Etydhv

Let $(M,g)$ be a Riemannian manifold with Levi-Civita connection $nabla$, and consider a one-form $alphainOmega^1(M;mathbbR)$ and a vector field $XinGamma(TM)$.

Since $alpha(X)in C^infty(M;mathbbR)$ we can consider it's differential $d(alpha(X))$. Is there a coordinate invariant way to computer this quantity? Maybe something like $d(alpha(X))(Y) = dalpha(X,Y) + alpha(nabla_YX)$? How does this generalize if we consider the differential of the smooth function $beta(X,Y)$ where $beta$ is now a 2-form and $X,Y$ are both vector fields?

Since $rm d(alpha(X))(Y) = Y(alpha(X))$, we indeed get from $$rm dalpha(X,Y) = X(alpha(Y)) - Y(alpha(X)) - alpha([X,Y])$$that $$rm d(alpha(X))(Y) = X(alpha(Y)) - alpha([X,Y]) - rm dalpha(X,Y),$$but this is not very satisfying as we still have $X(alpha(Y))$ on the right side. Since $nabla$ is torsion free, we may improve the right side to obtain $$rm d(alpha(X))(Y) = (nabla_Xalpha)(Y) + alpha(nabla_YX) - rm dalpha(X,Y),$$where $(nabla_Xalpha)(Y) = X(alpha(Y)) - alpha(nabla_XY)$ is the covariant derivative of $alpha$ in the direction of $X$. So if $iota_X$ denotes interior product, we can write $$ rm d(alpha(X)) = nabla_Xalpha + alpha circ nabla X - iota_X(rm dalpha), $$which is as coordinate-free as we can get. I'm not sure how this would actually be useful in practice, though. One can do the same and get more complicated formulas for $k$-forms with $k>1$ by using the same strategy, by using that if $omega$ is a $k$-form, then $nabla_Xomega$ is a $k$-form with $$(nabla_Xomega)(X_1,ldots, X_k) = X(omega(X_1,ldots,X_k)) - sum_i=1^k omega(X_1,ldots, nabla_XX_i,ldots, X_k).$$You will also have to use $$beginalign rm domega(X_0,ldots,X_k) &= sum_i=0^k (-1)^k X_i(omega(X_0,ldots, widehatX_i,ldots, X_k)) \ &qquad + sum_0leq i<jleq k (-1)^i+j omega([X_i,X_j],X_1,ldots, widehatX_i,ldots, widehatX_j,ldots, X_k). endalign$$

You can use Cartan's magical formula
$$
mathcalL_X omega = i_X domega + d(i_X omega),
$$
where $omega$ is an $n$-form, $mathcalL$ is the Lie derivative and $i$ stands for the interior product. Note that all these operations and the exterior derivative make sense on a differentiable manifold (not necessarily a Riemmannian manifold) so one does not need the Levi-Civita connection in order to get a coordinate free expression. However, I believe there are formulas that relate the Lie derivative with the Levi-Civita connection, so you can use these if you want expressions using the Levi-Civita connection.

For the $1$-form $alpha$ this gives:
$$
d(alpha(X)) = d(i_X alpha) = mathcalL_X alpha - i_X dalpha.
$$

For the $2$-form $beta$ one can repeatedly apply Cartan's formula:
$$
beginalign*
d(beta(X,Y)) &= d(i_Y i_X beta) \
&= mathcalL_Y(i_X beta) - i_Y d(i_X beta)\
&= mathcalL_Y(i_X beta) - i_Y mathcalL_X beta + i_Y i_X dbeta.
endalign*
$$