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 $dz_1\wedge dz_2\wedge d\bar{z_3}$?

  • $\begingroup$ Your guess is incorrect because $\ast$ does not distribute over the wedge product. $\endgroup$
    – Andrew
    Commented Aug 8, 2018 at 17:14
  • $\begingroup$ @Andrew so can I say that the action of $\bar{*}$ depends on the inner product? $\endgroup$
    – Danny
    Commented Aug 8, 2018 at 17:20
  • $\begingroup$ @Andrew Actually I wonder why is $\bar{*}:\epsilon^{p,q}\to\epsilon^{n-q,n-p}$ not $\bar{*}:\epsilon^{p,q}\to\epsilon^{n-p,n-q}$ $\endgroup$
    – Danny
    Commented Aug 8, 2018 at 17:24
  • $\begingroup$ It should map to $\epsilon^{n-p,n-q}$. This is because if $\alpha=dz_I d\bar{z}_J$ has $p$ of the dz's in it, then $\bar\ast\alpha$ needs to wedge with $\alpha$ to get to the volume form, and so $\bar\ast \alpha$ needs to have only the remaining $n-p$ $dz$'s in it. Same goes for the $d\bar{z}$'s. $\endgroup$
    – Andrew
    Commented Aug 8, 2018 at 17:38
  • 1
    $\begingroup$ And yes, the definition of $\bar\ast$ depends heavily on the choice of inner product. $\endgroup$
    – Andrew
    Commented Aug 8, 2018 at 17:39

1 Answer 1


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 $$ dz_1 \wedge d\bar{z}_3 \wedge d z_2 = - dz_1 \wedge dz_2 \wedge d\bar{z}_3. $$

  • $\begingroup$ There is something different with mathoverflow.net/questions/46844/… $\endgroup$
    – Danny
    Commented Aug 8, 2018 at 21:29
  • $\begingroup$ That is because they did not include conjugation. Without conjugation, $\ast: \Omega^{p,q}\to \Omega^{n-q, n-p}$. With conjugation it maps into $\Omega^{n-p, n-q}$. $\endgroup$
    – Andrew
    Commented Aug 8, 2018 at 23:56
  • $\begingroup$ Can you explain why without conjugation, $*:\Omega^{p,q}\to\Omega^{n-q,n-p}$? $\endgroup$
    – Danny
    Commented Aug 12, 2018 at 3:37
  • $\begingroup$ The proof I know is basically my Lemma 2, which proved that $\bar\ast$ maps into $\Omega^{n-p, n-q}$. Then lift the conjugate and you're done. $\endgroup$
    – Andrew
    Commented Aug 12, 2018 at 3:45
  • $\begingroup$ Indeed, I feel confused why via expanding by complex linearity, we have such a relation. $\endgroup$
    – Danny
    Commented Aug 12, 2018 at 3:49

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