Let $\theta$ a $1-$form. Why $d\theta(X,Y)=X\theta(y)-Y\theta(X)-\theta([X,Y])$? Let $\theta$ a $1-$form. Why $$\mathrm d\theta(X,Y)=X\theta(Y)-Y\theta(X)-\theta([X,Y])\ \ ?$$
I know that $\theta=\sum_{i=1}^n a\mathrm d x^i$ where $a\in \mathcal C^\infty (U)$ and $U$ an open. Then,
$$\mathrm d \theta=\sum_{i=1}^n \mathrm d a\wedge \mathrm d x^i.$$
Then,
$$\mathrm d \theta(X,Y)=\sum_{i=1}^n \mathrm d a\wedge \mathrm d x^i(X,Y)=\sum_{i=1}^n (\mathrm d a(X)\mathrm d x^i(Y)-\mathrm d a(Y)\mathrm d x^i(X))$$
$$=X(a)\sum_{i=1}^n\mathrm d x^i(Y)-Y(a)\sum_{i=1}^n\mathrm d x^i(X),$$
and I can't find the formula. Any idea ?
 A: Firstly observe that the equation is linear in $\theta$, so it suffices to consider the case $\theta = adx^k$ with some $a \in C^\infty(U)$. Thus,
$$d\theta = \sum_{i=1}^{n}\frac{\delta a}{\delta x^i}dx^i\wedge dx^k.$$
Let us write the vector fields in terms of coordinates, i.e.
$$X = \xi^1\frac{1}{\delta x^1} + \dots + \xi^n\frac{1}{\delta x^n},\ \ \ Y = \zeta^1\frac{1}{\delta x^1} + \dots + \zeta^n\frac{1}{\delta x^n},$$
with smooth functions $\xi^1,\dots,\xi^n,\zeta^1,\dots,\zeta^n$ on $U$.
We compute:
\begin{equation}\begin{split}
d\theta\cdot(X \wedge Y) & = (\sum_{i=1}^{n}\frac{\delta a}{\delta x^i}dx^i\wedge dx^k)(X \wedge Y)\\
& = \sum_{i=1}^{n}((\frac{\delta a}{\delta x^i}dx^i\wedge dx^k) (X \wedge Y))\\
& = \sum_{i=1}^{n}\frac{\delta a}{\delta x^i}(X(x^i))(Y(x^k))-\frac{\delta a}{\delta x^i}(Y(x^i))(X(x^k))\\
& =  \sum_{i=1}^{n}\frac{\delta a}{\delta x^i} (\xi^i\zeta^k - \xi^k\zeta^i),\\
X(\theta\cdot Y) & = X(adx^k(Y))\\
& = X(aY(x^k))\\
& = X(a\zeta^k)\\
& = \xi^1\frac{\delta(a\zeta^k)}{\delta x^1} + \dots + \xi^n\frac{\delta(a\zeta^k)}{\delta x^n}\\
& = \xi^1a\frac{\delta\zeta^k}{\delta x^1} + \xi^1\zeta^k\frac{\delta a}{\delta x^1} + \dots + \xi^na\frac{\delta\zeta^k}{\delta x^n} + \xi^n\zeta^k\frac{\delta a}{\delta x^n}\\
& = \sum_{i=1}^{n}(\xi^ia\frac{\delta\zeta^k}{\delta x^i} + \zeta^i\xi^k\frac{\delta a}{\delta x^i}),\\
Y(\theta\cdot X) & = \sum_{i=1}^{n}(\xi^ia\frac{\delta\zeta^k}{\delta x^i} + \zeta^i\xi^k\frac{\delta a}{\delta x^i}).
