In $\mathbb{R}^{3}$ consider the 3-valent graph

$$\Delta=\{(0,0,-t),\ t\geq 0\}\cup\{(0,-t,0),\ t\geq 0\}\cup\{(0,t,t),\ t\geq 0\}$$

and let $D$ be a tubular neighborhood of $\Delta$. Take $U=\mathbb{R}^{3}-D$ and assume we have a stitched Lagrangian fibration $f:X\rightarrow\mathbb{R}^{3}$ such that $U=f(X)$ and the seam is $Z=f^{-1}(\{b_{1}=0\}\cap U)$. Again we let $U^{+}=U\cap\{b_{1}\geq 0\}$, $U^{-}=U\cap\{b_{1}\leq 0\}$ and $\Gamma=U^{+}\cap U^{-}$. Also let $X^{\pm}=f^{-1}(U^{\pm})$. This time $\Gamma$ (hence $Z$) has three connected components

	$\displaystyle\Gamma_{c}$	$\displaystyle=$	$\displaystyle\{(0,t,s),\ t,s<0\}\cap U,$
	$\displaystyle\Gamma_{d}$	$\displaystyle=$	$\displaystyle\{(0,t,s),\ t>0,s<t\}\cap U,$
	$\displaystyle\Gamma_{e}$	$\displaystyle=$	$\displaystyle\{(0,t,s),\ s>0,t<s\}\cap U.$

Also denote by $Z_{c}$, $Z_{d}$ and $Z_{e}$ the corresponding connected components of $Z$ and by $\bar{Z}_{c}$, $\bar{Z}_{d}$ and $\bar{Z}_{e}$ their $S^{1}$ quotients.

Fix $b\in\Gamma_{c}$ and suppose that there is a basis $\{\gamma_{1},\gamma_{2},\gamma_{3}\}$ of $H_{1}(F_{b},\mathbb{Z})$ and generators $g_{1},g_{2},g_{3}$ of $\pi_{1}(U,b)$, satisfying $g_{1}g_{2}g_{3}=1$, with respect to which the monodromy transformations are

$$\mathcal{M}_{b}(g_{1})=T_{1}=\left(\begin{array}[]{ccc}1&-m_{1}&0\\ 0&1&0\\ 0&0&1\end{array}\right),\ \ \ \mathcal{M}_{b}(g_{2})=T_{2}=\left(\begin{array}[]{ccc}1&0&-m_{2}\\ 0&1&0\\ 0&0&1\end{array}\right)$$

(58)

and $\mathcal{M}_{b}(g_{3})=T_{3}=T_{2}^{-1}T_{1}^{-1}$, for non zero integers $m_{1}$ and $m_{2}$. We have that $\gamma_{1}$ is represented by the orbits of the $S^{1}$ action, since it is the only monodromy invariant cycle. Now, since $U-(\Gamma_{d}\cup\Gamma_{e})$ is contractible, $\{\gamma_{1},\gamma_{2},\gamma_{3}\}$ is a basis of $H_{1}(f^{-1}(U-(\Gamma_{d}\cup\Gamma_{e})),\mathbb{Z})$. Consider the diagrams:

or

induced by inclusions and restrictions. The map $j_{+}$ identifies $\{\gamma_{1},\gamma_{2},\gamma_{3}\}$ with a basis of $H_{1}(f^{-1}(U-(\Gamma_{c}\cup\Gamma_{d})),\mathbb{Z})$, which we call $\{\gamma_{1},\gamma_{2}^{+},\gamma_{3}^{+}\}$, while $j_{-}$ identifies it with another basis, which we call $\{\gamma_{1},\gamma_{2}^{-},\gamma_{3}^{-}\}$. Notice that the monodromy map $\mathcal{M}_{b}(g_{2})=j_{+}^{-1}\circ j_{-}$. We must have

$$\begin{cases}\gamma_{2}^{+}=\gamma_{2}^{-},\\ \gamma_{3}^{+}=m_{2}\gamma_{1}+\gamma_{3}^{-}.\end{cases}$$

(59)

Applying Proposition 6.5 to $f$ restricted to $f^{-1}(U-(\Gamma_{c}\cup\Gamma_{d}))$, we can consider the action coordinates map $\alpha$ on $U-(\Gamma_{c}\cup\Gamma_{d})$ computed with respect to $\{\gamma_{1},\gamma_{2}^{+},\gamma_{3}^{+}\}$ on $U^{+}$ and with respect to $\{\gamma_{1},\gamma_{2}^{-},\gamma_{3}^{-}\}$ on $U^{-}$. Let us denote these coordinates by $(b_{1}^{e},b_{2}^{e},b_{3}^{e})$. Similarly we can consider action coordinates on $U-(\Gamma_{d}\cup\Gamma_{e})$ with respect to the basis $\{\gamma_{1},\gamma_{2},\gamma_{3}\}$ of $H_{1}(f^{-1}(U-(\Gamma_{d}\cup\Gamma_{e})),\mathbb{Z})$. We denote them by $(b_{1}^{c},b_{2}^{c},b_{3}^{c})$. We have the identifications

$$\bar{Z}_{e}=T^{\ast}\Gamma_{e}\,/\,\langle db_{2}^{e},db_{3}^{e}\rangle_{\mathbb{Z}}$$

and

$$\bar{Z}_{c}=T^{\ast}\Gamma_{c}\,/\,\langle db_{2}^{c},db_{3}^{c}\rangle_{\mathbb{Z}}.$$

With respect to these coordinates we can compute the first order invariants $\ell_{1}^{e}$ and $\ell_{1}^{c}$ on $\bar{Z}_{e}$ and $\bar{Z}_{c}$ respectively. From Proposition 6.5 and identities (59) applied to $\ell_{1}^{c}$ and $\ell_{1}^{e}$ we obtain

$$\int_{[db_{2}^{c}]}\ell_{1}^{c}=\int_{[db_{3}^{c}]}\ell_{1}^{c}=0$$

and

$$\int_{[db_{2}^{e}]}\ell_{1}^{e}=0\ \ \text{and}\ \ \int_{[db_{3}^{e}]}\ell_{1}^{e}=m_{2}.$$

Similarly we construct the first order invariant $\ell_{1}^{d}$ on $\bar{Z}_{d}$. It will satisfy

$$\int_{[db_{2}^{d}]}\ell_{1}^{d}=m_{1}\ \ \text{and}\ \ \int_{[db_{3}^{d}]}\ell_{1}^{d}=0.$$

Again, monodromy is understood in terms of the difference in the cohomology class of the first order invariant. Example 5.5 is a special case of this situation, where $m_{1}=m_{2}=1$.