Misner-Thorne-Wheeler, p.92, Box 4.1, typo?

[gheremond]

In Misner-Thorne-Wheeler Gravitation, Chapter 4, Page 92, Box 4.1, at section 4, there is a formula for the contraction of a p-form and a p-vector. Now, it states that the contraction of a p-form basis with a p-vector basis gives the antisymmetrizer symbol, \left\langle {\omega ^{i_1 } \wedge \ldots \wedge \omega ^{i_p } ,e_{j_1 } \wedge \ldots \wedge e_{j_p } } \right\rangle = \delta ^{i_1 \ldots i_p } _{j_1 \ldots j_p } and there is a reference to exercises 3.13 and 4.12. I tried this part many many times and I always find the result to be p! times the antisymmetrizer. I also compared it for the case p=2 using the definition of the symbol from exercise 3.13, still the same result, I get an overall 2. Can anybody please explain what am I doing wrong here?

 

[javierR]

No typo. The symbol \delta^{ij}_{kl}\equiv \delta^{[i}_{k}\delta^{j]}_{l}\equiv \frac{1}{2!}\left(\delta^{i}_{k}\delta^{j}_{l}-\delta^{j}_{k}\delta^{i}_{l}\right) (which generalizes to n indices with a 1/n! factor), and basis 1-forms act on basis vectors as \omega^{i}(e_{j})=\delta^{i}_{j}.

 

[gheremond]

Thanks for the reply. However, if you look on page 88, you will see that the definition for \delta ^{\alpha \beta } _{\mu \nu } = \delta ^\alpha _\mu \delta ^\beta _\nu - \delta ^\alpha _\nu \delta ^\beta _\mu according to MTW does not carry the 1/2! factor that you mention. Furthermore, if you expand the wedge products into tensor products within the contraction symbol, you get \left\langle {\omega ^\alpha \wedge \omega ^\beta ,e_\mu \wedge e_\nu } \right\rangle = \left\langle {\omega ^\alpha \otimes \omega ^\beta - \omega ^\beta \otimes \omega ^\alpha ,e_\mu \otimes e_\nu - e_\nu \otimes e_\mu } \right\rangle

= \left\langle {\omega ^\alpha \otimes \omega ^\beta ,e_\mu \otimes e_\nu } \right\rangle - \left\langle {\omega ^\alpha \otimes \omega ^\beta ,e_\nu \otimes e_\mu } \right\rangle - \left\langle {\omega ^\beta \otimes \omega ^\alpha ,e_\mu \otimes e_\nu } \right\rangle + \left\langle {\omega ^\beta \otimes \omega ^\alpha ,e_\nu \otimes e_\mu } \right\rangle

= \delta ^\alpha _\mu \delta ^\beta _\nu - \delta ^\alpha _\nu \delta ^\beta _\mu - \delta ^\beta _\mu \delta ^\alpha _\nu + \delta ^\beta _\nu \delta ^\alpha _\mu = 2\left( {\delta ^\alpha _\mu \delta ^\beta _\nu - \delta ^\alpha _\nu \delta ^\beta _\mu } \right) = 2 \delta ^{\alpha \beta } _{\mu \nu }

again, using all the conventions of the book up to this point.