My question take source in D. Huybrechts Complex Geometry, p.262, in the lemma 6.1.9. He uses the wedge $\mathcal{A}^{0,p}(\Lambda^r \mathcal{T}_X)\times \mathcal{A}^{0,q}(\Lambda^s \mathcal{T}_X)\to \mathcal{A}^{0,p+q}(\Lambda^{r+s} \mathcal{T}_X) $, without defining it. But according to its use, it seems that some power of (-1) should be taken to obtain a right wedge. Why can't we say, for example in the case $r=s=1$, $(a\cdot d\overline{z}_{i_1}\wedge \dots \wedge d\overline{z}_{i_p}\otimes \frac{\partial}{\partial z_i})\wedge (b\cdot d\overline{z}_{j_1}\wedge d\overline{z}_{j_q}\otimes \frac{\partial}{\partial z_j}):=(d\overline{z}_I\wedge d\overline{z}_J)\otimes(a\cdot \frac{\partial}{\partial z_i}\wedge b\cdot \frac{\partial}{\partial z_j})$, while it seems that Huybrechts want $(a\cdot d\overline{z}_{i_1}\wedge \dots \wedge d\overline{z}_{i_p}\otimes \frac{\partial}{\partial z_i})\wedge (b\cdot d\overline{z}_{j_1}\wedge d\overline{z}_{j_q}\otimes \frac{\partial}{\partial z_j}):=(-1)^q(d\overline{z}_I\wedge d\overline{z}_J)\otimes(a\cdot \frac{\partial}{\partial z_i}\wedge b\cdot \frac{\partial}{\partial z_j})$ ?
Thanks for your answers.