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The projection of the element of two-dimensional surfaces in three dimensions on coordinate planes $x_\alpha x_\beta$ is an antisymmetric tensor of rank $2$

$$df_{\alpha\beta}=dx_{\alpha} dx_{\beta}'-dx_{\beta} dx'_{\alpha}$$

It is said in a textbook that as an element of integration over the two-dimensional surfaces in three dimensions is taken the dual of $df_{\alpha\beta}$

$$\ast df_{\alpha\beta}=\frac{1}{2}\varepsilon_{\alpha}\small^{\beta\gamma}df_{\beta\gamma}$$

How to understand intuitively the reason for switching from the projection to its dual?

Tursinbay
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  • This is the same identification we have with planes and normal vectors. – Ninad Munshi Jan 03 '20 at 19:56
  • Can you explain broadly? – Tursinbay Jan 04 '20 at 03:53
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    A plane requires a point and a normal vector. But let's say we were integrating over this plane. A translation in space does not affect it's differential element (differential form), so the only identifier it has left is its orientation. An area element also needs a size associated with it - in other words a magnitude and orientation, which can be encoded by another differential form. – Ninad Munshi Jan 04 '20 at 21:47

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