For any one-form (a linear form on the tangent space of each point) we have its exterior derivative $d\omega$ which is a two-form defined by $d\omega(X,Y)=D_X(\omega(Y))-D_Y(\omega(X))-\omega([X,Y])$
I'm trying to show that the value of $dw(X,Y)$ at a point $p$ is depndent only on the values of the vetcor fields $X,Y$ at that point.
I've just started diving into this subject, and I really don't have a good direction. I tried to think of all the tools I have so I thought of decomposing $X,Y$ to some orthonoraml base, or maybe even decompose $\omega$ by $\sum_i\omega(X_i)\omega^i$ where $\omega^i$ is the linear form associated with $X_i$ i.e for any $i,j$: $\omega^i(X_j)=\delta_{ij}$
I will be very glad for some help!
In addition, any general insight on this new definition of exterior derivative can be helpful. For now it seems quite peculiar.
Thanks!
First, we should define tensors. There are various types of tensors, and we focus on those which are relevant to your question. A tensor of type $(k,0)$ is a map$$T:\Gamma(TM)\times\ldots\times\Gamma(TM)\to C^\infty(M),$$which is $C^\infty(M)$-linear in every coordinate. In other words, $T$ eats $k$ vector fields and spits out a function. The linearity means that $T$ is additive in each coordinate, and for vector fields $X_1,\ldots,X_k$ and a function $f\in C^\infty(M)$ we have$$T(fX_1,X_2,\ldots,X_k)=\ldots=T(X_1,\ldots,X_{k-1},fX_k)=fT(X_1,\ldots,X_k).$$
The following is a fundamental observation about tensors:
Proposition: Let $T$ be a $(k,0)$ tensor. Then the value of $T(X_1,\ldots,X_k)$ at a point $p\in M$ depends only on the values of $X_1,\ldots,X_k$ at $p$. In other words, if $Y_1,\ldots,Y_k$ are also vector fields and we have $X_i(p)=Y_i(p),\quad i=1,\ldots k,$ then $T(X_1,\ldots,X_k)(p)=T(Y_1,\ldots,Y_k)(p)$.
Proofs of the proposition can be found in many textbooks. Specifically, I guess it is proved in doCarmo's Riemannian Geometry and in Lee's Smooth Manifolds.
To solve your question, you now need to show that $d\omega$ is a tensor. Namely, you should show $C^\infty(M)$-linearity. This is not so hard. You need Leibniz rule, and you need to know the expression for $[fX,Y]$, where $X$ and $Y$ are vector fields and $f$ is a function.