It seems that the correct version of (14.21) is

$$ \frac{1}{2}\,\langle d\alpha,u\wedge v\rangle=\partial_u\langle\alpha,v\rangle-\partial_v\langle\alpha,u\rangle-\langle\alpha,[u,v]\rangle $$ where $\alpha$ is a scalar valued 1-form. Similarly (14.22) should be

$$ \frac{1}{2}\,\langle dS,u\wedge v\rangle=\nabla_u\langle S,v\rangle-\nabla_v\langle S,u\rangle-\langle S,[u,v]\rangle $$ where $S$ is a tensor valued 1-form. To see this write $\alpha=\alpha_\mu dx^\mu$ and $d\alpha=\partial_\kappa\alpha_\mu dx^\kappa\wedge dx^\mu$. From

$$ dx^\kappa\wedge dx^\mu=dx^\kappa\otimes dx^\mu-dx^\mu\otimes dx^\kappa,\quad\quad u\wedge v=u\otimes v-v\otimes u,\quad\quad \langle dx^\mu,u\rangle=\langle dx^\mu,u^\nu e_\nu\rangle=u^\mu $$ it follows that

$$ \langle d\alpha,u\wedge v\rangle=2(\partial_\kappa\alpha_\mu) u^\kappa v^\mu-2(\partial_\kappa\alpha_\mu) u^\mu v^\kappa =2(\partial_u\alpha_\mu)v^\mu-2(\partial_v\alpha_\mu)u^\mu. $$ Using

$$ [u,v]=u^\mu(\partial_\mu v^\nu)e_\nu-v^\mu(\partial_\mu u^\nu)e_\nu $$ the RHS of (14.21) is

$$ \partial_u(\alpha_\mu v^\mu)-\partial_v(\alpha_\mu u^\mu)-\alpha_\nu u^\mu\partial_\mu v^\nu+\alpha_\nu v^\mu\partial_\mu u^\nu. $$ The last two terms are $-\alpha_\nu\partial_u v^\nu+\alpha_\nu\partial_v u^\nu$ so that we end up with $(\partial_u\alpha_\mu)v^\mu-(\partial_v\alpha_\mu)u^\mu$. This is half of the expression for $\langle d\alpha,u\wedge v\rangle$. $\quad\quad\Box$


It's a matter of definition.

Suppose $$ v_1 \wedge\ldots\wedge v_r \in \bigwedge^r(V), \\ v^{*1} \wedge\ldots\wedge v^{*r} \in \bigwedge^r(V^*). $$ Then the pairing between $\bigwedge^r(V)$ and $\bigwedge^r(V^*)$ is often defined by $$\tag{1} \langle v_1 \wedge \cdots \wedge v_r,\; v^{*1} \wedge \cdots \wedge v^{*r} \rangle = \det\bigl(\langle v_\alpha, v^{*\beta} \rangle\bigr). $$

If $\alpha \in V^*$, and $u,v \in V$, then $$\begin{align*} \langle \mathrm{d}\alpha,\; u \wedge v \rangle &= \alpha_{[j,i]} \langle \mathrm{d}x^i \wedge \mathrm{d}x^j,\; u \wedge v \rangle \\ &= \alpha_{[j,i]} \left| \begin{matrix} \langle \mathrm{d}x^i,\; u\rangle & \langle \mathrm{d}x^i,\; v\rangle \\ \langle \mathrm{d}x^j,\; u\rangle & \langle \mathrm{d}x^j,\; v\rangle \end{matrix} \right| \\ &= \alpha_{[j,i]} (\langle \mathrm{d}x^i,\; u\rangle \langle \mathrm{d}x^j,\; v\rangle - \langle \mathrm{d}x^i,\; v\rangle \langle \mathrm{d}x^j,\; u\rangle) \\ &= \alpha_{[j,i]} (\mathrm{d}x^i \wedge \mathrm{d}x^j)(u,v) \\ &= \mathrm{d}\alpha(u,v). \end{align*}$$ So from the famous formula $$ \mathrm{d}\alpha(u,v) = u(\alpha(v)) - v(\alpha(u)) - \alpha([u,v]) $$ we can get (14.21) of Misner, Thorne and Wheeler.

Some authors like to add an extra factor $r!$ to the right-hand side of the definition (1). This will result in an extra $\displaystyle\frac{1}{r!}$ in the left-hand side of (14.21).

  • $\begingroup$ Thanks for your comment. Of course. On p. 83 though, MTW define 2-forms and bivectors without extra factors. Their final result of Ch. 14, namely to express Riemann in terms of the Curvature-2-form , is not affected by that ``missing'' factor 1/2. That's because $\frac{1}{2}\langle d^2w,u\wedge v\rangle = (d^2w)(u,v)$ which can again be seen by expanding the 2-form $d^2w$ into wedge products and those wedge products and the bivector $u\wedge v$ into tensor products. $\endgroup$
    – Kurt G.
    Jun 20 at 8:03
  • $\begingroup$ To be precise: MTW show on p. 350 that $d^2v=e_\mu v^\nu (d{\omega^\mu}_\nu+{\omega^\mu}_\alpha\wedge{\omega^\alpha}_\nu)\,$. In the calculation of $\frac{1}{2}\langle d^2v,u\wedge v\rangle$ the $d{\omega^\mu}_\nu$ term can be dropped. One can easily see that the factor 1/2 cancels when going from $\frac{1}{2}\langle d^2v,u\wedge v\rangle$ to $(d^2v)(u,v)\,$ (2-form with the two slots filled in). $\endgroup$
    – Kurt G.
    Jun 20 at 8:10
  • $\begingroup$ Now I got it. Because of the vertical bars around $|j,i|$ in the definition of $\langle d\alpha,u\wedge v\rangle$ the summation is only over half of the index combinations than I thought. That's consistent with MTW's definition on p. 92. Your anwer is On Point ! Thanks ! $\endgroup$
    – Kurt G.
    Jun 20 at 9:01

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