Why does $dx\wedge dy$ give the area of the projection of a parallelogram onto the $xy$ plane?

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This article on differential forms says that

"$dx \wedge dy$, $dz \wedge dx$, and $dy \wedge dz$ are just $2$-forms that eat parallelograms and spit out the areas of their projections onto the $xy$, $zx$, and $yz$ planes respectively."

Why is that the case? As an example, why does $dx\wedge dy$ give the area of the projection of a parallelogram onto the $xy$ plane?

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Adding on User$203940$'s comment, we have

$$dx\wedge dy \ (v^1,v^2) := \sum_{\sigma\in\mathbb{S}_2} \text{sgn}(\sigma) dx\otimes dy \ (v^{\sigma(1)},v^{\sigma(2)}) = \sum_{\sigma\in\mathbb{S}_2} \text{sgn}(\sigma) \ v_x^{\sigma(1)}v_y^{\sigma(2)}$$

which is Leibniz' formula for determinant of the matrix $$\begin{pmatrix} v_x^1 & v_x^2\\ v_y^1 & v_y^2\\ \end{pmatrix}$$ i.e. the (signed) area of the projection of the paralellogram spanned by $v^1,v^2$ onto the $xy$ plane.