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The question is about the last sentence of the last corollary of Stabilizing the monodromy of an open book decomposition by Vicent Colin and Ko Honda. This question is also related to this other question of mine https://mathoverflow.net/posts/330596/edit.

The corollary of the aforementioned article says:

$\textbf{Corollary 3.1}$ Let $S$ be a surface with connected boundary $\partial S$ and let $h:S \to S$ be a pseudo-Anosov automorphism with fractional Dehn twist coefficient (at the only boundary component) greater than $2$. Then, any elementary positive stabilization along a non-boundary-parallel arc $\gamma \subset S$ is pseudo-Anosov.

The vast majority of the proof consists in showing that the stabilization $h'$ is not reducible. Then in the last sentence they conclude that it also cannot be periodic. Hence, by Nielsen-Thurston classification it must be pseudo-Anosov. However this last part is proven as follows in the last sentence of the paper:

To show that $h'$ is not periodic, consider $\delta = \partial S$. One easily verifies that the number of intersections with $a$ (the co-core of the handle used for stabilizing) increases with each iterate of $h'$.

They use all the hypothesis of the theorem to provide a proof for the first part and the way this last sentence is writen looks as if the last part was true in a more general setting.

The thing is that I cannot easily verify that. What I can easily verify is that the number of intersections of $h'(\delta)$ with $a$ is certainly $2$ (without using that $h$ is pseudo-anosov. I can also produce examples (see the edit below) of a periodic automorphism with connected boundary which, after a suitable positive stabilization, becomes a periodic automorphism aswell. So, in order to prove the last part we need to use that $h$ is pseudo-Anosov at some point, or the part on the fractional Dehn twist coefficient maybe. But it is not obvious to me why there are no relations in the surface that "un-wind" the twisting induced by the stabilization as in the examples that I can produce.

The rest of the question is an example that I claimed I could produce. It serves as a counter-example to Michael Lin's answer and it also serves as an example that the hypothesis on $h$ must be used. Also, since I have drawn some pictures, I hope this questions gets some more attention.

$\textbf{Example}$

For more details on how to obtain a description of the monodromies that I use as examples, you may have a look at Le groupe de monodromie du déploiement des singularités isolées de courbes planes I by Norbert A'Campo for example. Or any textbook on plane curve singularities probably contains a description of these monodromies.

The monodromy of the $A_2$ singularity $x^2+y^3: \mathbb{C}^2 \to \mathbb{C}$ consists of the composition of two right-handed Dehn twists around a "parallel" and a "meridian" of the torus as in the picture (that I quickly made) below that I just made. The monodromy is the composition of the right-handed Dehn twist around the blue curve and then around the green curve. You can see this in the paper of A'Campo for example or in any introductory text to Picard-Lefschetz theory. That this monodromy is freely periodic of order $6$ follows from the observation that a representative of the monodromy is $(x,y) \mapsto (e^{\pi i}x, e^{2\pi i/3}y)$.

Monodromy of the cusp

Losely speaking: in Singularity Theory "adding a Morse point" is more or less the same as "stabilizing". The singularity $x^2 +y^3$ "consists" of two Morse points. That is, after a generic perturbation it deforms to two Morse points. Also, its Dynkin diagram is the graph consisting of two vertices joint by a segment. Again, for more details on this, follow A'Campo's paper.

Now, the $A_3$ singularity $x^2+y^4$ has Milnor fiber the surface of genus $1$ and $2$ boundary components. And it can be obtain from the fiber of $A_2$ by attaching a $1$-handle to its boundary. See the picture below.

Monodromy of the stabilization of the cusp

The description of the monodromy is the composition of a Dehn twist around the blue curve, then the green curve and then the yellow curve (observe that the last two Dehn twists commute since their support is disjoint) This part can also be deduced from A'Campo's article. That the monodromy is freely periodic follows from the same reasoning as before. For instance, a representative is $(x,y) \mapsto (e^{\pi i}x, e^{\pi i/2}y)$. Which has order $4$.

So by taking $h$ the first monodromy and $h'$ the second we can see how $h'^4(\delta) \cap a$ is empty. Also $h'(\delta) \cap a$ consists of two points.

