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