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I am looking for an explicit presentation of the mapping class group of the annulus $\mathbb{A}^2$, after equipping it with $n$ interior punctures/marked points $\{x_1, \cdots, x_n\} \hookrightarrow \mathbb{A}^2$.

Here is how far I got:

The Birman-style exact sequence is (e.g. Massuyeau 2021, Thm. 3.13, p. 32)

$$ \cdots \to \mathrm{Homeos}^+(\mathbb{A}^2) \to \mathrm{Br}_n(\mathbb{A}^2) \longrightarrow \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \longrightarrow \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \to 1 $$

(on the left: homeomorphisms preserving the orientation and fixing the boundary pointwise),

where the generator of $\mathbb{Z}$ on the right is given by Dehn-twisting (DT) the annulus (Farb & Margalit 2012 Prop. 2.4, p. 68): holding the inner boundary fixed while making a full rotation of the outer boundary.

On the other hand, the annulus is one of the exceptions to the rule that $\mathrm{Homeos}^+(\mathbb{A}^2)$ tends to be trivial (Farb & Margalit 2012, Thm. 1.14) which means that it is not immediate how to truncate to a short exact sequence on the left.

But, at least, by Theorem 1 of

  • R. P. Kent & D. Pfeifer: "A Geometric and algebraic description of annular braid groups", Int. J. Algebra and Computation 12 01n02 (2002) 85-97 [doi:10.1142/S0218196702000997]

we have an explicit description of the annular braid group: it is the semidirect product

$$ \mathrm{Br}_n(\mathbb{A}^2) \,\simeq\, \mathrm{Br}_n^{\mathrm{aff}} \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}} $$

of the "affine braid group" (with a presentation just like Artin's, but for strands cyclically identified) with the free group generated by the operation "Cyc" of cyclically rotating all punctures (thought of as arranged equidistantly at equal radius in the annulus) "one step" in one direction.

If we assume that

$$ (\star) \;\;\;\;\; \mathrm{im}\big( \mathrm{Homeos}^+(\mathbb{A}^2) \to \mathrm{Br}_n(\mathbb{A}^2) \big) \overset{?}{\simeq} 1 $$

were trivial [edit: it is!, as pointed out in the comments below, according to Earle & Schatz 1970 p 170], then

$$ 1 \overset{?}{\to} \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}}\big) \longrightarrow \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \longrightarrow \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \to 1 \,. $$

would imply that

$$ \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \;\overset{?}{\simeq}\; \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}}\big) \rtimes \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \,, $$

which is claimed in Remark 1.2 (p. 6) of:

In fact, these authors also claim that the generators DT and Cyc commute --- which, I suppose, follows by noticing that both generators are given by radius-dependent rotations, and these all commute with each other.

But if, with assumption $(\star)$, we really have a short exact sequence as above, doesn't the Dehn twist DT actually conjugate the Artin generators to themselves, too (just shearing the form of the corresponding "whirl"-homeo), so that, under assumption $(\star)$, we'd actually have a direct product

$$ \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \;\overset{?}{\simeq}\; \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \mathbb{Z}\big) \times \mathbb{Z} $$

?

Finally: What's the image of the Dehn twist under the canonical map to the symmetric group? I gather the image is always the trivial permutation -- but I am not sure how to see this.

