# Curvature 2-forms: missing factor 1/2 in (14.21) and (14.22) of Misner, Thorne and Wheeler?

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 $$\langle d\alpha,u\wedge v\rangle$$. $$\quad\quad\Box$$

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).
• 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. Jun 20 at 8:03
• 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). Jun 20 at 8:10
• 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 ! Jun 20 at 9:01