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Mathematics
Differential Geometry
How Can Wedge Products Be Used in Differential Geometry?
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[QUOTE="Orodruin, post: 5838268, member: 510075"] An r-form is defined as a completely anti-symmetric type (0,r) tensor so indeed ##\omega_{\mu_1 \ldots \mu_r}## will be the components of that anti-symmetric tensor. Now, since the wedge product basis is anti-symmetric, adding something symmetric to the components will actually not matter at all (its contribution to the sum would be zero), but seen as just a type (0,r) tensor, its component would be anti-symmetric. Note that ##\omega = \omega_{\mu\nu} dx^\mu \otimes dx^\nu## is [I]not[/I] a 2-form, since it is not anti-symmetric and r-forms are anti-symmetric by definition. [/QUOTE]
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Differential Geometry
How Can Wedge Products Be Used in Differential Geometry?
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