1
$\begingroup$

enter image description here

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $E\cap T^{1,0}X = \{0\}$.

$\endgroup$
7
  • $\begingroup$ I feel a little confused, here $E$ is a complex space, but it seems you see $E$ as a real space. $\endgroup$
    – Danny
    Jul 31, 2018 at 12:59
  • $\begingroup$ Also, for my last part, the general case, the basis of the conjugate should be $\{ ...,1/2(x_j^*+iy_j^*,...\}$, then still we have $1/2(x_j^*+iy_j^*)(x_k+iy_k)=0$, thus the conjugate element acts trivially on $E$. $\endgroup$
    – Danny
    Jul 31, 2018 at 13:04
  • $\begingroup$ Yes I am thinking of $E$ as $2n$-dimensional real space and this is confusing. $E$ is complex so it has a complex structure $J$. We do not call it $i$, since that will cause confusion when you consider $E\otimes \mathbb C$. So your $x_j + i y_j$ is actually not in $E$ but only in $E\otimes \mathbb C$. $\endgroup$
    – user99914
    Jul 31, 2018 at 13:13
  • $\begingroup$ Now if $C^n=E=<d_1,..d_n>\cong <x_1, y_1,...,x_n,y_n>=R^{2n}$ and $F=<x_1^*,y_1^*,...,x_n^*,y_n^*>\oplus i<x_1^*,y_1^*,...,x_n^*,y_n^*>$, then how do elements of $F$ act on $E$? Can you give an example? $\endgroup$
    – Danny
    Jul 31, 2018 at 13:22
  • $\begingroup$ For example $(x_1^* + i y_1^*) (x_1) = x_1^* (x_1) + i y_1^*(x_1) = 1 + i\cdot 0 = 1$. $\endgroup$
    – user99914
    Jul 31, 2018 at 13:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .