Variant: twisted complex case [04BC]

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Variant: twisted complex case

The Floer theoretic obstruction for distinguished triangles can be easily generalized to involve many Lagrangians. Let $L^{\prime}$ be an exact immersed Lagrangian with bounding cochain built from the data of a twisted complex (17). We assume $L$ is isomorphic to $L^{\prime}$ in $D^{b}Fuk(X)$ , so we obtain a bordism current $\mathcal{C}$ with $\partial\mathcal{C}=L-L^{\prime}=L-\sum_{1}^{N}L_{i}$ . As before, all Lagrangians are assumed to be almost calibrated, and all intersections are transverse.

Theorem 3.26.

(Floer theoretic obstruction, multiple Lagrangian case) Assume the automatic transversality and the positivity condition hold for the bordism current $\mathcal{C}$ . Assume the destabilizing condition

\hat{\theta}_{1}>\hat{\theta}_{2}>\ldots>\hat{\theta}_{N},\quad\hat{\theta}_{i}=\arg\int_{L_{i}}\Omega.

Then the Lagrangian phase angle of $L$ has a lower bound on its oscillation:

\sup_{L}\theta_{L}\geq\hat{\theta}_{1},\quad\inf_{L}\theta_{L}\leq\hat{\theta}_{N},

(35)

and morever the J-volume of $L$ has a nontrivial lower bound

\text{Vol}_{J}(L)=\int_{L}e^{-i\theta}\Omega\geq\sum_{1}^{N}|\int_{L_{i}}\Omega|.

(36)

Proof.

Since most parts of the proof are identical to the distinguished triangle case, we will only sketch the main difference.

The Lagrangian boundaries on $\partial\Sigma$ are arranged in the clockwise order as $L,L_{N},L_{N-1},\ldots L_{1}$ . We construct the holomorphic function $F$ as in (33), and use it to produce complex valued volume forms on the $(n-1)$ -dimensional moduli spaces, such that for any $m=1,\ldots N$ ,

\int_{L}\Omega=\int_{\mathcal{M}}\tilde{\Omega}_{L},\quad\int_{L_{i}}\Omega=\int_{\mathcal{M}}\tilde{\Omega}_{L_{i}},\quad i=1,2,\ldots N.

As before, the real part of these complex volume forms are all non-negative, as a consequence of the positivity condition. The claim on the image curve $F(\Sigma)$ holds verbatim. Similarly to the distinguished triangle case, we produce nonnegatively weighted subsets $A_{1},\ldots A_{N}\subset L$ with $L=\sum A_{i}$ , such that

\text{Re}\int_{A_{m}}\Omega=\text{Re}\int_{L_{m}}\Omega>0,\quad\text{Im}\sum_{1}^{m}\int_{A_{i}}\Omega\geq\text{Im}\sum_{1}^{m}\ \int_{L_{i}}\Omega,\quad\int_{L}\Omega=\sum_{1}^{N}\int_{L_{i}}\Omega.

(37)

This implies

\arg\int_{A_{1}}\Omega\geq\arg\int_{L_{1}}\Omega,\quad\arg\int_{A_{N}}\Omega\leq\arg\int_{L_{N}}\Omega,

whence the phase inequality (35).

The J-volume can be bounded below by

\text{Vol}_{J}(L)=\int_{L}|\Omega|=\sum_{1}^{N}\int_{A_{i}}|\Omega|\geq\sum_{1}^{N}|\int_{A_{i}}\Omega|\geq\sum_{1}^{N}|\int_{L_{i}}\Omega|.

The last step uses the purely numerical Lemma 3.27 below. ∎

Lemma 3.27.

Let $z_{1},\ldots z_{N}$ be complex numbers, and $a_{1},\ldots a_{N}$ be fixed complex numbers with positive real parts, such that $\arg a_{1}>\arg a_{2}>\ldots>\arg a_{N}$ . Assume

\text{Re}(z_{i})=\text{Re}(a_{i}),\quad\text{Im}\sum_{1}^{m}z_{i}\geq\text{Im}\sum_{1}^{m}a_{i},\quad\sum_{1}^{N}z_{i}=\sum_{1}^{N}a_{i}.

Then $\sum_{1}^{N}|z_{i}|\geq\sum_{1}^{N}|a_{i}|$ .

Proof.

We argue by induction. The $N=2$ case is implied by Lemma 3.25. In general, we view $\sum_{1}^{N}|z_{i}|$ as a function of the imaginary parts of $z_{1},\ldots z_{N}$ subject to the constraints. Clearly this function achieves its minimum for some $(z_{i})$ . If $z_{1}=a_{1}$ , then we can conclude by induction. Otherwise $\text{Im}z_{1}>\text{Im}a_{1}$ . If $\arg z_{2}<\arg z_{1}$ , then we can fix $z_{1}+z_{2}$ and decrease $\sum_{1}^{2}|z_{i}|$ by Lemma 3.25, which would contradict minimality. Proceding with this argument, we are forced to have

\arg z_{1}\leq\arg z_{2}\leq\ldots\leq\arg z_{N},

whence

\arg\sum_{1}^{N}z_{i}\geq\arg z_{1}>\arg a_{1}>\arg\sum_{1}^{N}a_{i}

which contradicts $\sum_{1}^{N}z_{i}=\sum_{1}^{N}a_{i}$ . ∎