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Hodge star operator is an operator: $\bar{*}:\epsilon^{p,q}=\Gamma(X,\bigwedge^p\Omega^1_X\otimes\overline{\bigwedge^p\Omega^1_X})\to\epsilon^{n-q,n-p}$ with the relation $$\alpha\wedge\bar{*}({\beta})=<\alpha,\beta>vol$$

I am not sure about the vol in the complexified case, is it $dz_1\wedge...\wedge dz_n\wedge d\bar{z_1}\wedge...\wedge d\bar{z_n}$ or $dx_1\wedge... dx_n\wedge dy_1...\wedge dy_n$ or $dx_1\wedge dy_1\wedge...\wedge dx_n\wedge dy_n$, where $dx_i=\frac{dz_i+d\bar{z_i}}{2}$, $dy_i=\frac{dz_i-d\bar{z_i}}{2i}$?

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    $\begingroup$ reminds me of Hodge star. $\endgroup$ – Will Jagy Aug 8 '18 at 1:27
  • $\begingroup$ @WillJagy Do you have any idea? $\endgroup$ – Danny Aug 8 '18 at 1:53
  • $\begingroup$ I think the correct name is Hodge star, not Hedge star. $\endgroup$ – Will Jagy Aug 8 '18 at 1:55
  • $\begingroup$ @WillJagy Thanks, I have fixed the typo $\endgroup$ – Danny Aug 8 '18 at 1:57
  • $\begingroup$ I would suggest breaking this up into a few posts since you have so many questions. $\endgroup$ – Andrew Aug 8 '18 at 4:53
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The volume form is always going to be $dx_1 \wedge dy_1 \wedge ... \wedge dx_n \wedge dy_n$. To remember this I always remember how you give an inner product space $V$ with a complex structure $J$ a natural orientation: you start with a vector $x_1$, then set $y_1:= J(x_1)$. Next pick any $x_2$ orthogonal to $x_1$ and $y_1$, then set $y_2:=J(x_2)$, and so on. Thus, the orientation is $x_1, y_1, x_2, y_2, ...$, and hence you see the ordering on the volume form.

Now, you can prove the formula $$ \left(\frac{i}{2}\right)^n dz_1\wedge d\bar{z}_1 \wedge \dots \wedge dz_n\wedge d\bar{z}_n = dx_1 \wedge dy_1 \wedge \dots \wedge dx_n \wedge dy_n. $$ To see this, just calculate $dz\wedge d\bar{z}$: $$ dz\wedge d\bar{z} = (dx + i dy)\wedge(dx - idy) = 0 - i dx \wedge dy + i dy \wedge dx + 0 = (-2i) dx\wedge dy, $$ giving $$ \frac{i}{2} dz \wedge d\bar{z} = dx \wedge dy. $$ Then just multiply a bunch of these together.

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  • $\begingroup$ One more question, on the other hands, in the notation of $\{z_1,...,z_n,\bar{z_1},...,\bar{z_n}\}$, is the orientation $z_1,...,z_n,\bar{z_1},...,\bar{z_n}$? $\endgroup$ – Danny Aug 8 '18 at 17:04
  • $\begingroup$ And can you also help to take a look another question of mine from breaking up yesterday's question: math.stackexchange.com/questions/2875735/… $\endgroup$ – Danny Aug 8 '18 at 17:12
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    $\begingroup$ Not necessarily. According to my answer, the standard orientation would be $z_1, \bar{z}_1, \dots, z_n, \bar{z}_n$. You can check that the sign to permute this to $z_1, z_2, \dots, z_n, \bar{z}_1, \dots, \bar{z}_n$ is $(-1)^{n (n-1)/2}$, which isn't always $+1$. $\endgroup$ – Andrew Aug 8 '18 at 17:13

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