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    $\begingroup$ Is $\mathbb{A}^2$ the $n$-punctured annulus or the annulus itself? If the latter, I'm not certain that you get a short exact sequence as in your second paragraph: the annulus is explicitly excluded from 1.14, and since $\chi(\mathbb{A}^2)=0$ it's not covered by 9.1 either. $\endgroup$ Commented Mar 3 at 19:43
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    $\begingroup$ Re "It is implicitly claimed in Remark 1.2": any short exact sequence with cokernel $\mathbb{Z}$ splits, since $\mathbb{Z}$ is a free group. $\endgroup$ Commented Mar 3 at 19:44
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    $\begingroup$ @MarkGrant right, thanks, my bad. I am editing the question to fix this. $\endgroup$ Commented Mar 4 at 5:33
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    $\begingroup$ You might find it easier to proceed “the other way round”. The mapping class group of the n-punctured annulus is closely related to the $n+1$-strand braid group $B_{n+1}$ (whose presentation is surely well known, albeit not to me). Let $x_0$ be one of the punctures of the $(n+1)$-punctured disc on which $B_{n+1}$ acts up to homotopy. You need to pass to the stabiliser of $x_0$ and then “turn $x_0$ into a boundary component”, which corresponds to passing to a central extension. It remains to work out which central extension this is, which presumably requires some computation. $\endgroup$ Commented Mar 4 at 11:17
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    $\begingroup$ [I posted the above just as @Adrien posted his answer, which seems to advocate a similar approach.] $\endgroup$ Commented Mar 4 at 11:19

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$\newcommand{\A}{\mathbb A}$ Tbh I find this description of the annular braid group confusing: there is a standard trick that the $n$ strand annular braid group is isomorphic to the subgroup of the usual braid group on $n+1$ strands of braids where the first strand is fixed (the extra fixed strand represents the inner boundary component). In other words, shifting the indices and letting $\sigma_0,\dots,\sigma_{n-1}$ be the standard generators of $B_{n+1}$, the annular braid group is generated by $\tau=\sigma_0^2$ and the $\sigma_i$ for $i\geq 1$.

In addition to the standard braid relations for the $\sigma_i$, you just have the "reflection equation" $$\tau\sigma_1\tau\sigma_1=\sigma_1\tau\sigma_1\tau.$$

In this picture you can represent the Dehn twist along (a simple closed curve close to) the inner boundary component simply by putting a framing on the extra fixed strand, and I agree this is indeed the MCG you're looking for, and this is indeed just a direct product $B_n(\A^2)\times \mathbb Z$. In other words you can arrange for your Dehn twist to act as the identity on your marked point (in particular the induced permutation is the identity as well as you expect).

Edit: Btw I'm not sure which version Farb & Margalit refers to but the connected components of the group of diffeomorphisms (or homeomorphisms) of the annulus which fix the boundary pointwise are contractible so Birman's sequence is short in that case. See the theorem at the end of the introduction in Earle-Schatz Teichmüller theory for surfaces with boundary

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    $\begingroup$ Just as a side note: this is still an Artin group. $\endgroup$ Commented Mar 5 at 12:49
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    $\begingroup$ Yes, good point, indeed $B_n(A^2)$ is the braid group of Coxeter type B, and the presentation above is the "standard" one from that perspective. $\endgroup$ Commented Mar 5 at 15:39
  • $\begingroup$ @Adrien, thanks, and sorry for the slow reaction. The alternative presentation of the annular braid group that you mention is explained in Remark 1.1 of Gadbled et al that I cited (arxiv.org/pdf/1504.07596#page=4) Where you write "In this picture you can represent..." it sounds plausible, but I am not sure how this is a proof, maybe you could expand? $\endgroup$ Commented Mar 6 at 8:40
  • $\begingroup$ @Adrien Regarding your edit-remark: Theorem 1.14 on p. 43 here: euclid.nmu.edu/~joshthom/Teaching/MA589/farbmarg.pdf#page=60 excludes the closed annulus from the assumption. Hence it doesn't say that $\pi_1$ of homeos of the annulus is non-trivial, but of course it doesn't assert it's trivial, either. What's a reference for it being trivial? $\endgroup$ Commented Mar 6 at 8:40
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    $\begingroup$ Re: the final comment in this answer, unfortunately Farb and Margalit don't handle these kinds of "small" examples well. The proof that the mapping class group of a three-holed sphere is $\mathbb{Z}^3$ is incomplete, for instance. $\endgroup$ Commented Mar 6 at 13:38

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