I Wedge Product and r-form

  • Thread starter Silviu
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Hello! I am reading something about differential geometry and I have that for a manifold M and a point ##p \in M## we denote ##\Omega_p^r(M)## the vector space of r-forms at p. Then they say that any ##\omega \in \Omega_p^r(M)## can be expanded in terms of wedge products of one-forms at p i.e. ##\omega =\frac{1}{r!}\omega_{\mu_1 ...\mu_r}dx^\mu_1 \wedge dx^\mu_2 ... \wedge dx^{\mu_r}##, with ##\omega_{\mu_1 ...\mu_r}## completely antisymmetric. I am not sure why. If I have a 2-form, that would be ##\omega=\omega_{\mu\nu}dx^\mu dx^\nu##, but the ##\omega_{\mu\nu}## and ##\omega_{\nu\mu}## don't need to me in any relationship (equal or opposite), so how can I use the wedge product which would basically contain ##dx^\mu dx^\nu-dx^\nu dx^\mu## to obtain it? Thank you!
 

Orodruin

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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 not a 2-form, since it is not anti-symmetric and r-forms are anti-symmetric by definition.
 

martinbn

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...If I have a 2-form, that would be ##\omega=\omega_{\mu\nu}dx^\mu dx^\nu##...
It would be ##\omega=\omega_{\mu\nu}dx^\mu \wedge dx^\nu##
 

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