# How do I complete this proof involving the Hodge dual and inner product?

Here are the preliminaries: let $M$ be an $n$-dimensional differentiable manifold equipped with a metric $g$. Define two $p$-forms $\alpha,\beta\in \Omega^p(M)$ and an $n$-form $\eta=\sqrt{|g|}dx^1\wedge\cdots\wedge dx^n$ (the canonical volume form). The Hodge dual is the isomorphism $\star :\Omega^p(M)\rightarrow \Omega^{n-p}(M)$ and is defined by its action on a $p$-form $\omega$ like $\star\omega=\frac{1}{p!}\omega^{i_1\dotsc i_p} \iota(e_{i_p})\dotsb \iota(e_{i_1})\eta$ where $\iota(X)$ is the interior product wrt the vector field $X$, and is the antiderivation $\iota(X):\Omega^p(M)\rightarrow\Omega^{p-1}(M)$. $\{e_i\}$ is a set of local basis vector fields. The inner product $\langle .,.\rangle$ of two p-forms is defined as $\langle\alpha,\beta\rangle=\frac{1}{p!}\alpha_{i_1\dotsc i_p}\beta^{i_1\dotsc i_p}$. Given all this, I wish to verify $$\alpha\wedge\star\beta=\langle\alpha,\beta\rangle\eta$$ I have a pretty good idea of how to start. I can use the above formula for the Hodge dual to write $$\alpha\wedge\star\beta=\frac{1}{p!}\beta^{i_1\dotsc i_p}\alpha\wedge\big( \iota(e_{i_p})\dotsb \iota(e_{i_1})\eta\big)$$ Now I can to move the interior products from the right side of the wedge onto the left to get the components of $\alpha$, which are then contracted with the components of $\beta$. To this end I think I can use the product rule for antiderivations (for $\alpha$ a $p$-form and $y$ an antiderivation): $$y(\alpha\wedge\beta)=y\alpha\wedge\beta+(-)^p\alpha\wedge y\beta$$ So in the above equation I can move the $e_1$ interior product past the $e_p$ one and pick up $p-1$ $(-)$s (using $\iota(X)\iota(Y)=-\iota(Y)\iota(X)$). Since $\alpha$ is a $p$-form, I pick up $p$ $(-)$s, which multiplied by the $p-1$ $(-)$s is simply $-1$. In other words: $$\iota(e_{i_1})[\alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\eta]= \iota(e_{i_1})\alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\eta- \alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\iota(e_{i_1})\eta$$ So $$\iota(e_{i_1})\alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\eta= \alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\iota(e_{i_1}) \eta$$ iff $$\iota(e_{i_1})[\alpha\wedge \iota(e_{i_p})\dotsb \iota(e_{i_2})\eta]=0$$ (or at least symmetric, which I don't think it is). However, there is no particular reason for this to be zero as far as I can see.

Any help would be greatly appreciated.

• In these sorts of computations, I would always recommend working in an orthonormal basis, which simplifies things immensely. As long as you know everything you have defined is basis-independent, you may as well just use an orthonormal basis. – Phillip Andreae Oct 4 '14 at 21:45
• Are you saying that if the set of e's is orthonormal, then $\iota(e_{i_1})[\alpha\wedge \iota(e_{i_p})\cdots\iota(e_{i_2})\eta]=0$? – Ryan Unger Oct 4 '14 at 23:02
• Now that I think about it, you can see that it's zero just by counting degrees of forms: $\alpha$ is a $p$-form, and you're wedging it with an $(n-p+1)$-form, which gives you an $(n+1)$-form, which must be zero. So if that's all you need, it seems like that does it. But I still recommend working in an orthonormal basis to make your life easier in general! – Phillip Andreae Oct 5 '14 at 4:05
• >counting degrees of forms -- I'm upset that it's this easy. – Ryan Unger Oct 5 '14 at 4:09