Commutator of $J$-Tensor and Exterior Derivative Given the $J$-tensor on complex manifolds, I am trying to prove the following commutator relations for Kähler manifolds:
$$
[J,d] \;\; = \;\; d^c \hspace{3pc} [J,d^c] = -d
$$
where $d$ is the ordinary exterior derivative, and $d^c$ is the twisted exterior derivative written in local coordinates as
$$
d^c\omega \;\; =\;\; \sum_{i=1}^{2n} Je_i \wedge \nabla_{e_i}\omega.
$$
In the text I'm using the $J$ tensor can be extended to acting on $k$-forms by the following definition:
$$
J\omega \;\; =\;\; \sum_{i=1}^{2n} Je_i \wedge\left (e_i \lrcorner\;  \omega \right )
$$
As a first attempt I computed the following:
\begin{eqnarray*}
[J,d]\omega & = & J(d\omega) - d(J\omega) \\
& = & \sum_{i=1}^{2n}Je_i \wedge (e_i \lrcorner \; d\omega) - d\left (\sum_{i=1}^{2n} Je_i \wedge (e_i \lrcorner \; \omega) \right ) \\
& = & \sum_{i=1}^{2n}Je_i \wedge (e_i \lrcorner \; d\omega) - \sum_{i=1}^{2n} d(Je_i) \wedge (e_i \lrcorner \; \omega) + \sum_{i=1}^{2n} Je_i \wedge d(e_i \lrcorner \; \omega) \\
& = & \sum_{i=1}^{2n} Je_i \wedge \mathscr{L}_{e_i}\omega - \sum_{i=1}^{2n} d(Je_i) \wedge (e_i \lrcorner \; \omega)
\end{eqnarray*}
where $\mathscr{L}_{e_i}\omega$ is the Lie derivative of $\omega$ in the $e_i$-direction, which was derived from Cartan's magic formula.  I'm having quite a bit of trouble from this point onward.  For instance, how do we write this in the form above for the twisted exterior derivative?  How do I relate the Lie derivative to the Levi-Civita connection?  I also find the expression $d(Je_i)$ rather dubious.  Should I interpret $e_i$ as being a smooth section into the tangent bundle?  
 A: I do not see how the commutator relations could hold. $J$ acts on $(p, q)$- forms by multiplication by $i^{p-q} $. One can check that $[J, d](\alpha) = i^{p-q} ((i-1)\partial\alpha -(i+1)\bar\partial \alpha) $, which is not equal to $d^c\alpha=-i(\partial-\bar\partial) \alpha$. 
Similarly, for $(p, q)$ form $\alpha$, one compute $[J, d^c](\alpha) =i^{p-q} ((1+i)\partial\alpha+(1-i)\bar\partial\alpha)$. Again, which is not equal to $-d\alpha$.
What is known to be true is $J^{-1}dJ=d^c$. This can be seen by checking on $(p, q)$ forms:
 $$\begin{align}Jd^c\alpha &=J(-i\partial\alpha+i\bar\partial\alpha)\\& =-i^{p+2-q}\partial\alpha + i^{p-q} \bar\partial\alpha\\&=i^{p-q}d\alpha=dJ\alpha\end{align}$$

Edit: The following is to give an explanation on the equivalent definitions of $d^c$ etc.
In terms of complex local coordinates $z_1,...,z_n$, the real coordinates are $x_j+iy_j=z_j$. Then locally the (complexified) tangent space has complex basis $\{\frac{\partial}{\partial x_j}, \frac{\partial}{\partial y_j}\} $, and $J$ acts on these vectors by $J\left(\frac{\partial}{\partial x_j}\right) =\frac{\partial}{\partial y_j} $ and $J\left(\frac{\partial}{\partial y_j}\right) =-\frac{\partial}{\partial x_j} $.
