# Identifying a wedge-to-metric formula

In this question, the original poster wrote:

On every Riemannian manifold $$M$$, we can consider the Hodge $$*$$-operator, which is characterised by the following formula: $$a\wedge *b = (a,b)\nu.$$ Here $$a$$ and $$b$$ are smooth forms on $$M$$, $$(\ ,\ )$$ is a metric on $$\wedge T^*\!M$$ and $$\nu$$ is the volume form with respect to the Riemannian metric.

I'm looking to study this formula in particular, but it's difficult to search for because of the notation.

What are a couple webpages or books that discuss (or even derive) this formula? The simpler the better.

• Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $\mathbb{R}^3$ of $U \times V^* = \sin \alpha \|U\|\|V\|$ where $V^*$ is the vector directly orthogonal to V in TM and $\alpha$ the angle between $U$ and $V^*$. – Jean Marie Dec 5 '18 at 6:53
• @JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example? – Doubt Dec 5 '18 at 14:22
• I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form. – Jean Marie Dec 5 '18 at 15:42