# Hodge star operator on complex manifold

The Hodge star is defined as $$$$\alpha\wedge\overline{\star\beta} = \langle\alpha,\beta\rangle(\star1),\quad \alpha,\beta\in\Omega^{p,q}(M).$$$$ Here the inner product $$\langle\alpha,\beta\rangle$$ is defined as $$$$\langle\alpha,\beta\rangle = \frac1{p!q!}\sum_{I,\bar J}\alpha_{I,\bar J}\overline{\beta_{I,\bar J}}.$$$$

Applying this definition for $$\alpha = \beta = 1$$, $$1\wedge\overline{\ast1} = \langle1,1\rangle(\ast1) = (\ast1)$$. On the other hand, for $$\alpha = 1$$ and $$\beta = i$$ $$$$1\wedge\overline{\ast i} = \langle 1,i\rangle(\ast1) = -i(\ast1) = (-i)1\wedge\overline{\ast1} = 1\wedge\overline{i(\ast1)}.$$$$ Then, $$\ast i = i(\ast1)$$. Does it imply $$\ast$$ is a linear operator?

However, I found that $$\ast$$ is anti-linear here https://physics.stackexchange.com/a/162861/190058

My derivation above is wrong?

P.S. For the Hodge star, the transformation for basis is $$$$\label{1} \ast(\theta^I\bar\theta^{\bar J}) = \frac{(-i)^{n^2}}{(n-p)!(n-q)!}\varepsilon^{I\bar J}_{\quad I'\bar J'}\bar\theta^{I'}\theta^{\bar J'},$$$$ and $$$$\ast(\bar\theta^I\theta^{\bar J}) = \frac{i^{n^2}}{(n-p)!(n-q)!}\varepsilon^{I\bar J}_{\quad I'\bar J'}\theta^{I'}\bar\theta^{\bar J'},$$$$ where the last equation is the complex conjugation of the first one. ($$\because$$ $$\text{c.c.}(\ast\beta) = \text{c.c.}(\ast a + i\ast b) = \ast a - i\ast b = \ast(a - i\ast b) = \ast(\text{c.c.}(\beta))$$)

On the other hand, by changing the order of basis, \begin{align} \ast(\bar\theta^{\bar J}\theta^I) = \frac{(-i)^{n^2}}{(n-p)!(n-q)!}\varepsilon^{\bar JI}_{\quad\bar J'I'}\theta^{\bar J'}\bar\theta^{I'}. \end{align} For the first one and this one, which is correct?

• Without checking anything, this might be because physicists tend to use a different convention for inner products than mathematicians. In particular they typiclaly require that $\langle \cdot,\cdot\rangle$ is conjugate linear in the first variable, while the mathematicians' convention is to have the inner product be conjugate linear in the second variable. Commented Oct 24, 2021 at 20:09
• @kahen So if we use the physicists convention, $\star$ is antilinear while if we use the mathematicians convention $\star$ is linear?
– KoKo
Commented Oct 24, 2021 at 20:17
• Please do not use $i$ as an index when working with complex forms. Commented Oct 26, 2021 at 3:59
• Did I use $i$??
– KoKo
Commented Oct 26, 2021 at 9:06

There are a few conventions here. You can check whether it is linear as follows: $$\forall\alpha:\alpha\wedge \overline{\ast(k\beta)}=\langle \alpha,k\beta\rangle (\ast1)=\bar k\langle \alpha,\beta\rangle (*1)=\alpha\wedge\overline{k\ast(\beta)}\Longrightarrow \ast(k\beta)= k\ast(\beta)$$

On the other hand, it is also common in the literature to define the Hodge star operator by the equation $$\alpha\wedge\ast\beta=\langle \alpha,\beta\rangle(\ast1)$$. Likewise, we can check $$\forall\alpha:\alpha\wedge\ast(k\beta)=\langle\alpha,k\beta\rangle(\ast1)=\bar k\langle\alpha,\beta\rangle(\ast 1)=\alpha\wedge\bar k\ast(\beta)\Longrightarrow\ast(k\beta)=\bar k\ast(\beta)$$

This becomes antilinear. Note that the first one sends a $$(p,q)$$-form to $$(n-q,n-p)$$ form meanwhile the second definition sends a $$(p,q)$$-form to $$(n-p,n-q)$$ form.

• In my textbook the first convention is used. Here $\ast^2\alpha = (-1)^{p+q}\alpha$ for $(p,q)$-form $\alpha$. This property is independent of the convention?
– KoKo
Commented Oct 25, 2021 at 9:35
• Yes I think so. One property is that $\ast$ preserves the metric, whether it is linear or antilinear. (Really they are the same up to conjugation).
– lEm
Commented Oct 25, 2021 at 9:46
• In my textbook, the Hodge star operation is given by $$\overline{\ast\beta} = \frac{i^{n^2}}{(n-p)!(n-q)!p!q!}\beta_{I\bar J}\varepsilon^{I\bar J}_{\;\; I'\bar J'}\theta^{I'}\wedge\bar\theta^{J'}.$$ To check $\ast^2 = (-1)^{p+1}$, I would like to find the expression without complex conjugation. Is this the correct form? $$\ast(\theta^I\wedge\bar\theta^{\bar J}) = \frac{(-i)^{n^2}}{(n-p)!(n-q)!}\varepsilon^{I\bar J}_{\;\;\; I'\bar J'}\bar\theta^{I'}\wedge\theta^{\bar J'}.$$
– KoKo
Commented Oct 25, 2021 at 10:17