What Defines the Components of a 2-form?

[center o bass]

Hi I'm a bit confused about what actually define the components of a form. I just saw an argument where one found that

$$\underline{d \omega}^\rho (\vec e_\mu \wedge e_\nu) = - c_{\mu \nu}^\rho$$

and then the author wrote that this implied that

$$\underline{d \omega}^\rho = - \frac{1}2 c_{\mu \nu}^\rho \underline{\omega}^\mu \wedge \underline{\omega}^\nu$$

so if the components of a p-form is defined as

$$\underline{\alpha} = \frac{1}{p!} \alpha_{\mu_1 \ldots \mu_p} \underline{\omega}^{\mu_1} \wedge \ldots \wedge \underline{\omega}^{\mu_p}$$

where $\alpha_{\mu_1 \ldots \mu_p}$ are the components, it seems the argument above implies that one can find these by applying the p-form to the basis p-vectors.

However I tried this with a two form $\underline{\alpha} = 1/2 \alpha_{\mu \nu} \underline{\omega}^\mu \wedge \underline{\omega}^\nu$ using the definition of the wedge product

$$\underline{\alpha}(\vec{e}_\alpha \wedge \vec{e}_\beta) = \frac{1}{2} \alpha_{\mu \nu} \underline{\omega}^\mu \wedge \underline{\omega}^\nu (\vec{e}_\alpha \wedge \vec{e}_\beta) \\<br /> = \frac{1}{2} \alpha_{\mu \nu} 4 \underline \omega^{[\mu} \underline \omega^{\mu ]}( \vec e_{[\alpha} \vec e_{\beta]}) \\<br /> = 2\alpha_{\mu \nu} \delta^{[\mu}_{[\alpha} \delta^{\nu ]}_{\beta]} \\<br /> = 2 \alpha_{\alpha \beta}$$

Where I have used that $\underline \omega ^\mu \underline \omega^\nu = 2! \underline \omega^\mu \underline \omega^\nu$. But should I not get $\alpha_{\alpha \beta}$ here? Is the caculation wrong or is my assumption of what defines the components wrong?

 

[fzero]

Hi I'm a bit confused about what actually define the components of a form. I just saw an argument where one found that

$$\underline{d \omega}^\rho (\vec e_\mu \wedge e_\nu) = - c_{\mu \nu}^\rho$$

Your notation is a bit confusing, which might be the reason for the extra factor of 2 that you get later. The $e_\mu$ are a basis for vector fields, so we do not take a wedge product of them. It's typical to write that expression as

$$\underline{d \omega}^\rho ( e_\mu , e_\nu) = - c_{\mu \nu}^\rho,$$

where it's understood that

$$(\alpha \wedge \beta) (v,w) = \alpha(v) \beta(w) - \beta(w) \alpha(v)$$

pointwise. The pair of terms here is of course responsible for the factor of 2. However it's clear that in your expression

$$\underline{\alpha}(\vec{e}_\alpha \wedge \vec{e}_\beta) = \frac{1}{2} \alpha_{\mu \nu} \underline{\omega}^\mu \wedge \underline{\omega}^\nu (\vec{e}_\alpha \wedge \vec{e}_\beta) \\<br /> = \frac{1}{2} \alpha_{\mu \nu} 4 \underline \omega^{[\mu} \underline \omega^{\mu ]}( \vec e_{[\alpha} \vec e_{\beta]}) \\<br /> = 2\alpha_{\mu \nu} \delta^{[\mu}_{[\alpha} \delta^{\nu ]}_{\beta]} \\<br /> = 2 \alpha_{\alpha \beta}$$

you're introducing an extra factor of 2 from the erroneous manipulation of the vector fields.

 

[center o bass]

Your notation is a bit confusing, which might be the reason for the extra factor of 2 that you get later. The $e_\mu$ are a basis for vector fields, so we do not take a wedge product of them. It's typical to write that expression as

$$\underline{d \omega}^\rho ( e_\mu , e_\nu) = - c_{\mu \nu}^\rho,$$

where it's understood that

$$(\alpha \wedge \beta) (v,w) = \alpha(v) \beta(w) - \beta(w) \alpha(v)$$

pointwise. The pair of terms here is of course responsible for the factor of 2. However it's clear that in your expression



you're introducing an extra factor of 2 from the erroneous manipulation of the vector fields.

Check out page 59 in

http://www.google.no/url?sa=t&rct=j...uSWMaT_Vs9fO_AucIQGIMOA&bvm=bv.43287494,d.Yms

Here 'p-vectors' are introduced where wedge products of the vector basis is taken.
The argument I referred to is at page 131. Check out the equations (6.175) and (6.176).

The that argument then fallacious?