We can define a new basis for the complexified tangent space by taking $\frac{\partial}{\partial z_j} :=\frac12\left(\frac{\partial}{\partial x_j} - i\frac{\partial}{\partial y_j} \right)$ and $\frac{\partial}{\partial \bar z_j} =\frac12\left(\frac{\partial}{\partial x_j} +i\frac{\partial}{\partial y_j} \right) $. Then one can verify that $\frac{\partial}{\partial z_j}$ and $\frac{\partial}{\partial \bar z_j}$ are eigenvectors of $J$ with eigenvalues $i$ and $-i$ respectively. With respect to this new basis, the dual basis on the cotangent space is given by $dz_j= dx_j+idy_j$ and $d\bar z_j=dx_j-idy_j$.
A $(p, q)$-form is then defined to be any complex differential form so that when expressed in local coordinates, it is a sum of wedge products consisting of $p$ terms of type $dz_j$ and $q$ terms type $d\bar z_j$ (it suffices to check on an open cover of charts, why?). We also see that any $k$ forms can be written as a sum of $(p, q) $ where $p+q=k$. So to specify the operators $d, \partial, \bar\partial$, etc, we can check their actions on $(p, q) $ forms and extend by linearity. 
Hence if we take exterior derivative of the local $(p, q)$ form $\alpha = fdz_{i_1}\wedge... \wedge dz_{i_p} \wedge d\bar z_{j_1}\wedge...\wedge d\bar z_{j_q} $, we get $$d\alpha =\sum_{k=1}^n\left( \frac{\partial f}{\partial z_k} dz_k \wedge...\right) +\sum_{k=1}^n\left( \frac{\partial f}{\partial \bar z_k} d\bar z_k\wedge... \right) :=\partial\alpha+\bar\partial\alpha $$. Here one defines $\partial\alpha$ and $\bar\partial\alpha$ to be the $(p+1,q)$ and $(p, q+1)$ parts of $d\alpha$ respectively. Again, one can verify that this definition is independent of the local coordinates chosen.
As described in the notes you mentioned in the comment, if we write $e_k =\frac{\partial}{\partial x_k} $ and $e_{n+k}=\frac{\partial}{\partial y_k} $ for $k\leq n$. We have $Je^k=e^{n+k}$ and $Je^{n+k} =-e^k$ for $k\leq n$. Then we will verify that $\partial$ and $\bar\partial$ are given by $$\partial\alpha=\sum_{k=1}^n \frac12 dz_k\wedge \nabla_{e_k} \alpha-i\sum_{k=1}^n \frac12 dz_k\wedge \nabla_{e_{n+k}} \alpha$$ and $$\bar\partial\alpha=\sum_{k=1}^{n}\frac12 d\bar z_k\wedge \nabla_{e_k} \alpha+i\sum_{k=1}^n \frac12 d\bar z_k \wedge \nabla_{e_{n+k}}\alpha. $$ This follows from noticing $dz_k(e_k) =1$, $dz_k(e_{n+k}) =i$, $d\bar z_k(e_k)=1$ and $d\bar z_k(e_{n+k}) =-i$. Using the above formula for $d\alpha$, one has $$\begin{align} & \frac12 dz_k\wedge\nabla_{e_k} \alpha - \frac i2 dz_k\wedge\nabla_{e_{n+k}}\alpha\\& =  \frac12\left(\frac{\partial f} {\partial z_k}dz_k \wedge... +\frac{\partial f}{\partial\bar z_k} dz_k\wedge...\right) +\frac12\left((-i)i \frac{\partial f} {\partial z_k} dz_k+(-i)(-i) \frac{\partial f} {\partial\bar z_k}\right)\\& =\frac{\partial f} {\partial z_k} \wedge dz_k\wedge... \end{align} $$
A similar computation for $\bar\partial$ verifies the claim. So finally, we have $$i(\partial\bar\alpha - \partial\alpha) =i\left(\sum_{k=1}^n - ie^{n+k} \wedge\nabla_{e_k}\alpha+i\sum_{k=1}e^k\wedge\nabla_{e_{n+k}}\alpha\right)=\sum_{k=1}^{2n}Je^k\wedge\nabla_{e_k}\alpha= d^c\alpha. $$
We are finally done! 
