# Hodge star operator $\bar{*}$ and volume form

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 $dx_i=\frac{dz_i+d\bar{z_i}}{2}$, $dy_i=\frac{dz_i-d\bar{z_i}}{2i}$?

• reminds me of Hodge star. – Will Jagy Aug 8 '18 at 1:27
• @WillJagy Do you have any idea? – Danny Aug 8 '18 at 1:53
• I think the correct name is Hodge star, not Hedge star. – Will Jagy Aug 8 '18 at 1:55
• @WillJagy Thanks, I have fixed the typo – Danny Aug 8 '18 at 1:57
• I would suggest breaking this up into a few posts since you have so many questions. – Andrew Aug 8 '18 at 4:53

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.
• One more question, on the other hands, in the notation of $\{z_1,...,z_n,\bar{z_1},...,\bar{z_n}\}$, is the orientation $z_1,...,z_n,\bar{z_1},...,\bar{z_n}$? – Danny Aug 8 '18 at 17:04
• 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 $+1$. – Andrew Aug 8 '18 at 17:13