Exterior differentiation of one form on a smooth manifold Prove : $d$$\omega$$(V,W)$=$V \omega (W) - W \omega(V) -\omega([V,W])$
where $\omega$ is a one-form and V,W are vector fields on a smooth manifold M.
I'm fine with the right side of the equation, but I don't know what $d$$\omega$$(V,W)$ means or looks like here.
Can you please explain?
Thanks.
 A: Let $w=fdg$ where $f, g$ are $0$-forms
Evaluate the first term
$$dw(V, W)=df\wedge dg(V, W)=df(V)dg(W)-df(W)dg(V)=V(f)W(g)-V(g)W(f)$$
the second term
$$Vω(W)−Wω(V)−ω([V,W])$$
$$=V(fdg(W))-W(fdg(V))-fdg([V,W])$$
$$=V(fW(g))-W(fV(g))-f(VW(g)-WV(g))$$
$$=V(f)W(g)-V(g)W(f)$$
are the same.
A: $df(X):=X(f)$. And $$ \alpha:=gdf_1\wedge \cdots \wedge df_k
\Rightarrow d(\alpha):= dg\wedge df_1\wedge \cdots \wedge df_k
$$
Hence if $E_i$ is a local coordinate vector field and $E_i^\ast $
  is a dual, then $$  df(E_i)=E_i(f) \Rightarrow df=E_i(f)E_i^\ast
  $$
Hence $$ d(df)=0 \Rightarrow d(E_i(f))\wedge E_i^\ast + E_i(f)
  dE_i^\ast =0 $$
Here $$ [E_m,E_i]=0 \Rightarrow d(E_i(f))\wedge E_i^\ast =
E_m(E_i(f)) E_m^\ast \wedge E_i^\ast =0 $$
That is $$ dE_i^\ast =0 $$
Returning to OP : If $\omega=\omega_iE_i^\ast,\ V=V_i E_i,\ W=W_kE_k
$ then
$$ d\omega (V,W)= E_m(\omega_i) E_m^\ast \wedge E_i^\ast  (V,W)=
E_m(\omega_i) [V_mW_i-W_mV_i]
$$
$$ V(\omega (W))=V(\omega_iW_i)=V_mE_m(\omega_i)W_i +
\omega_i V(W_i) $$
Now a proof is followed from simple calculation.
