# Hodge star operator and exterior calculation

I am learning complex geometry by D. Huybrechts. Here is a formula that I can't understand $$\omega \wedge \beta\wedge \star \alpha=\beta\wedge(\omega \wedge \star \alpha)\tag 1$$

Here $$\omega$$ is the fundamental form which is a $$2$$-form actually. And I try to expand both sides by the definition $$\alpha \wedge \star \beta= \langle \alpha, \beta \rangle \cdot vol$$

For LHS, we have $$\langle \omega \wedge \beta,\alpha \rangle \cdot vol$$ and for RHS, we have $$\beta \wedge \langle \omega, \alpha \rangle \cdot vol$$ But here is the things I don't understand:

(1) And let's suppose $$\alpha,\beta \in \land^k V^{*}$$. For $$\langle \omega \wedge \beta,\alpha \rangle$$, here $$\omega \wedge \beta$$ is a $$(k+2)$$- form so how can an inner product operated by a $$k+2$$-form and a $$k$$-form ? So does for RHS.

(2) How to prove $$(1)$$ by expanding in my way or the other methods?

As $$\omega$$ is a $$2$$-form, it commutes for the $$\wedge$$-product with other $$p$$-forms, (because if $$\alpha$$ and $$\beta$$ are $$p$$ and $$q$$-forms, $$\alpha\wedge \beta = (-1)^{pq} \beta\wedge \alpha$$). Moreover, this product is associative. Forget about the $$\star$$ in $$(\star \alpha)$$ and consider only $$(\star\alpha)$$ as a differential form. Then : \begin{align} \omega \wedge \beta \wedge \star \alpha &= \left(\omega \wedge \beta\right) \wedge \star \alpha \\ &= \left(\beta \wedge \omega \right)\wedge \star \alpha \\ &= \beta \wedge \left( \omega \wedge \star \alpha\right) \end{align}

• Oh, how careless I am ,I forgot $\omega$ is a two form. ..But I wonder can a p-form and a q-form operated on inner product? I learned that the inner pruduct is only defined for the same degree.
– LSY
Commented May 24, 2020 at 12:28
• Let $n$ be the dimension of your vector space / manifold. If $\alpha$ is a $k$- form, then $\star \alpha$ is a $(n-k)$ -form. It is defined to be the unique for for which, for all $k$-forms $\theta$, $\theta \wedge \star \alpha = <\theta,\alpha> \mathrm{dvol}$. Here $\theta$ and $\alpha$ do have the same degree ! It is just the way to define the $(n-k)$-form $\star \alpha$. In your example, if $\omega$, $\alpha$ and $\beta$ are of degree $2$, $p$ and $q$ respectively and if $n$ is the dimension of the vector space, then $\omega\wedge \beta \wedge \star\alpha$ is of degree $2 + q + (n-p)$ Commented May 24, 2020 at 12:32
• I think I find what am I missing. The operation priority of Hodge $\star$ operator is less than the wedge operation !
– LSY
Commented May 24, 2020 at 12:43
• It has to be read as $\omega \wedge \beta \wedge (\star \alpha)$. I don't get what priority you are talking about! Commented May 24, 2020 at 12:45
• I originally believe we first do the wedge operation $(\omega \wedge \beta)$ ,and form a $(p+2)$- form and then we apply the Hodge star operator, so $(\star \alpha)$ is actually $n-(p+2)$-form. so that the inner product have the same degree and it makes sense.That's what I originally mean by the "operation priority"... I made some mistskes.
– LSY
Commented May 24, 2020 at 12:59