Why is $\epsilon^{p,q}(X):=\Gamma(X,\bigwedge^p\Omega^1_X\otimes\overline{\bigwedge^p\Omega^1_X})$? Here $\Omega^1_X=(T^*X)^{1,0}$, from : this notes we have
$\epsilon^{p,q}(X):=\Gamma(X,\bigwedge^p\Omega^1_X\otimes\overline{\bigwedge^p\Omega^1_X})$
But I wonder why it's tensor not wedge? Shouldn't it be $\epsilon^{p,q}(X):=\Gamma(X,\bigwedge^p\Omega^1_X\wedge\overline{\bigwedge^p\Omega^1_X})$?
See O'well's book (p.33) or here
 A: Let us reduce ourselves to the fiber. If $V$ and $\bar V$ are two different $\Bbb R$-spaces with $W=V\oplus \bar V$, and assume there is an isomorphism over $\Bbb R$, $v\to\bar v$, between the two summands. 
We fix a basis $B$ of $V$. Thus also one, $\bar B$, for $\bar V$.
Now let us consider an element in $$\wedge^\cdot W\ .$$ It can be written uniquely "adapted" w.r.t. the basis $B\sqcup \bar B$ of $W$ over $\Bbb R$. Adapted means the following. We order $B$ and correspondingly $\bar B$. Each wedge product in $\wedge^\cdot V$ is rewritten so that we have first a wedge in $\wedge^\cdot V$, and then one in $\wedge^\cdot \bar V$.
This gives a well defined map
$$ \wedge^\cdot W\to \wedge^\cdot V\otimes \wedge^\cdot \bar V\ ,$$
which is inverse to the obvious map
$$ \wedge^\cdot W\leftarrow \wedge^\cdot V\otimes \wedge^\cdot \bar V\ ,$$
in opposite direction.
So why $X\otimes Y$ and not $X\wedge Y$ for $X=\wedge^\cdot V$, $Y=\wedge^\cdot \bar V$?
Because these are (seen as) different spaces from the start. (And there is no alternation in general, for two different spaces, $X,Y$. There is a space $X\otimes Y$, and a space $Y\otimes X$, defined using universal constructions, but there is no $X\wedge Y$. We would have to identify the two tensor products in a common world, but in our case we definitively not want to identify, for instance $dz_k$ with $d\bar z_k$ in the tangential space, we did our best to have them different.)
