We now calculate the conditions on the functions $u(x,y),v(x,y):\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}}$ for the 3-fold $N$ of (30) to be special Lagrangian.

Now if $z_{1}=z_{2}=0$ then $|z_{1}|^{2}-|z_{2}|^{2}$ has zero derivative at $\bf z$. Thus points of the form $(0,0,z_{3})$ in $N$ will be singular. Clearly, these occur exactly when $z_{3}=x+iv(x,0)$ for $x\in\mathbin{\mathbb{R}}$ with $u(x,0)=0$. Also, as $|z_{1}|^{2}-|z_{2}|^{2}=0$, such points occur in $N$ only when $a=0$. We shall see that these are the only singular points in $N$, provided $u$ and $v$ are differentiable.

To prove part (a) we need to show that each ${\bf z}\in N$ not of the form $(0,0,z_{3})$ is a nonsingular point of $N$, and the tangent space $T_{\bf z}N$ is a special Lagrangian 3-plane $\mathbin{\mathbb{R}}^{3}$ in $\mathbin{\mathbb{C}}^{3}$. As $N$ is $\mathbin{\rm U}(1)$-invariant, it is enough to prove this for one point in each orbit of the $\mathbin{\rm U}(1)$-action (29). Since $|z_{1}|=|z_{2}|$ on $N$, each $\mathbin{\rm U}(1)$-orbit in $N$ contains one or two points $(z_{1},z_{2},z_{3})$ with $z_{1}=z_{2}$.

Thus it is enough to show that $T_{\bf z}N$ exists and is special Lagrangian for points ${\bf z}=(z_{1},z_{1},z_{3})$ in $N$ with $z_{1}\neq 0$. In our next lemma we identify $T_{\bf z}N$ at such a point. The proof is elementary, and is left as an exercise.

Now define $\times:\mathbin{\mathbb{C}}^{3}\times\mathbin{\mathbb{C}}^{3}\rightarrow\mathbin{\mathbb{C}}^{3}$ as in (2), and apply Proposition 2.4 with ${\bf r}={\bf p}_{1}$ and ${\bf s}={\bf p}_{2}$. Clearly ${\bf p}_{1}$ and ${\bf p}_{2}$ are linearly independent, and $\omega({\bf p}_{1},{\bf p}_{2})=0$. So Proposition 2.4 shows that $\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{1}\times{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}}$ is the unique SL 3-plane in $\mathbin{\mathbb{C}}^{3}$ containing $\langle{\bf p}_{1},{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}}$.

Therefore $\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{3}\rangle_{\scriptscriptstyle\mathbb{R}}$ is an SL 3-plane if and only if ${\bf p}_{3}\in\langle{\bf p}_{1},{\bf p}_{2},{\bf p}_{1}\times{\bf p}_{2}\rangle_{\scriptscriptstyle\mathbb{R}}$. Combining equations (2), (34) and (35) gives

So suppose ${\bf p}_{3}=\alpha{\bf p}_{1}+\beta{\bf p}_{2}+\gamma{\bf p}_{1}\times{\bf p}_{2}$. As the first two coordinates are equal in ${\bf p}_{2},{\bf p}_{3}$ and ${\bf p}_{1}\times{\bf p}_{2}$ but not in ${\bf p}_{1}$, we see that $\alpha=0$. Taking real parts in the third coordinate gives $\beta=0$. And comparing real multiples of $i\bar{z}_{1}$ in the first coordinate shows that $\gamma={\textstyle\frac{1}{2}}|z_{1}|^{-2}$.

