(S, \partial S)$

and

$\PMod(S_{g,n})$ the pure mapping class group of $S_{g,n}$ ($n$ is the number of punctures and boundary components) is the subgroup of $\Mod(S_{g,n})$ consisting of elements that fix each puncture and fix each boundary component setwise (send each puncture/or boundary component to itself)?

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  • $\begingroup$ Did you ask a similar question earlier? $\endgroup$
    – Will Sawin
    Apr 2 at 14:44
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    $\begingroup$ Yes it is related to these questions but I still don't understand the current one mathoverflow.net/questions/144857/… mathoverflow.net/questions/417191/… $\endgroup$
    – user479845
    Apr 2 at 14:47
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    $\begingroup$ fixing a boundary and a puncture are different things, because fixing a boundary pointwise prevents you from doing a rotation around it, whereas if you "just" preserve a puncture you can do it $\endgroup$
    – alesia
    Apr 2 at 17:36
  • $\begingroup$ Q1: no. Q2: yes. Well, provided you fix the boundary. Dehn twists can't permute the boundary components. Q3: You can make any definition you want, the important thing is to clearly tell people what you are talking about when you discuss things. $\endgroup$ Apr 3 at 2:42
  • $\begingroup$ Why is the answer "no" for Q1? Thank you. $\endgroup$
    – user479845
    Apr 3 at 11:51

1 Answer 1

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This really is very similar to your previous question.

Answer 1: If $S$ is compact and without boundary then the two groups are isomorphic. (However, be aware that there are yet other definitions of the pure mapping class group where the latter is finite index in the former. See Ivanov's book.)

This is also the case when $S$ is compact and with boundary in the following sense. The group $\mathrm{Mod}(S, \partial S)$ (of isotopy classes fixing boundary points pointwise) is isomorphic to the pure mapping class group $\mathrm{PMod}(S)$, However, neither of these is isomorphic to $\mathrm{PMod}(S - \partial S)$. This is because in the former groups we have Dehn twists about boundary components.

Answer 2: Yes, if you require boundary points to be fixed (and you require boundary points to exist and orientations to be preserved - consider the two-sphere).

Answer 3: The difference between boundary components and punctures is that you can Dehn twist about the former and get something not isotopic to the identity (when fixing boundary components pointwise).

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  • $\begingroup$ It does make sense - as there are no punctures, there is no restriction, and all mapping classes are pure. $\endgroup$
    – Sam Nead
    Apr 3 at 20:27

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