If $\dim_{\mathbb{R}}S<n$, then

$$H^{n}(X,\mathbb{C})=H^{n}(T_{r}(S),\mathbb{C})=H^{n}(S,\mathbb{C})=\{0\}$$

for any $r>0$, where $T_{r}(S)=\{p\in X|dist_{g_{0}}(p,S)\leq r\}$. Let $r_{0}>r_{1}>\sigma$ such that $T_{r_{1}}(S)\subset B_{g_{0}}(x_{0},r_{0})$, and $F_{r_{0},k}(T_{r_{1}}(S))\supset B_{\tilde{g}_{k}}(x_{k},\sigma)$ for $k\gg 1$. Then the inclusion maps induce homeomorphisms on cohomology groups

$$H^{n}(M,\mathbb{C})\longrightarrow H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})\longrightarrow H^{n}(B_{\tilde{g}_{k}}(x_{k},\sigma),\mathbb{C}),$$

$${\rm and}\ \ \ [\tilde{\Omega}_{k}]\mapsto[\tilde{\Omega}_{k}|_{F_{r_{0},k}(T_{r_{1}}(S))}]\mapsto[\tilde{\Omega}_{k}|_{B_{\tilde{g}_{k}}(x_{k},\sigma)}]\neq 0.$$

Thus $[\tilde{\Omega}_{k}|_{F_{r_{0},k}(T_{r_{1}}(S))}]\neq 0$ in $H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})$, which contradicts to

$$H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})\cong H^{n}(T_{r_{1}}(S),\mathbb{C})=\{0\}.$$

∎

If $\omega_{E}|_{\bar{S}}\neq 0$, and thus $\omega_{0}|_{S}\neq 0$, then there are two vectors $v_{1},v_{2}\in\bar{S}$ such that $\omega_{E}(v_{1},v_{2})>0$. By perturbing $v_{1}$ and $v_{2}$ a little bit if necessary, we have that $\tilde{\Sigma}=\mathfrak{q}(\{t_{1}v_{1}+t_{2}v_{2}|t_{i}\in\mathbb{R}\})$ is a closed 2-torus in $\tilde{S}$, i.e. a closed 2-parameters subgroup. Thus $\Sigma=\pi_{h}(\tilde{\Sigma})$ is a closed oriented surface in $S$, which satisfies

$$\int_{\Sigma}\omega_{0}\geq\frac{1}{\lambda(n)}\int_{\tilde{\Sigma}}\pi_{h}^{*}\omega_{0}\geq\frac{\omega_{E}(v_{1},v_{2})}{\lambda(n)\|v_{1}\wedge v_{2}\|_{h_{E}}}V_{E}>0,$$

where $V_{E}$ denotes the Euclidean area of the intersection of $\{t_{1}v_{1}+t_{2}v_{2}|t_{i}\in\mathbb{R}\}$ with the fundamental domain of the quotient map $\mathfrak{q}$. From the smooth convergence of $(M_{k},\tilde{\omega}_{k},\tilde{g}_{k})$,

$$\lim_{k\longrightarrow\infty}i^{-2}_{g_{k}}(p_{k})\int_{F_{r,k}(\Sigma)}\omega_{k}=\lim_{k\longrightarrow\infty}\int_{F_{r,k}(\Sigma)}\tilde{\omega}_{k}=\lim_{k\longrightarrow\infty}\int_{\Sigma}F_{r,k}^{*}\tilde{\omega}_{k}=\int_{\Sigma}\omega_{0},$$

for $r\gg 1$ such that $\Sigma\subset B_{g_{0}}(p_{0},r)$. Thus

$$0<\frac{1}{2}i^{2}_{g_{k}}(p_{k})\int_{\Sigma}\omega_{0}\leq\int_{F_{r,k}(\Sigma)}\omega_{k}\leq 2i^{2}_{g_{k}}(p_{k})\int_{\Sigma}\omega_{0}\leq 2k^{-2}\int_{\Sigma}\omega_{0}<1,$$

for $k\gg 1$. Since $[F_{r,k}(\Sigma)]\in H_{2}(M_{k},\mathbb{Z})$ and $[\omega_{k}]\in H^{2}(M_{k},\mathbb{Z})$, we obtain

$$\int_{F_{r,k}(\Sigma)}\omega_{k}\in\mathbb{Z},$$

which is a contradiction. Hence $\omega_{E}|_{\bar{S}}\equiv 0$ and $\omega_{0}|_{S}\equiv 0$, which implies that $S$ is a lagrangian submanifold $(X,\omega_{0})$ by combining Lemma 3.1.

Since $\bar{S}$ is a lagrangian linear subspace of $(\mathbb{C}^{n},\omega_{E})$, there is a $\theta_{0}\in\mathbb{R}$ such that ${\rm Im}e^{\sqrt{-1}\theta_{0}}\Omega_{E}|_{\bar{S}}=0$. This implies that $\bar{S}$ is a special lagrangian linear subspace of phase $\theta_{0}$ in $(\mathbb{C}^{n},\omega_{E},\Omega_{E})$. Thus $S$ is a special lagrangian submanifold of phase $\theta_{0}$ in $(X,\omega_{0},\Omega_{0})$, i.e.