\end{split}\end{equation}
Next, you can show that the Lie bracket of $\eta_1\frac{\delta}{\delta x^i}$ and $\eta_2\frac{\delta}{\delta x^j}$ for any smooth $\eta_1,\eta_2$ is $\eta_1\frac{\delta\eta_2}{\delta x^i}\frac{\delta}{\delta x^j} - \eta_2\frac{\delta\eta_1}{\delta x^j}\frac{\delta}{\delta x^i}$, moreover the Lie bracket as a function is linear in both entries, which gives us the Lie bracket of $X$ and $Y$. It follows that
\begin{equation}\begin{split}
\theta\cdot[X,Y] & = \theta\cdot(\sum_{i,j=1}^n \xi^i\frac{\delta\zeta^j}{\delta x^i}\frac{j}{\delta x^j} - \zeta^j \frac{\delta\xi^i}{\delta x^j}\frac{\delta}{\delta x^i})\\
& = \theta\cdot(\sum_{i=1}^n (\sum_{j=1}^n \xi^j \frac{\delta\zeta^i}{\delta x^j} - \zeta^j \frac{\delta\xi^i}{\delta x^j})\frac{\delta}{\delta x^i})\\
& = adx^k (\sum_{i=1}^n (\sum_{j=1}^n \xi \frac{\delta\zeta^i}{\delta x^j} - \zeta^j \frac{\delta\xi^i}{\delta x^j})\frac{\delta}{\delta x^i})\\
& = a (\sum_{j=1}^n \xi^j \frac{\delta\zeta^k}{\delta x^j} - \zeta^j\frac{\delta\xi^k}{\delta x^j})\\
& = \sum_{j=1}^n (a\xi^j\frac{\delta\zeta^k}{\delta x^j} - a\zeta^j\frac{\delta\xi^k}{\delta x^j}).
\end{split}\end{equation}
Now putting all terms together yields
\begin{equation}\begin{split}
X(\theta\cdot Y) - Y(\theta\cdot X) - \theta\cdot[X,Y] & = (\sum_{i=1}^n \xi^ia\frac{\delta\zeta^k}{\delta x^i} + \xi^i\zeta^k\frac{\delta a}{\delta x^i}) - (\sum_{i=1}^n \zeta^ia\frac{\delta\xi^k}{\delta x^i} + \zeta^i\xi^k\frac{\delta a}{\delta x^i})\\
& - (\sum_{i=1}^n a\xi^i\frac{\delta\zeta^k}{\delta x^i} - a\zeta^i\frac{\delta\xi^k}{\delta x^i})\\
& = \sum_{i=1}^n (\xi^ia\frac{\delta\zeta^k}{\delta x^i} + \xi^i\zeta^k\frac{\delta a}{\delta x^i} - \zeta^ia\frac{\delta\xi^k}{\delta x^i} - \zeta^i\xi^k\frac{\delta a}{\delta x^i} - a\xi^i\frac{\delta\zeta^k}{\delta x^i} + a\zeta^i\frac{\delta\xi^k}{\delta x^i})\\
& = \sum_{i=1}^n (\xi^i\zeta^k\frac{\delta a}{\delta x^i} - \zeta^i\xi^k\frac{\delta a}{\delta x^i})\\
& = \sum_{i=1}^n (\xi^i\zeta^k - \xi^k\zeta^i)\frac{\delta a}{\delta x^i}\\
& = d\theta\cdot(X \wedge Y).
\end{split}\end{equation}
As already mentioned, the result extends linearly to all 1-forms $\theta$.
A: Write $X=\sum_iX_i{\partial\over{\partial x_i}}$
$\theta(X)=\sum_i a_iX_i$, $Y.\theta(X)=\sum_ida_i(Y)X_i+a_idX_i(Y)$.
This implies that $X.\theta(Y)-Y.\theta(X)-\theta([X,Y])=\sum_ida_i(X)Y_i+a_idY_i(X)-da_i(Y)X_i-a_idX_i(Y)-\sum_ia_i(dY_i(X)-dX_i(Y))$
$=\sum_ida_i(X)Y_i-da_i(Y)X_i.$
