Regarding wells' p.156 conjugate liner 1-form

Question

I am kind of confused with the conjugate linear 1-form, because it seems that to any conjugate linear 1-form $x$ and any $v\in E$, $x(v)=0$, in this sense the conjugation doesn't make sense.

I am considering an example: Let $X$ be a complex manifold of dim $n$ and $E$ be the tangent space of $X$ at $a$: $T_a^{1,0}X=<...,\frac{\partial}{\partial x_j}-i\frac{\partial}{\partial y_j},...>=<...,\frac{\partial}{\partial z_j},...>$, where $\frac{\partial}{\partial z_j}=1/2(\frac{\partial}{\partial x_j}-i\frac{\partial}{\partial y_j})$,

Now with this notation, we have $F$ above should be $(T_a^*X_0)_{\mathbb{C}}=(T_a^*X)^{1,0}\oplus (T_a^*X)^{0,1}=<...,dz_j,...>\oplus <...,d\bar{z_j},...>$, s.t. $dz_i(\frac{\partial}{\partial z_j})=\delta_{ij}$ and $d\bar{z_i}(\frac{\partial}{\partial {z_j}})=0$. And $\wedge^{1,0}F=(T_a^*X)^{1,0}$, $\wedge^{0,1}F=(T_a^*X)^{0,1}$.

But by $d\bar{z_i}(\frac{\partial}{\partial {z_j}})=0$, to any $v\in E=T_a^{1,0}X=<...,\frac{\partial}{\partial z_j},...>$, we have $d\bar{z_i}(v)=0$. But $d\bar{z_i}$ is the conjugate of $d{z_i}$, and $d{z_i}(v)$ might not have to be $0$, thus there is a contradiction. Where did I go wrong?

More generally, let $E=<d_1,..d_n>=<x_1+iy_1,...,x_n+iy_n>$, then $F=<x_1^*,y_1^*,...,x_n^*,y_n^*>\oplus i<x_1^*,y_1^*,...,x_n^*,y_n^*>$, then the $\wedge^{1,0}F$ as the set of complex linear 1-forms should be $<...,1/2(x_j^*-iy_j^*),...>$, then what's the conjugate? It seems that it has a similar problem.

Answer 1

The confusion is at the line $E = T^{1,0}X$. It is true that there is an (complex isomorphism between $(E, J)$ and $(T^{1,0}X, \sqrt{-1})$ given by

$$ v\in E\mapsto v - \sqrt{-1} Jv \in T^{1,0}X,$$

but if you think of $E \cong E\otimes \{1\}$ in $E\otimes \mathbb C$, then $E\cap T^{1,0}X = \{0\}$.