Proof. [03KI]

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Proof. Explicit calculation using (2) shows that

	$\displaystyle g({\bf r},{\bf r}\times{\bf s})=g({\bf s},{\bf r}\times{\bf s})=0,$		(3)
	$\displaystyle\omega({\bf r},{\bf r}\times{\bf s})=\omega({\bf s},{\bf r}\times{\bf s})=0,$		(4)
	$\displaystyle\|{\bf r}\times{\bf s}\|^{2}=\|\bf r\|^{2}\|\bf s\|^{2}-g({\bf r},{\bf s})^{2}-\omega({\bf r},{\bf s})^{2},$		(5)
	$\displaystyle\text{and}\quad(\mathop{\rm Im}\Omega)({\bf r},{\bf s},{\bf r}\times{\bf s})=0,$		(6)

for all ${\bf r},{\bf s}\in\mathbin{\mathbb{C}}^{3}$ . When ${\bf r},{\bf s}$ are linearly independent and $\omega({\bf r},{\bf s})=0$ , equation (3) shows that ${\bf r}\times{\bf s}$ is orthogonal to ${\bf r},{\bf s}$ , and (5) that $|{\bf r}\times{\bf s}|\neq 0$ . Therefore ${\bf r},{\bf s}$ and ${\bf r}\times{\bf s}$ are linearly independent.

Also we have $\omega({\bf r},{\bf s})=\omega({\bf r},{\bf r}\times{\bf s})=\omega({\bf s},{\bf r}\times{\bf s})=0$ by (4), so that $\langle{\bf r},{\bf s},{\bf r}\times{\bf s}\rangle_{\scriptscriptstyle\mathbb{R}}$ is a Lagrangian 3-plane. Then (6) shows that $\langle{\bf r},{\bf s},{\bf r}\times{\bf s}\rangle_{\scriptscriptstyle\mathbb{R}}$ is a special Lagrangian 3-plane, by Proposition 2.7. It is easy to see that this is the only SL 3-plane in $\mathbin{\mathbb{C}}^{3}$ containing $\langle{\bf r},{\bf s}\rangle_{\scriptscriptstyle\mathbb{R}}$ . $\square$