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  • $\begingroup$ Your example has $FDTC = 1/6$. The upshot of $FDTC \geq 2$ is a guarantee that each time you apply $h$ (and before applying the new Dehn twist $T_b$ to get $h' = T_bh$), your curve (the iterated image of $\delta$) intersects the stabilising curve $b$. $\endgroup$ – magicker72 Jun 7 at 4:41
  • $\begingroup$ Sure, this is what I suspect. Obviously (because of my example) you have to use either that $h$ is pseudo-anosov or something about its FDTC. Still I don't see why this is a guarantee that the iterated image of $\delta$ will always intersect $\gamma$. For example, take $g$ the automorphism that is "like" $h$ but with $0 \leq FDTC <1$. Then $h = T^n g$ where T is a boundary Dehn twist. Then, it could happen that when you apply $g$ first the iterated image of $\delta$ does not intersect $\gamma$. Then the boundary Dehn twists dont apply. $\endgroup$ – Paul Jun 7 at 4:46
  • $\begingroup$ Just for the record: I'm not claiming that the statement is incorrect. I just don't see it completely clear. There are a lot of slipery slopes in mapping class group and the devil is in the details. If you use that FDTC >2 and not that $h$ is pseudo-anosov, it would be worth writing that out, don't you agree? $\endgroup$ – Paul Jun 7 at 4:48
  • $\begingroup$ I deduce from your comment that you suggest that the following is true: Given $h:S \to S$ any automorphism on a surface with connected boundary and such that FDCT >2. Then, any elementary stabilization (along a non-boundary-parallel arc) contains a pseudo-anosov piece in its Nielsen-Thurston decomposition. $\endgroup$ – Paul Jun 7 at 4:50
  • $\begingroup$ Your deduction sounds reasonable. For your other comments, not that $h$ acts trivially on $\delta$, so the image under $h'$ runs over the 1-handle. From then on, you can break it up into arcs: those inside $\Sigma$ and those going over the 1-handle. The arcs inside $\Sigma$ will always intersect $b$, due to the FDTC condition, and the other arcs should be fixed (or at least, that can be arranged by an isotopy). $\endgroup$ – magicker72 Jun 7 at 5:44
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You don't have to use any of the hypotheses on $h$. $a$ is the co-core of the band, which intersects the twisting curve $\gamma$ exactly once, so $|h'(\delta) \cap a| = 1$. Examining the image of $\delta$ under iterates of $h'$ the along the band, you can see that the action of $h$ doesn't do anything (is the identity) and the Dehn twist causes $\delta$ to run along the band an additional time.

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  • $\begingroup$ Yes, that happens the first time. After that, there is part of $h'(\delta)$ which is inside $S$. And now $h$ does something on it. There's nothing that tells me that after some iterations the number of intersections with $a$ of $h'^n(\delta)$ decreases back to $0$. $\endgroup$ – Paul Jun 5 at 22:34
  • $\begingroup$ And also, $|h'(\delta) \cap a |= 2$ because the stabilization curve intersects twice $\delta$. $\endgroup$ – Paul Jun 5 at 22:35
  • $\begingroup$ Actually I have proven that you do have to use the hypothesis on $h$. Take the monodromy of $x^2+y^3$ as $h$. This automorphism is a periodic automorphism of order $6$ on the torus minus a disk. The map consists of the composition of two right-handed Dehn twists along curves that intersect once. Now, there is a stabilization of this monodromy to obtain the monodromy of $x^2 + y^4$. In this case, the monodromy is the composition of the previous one with a right-handed Dehn twist around a curve that intersects once only one of the previous two curves. And this monodromy has order $4$. $\endgroup$ – Paul Jun 5 at 22:47
  • $\begingroup$ I think you may be confusing co-core with core? $\endgroup$ – Michael Lin Jun 5 at 22:54
  • $\begingroup$ No. The co-core is the segment that does not touch the boundary of the original surface and intersects once the core of the handle being attached. $\delta$ intersects twice the twisting curve $\gamma.$ $\endgroup$ – Paul Jun 5 at 22:56

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