# Clarification for arXiv:1606.08656

[2018-01-31 Wed]

The paper Ivan Smith and I wrote on Lagrangian pinwheels in $$\mathbf{CP}^2$$ and Markov numbers has recently appeared in Geometry and Topology. Shortly thereafter, Yong-Geun Oh contacted us with some excellent questions where our exposition was less than clear. In case anyone else has the same questions, I thought I would write a blog post clarifying these points.

• Lemma 2.16(c) should say $$T=\mathbb{Z}/(m)$$ rather than $$\mathbb{Z}/(\Delta/m)$$.
• There is a discrepancy between what we write in Section 2.3:
• The generic orbit is homologous to $$p^2e/g$$.'' (correct)
and what we write in the final paragraph of Remark 3.15:
• the generic Reeb orbits are homologically trivial in $$\Sigma_{p,q}$$'' (not true in general)
Indeed, since $$H_1(\Sigma_{p,q};\mathbb{Z})=\mathbb{Z}/(p^2)$$, this only holds when $$g=1$$, which happens if and only if $$p$$ is odd (see Remark 2.12).

This is important, because Remark 3.15 is used to justify why certain holomorphic curves asymptote to nongeneric Reeb orbits. As stated, the argument we give there holds only for odd $$p$$.

I will now explain the two situations in which we needed this fact and how it generalises to arbitrary values of $$p$$. In the case when $$p$$ is odd (and $$g=1$$), this reduces to what we wrote.

1. In Theorem 4.16, we have a holomorphic cylinder which we have extracted from an SFT neck-stretching argument. One of its ends is asymptotic to a Reeb orbit on a component of the boundary called $$\Sigma_{p_2,q_2}$$ (a contact lens space). We have already established that the Reeb orbits fall into two categories: exceptional'' and generic'' (the Reeb orbits give a Seifert fibration of the lens space). We want to show that this asymptote is to an exceptional orbit.

The homology class of the generic orbit is $$p^2/g$$ times the class of the exceptional orbit. The relative homology class of the cylinder is (in notation from the paper) $$D\mathcal{E}$$ and the connecting homomorphism to $$H_1(\Sigma_{p_2,q_2};\mathbf{Z})=\mathbf{Z}/(p_2^2)$$ is to reduce $$D$$ modulo $$p_2^2$$. We know that $$D\leq p_2/3$$ by Lemma 4.11(B). If the asymptote were generic then $$D$$ would reduce modulo $$p_2^2$$ to something divisible by $$p_2^2/g$$ (where $$g$$ is $$1$$, $$2$$ or $$4$$, depending on $$p_2$$ by Remark 2.12). But then we get $$p_2^2/g\leq p_2/3$$, which implies $$p_2\leq g/3\leq 4/3$$ so we don't need to worry about this case as $$p_2$$ is at least $$2$$.

2. In the other situation, if the SFT limit is a plane (the case $$|Z|=1$$ in the paper) and the asymptote is generic then the homology class of the plane is, again, divisible by $$p^2/g$$ and this contradicts the inequality $$D\leq 2p/3$$ from Lemma 4.8.

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.