Thus $T_{\bf z}N$ is special Lagrangian if and only if ${\bf p}_{1}\times{\bf p}_{2}=2|z_{1}|^{2}{\bf p}_{3}$. By (36) and (37), this reduces to

But $u=\mathop{\rm Re}(z_{1}^{2})$ and $y=\mathop{\rm Im}(z_{1}^{2})$ by (31), so that $|z_{1}|^{4}=u^{2}+y^{2}$, and $|z_{1}|^{2}=(u^{2}+y^{2})^{1/2}$. Substituting this into (38) gives equation (32), which proves part (a) of Proposition 6.1. Part (b) is left to the reader. $\square$

Equations (32) and (33) are nonlinear versions of the Cauchy–Riemann equations. For if we replace the factors $2(u^{2}+y^{2})^{1/2}$ and $(4u^{2}+4y^{2}+a^{2})^{1/2}$ in (32) and (33) by 1, the equations become

which are the conditions for $u-iv$ to be a holomorphic function of $x+iy$. We may therefore expect the solutions of (32) and (33) to have qualitative features in common with solutions of the Cauchy–Riemann equations.

Note that (32) is the case $a=0$ of equation (33), so we will often use (33) to refer to both, without assuming $a\neq 0$. Following Harvey and Lawson [9, Th. III.2.7], who use results of Morrey, we may prove:

Now holomorphic functions on $\mathbin{\mathbb{C}}$ are determined uniquely by their values on $\mathbin{\mathbb{R}}$. In the same way, solutions of (33) on $\mathbin{\mathbb{R}}^{2}$ are determined by their values on the $x$-axis. We state this in the following proposition, which may be proved using the Cauchy–Kowalevksy Theorem [17, p. 234].

Next we show that solutions $u,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}}$ of (33) may be written in terms of a single potential $f:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}}$.

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.

The functions $u,v$ of Propositions 6.7 and 6.8 provide examples of explicit solutions of equations (32) and (33). Here are some more examples of solutions to (32) and (33). The author constructed them by choosing a particular form for $u,v$ involving arbitrary functions of only one variable, and solving the resulting o.d.e.s.

The next example uses the idea that if $u(x)={\textstyle\frac{1}{2}}y^{2}g(x)-(2g(x))^{-1}$ for some function $g>0$, then $(u^{2}+y^{2})^{1/2}={\textstyle\frac{1}{2}}y^{2}g(x)+(2g(x))^{-1}$. This simplifies (32).

The following example assumes that $u(x,y)=y\,g(x)$ for some nonzero $g$.

We shall now focus on the behaviour of solutions of (32) near points $(x,0)$ with $u(x,0)=0$.

We call a point $(x,0)$ in $U$ with $u(x,0)=0$ a singularity of the solution $u,v$. We call a singularity $(x,0)$ isolated if there exists $\epsilon>0$ such that the open disc $B_{\epsilon}(x,0)$ of radius $\epsilon$ about $(x,0)$ lies in $U$, and the only point $(x^{\prime},y^{\prime})$ in $B_{\epsilon}(x,0)$ with $u(x^{\prime},y^{\prime})=0$ and $v(x^{\prime},y^{\prime})=v(x,0)$ is $(x,0)$.

Let $(x,0)$ be an isolated singularity of $u,v$, and let $\epsilon$ be as above. Consider the map $\gamma:{\mathcal{S}}^{1}\rightarrow\mathbin{\mathbb{C}}$ given by

As $(x,0)$ is isolated we see that $\gamma$ is smooth and maps ${\mathcal{S}}^{1}\rightarrow\mathbin{\mathbb{C}}\setminus\{0\}$. Define the order of the isolated singularity $(x,0)$ to be the winding number of $\gamma$ about 0 in $\mathbin{\mathbb{C}}$. It is easy to show that the order is independent of $\epsilon$, provided $\epsilon>0$ is sufficiently small.

Not all singularities are isolated. For instance, if we put $u(x,y)=\alpha y$ and $v(x,y)=\alpha x+c$ for $\alpha,c\in\mathbin{\mathbb{R}}$, as in Example 6.3, then $(x,0)$ is a nonisolated singularity for all $x$. However, the author believes that nonisolated singularities are rather nongeneric, and so not of much interest in this paper. Also, by analogy with the Identity Theorem of complex analysis, the author conjectures that if $(x,0)$ is a singularity of $u,v$ and $x$ is an isolated zero of $u(x^{\prime},0)$, then $(x,0)$ is isolated.