Weighted Sobolev space with exponential growth [04A7]

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Weighted Sobolev space with exponential growth

We now discuss solutions to linearized Cauchy-Riemann equations in weighted Sobolev spaces $W^{1,2;\mu}$ (cf. [70, section 2]). These spaces agree with their unweighted counterparts along the strip like input ends, but at the strip like output end $s\gg 0$ , a vector field $v\in W^{l,2;\mu}$ means that $\exp(-\mu s)v$ lies in $W^{l,2}$ . Generally we choose $\mu$ to avoid a discrete set of indicial values. The main point of these weighted Sobolev spaces is that they allow for holomorphic vector fields with prescribed exponential growth along the output end, which is conceptually similar to allowing for meromorphic functions in Riemann surface theory. If we think of the strip like end $q$ as the infinity (resp. the origin) in the upper half plane model of $\Sigma$ , then the natural coodinate is $z=e^{\pi(s+it)}$ (resp. $z=e^{-\pi(s+it)}$ ), and the exponential growth $o(e^{\mu s})$ becomes $o(|z|^{\mu/\pi})$ (resp. $o(|z|^{-\mu/\pi})$ .

For larger $\mu$ more vector fields are included in the Sobolev space, and the index increases by one each time $\mu$ crosses an indicial value (counted with multiplicity). In our problem, the indicial values are

\phi_{1}+\pi\mathbb{Z},\quad\phi_{2}+\pi\mathbb{Z},\ldots,\phi_{n}+\pi\mathbb{Z},

where $\phi_{1},\ldots\phi_{n}$ are the characterizing angles at the Lagrangian intersection point $q$ at the output end. Then the index for the linearized Cauchy-Riemann operator $W^{1,2;\mu}\to L^{2,\mu}$ is

\deg q-\sum_{1}^{k}\deg p_{i}+\text{number of indicial values between $0$ and $\mu$},

(25)

where $k$ is the number of input ends. In particular, for holomorphic strips with $\deg p=\deg q$ (resp. $\deg q-\deg p=1$ ), then the index for $\mu=\pi$ is equal to $n$ (resp. $n+1$ ). In contrast, the ordinary index (for the $\mu=0$ case) is zero, and the moduli space obtained by taking $\mathbb{R}$ -quotient has virtual dimension $-1$ (resp. zero). There are in fact sufficient conditions to rule out the negative dimension moduli spaces, and constrain the zero dimensional moduli spaces:

Lemma 3.9.

In the holomorphic strip case, assume $v_{1},\ldots v_{n}$ are in the kernel of the linearized Cauchy-Riemann operator on $W^{1,2;\mu=\pi}$ , such that $\Omega(v_{1},\ldots v_{n})$ does not vanish identically. Then $\deg q-\deg p\geq 1$ . When the equality is achieved, the holomorphic strip is an immersion up to the boundary with minimal vanishing at the corner, and the zero dimensional moduli space is regular.

Proof.

We modify the proof of Lemma 3.5 and Cor. 3.6. We think of the corner $q$ as the origin in the upper half plane model. Without loss of generality $v_{n}$ is the $\mathbb{R}$ -translation vector field of the holomorphic strip. Then the leading order asymptotic is

v_{k}=(a_{k1}z^{\phi_{1}/\pi-1},\ldots,a_{kn}z^{\phi_{n}/\pi-1})+O(1),\quad k=1,2,\ldots n-1,

and

v_{n}=(a_{n1}z^{\phi_{1}/\pi},\ldots,a_{nn}z^{\phi_{n}/\pi})+O(z).

hence

\Omega(v_{1},\ldots v_{n})=z^{(\sum\phi_{k})/\pi-n+1}(\det(a_{kj})+o(1)),

The excess vanishing order is $\geq 1-n$ , where negative order stands for poles. By the index formula (23) for the ordinary linearized Cauchy-Riemann equation, we have

