that $J_1=J_2J_3=-J_3J_2$ and $J_2=J_3J_1=-J_1J_3$, and the corresponding Kähler forms are $\omega=\omega_1$, $\omega_2$, $\omega_3$ which are orthogonal and the same length with $\Upsilon=\omega_2+i\omega_3$.

This shows we should only consider complex dimension 3 and higher to find new calibrated submanifolds.

Let $f:\R^n\rightarrow \R^n$ be a smooth function and let $N=\text{Graph}(f)\subseteq\R^{2n}=\C^n$. We want to see when $N$ is special Lagrangian. We see that tangent vectors to $N$ are given by $$e_1+i\nabla_{e_1}f,\ldots,e_n+i\nabla_{e_n}f.$$ Hence $N$ is Lagrangian if and only if $$\omega(e_j+i\nabla_{e_j}f,e_k+i\nabla_{e_k}f)=\nabla_{e_k}f_j-\nabla_{e_j}f_k=0$$ for all $j,k$. Since $\R^n$ is simply connected, this occurs if and only if there exists $F$ such that $f_j=\nabla_{e_j}F$; i.e. $f=\nabla F$.

Recall that $\Upsilon=\d z_1\wedge\ldots\wedge \d z_n$. We know that $N$ is special Lagrangian if and only if $N$ is Lagrangian and $\operatorname{Im}\Upsilon$ vanishes on $N$. Now $$\Upsilon(a_1+ib_1,\ldots,a_n+ib_n)=\operatorname{det}_{\C}(A+iB)$$ where $A,B$ are the matrices with columns $a_i,b_j$ respectively. Hence $$\Upsilon(e_1+i\nabla_{e_1}\nabla F,\ldots,e_n+i\nabla_{e_n}\nabla F)=\operatorname{det}_{\C}(I+i\text{Hess}\,F),$$ where $\operatorname{Hess}F=(\frac{\partial^2 F}{\partial x_i\partial x_j})$.

Therefore $N=\text{Graph}(f)$ is special Lagrangian (up to a choice of orientation) if and only if $f=\nabla F$ and $$\operatorname{Im}\operatorname{det}_{\C}(I+i\operatorname{Hess}F)=0.$$

If $n=2$, $$I+i\text{Hess}\,F=\left(\begin{array}{cc} 1+iF_{xx} & iF_{xy} \\ iF_{yx} & 1+iF_{yy}\end{array}\right).$$ Therefore, the determinant gives $$1-F_{xx}F_{yy}+F_{xy}^2+i(F_{xx}+F_{yy}),$$ then the imaginary part is $F_{xx}+F_{yy}$. Therefore, $N$ is special Lagrangian if and only if $\Delta F=0$.

As we know, a 