Consider the holomorphic bundle $E=\Omega_X^p$ over complex manifold $X$. $\bar{\partial}^*$ is the formal adjoint of $\bar{\partial}$ with respect to inner product of differential forms.
As in Hodge Theory and Complex Algebraic Geometry by Claire Voisin, the end of 5.1.3, remark 5.11, p.123, she says $(-1)^p\frac{1}{2}\bar{\partial}_E^* = \bar{\partial}^*$.
However I think $2^p\langle \cdot ,\cdot \rangle_{A^{0,q}\ (\Omega_X^p\ )} = \langle \cdot ,\cdot \rangle_{A^{p,q}}$. And \begin{align} \langle\bar{\partial}_E u,v\rangle_{A^{0,q}\ (\Omega_X^p)} &= \langle u,\bar{\partial}^*_E v\rangle_{A^{0,q}\ (\Omega_X^p)} \\ &= (-1)^p2^{-p} \langle\bar{\partial} u,v\rangle_{A^{p,q}} \\ &= (-1)^p2^{-p} \langle u,\bar{\partial}^* v\rangle_{A^{p,q}}\\ & = \langle u,(-1)^p\bar{\partial}^* v\rangle_{A^{0,q}\ (\Omega_X^p)} \end{align} and therefore $(-1)^p\bar{\partial}_E^* = \bar{\partial}^*$.
My question is where does the $\frac{1}{2}$ in the book come from?
I realize that the main point is how the inner product is defined. Maybe for the inner product in $A^{0,q}\ (\Omega_X^p\ )$, the inner product of antiholomorphic differential form $d\bar{z}$ is induced by the Hermitian metric in $T_X$, while the inner product in $A^{p,q}$ is induced by Euclidean metric of $(T_X\otimes C)^*$ where the Euclidean metric is induced by hermitian metric of $T_X$. Am I right?