\deg q-\deg p\geq 1-n+n=1,

and equality forces $\Omega(v_{1},\ldots v_{n})$ to have no interior zero, no boundary zero, minimal zero at $p$ , and $\det(a_{kj})\neq 0$ at $q$ . The argument in Cor. 3.6 shows $v_{1},\ldots v_{n},\frac{v_{n}}{z}$ span the real vector space of first order deformations in $W^{1,2;\pi}$ . In particular, the only first order deformation which decays at $q$ is the $\mathbb{R}$ -translation vector field. Thus the cokernel to the ordinary linearized Cauchy-Riemann operator vanishes, and the moduli space is regular. ∎

A very analogous statement holds in the polygon case, and is left to the reader:

Lemma 3.10.

In the holomorphic polygon case, assume $v_{1},\ldots v_{n-1}$ are in the kernel of the extended linearized Cauchy-Riemann operator on $W^{1,2;\mu=\pi}$ , such that $\Omega(\cdot,v_{1},\ldots v_{n-1})$ does not vanish identically as a 1-form on $\Sigma$ . Then $\deg q-\sum_{1}^{k}\deg p_{k}+k-2\geq 0$ . When the equality is achieved, the holomorphic polygon is an immersion up to the boundary with minimal vanishing at the corner, and the zero dimensional moduli space of holomorphic polygons is regular at $u:\Sigma\to X$ .

A similar statement applies to teardrop curves:

Lemma 3.11.

(Regularity of teardrops) Let $u:\Sigma\to X$ be a teardrop curve with a unique output $q$ and no input ends. Assume $v_{1},\ldots v_{n-1}$ are in the kernel of the linearized Cauchy-Riemann operator on $W^{1,2;\mu=\pi}$ , such that $\Omega(\cdot,v_{1},\ldots v_{n-1})$ does not vanish identically as a 1-form on $\Sigma$ . Then $\deg q\geq 2$ . When the equality is achieved, the teardrop curve is an immersion up to the boundary with minimal vanishing at the corner, and the kernel of the ordinary Cauchy-Riemann operator is spanned as a real vector space by the Möbius vector fields on $\Sigma$ fixing the $q$ corner, and the cokernel vanishes.

Proof.

We modify the proof of Lemma 3.9. We think of the corner $q$ as the origin in the upper half plane model, and take $v_{n}$ instead to be the Möbius vector field $z^{2}\partial_{z}$ on $\Sigma$ . This has one higher order of vanishing:

v_{n}=(a_{n1}z^{\phi_{1}/\pi+1},\ldots a_{nn}z^{\phi_{n}/\pi+1})+O(z^{2}).

This leads to

\Omega(v_{1},\ldots v_{n})=z^{(\sum\phi_{k})/\pi-n+2}(\det(a_{kj})+o(1)),

so the excess vanishing order at $q$ is $\geq 2-n$ . The index of the ordinary linearized Cauchy-Riemann operator is

\deg q=2\sum(\text{interior zeros})+\sum(\text{boundary zeros})+\sum(\text{excess corner zeros})+n,

whence $\deg q\geq 2$ .

When the equality is achieved, then there is no interior or boundary zero, and $\det(a_{kj})\neq 0$ at the corner, hence the immersion claim. The argument in Cor. 3.6 shows that $v_{1},\ldots v_{n},\frac{v_{n}}{z},\frac{v_{n}}{z^{2}}$ span the real vector space of first order deformations in $W^{1,2;\pi}$ . In particular, the only first order deformation which decays at $q$ are spanned by $v_{n}$ and $z^{-1}v_{n}$ , namely the Möbius generators. Since the index of the ordinary Cauchy-Riemann operator is two, the cokernel must have dimension zero, namely the obstruction vanishes. ∎

The above lemma describes the optimal case for teardrop curves. Such $\deg q=2$ teardrop curves arise in isolated zero dimensional moduli spaces after taking the $Aut(D^{2},q)$ quotient, and the counting contribution to $m_{0}$ are $\pm 1$ depending on the spin structure and the orientation issues.