From the way $f$ is defined in Example 5.8, we can assume

$$\Sigma_{h,M}=\{t=0,\ u_{1}=0,\ |u_{2}|^{2}<\epsilon/4\},$$

where $\epsilon$ is as in (46) and $M=\log(\sqrt{\epsilon}/2)$. For any $\tau>0$ denote open sets

$$N^{\tau}=\{(t,u_{1},u_{2})\ |\ \max(|u_{1}|,|u_{2}|^{2})<\tau\}.$$

From now on we assume $f$ is restricted to $N^{\epsilon/2}$. As one can easily see from the construction, the map $G_{t}$ defining $f$, restricted to $N^{\epsilon/2}$ is

$$G_{t}(u_{1},u_{2})=\left(\log|u_{2}|,\log\left|\frac{u_{1}}{\sqrt{|t|+\sqrt{t^{2}+|u_{1}|^{2}}}}-1\right|\right).$$

(70)

This is the map that we want to perturb, but just on a smaller neighborhood. We do it applying the idea already anticipated at the end of Example 5.7. In fact we notice that $G_{t}$ is invariant with respect to the $S^{1}$ action

$$e^{i\theta}(u_{1},u_{2})=(u_{1},e^{2i\theta}u_{2}),$$

which is also Hamiltonian with respect to the reduced symplectic form $\omega_{t}$ given in (28). The moment map is

$$(u_{1},u_{2})\mapsto|u_{2}|^{2}.$$

So, if $g$ is a real function depending only on $u_{1},t$ and $s=|u_{2}|^{2}$, then

$$(u_{1},u_{2})\mapsto(\log|u_{2}|,\ g(u_{1},t,|u_{2}|^{2}))$$

is a Lagrangian fibration with respect to $\omega_{t}$, provided the level sets of $u_{1}\mapsto g(u_{1},t,s)$ are one dimensional submanifolds for every $s$ and $t$. For example, consider a real non-negative function $\rho$ defined on $\mathbb{R}^{3}$ such that, for every fixed $(t,s)\in\mathbb{R}^{2}$, the map

$$u\mapsto\frac{u}{\rho(|u|^{2},t,s)}$$

(71)

is a local homeomorphism of a neighborhood of $u=0$, then $g=\log|\frac{u}{\rho}-1|$ defines a Lagrangian fibration (at least in a neighborhood of $0$). In particular

$$\rho_{0}(r,t)=\sqrt{|t|+\sqrt{t^{2}+r}},$$

with $(r,t)\in\mathbb{R}^{2}$ gives the map $G_{t}$ in (70), but it is not smooth. It is easy to see that if $\rho$ is smooth on $\mathbb{R}^{3}$ and satisfies

$$\rho>\rho_{0}$$

(72)

then the map (71) is an orientation preserving diffeomorphism (at least near $u=0$). So let us choose a smooth $\rho_{1}$, defined on $\mathbb{R}^{2}$ and satisfying $\rho_{1}>\rho_{0}$, and let

$$g_{j}=\log\left|\frac{u_{1}}{\rho_{j}(|u_{1}|^{2},t)}-1\right|,$$

for $j=0,1$. We wish to find a $g$ which interpolates between $g_{0}$ and $g_{1}$. More precisely, we want $g$ to be equal to $g_{0}$ outside $N^{3\epsilon/8}$ and to $g_{1}$ on some smaller open neighborhood of $\Sigma_{h,M}$. Clearly $(u_{1},u_{2})\in N^{3\epsilon/8}$ if and only if $(|u_{1}|^{2},|u_{2}|^{2})$ is in the rectangle

$$S_{0}=[-9\epsilon^{2}/64,9\epsilon^{2}/64]\times[-3\epsilon/8,3\epsilon/8].$$

Now let $S_{1}$ be a closed neighborhood of $0$ in $\mathbb{R}^{2}$ which is contained in the interior of $S_{0}$, e.g. a smaller rectangle. Taking a $\sigma\in C^{\infty}(\mathbb{R}^{2})$, which is $0$ outside $S_{0}$ and $1$ on $S_{1}$, let us define

$$\rho(r,t,s)=(1-\sigma(r,s))\rho_{0}(r,t)+\sigma(r,s)\rho_{1}(r,t),$$

so that $\rho$ is equal to $\rho_{0}$ outside $S_{0}$ and it is equal to $\rho_{1}$ on $S_{1}$. Clearly $\rho>\rho_{0}$. We leave it to the reader to check that choices can be made so that with this $\rho$, (71) is indeed a homeomorphism. Now define

$$g=\log\left|\frac{u_{1}}{\rho(|u_{1}|^{2},t,|u_{2}|^{2})}-1\right|.$$

Clearly $g$ is equal to $g_{0}$ outside $N^{3\epsilon/8}$ and to $g_{1}$ on

$$N_{h,M}=\{(|u_{1}|^{2},|u_{2}|^{2})\in S_{1}\}$$

which, with a suitable choice of $S_{1}$, is a neighborhood of $\Sigma_{h,M}$. Moreover $u\mapsto g(u,t,s)$ has $1$-dimensional level sets. We can therefore replace the second component of $G_{t}$ in (70) with $g$ and redefine

$$G_{t}(u_{1},u_{2})=(\log|u_{2}|,g),$$

which is smooth on $N_{h,M}$. This proves the lemma for $\Sigma_{h,M}$. A schematic picture of this smoothing is described in Figure 12. The vertical lines represent fibres of $f$ over the horizontal leg. The base of the fibration is represented by the horizontal line on the bottom of the picture; the bold segment on the right represents the region where the codimension one part of $\Delta$ begins. The shaded region represents the locus where $f$ is not smooth. The dashed region is $N_{h,M}$.

