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 $$\alpha\wedge\bar{*}({\beta})=\langle\alpha,\beta \rangle \text{vol}$$
Can you give an example to illustrate a specific formula about how is $\bar{*}$ defined, for example, if $\alpha=dz_2\wedge d\bar{z_1}\wedge d\bar{z_4}$ and $n=4$, then what's $\bar{*}\alpha$?
My guess is $\bar{*}\alpha=\overline{(*dz_2)\wedge(*d\bar{z_1}\wedge d\bar{z_4})}=\overline{
(-dz_1\wedge dz_3\wedge dz_4)\wedge d\bar{z_2}\wedge d\bar{z_3}}=-dz_2\wedge dz_3\wedge d\bar{z_1}\wedge d\bar{z_3}\wedge d\bar{z_4}$
By the way,above I assume that if $\alpha=dz^I\wedge d\bar{z}^J$, then is $\bar{\alpha}=dz^J\wedge d\bar{z}^I$, is that correct?
Moreover, is there is formula to illustrate the relation between $dz_1\wedge d\bar{z_3}\wedge dz_2$ and $dz_1\wedge dz_2\wedge d\bar{z_3}$?
Best Answer
To make notation easier I will write $z, x, y$ instead of $dz, dx, dy.$ Furthermore, this all boils down to linear algebra, so I will work with a real vector space $V$ with an inner product and compatible complex structure $J$ where $y_i = J(x_i)$. Further assume that the $x_i$'s and $y_i$'s are orthogonal. Remember how the inner product on $V$ extends to $\bigwedge^k V$: we declare that monomials in the $x$ and $y$'s of length $k$ are orthonormal. Set $z_j = x_j + i y_j$.
I will call a $(p,q)$ form a monomial if it is of the form $z_I \wedge \bar{z}_J$. I will often omit wedge product symbols and write $z_I \bar z_J$.
I will record some observations:
Lemma $1$. $\bar\ast$ takes complex $k$-forms into $(n-k)$-forms.
Proof This is the (conjugate) of the complexification of the real Hodge star, which has this same property.
Lemma $2$. $\bar\ast(z_I\bar{z}_J) = b z_{I^c} \bar{z}_{J^c}$ where $I^c$ is the complementary set of indices, and $b$ is some nonzero constant.
Proof. Say $z_I \bar{z}_J$ is a $(p,q)$-form. Write $\bar\ast (z_I \bar{z}_J) = \sum_{K,L} b_{K,L} z_K\bar{z}_L$, where $|K| + |L| = 2n-(p+q)$ (this implicitly uses lemma $1$ to get the sizes of the multi-indices right). Since $z_I\bar{z}_J \wedge \bar\ast(z_I\bar{z}_J) = ||z_I \bar{z}_J||^2 \cdot \text{vol}$, we note that $\sum_{K,L} b_{K,L} z_I\bar{z}_J z_K\bar{z}_L$ must equal a positive multiple of the volume form. But $z_I \bar{z}_J z_K\bar{z}_L=0$ unless $K= I^c$ and $L=J^c$. You can think about why this last step is true; the reasoning should morally be that there will either be more than $n$ $z$'s or more than $n$ $\bar{z}$'s, killing the product $z_I \bar{z}_Jz_K\bar{z}_L$. The constant $b := b_{I^c, J^c}$ is nonzero because we need to have $$ b z_I\bar{z}_J z_{I^c}\bar{z}_{J^c} = ||z_I\bar{z}_J||^2 \cdot \text{vol} \neq 0. $$
Computing $\bar\ast\alpha$ for $\alpha = z_2\bar{z}_1\bar{z}_4$, by lemma $2$ we already know that $$ \bar\ast\alpha = b \cdot z_1 z_3 z_4 \bar{z}_2\bar{z}_3, $$ and it remains to only calculate the value of $b$. We'll need $||\alpha||$ for this.
Expand $\alpha$ into $x$ and $y$'s: $$\alpha = z_2\bar{z}_1\bar{z}_4 = x_2x_1x_4 -i x_2y_1x_4 + i y_2x_1x_4 + y_2y_1y_4 -ix_2x_1y_4 - x_2y_1y_4 + y_2x_1y_4 - iy_2y_1y_4.$$ Since these monomials in $x$ and $y$ are all orthonormal and distinct, we can see that $||\alpha||^2 = 8$.
So, $$ z_2 \bar{z}_1 \bar{z}_4 \wedge b z_1 z_3 z_4\bar{z}_2\bar{z}_3 = 8 \cdot \text{vol}. $$ Next, you can check that $$ z_2 \bar{z}_1 \bar{z}_4 z_1 z_3 z_4\bar{z}_2\bar{z}_3 = - z_1\bar{z}_1z_2\bar{z}_2 z_3\bar{z}_3z_4\bar{z}_4 = -\left(\frac{2}{i}\right)^4 \cdot \text{vol} $$ and so we find that $$ -b \left(\frac{2}{i}\right)^4 = 8, $$ or $b=-\frac{1}{2}$. Therefore $$ \bar\ast(z_2\bar{z}_1\bar{z}_4) = -\frac{1}{2} z_1z_3z_4\bar{z}_2\bar{z}_3. $$
Your second question: no. If $\alpha = z_I \bar{z}_J$, then $\bar{\alpha} = \bar{z}_I z_J \neq z_J \bar{z}_I$ in general. You would have to figure out the number of permutations needed and count the right number of signs.
I implicitly answered your last question above, but I'll state it clearly here: you permute $dz$ and $d\bar{z}$'s like ordinary $1$-forms, so $$ dz_1 \wedge d\bar{z}_3 \wedge d z_2 = - dz_1 \wedge dz_2 \wedge d\bar{z}_3. $$