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At a point $p$, write $W=T_{p}L$ and take the orthogonal complement with respect to the Riemannian metric $g$. The compatible triple identity givesIf $L$ is a Lagrangian submanifold, then $\omega (u,v)=0$ for $u,v\in W$, so $JW\subset W^{\perp }$. Both have dimension $\frac{1}{2}\dim M$, hence equality.

Conversely, if $JW=W^{\perp }$, equality of dimensions gives $\dim W=\frac{1}{2}\dim M$, and $g(Ju,v)=0$ for $u,v\in W$ gives $\omega {|}_W=0$. ThereforeA positive-definite inner product has $W\cap W^{\perp }=\{0\}$. Hence $W\cap JW=\{0\}$, proving that every Lagrangian submanifold is a totally real submanifold for a compatible almost complex structure.

Use ${\mathbb{C}}^2$ with the standard symplectic form ${\omega }_0=d{x}_1\wedge d{y}_1+d{x}_2\wedge d{y}_2$ and the compatible almost complex structure given by multiplication by $i$. Consider the embedded submanifoldIn coordinates $(x_1,y_1,x_2,y_2)$, its tangent space is $W=\{(a,b,b,0):a,b\in \mathbb{R}\}$. Applying $J$ gives $J(a,b,b,0)=(-b,a,0,b)$. If this is again in $W$, its fourth coordinate forces $b=0$, and its second and third coordinates then force $a=0$. Thus $W\cap JW=0$ and ${\dim }_{\mathbb{R}}L=2$: $L$ is a totally real submanifold.

However, the pullback of a differential form under the parametrization isThus this totally real submanifold is not a Lagrangian submanifold.