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. $\square$