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I was reading The symplectic Floer homology of a Dehn twist by P. Seidel, which you can find here.

In Lemma 3(ii) the following topological property of Dehn twists is stated without proof:

Let $\Sigma$ be an oriented compact surface. Let $C_1,\dots,C_n \subset \Sigma$ be noncontractible disjoint circles contained in $\Sigma$, and for every $i=1,\dots,n$ let $T_i$ be a positively oriented Dehn twist around $C_i$. Also, for every $i=1,\dots,n$ choose a sign $\sigma_i \in \{\pm 1\}$. Assume that if a component of $\Sigma - \bigcup C_i$ is a cylinder bounded by $C_i$ and $C_j$, then $\sigma_i = \sigma_j$. Let $T := T_1^{\sigma_1} \circ \dots \circ T_n^{\sigma_n}$.
If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$, and $T \circ \gamma$ is homotopic to $\gamma$ relative to the endpoints, then $\gamma$ is homotopic relative to the endpoints to a path $\gamma'$ contained in $\mathrm{Fix} T$.

I guess this is an (almost) trivial fact for experts on MCG of surfaces, hence the lack of proof, but could someone provide me a reference where this is proven?

Remark. For his purposes, Seidel assumes that $\Sigma$ has no boundary and that its genus is greater than 1, but I think the claim is still true in this generality.

Thank you in advance for your time.

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1 Answer 1

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The Primer has "model proofs" for various steps of the proof. Bleiler's notes on Casson's lectures are also very good on these topics. (The notes cover far less material than the Primer, so they are much much shorter.)


I find the given statement confusing, because the conclusion is not as strong as it should be. I think the conclusion should be as follows:

If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$, and if $T \circ \gamma$ is homotopic to $\gamma$ (rel endpoints), then there is a homotopy of $\gamma$ (rel endpoints) making $\gamma$ disjoint from the union $\cup C_i$.

Proof: We prove the contrapositive. Homotope $\gamma$ (rel endpoints) to have minimal intersection number with $\cup C_i$. Suppose that the result still meets $\cup C_i$. So $\gamma$ crosses one of the curves, say $C$. Let $(B_j)_j$ be the subcollection of $(C_i)_i$ which are all parallel to $C$. Let $X \subset \Sigma$ be an annulus that contains all of the $B_j$ as essential curves in its interior. We may arrange matters so that $\gamma$ has minimal intersection number with $\partial X$. Let $\delta$ be a component of $\gamma \cap X$ which meets some of the $B_j$. We may (and do) assume that the endpoints of $\delta$ are fixed by all of the $T_i$.

Let $\Sigma^C$ be the cover of $\Sigma$ which is homeomorphic to an annulus and where $C$ does not unwrap. We use the initial point of $\delta$ as our base-point; let $\gamma^C$ be the resulting lift of $\gamma$ to $\Sigma^C$. Let $\delta^C \subset \gamma^C$ be the resulting lift of $\delta$. The arc $\delta^C$ crosses the lifts of (the correct) $B_j$ due to our choice of base-point.

We now lift $T(\gamma)$ using the same base-point. It lifts (up to homotopies in $\Sigma^C$ supported in bigons disjoint from $C$) to a copy of $T_C^n(\gamma^C)$. Here $n$ is the number of parallel copies of $C$ crossed by $\delta$ (and the sign depends on the sign of the twists along the $B_j$). Since $T_C^n(\gamma^C)$ has different winding number (than $\gamma^C$) with the curves $B_j$, it follows that they are not homotopic (rel endpoints) in $\Sigma^C$. Thus (by homotopy lifting) $T(\gamma)$ and $\gamma$ are not homotopic (rel endpoints) in $\Sigma$, as desired.

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    $\begingroup$ Thank you for your answer! You're also right about the conclusion of the claim; that's what Seidel claims in Lemma 3(ii). At the beginning of Lemma 4 he uses the particular case I wrote, hence my mistake when writing the post. Let me edit the question so there's no confusion for future readers. $\endgroup$
    – Don
    Commented Jul 29, 2023 at 15:18

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