This question was asked in my mid term of smooth manifolds and I couldn't solve it in exam time. I tried it again at home and I think I need help.
Question: Let w be a 1-form on $\mathbb{R}^n$. If X and Y are smooth vector fields on $\mathbb{R}^n$, show that dw(X,Y) = X(w(Y))-Y(w(X)) - w([X,Y]).
1-form is equivalent to saying linear functional. But I am not able to make any progress due as I am not sure which result should I use.
Can you please help me with this proof by giving a few hints?
Thanks!
Let $X = \sum_i a_i \frac{\partial }{\partial x_i}$, $Y = \sum_j b_j \frac{\partial}{\partial x_j}$ so that $[X,Y] = \sum_{k=1}^n (\sum_i a_i \frac{\partial b_k}{\partial x_i} - b_i \frac{\partial a_k}{\partial x_i}) \frac{ \partial }{\partial x_k}$.
Let $w = \sum f_i dx^i$ so that $dw = \sum_{i,j} \frac{\partial f_i}{\partial x_j} dx^j \wedge dx^i$. Recall that $X$ acts on functions $f$ via $Xf(p) = \sum a_i(p) \frac{\partial f}{\partial x_i} (p)$.
We have $w(Y) = \sum_j f_j \cdot b_j$ and $w(X) = \sum_i f_i \cdot a_i$ so that
$\frac{\partial w(Y)}{\partial x_i} = \sum_{i,j} \frac{\partial f_j}{\partial x_i} b_j + f_j \cdot \frac{\partial b_j}{\partial x_i} $
and
$\frac{\partial w(X)}{\partial x_j} = \sum_{i,j} \frac{ \partial f_i}{\partial x_j} \cdot a_i + f_i \cdot \frac{\partial a_i}{\partial x_j}.$
Now the left-hand side gives
\begin{align*} dw(X,Y) = \sum_{i,j} \frac{\partial f_i}{\partial x_j} dx^j \wedge dx^i (X,Y) \\ = \sum_{i,j} \frac{\partial f_i}{\partial x_j} (p) \cdot (a_jb_i - a_ib_j)(p) \end{align*}
and the right-hand side $X(w(Y)) - Y(w(X)) - w([X,Y])$ gives
$$ \sum_{i=1}^n a_i(p) \frac{\partial w(Y)}{\partial x_i}(p) - \sum_{j=1}^n b_j (p) \frac{\partial w(X)}{\partial x_j} - \sum_{k=1}^n f_k(p) \cdot ( \sum_i a_i \frac{\partial b_k}{\partial x_i} - b_i \frac{\partial a_k}{\partial x_i})(p) = ... $$ $$...= \sum_{i,j}^n a_i \cdot ( \frac{\partial f_j}{\partial x_i} b_j + f_j \cdot \frac{\partial b_j}{\partial x_i}) - \sum_{i,j} b_j \cdot ( \frac{ \partial f_i}{\partial x_j} \cdot a_i + f_i \cdot \frac{\partial a_i}{\partial x_j}) - \sum_{k=1}^n f_k \cdot (\sum_i a_i \frac{\partial b_k}{\partial x_i} - b_i \frac{\partial a_k}{\partial x_i})$$ $$ = \sum_{i,j} \frac{\partial f_i}{\partial x_j} (p) \cdot (a_jb_i - a_ib_j)(p) $$
which is the left-hand side above.