The case of the vertical leg is done in the same way. At first sight it is not so obvious that also the diagonal leg can be treated in the same way. So let us give some explanation. When $|u_{2}|^{2}\geq M$, the map $G_{t}$ becomes

$$G_{t}(u_{1},u_{2})=\left(\log\left|\frac{u_{1}}{\rho_{0}(|u_{1}|^{2},t)}-u_{2}\right|,\,\log\left|\frac{u_{1}}{\rho_{0}(|u_{1}|^{2},t)}+u_{2}\right|\right).$$

(73)

The first observation is that this map is invariant under the $S^{1}$-action

$$e^{i\theta}(u_{1},u_{2})=(e^{i\theta}u_{1},e^{i\theta}u_{2}).$$

(74)

After the following change of coordinates on the base

$$(x_{1},x_{2})\mapsto\left(\frac{e^{2x_{1}}+e^{2x_{2}}}{2},x_{1}-x_{2}\right)$$

this becomes

$$G_{t}(u_{1},u_{2})=\left(\frac{\sqrt{t^{2}+|u_{1}|^{2}}+|u_{2}|^{2}}{2}-\frac{|t|}{2},\,\log\frac{\left|u_{1}/\rho_{0}-u_{2}\right|}{\left|u_{1}/\rho_{0}+u_{2}\right|}\right).$$

(75)

One can check that for every fixed $t\in\mathbb{R}$ the map

$$(u_{1},u_{2})\mapsto\frac{\sqrt{t^{2}+|u_{1}|^{2}}+|u_{2}|^{2}}{2},$$

is the moment map of the $S^{1}$-action (74), with respect to the reduced symplectic form $\omega_{t}$. Moreover, if one replaces $u_{1}=z_{1}z_{2}$, $u_{2}=z_{3}$ and $t=\frac{|z_{1}|^{2}-|z_{2}|^{2}}{2}$, then the above map becomes

$$\nu:(z_{1},z_{2},z_{3})\mapsto\frac{|z_{1}|^{2}+|z_{2}|^{2}}{4}+\frac{|z_{3}|^{2}}{2}$$

which is a smooth map on the total space. Let us denote

$$s=\frac{\sqrt{t^{2}+|u_{1}|^{2}}+|u_{2}|^{2}}{2}.$$

The second component of (75) can be rewritten as

$$g_{0}(u_{1},u_{2})=\log\left|\frac{2u_{1}/\rho_{0}}{u_{1}/\rho_{0}+u_{2}}-1\right|.$$

We can now apply the same strategy we used in the case of the horizontal leg. We observe that we could replace this $g_{0}$ with any other $S^{1}$-invariant function $g$. In particular we could replace $\rho_{0}$, which is $S^{1}$-invariant, with another smooth $S^{1}$-invariant $\rho_{1}$. As before, we then interpolate $\rho_{0}$ and $\rho_{1}$ with a cut off function $\sigma$ depending on $|u_{1}|^{2}$ and $s$. We avoid writing the details here, as they just follow the same argument as before.

In the end we obtain that, in a small neighborhood of $\Sigma_{d,M}$, $G_{t}$ can be written as:

$$G_{t}=\left(s-\frac{|t|}{2},\log\left|\frac{2u_{1}/\rho_{1}}{u_{1}/\rho_{1}+u_{2}}-1\right|\right),$$

where now the second component is smooth. The first component is not quite smooth yet. We saw that $s$ is smooth when lifted to the total space, but $|t|$ isn’t. The total fibration becomes of the type

$$f(z_{1},z_{2},z_{3})=\left(\mu,\nu-\frac{|\mu|}{2},g(z_{1}z_{2},z_{3},\mu,\nu)\right),$$

where $g$ is smooth. We see that after a change of coordinates on the base of the type

$$(b_{1},b_{2},b_{3})\mapsto\left(b_{1},b_{2}+\frac{|b_{1}|}{2},b_{3}\right)$$

(76)

this fibration becomes

$$f(z_{1},z_{2},z_{3})=\left(\mu,\nu,g(z_{1}z_{2},z_{3},\mu,\nu)\right),$$

which is smooth. One can find a global change of coordinates on the base which acts like (76) only in a neighborhood of the end of the diagonal leg and is the identity elsewhere. This ends the proof of the Lemma. ∎