How do we wedge the complex differentials $\mathrm{d}z^i$ and $\mathrm{d}\bar z^{\bar j}$? By the standard definition of the wedge product as an alternated tensor product, I would think we have
$$\tag{1}\mathrm{d}z^i\wedge\mathrm{d}\bar z^{\bar j}=\mathrm{d}z^i\otimes\mathrm{d}\bar z^{\bar j}-\mathrm{d}\bar z^{\bar j}\otimes\mathrm{d}z^i$$
As I understand it, the left side should be a section of exterior algebra $\Lambda^{1,1}M$. However, in the lecture notes I'm reading (arXiv), this exterior algebra is defined on page 10 as $$\Lambda^{1,1}M:=\Lambda^1T^*M\otimes\Lambda^1\bar T^*M$$
where $T^*M$ is the holomorphic cotangent bundle and $\bar T^*M$ its antiholomorphic counterpart. By inspection, however, the rhs. of (1) can't be a section of this product space because the first term is a section of $\Lambda^1T^*M\otimes\Lambda^1\bar T^*M$, the second factor a section of $\Lambda^1\bar T^*M\otimes\Lambda^1 T^*M$ and the tensor product is not generally commutative.
So how does this actually work? Any help would be greatly appreciated.
 A: Ref. 1 writes on p.10:

$$
\Lambda^k T^{\ast}_{\mathbb{C}} M ~=~ \bigoplus_{j=0}^k \Lambda^{j,k-j} M,\tag{1.11}$$
where we defined
$$\Lambda^{p,q} M ~:=~ \Lambda^pT^{*(1,0)}M\otimes\Lambda^{q}T^{*(0,1)}M.\tag{1.11b}$$

Here $M$ is a $2n$-dimensional real manifold with a complex structure $J$; the symbol $\otimes$ denotes the standard (un-antisymmetrized) tensor product over $\mathbb{C}$; the symbol $\wedge$ denotes the antisymmetrized tensor product/wedge product/exterior product over $\mathbb{C}$; and $T^{\ast}_{\mathbb{C}} M:=\mathbb{C}\otimes T^{\ast} M$ is the complexified cotangent space.
OP got a point that it would be more natural to define
$$\begin{align}\Lambda^{p,q} M ~:=~& \Lambda^pT^{*(1,0)}M\wedge\Lambda^{q}T^{*(0,1)}M\cr ~:=~&\Phi(\Lambda^pT^{*(1,0)}M\otimes\Lambda^{q}T^{*(0,1)}M) ~\subset~ \Lambda^{\bullet} T^{\ast}_{\mathbb{C}} M \end{align}\tag{*}$$
rather than the definition (1.11b). Here the definition (*) is canonically isomorphic to the definition (1.11b). Explicitly, the canonical isomorphism $\Phi$ reads
$$\begin{align}\sum_i\eta_i \otimes \bar{\omega}_i\quad\stackrel{\Phi}\mapsto\quad&\sum_i\eta_i \wedge \bar{\omega}_i, \cr 
\eta_i~\in~\Lambda^pT^{*(1,0)}M,\qquad & \bar{\omega}_i~\in~\Lambda^{q}T^{*(0,1)}M, \end{align}\tag{**}$$
using a hopefully clear notation.
Example:
$$\begin{align}\mathrm{d}z^i\otimes\mathrm{d}\bar{z}^{\bar{\jmath}}~\stackrel{\Phi}\mapsto~
\mathrm{d}z^i\wedge\mathrm{d}\bar{z}^{\bar{\jmath}}
~=~&\frac{1}{2}\left(\mathrm{d}z^i\wedge\mathrm{d}\bar{z}^{\bar{\jmath}} 
-\mathrm{d}\bar{z}^{\bar{\jmath}}\wedge\mathrm{d}z^i\right)\cr
~=~&-\mathrm{d}\bar{z}^{\bar{\jmath}}\wedge\mathrm{d}z^i~\in~\Lambda^{\bullet} T^{\ast}_{\mathbb{C}} M .\end{align}\tag{***} $$
Alternatively, if one uses the definition (1.11b), then the equality symbol "$=$" in eq. (1.11) should be interpreted as an isomophism symbol "$\cong$".
References:

*

*V. Bouchard, Lectures on complex geometry, Calabi-Yau manifolds and toric geometry,  arXiv:hep-th/0702063.

