How Can Wedge Products Be Used in Differential Geometry?

[Silviu]

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]

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]

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

It would be $\omega=\omega_{\mu\nu}dx^\mu \wedge dx^\nu$