Consider a fibration $f$ satisfying the following analogues of Assumptions 7.1–7.1.

These assumptions are a kind of toy model, designed to illustrate some aspects of how the fibrations of §5 and §7 might fit together in a Calabi–Yau 3-fold, and to perform a topological calculation. We are not making the conjecture that a fibration actually exists satisfying these assumptions.

In Figure 3 we sketch the functions $u_{0,b}(x,0)$ for several values of $b$, on the same graph, to display the general features we expect of these functions. The basic idea is that $u_{0,b}(x,0)$ should look a bit like $b+\cos x$, in that it has the same periodic behaviour, is zero at the same points, and is increasing and decreasing in the same regions. But at its zeros $u_{0,b}(x,0)$ has gradient zero, whereas $b+\cos x$ generally does not.

Let us describe the fibres $N_{a,b,c}$ of $f$. From §6, $N_{a,b,c}$ is singular if and only if $a=0$ and $u_{0,b}(x,0)=0$ for some $x$. But part (iii) of Assumption 8.1 shows that $u_{0,b}(x+2\pi\mathbin{\mathbb{Z}},0)=0$ if and only if $\cos x=-b$. This has solutions when $b\in[-1,1]$. Therefore $N_{a,b,c}$ is singular if and only if $a=0$ and $b\in[-1,1]$.

It is easy to show that the nonsingular fibres of $f$ are all diffeomorphic to $T^{2}\times[-\pi,\pi]$. The singular fibres $N_{0,b,c}$ for $b\in(-1,1)$ are diffeomorphic to $T^{2}\times[-\pi,\pi]$ with two homologous circles collapsed to two points, and the singular fibres $N_{0,\pm 1,c}$ are diffeomorphic to $T^{2}\times[-\pi,\pi]$ with one circle collapsed to a point.

The discriminant $\Delta_{f}$ of $f$ is the set of $(a,b,c)\in\mathbin{\mathbb{R}}^{3}$ such that $f^{-1}(a,b,c)$ is singular. From above we see that

$$\Delta_{f}=\bigl\{(0,b,c):b\in[-1,1],\quad c\in\mathbin{\mathbb{R}}\bigr\}.$$

(60)

We can think of $\Delta_{f}$ as a ‘ribbon’ in $\mathbin{\mathbb{R}}^{3}$. To understand the topology of a fibration $f:X\rightarrow B$ with singularities, it is often helpful to calculate the monodromy around nontrivial loops in $B\setminus\Delta_{f}$, as in Ruan [18, §4] or Gross [7, §1], for instance.

In our case, $\pi_{1}\bigl(\mathbin{\mathbb{R}}^{3}\setminus\Delta_{f}\bigr)$ is isomorphic to $\mathbin{\mathbb{Z}}$, and generated by the circle $\gamma:{\mathcal{S}}^{1}\rightarrow\mathbin{\mathbb{R}}^{3}\setminus\Delta_{f}$ given by $\gamma({\rm e}^{i\theta})=(2\cos\theta,2\sin\theta,0)$. So we would like to understand the monodromy around $\gamma$. It turns out that the monodromy action on the homology $H_{1}\bigl(T^{2}\times[-\pi,\pi];\mathbin{\mathbb{Z}}\bigr)\cong\mathbin{\mathbb{Z}}^{2}$ is trivial.

This is because the topologically interesting transformations happen in the $[-\pi,\pi]$ directions, which $H_{1}\bigl(T^{2}\times[-\pi,\pi];\mathbin{\mathbb{Z}}\bigr)$ does not detect. To get round this we will identify the two boundary components $T^{2}\times\{\pi\}$ and $T^{2}\times\{-\pi\}$ of each fibre $N_{a,b,c}$, so that the nonsingular fibres become 3-tori $T^{3}$, and then evaluate the monodromy around $\gamma$ on $H_{1}(T^{3};\mathbin{\mathbb{Z}})$, which is nontrivial.

