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I wanted to maybe extend Hodge star/ Technical question to a new question so others could benefit from the idea.

So there we discussed that when the $\star$ is Hodge duality star then it is real-linear, by that, if say $\omega$ is a 1-form, then $$\star(c\omega)=c\star(\omega)$$ c being a real number or real function.

I now ask what happens if we have a complex function now outside the $\star$, that is can we treat this the same if our Hodge duality is real-linear?

In other words, if $\alpha$ is complex function, can we say that

$$\star(\alpha\omega)=\alpha\star(\omega)?$$

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PhilosophicalPhysics
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Let $V$ be a complex vector space equipped with an inner product $\langle\, , \rangle$. There is an induced inner product on $(p,q)$-forms which I will also denote by $\langle\, , \rangle$. The Hodge dual is defined intrinsically as follows: for $(p, q)$-forms $\alpha$ and $\beta$,

$$\alpha\wedge\overline{\star\beta} = \langle\alpha, \beta\rangle dV.$$

Defined in this way, the Hodge dual is complex linear, i.e. $\star(c\beta) = c(\star\beta)$.

Rarely (one may even argue, incorrectly), the Hodge dual is defined without the complex conjugation on the left hand side. In this case, the Hodge dual will be conjugate linear, i.e. $\star(c\beta) = \bar{c}(\star\beta)$.

If the Hodge dual is being used in the complex setting without a definition stated beforehand, I would understand that to mean the map given by the first definition. I would consider the map in the second definition the Hodge dual composed with conjugation.

Michael Albanese
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