Let $M$ be a complex manifold of dimension $n$. It means that $M$ is a real smooth smooth manifold of dimension $2n$. Suppose that the real tangent bundle of $M$ has a local basis:
$$\left\{\frac{\partial }{\partial x_1},\ldots,\frac{\partial}{\partial x_n},\frac{\partial }{\partial y_1},\ldots,\frac{\partial}{\partial y_n}\right\}$$
and the real cotangent bundle has local basis
$$\left\{dx_1,\ldots,dx_n,dy_1,\ldots dy_n\right\}$$
Then we put $$\frac{\partial}{\partial z_j}:=\frac{1}{2}\left(\frac{\partial }{\partial x_j}-i \frac{\partial }{\partial y_j}\right)$$
$$\frac{\partial}{\partial \bar z_j}:=\frac{1}{2}\left(\frac{\partial }{\partial x_j}+i \frac{\partial }{\partial y_j}\right)$$
$$dz_j:=dx_j+idy_j$$
$$d\bar z_j:=dx_j-idy_j$$
Now consider the complexified cotangent bundle $(T^\ast M)_{\mathbb C}$, it has a decomposition:
$$(T^\ast M)_\mathbb C:=T^\ast M^{1,0}\oplus T^\ast M^{0.1}$$
where
$$T^\ast M^{(1,0)}=\left<dz_j:j=1,\ldots n\right>$$
$$T^\ast M^{(1,0)}=\left<d\bar z_j:j=1,\ldots n\right>\,.$$
At this point one defines the algebra of differential $(p,q)$-forms on $M$ as:
$$\bigwedge^{p,q}M:=\bigwedge^{p}T^\ast M^{1,0}\otimes \bigwedge^{q}T^\ast M^{0,1}$$
So locally any $(p,q)$-differential form can be written as
$$\omega=\sum_{i_1<\ldots<i_p} \alpha_{i_1,\ldots,i_p}dz_{i_1}\wedge\ldots\wedge dz_{i_p}\otimes \sum_{j_1<\ldots<j_q}\beta_{i_1,\ldots,i_q}d\bar z_{j_1}\wedge\ldots\wedge d\bar z_{j_q}\,.$$
So why in every textbook a $(p,q)$-dffierential form is written simply as:
$$\omega=\sum \alpha_{i_1,\ldots,i_p}dz_{i_1}\wedge\ldots\wedge dz_{i_p}\wedge d\bar z_{j_1}\wedge\ldots\wedge d\bar z_{j_q}\;\;?$$
Where is the tensor product?
Best Answer
If $V$ and $W$ are complex vector spaces, then you have canonical isomorphisms $$ \bigoplus_{j+k=m} \bigwedge^j V \otimes \bigwedge^k W \to \bigwedge^m V \oplus W $$ given on each direct summand $\bigwedge^j V \otimes \bigwedge^k W$ with $j+k=m$ by $$ (v_1 \wedge \cdots \wedge v_j) \otimes (w_1 \wedge \cdots w_k) \mapsto v_1 \wedge \cdots \wedge v_j \wedge w_1 \wedge \cdots \wedge w_k. $$ Applied fibrewise in your case, this gives isomorphisms of complex vector bundles $$ \bigoplus_{p+q=m} \bigwedge^p T^\ast M^{(1,0)} \otimes \bigwedge^q T^\ast M^{(0,1)} \to \bigwedge^m T^\ast M_{\mathbb{C}}; $$ it's completely standard (and generally convenient) to identify $$ \bigwedge^{p,q} T^\ast M_{\mathbb{C}} := \bigwedge^p T^\ast M^{(1,0)} \otimes \bigwedge^q T^\ast M^{(0,1)} $$ with its image in $\bigwedge^{p+q}T^\ast M_{\mathbb{C}}$ under this map, which is what you're seeing whenever textbooks leave out the tensor product signs you're asking about.