Let $N_{a,b,c}$ be a fibre of $f$. We need a way to identify the two components of $\partial N_{a,b,c}$. Here is one way to do it. Let ${\bf z}=(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})$ and ${\bf z}^{\prime}=(z_{1}^{\prime},z_{2}^{\prime},z_{3}^{\prime}+2\pi\mathbin{\mathbb{Z}})$ lie in $N_{a,b,c}$ with $\mathop{\rm Im}(z_{1}z_{2})=\pi$ and $\mathop{\rm Im}(z_{1}^{\prime}z_{2}^{\prime})=-\pi$, so that ${\bf z}$ and ${\bf z}^{\prime}$ lie in different boundary components. We identify $\bf z$ and ${\bf z}^{\prime}$ if

$$\mathop{\rm Re}(z_{3})+2\pi\mathbin{\mathbb{Z}}=\mathop{\rm Re}(z_{3}^{\prime})+2\pi\mathbin{\mathbb{Z}}\quad\text{and}\quad\arg(z_{1})+2\pi\mathbin{\mathbb{Z}}=\arg(z_{1}^{\prime})+2\pi\mathbin{\mathbb{Z}},$$

(61)

where $\arg(r{\rm e}^{i\theta})=\theta$ for $r>0$ and $\theta\in[0,2\pi)$ is the argument of a nonzero complex number. As $\mathop{\rm Im}(z_{1}z_{2})=\pi$ and $\mathop{\rm Im}(z_{1}^{\prime}z_{2}^{\prime})=-\pi$ both $z_{1}$ and $z_{1}^{\prime}$ are nonzero, and so $\arg(z_{1})$ and $\arg(z_{1}^{\prime})$ are well-defined.

Let $\tilde{N}_{a,b,c}$ be $N_{a,b,c}$ with its boundary components identified as above. It is easy to show that (61) defines a diffeomorphism between the two components of $\partial N_{a,b,c}$, and that $\tilde{N}_{a,b,c}$ is a copy of $T^{3}$ when $N_{a,b,c}$ is nonsingular.

To calculate the monodromy we will cover $\mathbin{\mathbb{R}}^{3}\setminus\Delta_{f}$ by two closed sets $U^{\pm}$, and define a trivialization of $f$ over each set. Let

$$U^{+}=\bigl\{(a,b,c)\in\mathbin{\mathbb{R}}^{3}:\text{$a\geqslant 0$, and $b\notin[-1,1]$ if $a=0$}\bigr\}.$$

(62)

For each $(a,b,c)\in U^{+}$, define $\Phi^{+}_{a,b,c}:\tilde{N}_{a,b,c}\rightarrow(\mathbin{\mathbb{R}}/2\pi\mathbin{\mathbb{Z}})^{3}$ by

$$\begin{split}&\Phi^{+}_{a,b,c}:(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})\mapsto\\ &\bigl(\arg(z_{1})+2\pi\mathbin{\mathbb{Z}},\mathop{\rm Re}(z_{3})+2\pi\mathbin{\mathbb{Z}},\mathop{\rm Im}(z_{1}z_{2})+2\pi\mathbin{\mathbb{Z}}\bigr).\end{split}$$

(63)

There are two issues involved in proving $\Phi_{a,b,c}^{+}$ is well-defined. Firstly, we must show that $z_{1}\neq 0$ in $\tilde{N}_{a,b,c}$, so that $\arg(z_{1})$ exists. This is true because $z_{1}$ and $z_{2}$ cannot both be zero, as then $(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})$ would be a singular point, contradicting $(a,b,c)\notin\Delta_{f}$. But $|z_{1}|^{2}-|z_{2}|^{2}=a\geqslant 0$, so that if $z_{1}=0$ then $z_{2}=0$. Thus $z_{1}\neq 0$ in $\tilde{N}_{a,b,c}$. The second issue is that points identified in $N_{a,b,c}$ must have the same image under $\Phi_{a,b,c}^{+}$. This follows from (61).

The other closed set in $\mathbin{\mathbb{R}}^{3}\setminus\Delta_{f}$ is

$$U^{-}=\bigl\{(a,b,c)\in\mathbin{\mathbb{R}}^{3}:\text{$a\leqslant 0$, and $b\notin[-1,1]$ if $a=0$}\bigr\}.$$

(64)

We cannot use (63) to trivialize $f$ over this set, because $z_{1}$ will become zero in some $N_{a,b,c}$, and so $\arg(z_{1})$ may not be well-defined. Instead we must do something more complicated.

Let $\eta:[-\pi,\pi]\rightarrow[0,1]$ be smooth with $\eta(y)=0$ for $|y|\leqslant 1$ and $\eta(y)=1$ for $|y|\geqslant 2$. For each $(a,b,c)\in U^{-}$, define $\Phi^{-}_{a,b,c}:\tilde{N}_{a,b,c}\rightarrow(\mathbin{\mathbb{R}}/2\pi\mathbin{\mathbb{Z}})^{3}$ by

$$\begin{split}\Phi^{-}_{a,b,c}:(z_{1}&,z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})\mapsto\bigl(\eta\bigl(\mathop{\rm Im}(z_{1}z_{2})\bigr)\arg(z_{1}z_{2})\\ &-\arg(z_{2})+2\pi\mathbin{\mathbb{Z}},\mathop{\rm Re}(z_{3})+2\pi\mathbin{\mathbb{Z}},\mathop{\rm Im}(z_{1}z_{2})+2\pi\mathbin{\mathbb{Z}}\bigr).\end{split}$$

(65)

Here for $a<0$ we may have $z_{1}z_{2}=0$ at some points in $\tilde{N}_{a,b,c}$, so that $\arg(z_{1}z_{2})$ is undefined. But then $\eta\bigl(\mathop{\rm Im}(z_{1}z_{2})\bigr)=0$, so we take this term to be zero.

Also, $\arg(z_{1}z_{2})$ changes discontinuously when $z_{1}z_{2}$ is real and positive, but the $\eta$ term is zero here, so that $\Phi^{-}_{a,b,c}$ is continuous. The term $\arg(z_{2})$ in (65) always exists by the argument above. When $|\mathop{\rm Im}(z_{1}z_{2})|\geqslant 2$ we have

$$\eta\bigl(\mathop{\rm Im}(z_{1}z_{2})\bigr)\arg(z_{1}z_{2})-\arg(z_{2})+2\pi\mathbin{\mathbb{Z}}=\arg(z_{1})+2\pi\mathbin{\mathbb{Z}},$$

so that (65) agrees with (63). Therefore points identified in $N_{a,b,c}$ have the same image under $\Phi_{a,b,c}^{-}$, as above. So $\Phi^{-}_{a,b,c}$ is well-defined.

Now we can calculate the monodromy around $\gamma$. Starting at ${\rm e}^{i\theta}=1$, and going round $\gamma$ once in the positive direction, we first cross over from $U^{+}$ to $U^{-}$ at ${\rm e}^{i\theta}=i$. The transition map between trivializations is $\Phi^{-}_{0,2,0}\circ(\Phi^{+}_{0,2,0})^{-1}$. Then we cross from $U^{-}$ back to $U^{+}$ at ${\rm e}^{i\theta}=-i$, with transition map $\Phi^{+}_{0,-2,0}\circ(\Phi^{-}_{0,-2,0})^{-1}$. So the overall monodromy transformation round $\gamma$ is

$$\Phi^{+}_{0,-2,0}\circ(\Phi^{-}_{0,-2,0})^{-1}\circ\Phi^{-}_{0,2,0}\circ(\Phi^{+}_{0,2,0})^{-1}.$$

(66)

Let $(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})\in N_{0,2,0}$ and write

	$\displaystyle\Phi^{+}_{0,2,0}(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})$	$\displaystyle=(\theta^{+}+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}})$
	$\displaystyle\text{and}\quad\Phi^{-}_{0,2,0}(z_{1},z_{2},z_{3}+2\pi\mathbin{\mathbb{Z}})$	$\displaystyle=(\theta^{-}+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}).$

Then $\theta^{+}=\arg(z_{1})$ and $\theta^{-}=\eta\bigl(\mathop{\rm Im}(z_{1}z_{2})\bigr)\arg(z_{1}z_{2})-\arg(z_{2})$ by (63) and (65), so that

$$\theta^{+}-\theta^{-}=\bigl(1-\eta\bigl(\mathop{\rm Im}(z_{1}z_{2})\bigr)\bigr)\arg(z_{1}z_{2})=\bigl(1-\eta(y)\bigr)\arg\bigl(u_{0,2}(x,y)+iy\bigr).$$

But $u_{0,2}(x,0)>0$ for all $x$ by Assumption 8.1. It follows that

$$\arg\bigl(u_{0,2}(x,y)+iy\bigr)=\begin{cases}0,&y=0,\\ \cot^{-1}\bigl(y^{-1}u_{0,2}(x,y)\bigr),&y>0,\\ \pi+\cot^{-1}\bigl(y^{-1}u_{0,2}(x,y)\bigr),&y<0,\\ \end{cases}$$

as $\arg$ maps into $[0,2\pi)$, where $\cot^{-1}$ maps $\mathbin{\mathbb{R}}\rightarrow(0,\pi)$. Combining the last four equations gives

		$\displaystyle\Phi^{+}_{0,2,0}\circ(\Phi^{-}_{0,2,0})^{-1}:(\theta+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}})\mapsto$
		$\displaystyle\begin{cases}\bigl(\theta+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},2\pi\mathbin{\mathbb{Z}}\bigr),&y=0,\\ \bigl(\theta+(1-\eta(y))\cot^{-1}(y^{-1}u_{0,2}(x,y))+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}\bigr),&y>0,\\ \bigl(\theta+(1-\eta(y))\bigl(\pi\!+\!\cot^{-1}(y^{-1}u_{0,2}(x,y))\bigr)\!+\!2\pi\mathbin{\mathbb{Z}},x\!+\!2\pi\mathbin{\mathbb{Z}},y\!+\!2\pi\mathbin{\mathbb{Z}}\bigr),&y<0.\end{cases}$

Similarly we find that

$$\arg\bigl(u_{0,-2}(x,y)+iy\bigr)=\begin{cases}\pi,&y=0,\\ \cot^{-1}\bigl(y^{-1}u_{0,-2}(x,y)\bigr),&y>0,\\ \pi+\cot^{-1}\bigl(y^{-1}u_{0,-2}(x,y)\bigr),&y<0,\\ \end{cases}$$

as $u_{0,-2}(x,0)<0$ for all $x$, so that

		$\displaystyle\Phi^{-}_{0,-2,0}\circ(\Phi^{+}_{0,-2,0})^{-1}:(\theta+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}})\mapsto$
		$\displaystyle\begin{cases}\bigl(\theta+\pi+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},2\pi\mathbin{\mathbb{Z}}\bigr),&y=0,\\ \bigl(\theta+(1-\eta(y))\cot^{-1}(y^{-1}u_{0,-2}(x,y))+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}\bigr),&y>0,\\ \bigl(\theta+(1-\eta(y))\bigl(\pi\!+\!\cot^{-1}(y^{-1}u_{0,-2}(x,y))\bigr)\!+\!2\pi\mathbin{\mathbb{Z}},x\!+\!2\pi\mathbin{\mathbb{Z}},y\!+\!2\pi\mathbin{\mathbb{Z}}\bigr),&y<0.\end{cases}$

From (66) and the equations above we see that the monodromy is

	$\displaystyle(\theta+$	$\displaystyle 2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}})\mapsto$
		$\displaystyle\begin{cases}\bigl(\theta+(1-\eta(y))\pi+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}\bigr),&y=0,\\ \begin{gathered}\bigl(\theta+(1-\eta(y))\bigl(\cot^{-1}(y^{-1}u_{0,2}(x,y))\qquad\qquad\qquad\qquad\qquad\\ -\cot^{-1}(y^{-1}u_{0,-2}(x,y))\bigr)+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}\bigr),\end{gathered}&y\neq 0.\end{cases}$

Note that although this is continuous as a map to $(\mathbin{\mathbb{R}}/2\pi\mathbin{\mathbb{Z}})^{3}$, the expression

$$\bigl(1-\eta(y)\bigr)\bigl(\cot^{-1}(y^{-1}u_{0,2}(x,y))-\cot^{-1}(y^{-1}u_{0,-2}(x,y))\bigr)$$

decreases discontinuously by $-2\pi$ at $y=0$, as $u_{0,2}(x,0)>0$ and $u_{0,-2}(x,0)<0$ for all $x$. It follows that the monodromy is homotopic to

$$(\theta+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}})\mapsto(\theta+y+2\pi\mathbin{\mathbb{Z}},x+2\pi\mathbin{\mathbb{Z}},y+2\pi\mathbin{\mathbb{Z}}),$$

which is a transformation of $T^{3}$ with monodromy

$$\begin{pmatrix}1&0&1\\ 0&1&0\\ 0&0&1\end{pmatrix}$$

(67)

with respect to the obvious basis of $H_{1}(T^{3};\mathbin{\mathbb{Z}})$.

This is important, and is what gives the calculation above topological meaning. If we had defined the fibres $\tilde{N}_{a,b,c}$ only for nonsingular $N_{a,b,c}$, without the requirement that the identification of boundary components should extend continuously over the singular fibres, then we would have had more topological freedom in choosing how to identify the boundaries of the nonsingular $N_{a,b,c}$ to get $\tilde{N}_{a,b,c}$, and the monodromy matrix would depend on this choice.