Are these two definitions of exterior derivative equivalent? I saw two definition of the exterior derivative of a $k$-form $\omega$.
First definition: $$(d\omega)_p(v_0,\ldots,v_k)= \sum_{i=0}^k(-1)^id(\omega(v_0,\ldots,\hat{v_i},\ldots,v_k))_p(v_i)$$
Second definition: $$(d\omega)_p(v_0,\ldots,v_k)=\sum_{i=0}^k(-1)^iv_i(p)(\omega\circ (v_0,\ldots,\hat{v_i},\ldots,v_k))+  \sum_{i<j}(-1)^{i+j}\omega_p\circ([v_i,v_j],v_0,\ldots,\hat{v_i},.\hat{v_j}.,v_k)$$
Are they same? How to prove that they are same - can I get a proof of these definitions being equivalent or not equivalent?
 A: These definitions are the same. In fact, there is a third, more common definition that they are equivalent to. Namely, if we take a local coordinate system $(x^{1}, \cdots, x^{n})$, we get $dx^1 , \cdots , dx^n$ as a basis for the cotangent space of a point in the chart. If we take a multi-index $I = (i_{1}, \cdots, i_{k})$, then we can consider a $k$-form of the form
$$\omega = f_{I}dx^{I}$$
and define
$$d\omega = \sum_{i=1}^{n}\frac{\partial f_{I}}{x^{i}}dx^{i} \wedge dx^{I} \,\,\,\,\,\, (\dagger)$$
This definition extends to any $k$-form by linearity. It isn't too hard to see that this is equivalent to your first definition, since the derivative of $\omega(v_{0}, \cdots, \hat{v_{i}}, \cdots v_{k})$ corresponds to
$$\frac{\partial f_{I}}{x^{i}}\wedge dx^{I}$$
And then the evaluation at $v_i$ is like $\wedge dx^{i}$, where you need to include a power of $(-1)^i$ to compensate for how you are shuffling around the basis. Note that here you need to take the derivative of a real-valued function defined on the manifold, so you need to choose a coordinate system. The second definition has the advantage of being defined without a coordinate system. The proof that it is equivalent to definition $(\dagger)$ is a little messy.
First you show that the second definition enjoys "function linearity", so that if you multiply one of the vector fields by a $C^\infty$ function $f$, it pulls out of the sum. This is a little messy, but the details can be sorted through with enough determination. Note that definition $(\dagger)$ is clearly function linear, since we are not taking "derivatives" of vector fields, such as when taking the Lie bracket. Also, both definitions are linear. Thus it suffices to show equality on coordinate vector fields, and again you get some slightly messy casework. The details are done in Spivak's text "A comprehensive introduction to Differential Geometry, Vol. I", on pages 213-214.