$$\omega_{0}|_{S}=0,\ \ \ {\rm Im}e^{\sqrt{-1}\theta_{0}}\Omega_{0}|_{S}=0.$$

∎

By the smooth convergence of $\tilde{\omega}_{k}$ and Lemma 3.2,

$$\lim_{k\longrightarrow\infty}\int_{F_{r,k}(A)}\tilde{\omega}_{k}=\int_{A}\omega_{0}=0,$$

for any cycle $A\in H_{2}(S,\mathbb{Z})$. For $k\gg 1$, we have

$$|\int_{F_{r,k}(A)}\omega_{k}|=i^{2}_{g_{k}}(p_{k})|\int_{F_{r,k}(A)}\tilde{\omega}_{k}|<\frac{1}{2k^{2}}<1.$$

Since $[F_{r,k}(A)]\in H_{2}(M_{k},\mathbb{Z})$ and $[\omega_{k}]\in H^{2}(M_{k},\mathbb{Z})$, we obtain $\int_{F_{r,k}(A)}\omega_{k}\in\mathbb{Z}$. This implies that $|\int_{F_{r,k}(A)}\omega_{k}|=0$, and we obtain the conclusion

$$\int_{A}F_{r,k}^{*}\tilde{\omega}_{k}=0.$$

∎

We choose coordinates $x_{1},\cdots,x_{n}$ on $\bar{S}$ and $y_{1},\cdots,y_{n}$ on $\bar{S}^{\perp}$ such that

$$g_{E}=\sum(dx_{j}^{2}+dy_{j}^{2}),\ \ \ \omega_{E}=\sum dx_{j}\wedge dy_{j},\ \ \ e^{\sqrt{-1}\theta_{0}}\Omega_{E}=\bigwedge_{j=1}^{n}(dx_{j}+\sqrt{-1}dy_{j}).$$

If $\mathcal{G}$ is a subgroup of the fundamental group $\pi_{1}(X)=\pi_{1}(S)$, then $\mathcal{G}$ acts on $\mathbb{C}^{n}$ preserving $g_{E}$, $\omega_{E}$ and $\Omega_{E}$, and $\bar{S}$ is a invariant subspace. For any $\gamma\in\mathcal{G}$, we have $\gamma\cdot(v+w)=G_{\gamma}(v+w)+b_{\gamma}$, where $G_{\gamma}\in U(\mathbb{C}^{n})$, $b_{\gamma}\in\bar{S}$, $v\in\bar{S}$ and $w\in\bar{S}^{\perp}$. Since $\bar{S}$ is invariant, we obtain then $G_{\gamma}(v+w)=A_{\gamma}v+B_{\gamma}w+C_{\gamma}w$ where $A_{\gamma}\in SO(\bar{S})$, $B_{\gamma}\in SO(\bar{S}^{\perp})$, and $C_{\gamma}\in Hom(\bar{S}^{\perp},\bar{S})$. Moreover, $G_{\gamma}\in SO(\mathbb{R}^{2n})$ implies $C_{\gamma}=0$. Since $\omega_{E}(G_{\gamma}(v+w),G_{\gamma}(v+w))=\omega_{E}(v+w,v+w)$, we have $B_{\gamma}=A_{\gamma}^{-1,T}=A_{\gamma}$, and $\gamma\cdot(v+w)=A_{\gamma}(v+w)+b_{\gamma}$. Thus $\pi_{1}(\tilde{X})\cong\Lambda$ acts on $\mathbb{C}^{n}$ given by $\gamma\cdot(v+w)=v+w+b_{\gamma}$, $b_{\gamma}\in\Lambda$, for any $v\in\bar{S}$ and $w\in\bar{S}^{\perp}$. This implies that $\tilde{X}=\mathbb{C}^{n}/\pi_{1}(\tilde{X})\cong\bar{S}/\Lambda\times\bar{S}^{\perp}=\tilde{S}\times\bar{S}^{\perp}$, and $\pi^{*}g_{0}=h+h_{E}$ where $h_{E}=g_{E}|_{\bar{S}^{\perp}}$, and $h$ is the standard flat metric on $\tilde{S}$ induced by $g_{E}|_{\bar{S}}$.

The $\pi_{1}(X)$-action on $\mathbb{C}^{n}$ descents to a $\Gamma$-action on $\tilde{X}$, which is a product action since the $\pi_{1}(X)$-action is so. Moreover, $\tilde{S}\times\{0\}$ is a invariant set as $\bar{S}\times\{0\}$ is invariant under the $\pi_{1}(X)$-action. If we denote the quotient map $\mathfrak{q}_{1}:\mathbb{C}^{n}\longrightarrow\mathbb{C}^{n}/\Lambda=\tilde{X}$, then $\bar{\pi}=\pi\circ\mathfrak{q}_{1}$, $g_{E}=\mathfrak{q}_{1}^{*}\pi^{*}g_{0}$, $\omega_{E}=\mathfrak{q}_{1}^{*}\pi^{*}\omega_{0}$, and $\Omega_{E}=\mathfrak{q}_{1}^{*}\pi^{*}\Omega_{0}$. Since $\omega_{E}|_{\bar{S}\times\{y\}}=0$ and $e^{\sqrt{-1}\theta_{0}}\Omega_{E}|_{\bar{S}\times\{y\}}=0$ for $y\in\bar{S}^{\perp}$, we obtain that

$$\pi^{*}\omega_{0}|_{T^{n}\times\{y\}}\equiv 0,\ \ {\rm and}\ \ \pi^{*}{\rm Im}e^{\sqrt{-1}\theta_{0}}\Omega_{0}|_{T^{n}\times\{y\}}\equiv 0,$$

for a constant $\theta_{0}\in\mathbb{R}$. ∎