In Corollary 4.2 and Theorems 5.2 and 5.4 we defined examples of special Lagrangian fibrations $f:\mathbin{\mathbb{C}}^{3}\rightarrow\mathbin{\mathbb{R}}^{3}$. We shall now show that each fibre $f^{-1}(a,b,c)$ of these fibrations may be written in the form (30). For Corollary 4.2 this is trivial:

Next we show that the fibres $f^{-1}(a,b,c)=N_{a,b+ic}$ of the fibration $f$ of Theorem 5.2 may be written in the form (30).

We will use equations (42)–(44) to prove parts (b) and (c) of the proposition. First suppose $x=b$. Then (43) gives $u(v-c)=0$, so $u=0$ or $v=c$. If $v=c$ then (42) gives $u^{2}+y^{2}=0$, so $u=y=0$. Thus $x=b$ implies $u=0$. By a similar argument $u=0$ implies $x=b$, so $u=0$ if and only if $x=b$, as in part (b). In the same way, $v=c$ if and only if $y=0$, as in part (c).

We claim that the two terms $u(x-b)$ and $-y(v-c)$ in (44) are both nonnegative. If one is zero this is obvious. So suppose both are nonzero, so that $x-b,y,u$ and $v-c$ are all nonzero. From (43), the signs of three of these terms determine the sign of the fourth. It is easy to verify that for all eight sign possibilities, $u(x-b)$ and $-y(v-c)$ have the same sign. So both are nonnegative by (44). Hence $u(x-b)\geqslant 0$, and $u=0$ if and only if $x=b$. Clearly, this proves part (b). Part (c) follows in the same way.

Next we shall show that for each pair $(x,y)$, there is exactly one pair $(u,v)$ satisfying (42)–(44). Multiplying (42) by $(v-c)^{2}$ and replacing $u^{2}(v-c)^{2}$ by $y^{2}(x-b)^{2}$ using (43), we get

This is a sextic in $v$, independent of $u$. Putting $\alpha=(v-c)^{2}$, it becomes

Thus $(v-c)^{2}$ is a real, nonnegative root of the cubic $P$. Divide into cases

• (i)

  $x\neq b$, $y\neq 0$ and $P$ has three real roots $\gamma_{1},\gamma_{2},\gamma_{3}$, not necessarily distinct;

• (ii)

  $x\neq b$, $y\neq 0$ and $P$ has one real root $\gamma$ and a complex conjugate pair of non-real roots $\delta,\bar{\delta}$;

• (iii)

  $y=0$; and (iv) $x=b$ and $y\neq 0$.

We shall show that in cases (i)–(iii), the cubic $P$ has exactly one real nonnegative root, giving a unique value of $(v-c)^{2}$. In case (iv) there are two nonnegative roots, but one can be excluded.

In case (i) we have $\gamma_{1}+\gamma_{2}+\gamma_{3}=-2(x-b)^{2}<0$, so at least one $\gamma_{j}$ is negative. But $\gamma_{1}\gamma_{2}\gamma_{3}=y^{2}(x-b)^{2}>0$, so an even number of $\gamma_{j}$ are negative and an odd number positive. The only possibility is that one $\gamma_{j}$ is positive and two negative. So $P$ has exactly one nonnegative root. In case (ii) we have $\gamma|\delta|^{2}=y^{2}(x-b)^{2}>0$, proving that $\gamma>0$, so $P$ has exactly one nonnegative root. In case (iii) we have $P(\alpha)=\alpha\bigl(\alpha+(x-b)^{2}\bigr)^{2}$, with roots 0 and $-(x-b)^{2}$ (twice), so the only nonnegative root is 0.

In case (iv) we have $P(\alpha)=\alpha^{3}-y^{2}\alpha$, with roots $y,0$ and $-y$. Thus there are two nonnegative roots, $|y|$ and 0. However, if $\alpha=0$ then $(v-c)^{2}=0$, and $(x-b)^{2}=0$ by assumption, so the right hand side of (42) is zero. But $y\neq 0$, so the left hand side is positive, a contradiction. So $\alpha\neq 0$, and there is one allowable value for $\alpha$, which is $|y|$.

We have shown that (42) and (43) determine $(v-c)^{2}$ uniquely, and that there is a solution $(v-c)^{2}$ for all $x,y$. This yields $(v-c)$ up to sign. But part (c) gives the sign of $v-c$, so $v$ is determined uniquely. If $v\neq c$, equation (43) determines $u$. If $v=c$ then $y=0$ by (c), so (42) gives $u^{2}=(x-b)^{2}$, and $u=\pm(x-b)$. The sign of $u$ is given by (b). Therefore for all pairs $x,y$, there are unique solutions $u,v$ to (42)–(44).

Let us review what we have proved so far. If $(z_{1},z_{2},z_{3})\in N_{0,b+ic}$ and $x,y,u,v$ are defined by (41), then they satisfy (42)–(44). Also, given any $x,y$ there exist unique $u,v$ satisfying (42)–(44). This defines the functions $u(x,y),v(x,y)$ in the proposition uniquely, and it shows that $N_{0,b+ic}$ is a subset of the 3-fold $N$ of (40). The converse, that $N\subseteq N_{0,b+ic}$, follows easily by reversing the argument above, since if $(z_{1},z_{2},z_{3})\in N$ then (42)–(44) are equivalent to the equations defining $N_{0,b+ic}$. Hence $N=N_{0,b+ic}$.

It remains to prove parts (a), (d) and (e). The smoothness in (a) follows directly from (42)–(44), or indirectly from the fact that $N_{0,b+ic}$ is smooth except at $(0,0,b+ic)$, and $u,v$ satisfy (33) where they are smooth by Proposition 6.1. For part (d), set $y=0$. Then $v=c$ by (c), so (42) gives $u^{2}=(x-b)^{4}$. So $u(x,0)=\pm(x-b)^{2}$, and the sign is determined by (b). Part (e) follows in the same way. This completes the proof for $a=0$.

When $a\neq 0$, equation (42) must be replaced by

but the rest of the proof is more-or-less unchanged. $\square$

Here is the analogue of this for the fibration $f^{\prime}$ of Theorem 5.4.

The last three results show that the fibres of the fibrations of Corollary 4.2 and Theorems 5.2 and 5.4 may all be written in the form (30). This will be important to us in §7, where we shall discuss fibrations which mix the properties of these three fibrations, and we will use the coordinate system (30) to define the fibres.