Holomorphic polygon [04A1]

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Holomorphic polygon

We now move on to holomorphic polygons with $k+1$ corner points for $k\geq 2$ . The extended linearized Cauchy-Riemann equation (cf. [69, Chapter 9]) involves a vector field $v\in C^{\infty}(\Sigma,u^{*}TX)$ decaying at the ends, and $\rho\in\Omega^{0,1}(\Sigma,T\Sigma)$ representing a tangent vector of the Stasheff associahedron (i.e. the deformation of Riemann surface structure on the domain $\Sigma$ ), satisfying

\bar{\partial}v+\frac{1}{2}J_{X}\circ du\circ\rho=0.

where $J_{X}$ is the complex structure on $X$ . Here $\rho$ can be taken to be compactly supported, so $\bar{\partial}v=0$ near the corners. It immediately follows that

Lemma 3.7.

Given first order deformation vector fields $v_{1},\ldots v_{n-1}$ , then the (1,0)-form $\Omega(\cdot,v_{1},\ldots v_{n-1})$ on $\Sigma$ is holomorphic.

Remark 3.6.

Adding vector fields on $\Sigma$ valued in $T\Sigma$ to $v_{1},\ldots v_{n-1}$ does not affect $\Omega(\cdot,v_{1},\ldots v_{n-1})$ as a 1-form on $\Sigma$ . Thus this 1-form is insensitive to how one represents the Riemann surface structures on the abstract polygon.

We impose that the input at one of the $k$ corners maps to an intersection point in $CF^{0}(L,L^{\prime})$ , and the other inputs map to degree one self intersections of $L$ or $L^{\prime}$ . The output maps to $q\in CF^{*}(L,L^{\prime})$ .

Proposition 3.8.

(Automatic transversality, polygon case) Suppose $v_{1},\ldots v_{n-1}$ are linearly independent first order deformation vector fields. Either $\Omega(\cdot,v_{1},\ldots v_{n-1})$ vanishes identically as a 1-form on $\Sigma$ , or we must have $\deg q\geq n$ , and when the equality holds then $\Omega(\cdot,v_{1},\ldots v_{n-1})$ only vanishes at corners. In this case, all first order defomation vector fields are spanned by $v_{1},\ldots v_{n-1}$ , the holomorphic polygon $u:\Sigma\to X$ is an immersion up to the boundary, the obstruction of the extended linearized operator vanishes, and the moduli space is smooth at $u:\Sigma\to X$ .

Proof.

By viewing the domain of the polygon as a strip with extra boundary punctures, we produce a holomorphic vector field $v_{n}$ as the $\mathbb{R}$ -translation vector field. However, unlike in the strip case, at the degree one self intersection corners $v_{n}$ does not typically have the required decay to be admitted as a deformation vector field. Indeed, by thinking about such a corner point as the origin in the upper half plane model of $\Sigma$ with local coordinate $z$ , then $zv_{n}$ decays at the corner, but not necessarily $v_{n}$ itself.

Now $v_{1}\wedge\ldots v_{n}$ is a section of $\Lambda^{n}TX$ with boundary value on $\Lambda^{n}TL$ , and $\Omega(v_{1},\ldots v_{n})$ is a holomorphic function on $\Sigma$ . We assume from now on that it is not identically zero. The index of the ordinary Cauchy-Riemann operator is

\deg q-\sum_{1}^{k}\deg p_{k}=\deg q-k+1.

Invoking (23) this is computable from the vanishing orders of $\Omega(v_{1},\ldots v_{n})$ :

\deg q-k+1=2\sum(\text{interior zeros})+\sum(\text{boundary zeros})+\sum(\text{corner zeros})+n.

The interior and boundary vanishing orders are non-negative. Since the $v_{k}$ are holomorphic near the corners without correction, the proof of Lemma 3.5 shows that the excess vanishing order at the $CF^{0}(L,L^{\prime})$ corner and the $q$ corner are both non-negative. At the degree one self intersection corners, the excess vanishing order of $\Omega(v_{1},\ldots v_{n-1},zv_{n})$ is nonnegative by the same previous arguments, so $\Omega(v_{1},\ldots v_{n-1},v_{n})$ itself has excess vanishing order $\geq-1$ . Hence $\deg q-k+1\geq n-k+1,$ namely $\deg q\geq n$ .

When the equality is achieved, then all bounds are saturated. In particular, $\Omega(v_{1},\ldots v_{n})$ can only vanish at the corners, so $u:\Sigma\to X$ is an immersion up to boundary. At the corners, the same arguments in Corollary 3.6 shows the failure of immersion is minimal.

If $v$ is the deformation vector field corresponding to an arbitrary kernel element of the extended linearized operator, then after subtracting off a constant linear combination of $v_{1},\ldots v_{n-1}$ , we may assume $v$ is tangent to $\Sigma$ at any chosen point on $\partial\Sigma$ . The same argument in Corollary 3.6 shows $v$ is tangent to the image of $\Sigma$ . The immersion property allows us to lift $v$ to the domain $\Sigma$ . There is no room to deform the complex structure of $\Sigma$ , nor is there any automorphism of $\Sigma$ , so in fact $v$ vanishes identically. This shows that $v_{1},\ldots v_{n-1}$ span all first order deformations. But $\deg q=n$ implies that the index of the extended linearized operator is

\deg q-\sum_{1}^{k}p_{k}+k-2=n-1

Thus the cokernel dimension is zero, namely the obstruction space vanishes. Consequently, the moduli space of such holomorphic polygons is smooth. ∎

Remark 3.7.

Using the Floer degree formula (63), the asymptotic behaviour of $\Omega(v_{1},\ldots v_{n})$ at the corners can be extracted from the above proof: at the $CF^{0}(L,L^{\prime})$ corner point $p$

\Omega(v_{1},\ldots v_{n})=a_{p}z^{(\theta_{L^{\prime}}-\theta_{L})(p)/\pi}(1+O(z)),\quad a_{p}\neq 0,\quad\arg a_{p}=\theta_{L}(p)\mod\pi\mathbb{Z}.

At the degree one self intersections $p_{l}\in CF^{1}(L_{+},L_{-})$ on $L$ or $L^{\prime}$ ,

\Omega(v_{1},\ldots v_{n})=a_{l}z^{(\theta_{L_{-}}-\theta_{L_{+}})(p_{l})/\pi}(1+O(z)),\quad a_{l}\neq 0,\quad\arg a_{l}=\theta_{L_{+}}(p_{l})\mod\pi\mathbb{Z}.

At the degree $n$ output $q$ ,

\Omega(v_{1},\ldots v_{n})=a_{q}z^{(\theta_{L}-\theta_{L^{\prime}})(q)/\pi}(1+O(z)),\quad a_{q}\neq 0,\quad\arg a_{q}=\theta_{L^{\prime}}(q)\mod\pi\mathbb{Z}.