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I am trying to understand the proof of Theorem 1.7 on page 88 of Harvey-Lawson's Calibrated Geometries. I do not understand how they conclude that $(dz_1 \wedge \dots \wedge dz_n, A(e_1\wedge \dots \wedge e_n)) = \det_{\mathbb{C}}A$. I will describe the problem in more detail below.

Consider an n-plane $\zeta \in G(n.2n)$, and pick a basis $\varepsilon_1, \dots, \varepsilon_n$ for $\zeta$. Define the A as the matrix mapping the standard basis $e_1,\dots,e_n,Je_1,\dots,Je_n$ on $\mathbb{R}^{2n} \cong \mathbb{C}^n$, respectively to the vectors $\varepsilon_1, \dots, \varepsilon_n, J\varepsilon_1, \dots, J\varepsilon_n$. Then we have $\varepsilon_1\wedge \dots\wedge \varepsilon_n = A(e_1\wedge \dots\wedge e_n)$. They claim that $$(dz_1 \wedge \dots \wedge dz_n, A(e_1\wedge \dots \wedge e_n)) = \rm{det}_{\mathbb{C}}A .$$

I did the following calculation. In differential geometric language we have $$\frac{\partial}{\partial x_1}\wedge \dots \wedge \frac{\partial}{\partial x_n} = (1+\sqrt{-1})Id\left(\frac{\partial}{\partial z_1}\wedge \dots \wedge \frac{\partial}{\partial z_n}\right)$$ where Id is the identity matrix. And so $$\varepsilon_1\wedge \dots\wedge \varepsilon_n = A\left(\frac{\partial}{\partial x_1}\wedge \dots \wedge \frac{\partial}{\partial x_n}\right) = (1+\sqrt{-1})A\left(\frac{\partial}{\partial z_1}\wedge \dots \wedge \frac{\partial}{\partial z_n}\right).$$ So I think $$(dz_1 \wedge \dots \wedge dz_n, \varepsilon_1\wedge \dots\wedge \varepsilon_n) = \rm{det}_{\mathbb{C}}(A(1+\sqrt{-1})) = (1+\sqrt{-1})^n\rm{det}_{\mathbb{C}} A.$$ I am struggling to understand what my mistake here is. Could you please help me out.

Arctic Char
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user573253
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  • Where is the inequality? – Arctic Char Jul 26 '21 at 15:34
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    @arctic-char The section in the paper Calibrated Geometries is titled the special Lagrangian inequality. But I realise now my question is not actually related to the inequality, so the title is misleading. I am sorry about that. They state the inequality in the following page as $|dz(\zeta)|^2 \le |\zeta|^2$ with equality if and only if $\zeta$ is Lagrangian. – user573253 Jul 26 '21 at 16:11
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    It is not true that $\partial_{x_1}\wedge\dots\wedge\partial_{x_n}=(1+\sqrt{-1})(\partial_{z_1}\wedge\dots\wedge\partial_{z_n})$ --- note that $\partial_{x_i}\neq(1+\sqrt{-1})\partial_{z_i}$ in $\mathbb{R}^{2n}$. – user10354138 Jul 26 '21 at 17:02

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