5
$\begingroup$

let $(M,g)$ be an oriented Riemannian manifold. Let $*$ be the hodge operator, I want to prove that $$*\mathrm{vol}_g =1$$ where $\mathrm{vol}_g$ is the associate volume form $\sqrt{g} e^1\wedge \cdots \wedge e^n$

In these notes it's given as a fact, but when I started to prove it, I got confused by the underlying scalar product and how the pairing works, hence can someone show me some hints on how to prove that? I don't want a full solution, only a clarification.

In fact using the definition of hodge star I think that it's enough to prove that $g(\mathrm{vol}_g,\mathrm{vol}_g)=1$

but I don't know how to prove it.

Given this, then is it true that, for a given top-form ($n$-dim differential form), $w=(*w)\mathrm{vol}_g$? Because I'd argue in this way: $w=\tilde{w}(x)\mathrm{vol}_g$ because we are working in a one dimensional vector space, then $*w={*\tilde{w}}(x)\mathrm{vol}_g=\tilde{w}(x)\,{*\mathrm{vol}}_g=\tilde{w}(x)$.

Is it correct?

ADDENDUM In this setting, given $\eta \in \Omega^p(M)$, we define $*\eta$ as the unique element such that, for every $w \in \Omega^{p}(M)$ we have $$ w \wedge *\eta = (\bigwedge^p g)(w,\eta) \mathrm{vol}_g$$

$\endgroup$
4
  • 1
    $\begingroup$ There are many different starting definitions of "Hodge star operator." Some definitions will make $*\text{vol}_g = 1$ immediately obvious, and others will require more work. You need to tell us which one you're using. $\endgroup$ Commented Jan 1, 2015 at 21:57
  • $\begingroup$ @JesseMadnick I've update my question, hope everything is clarified $\endgroup$
    – Luigi M
    Commented Jan 1, 2015 at 22:10
  • $\begingroup$ I'm assuming $\bigwedge^p g$ is just meaning the induced Riemannian metric on the exterior $p$-forms. Correct? $\endgroup$ Commented Jan 1, 2015 at 22:40
  • $\begingroup$ @RobinGoodfellow yes, correct $\endgroup$
    – Luigi M
    Commented Jan 1, 2015 at 22:55

1 Answer 1

4
$\begingroup$

In these contexts I often find it useful to work in an orthonormal basis. Let $e^1, \dots, e^n$ be an orthonormal basis for $T_x^\ast M$ (this is just a pointwise linear algebra computation, so we can fix a point $x \in M$). Note that $\sqrt{g} = 1$, so $\text{vol}_g = e^1 \wedge \dots \wedge e^n$. The key thing is that if we start with an orthonormal basis for $T_x^\ast M$, the set of forms of the form $e^{i_1} \wedge \dots \wedge e^{i_p}$, where $1 \leq i_1 < \dots < i_p \leq n$, form an orthonormal basis for $\Lambda^p T_x^\ast M$. From this it is immediate that $g(\text{vol}_g, \text{vol}_g) = 1$. Do you see how this implies that $\ast \text{vol}_g = 1$?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .