Let $U\subset\mathbb{R}^{2}$ be an open annulus in $\mathbb{R}^{2}$ centered at the origin. As usual denote $U^{+}=U\cap\{b_{1}\geq 0\}$, $U^{-}=U\cap\{b_{1}\leq 0\}$ and $\Gamma=U^{+}\cap U^{-}$. This time $\Gamma$ is disconnected. We let $\Gamma_{u}=\Gamma\cap\{b_{2}\geq 0\}$ and $\Gamma_{d}=\Gamma\cap\{b_{2}\leq 0\}$ be the upper and lower parts of $\Gamma$ respectively. Now let $f:X\rightarrow\mathbb{R}^{2}$ be a stitched Lagrangian fibration such that $f(X)=U$. Observe that the seam $Z$ has two connected components: $Z_{u}=f^{-1}(\Gamma_{u})$ and $Z_{d}=f^{-1}(\Gamma_{d})$. Denote by $\bar{Z}_{u}$ and $\bar{Z}_{d}$ the respective $S^{1}$ quotients, i.e. the connected components of $\bar{Z}$. Let $b\in\Gamma_{u}$ and choose as generator of $\pi_{1}(U,b)$ an anti-clock-wise oriented curve starting at $b$ and going once around $0$. Suppose that with respect to a basis $\{\gamma_{1},\gamma_{2}\}$ of $H_{1}(F_{b},\mathbb{Z})$ the monodromy is

$$\left(\begin{array}[]{cc}1&-m\\ 0&1\end{array}\right),$$

(57)

for some integer $m\neq 0$. In this case we must have that $\gamma_{1}$ is represented by the orbits of the $S^{1}$ action. As usual let $X^{\pm}=f^{-1}(U^{\pm})$. Since $U-\Gamma_{d}$ is contractible we can think of $\{\gamma_{1},\gamma_{2}\}$ as a basis of $H_{1}(f^{-1}(U-\Gamma_{d}),\mathbb{Z})$. Consider the diagrams:

or

induced by inclusions and restrictions. The map $j_{+}$ identifies $\{\gamma_{1},\gamma_{2}\}$ with a basis $\{\gamma_{1},\gamma_{2}^{+}\}$ of $H_{1}(f^{-1}(U-\Gamma_{u}),\mathbb{Z})$, whereas $j_{-}$ with a basis $\{\gamma_{1},\gamma_{2}^{-}\}$. Notice that monodromy is given by $j_{+}^{-1}\circ j_{-}$. Therefore we must have $\gamma_{2}^{+}=m\gamma_{1}+\gamma_{2}^{-}$. Hence $\{\gamma_{1},\gamma_{2}^{+}\}$ and $\{\gamma_{1},\gamma_{2}^{-}\}$ satisfy conditions (a) and (b) in the previous section. Applying Proposition 6.5 to $f$ restricted to $f^{-1}(U-\Gamma_{u})$ we can consider the action coordinates map $\alpha$ constructed by taking action coordinates with respect to $\{\gamma_{1},\gamma_{2}^{+}\}$ on $U^{+}$ and with respect to $\{\gamma_{1},\gamma_{2}^{-}\}$ on $U^{-}$. Denote by $(b_{1}^{d},b_{2}^{d})$ such coordinates. Similarly on $U-\Gamma_{d}$ we can consider action angle coordinates with respect to the basis $\{\gamma_{1},\gamma_{2}\}$. Denote by $(b_{1}^{u},b_{2}^{u})$ these coordinates. In particular we have the identifications

$$\bar{Z}_{d}=T^{\ast}\Gamma_{d}\,/\,\langle db_{2}^{d}\rangle_{\mathbb{Z}}$$

and

$$\bar{Z}_{u}=T^{\ast}\Gamma_{u}\,/\,\langle db_{2}^{u}\rangle_{\mathbb{Z}}.$$

With respect to this choice of coordinates we can compute the first order invariants of $f$, $\ell_{1}^{u}$ and $\ell_{1}^{d}$ on $\bar{Z}_{u}$ and $\bar{Z}_{d}$, respectively. Then (49) should hold, therefore we obtain

$$\int_{[db_{2}^{u}]}\ell_{1}^{u}=0\ \ \text{and}\ \ \int_{[db_{2}^{d}]}\ell_{1}^{d}=m.$$

This tells us that monodromy can be read from a jump in cohomology class of the first order invariant associated to action coordinates.