The Hodge star is defined as \begin{equation} \alpha\wedge\overline{\star\beta} = \langle\alpha,\beta\rangle(\star1),\quad \alpha,\beta\in\Omega^{p,q}(M). \end{equation} Here the inner product $\langle\alpha,\beta\rangle$ is defined as \begin{equation} \langle\alpha,\beta\rangle = \frac1{p!q!}\sum_{I,\bar J}\alpha_{I,\bar J}\overline{\beta_{I,\bar J}}. \end{equation}
Applying this definition for $\alpha = \beta = 1$, $1\wedge\overline{\ast1} = \langle1,1\rangle(\ast1) = (\ast1)$. On the other hand, for $\alpha = 1$ and $\beta = i$ \begin{equation} 1\wedge\overline{\ast i} = \langle 1,i\rangle(\ast1) = -i(\ast1) = (-i)1\wedge\overline{\ast1} = 1\wedge\overline{i(\ast1)}. \end{equation} 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 \begin{equation}\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'}, \end{equation} and \begin{equation} \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'}, \end{equation} 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?
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.