# Understanding construction of the Hodge dual

I'm looking at the following construction of the Hodge dual on Wikipedia: https://en.wikipedia.org/wiki/Hodge_dual

I'll summarize what Wikipedia says. We let $V$ be an $n$-dimensional vector space over the reals equipped with an inner product. We let $\{e_1,\dots,e_n\}$ be an orthogonal basis, and $\omega=e_1\wedge\cdots\wedge e_n$. The hodge dual by definition should be an $\Bbb R$-linear map $*:\Lambda^k(V)\to\Lambda^{n-k}(V)$ for any $0\le k\le n$ such that for all $\alpha,\beta\in \Lambda^k(V)$, $\alpha\wedge*\beta=(\alpha,\beta)\omega$.

Here is the construction:

If we fix $\lambda\in \Lambda^k(V)$, then for every $\theta\in \Lambda^{n-k}(V)$ there is a real number $f_{\lambda}(\theta)$ such that $\lambda\wedge\theta=f_{\lambda}(\theta)\omega$. This defines a bounded linear operator $f_{\lambda}:\Lambda^{n-k}(V)\to\Bbb R$, so by the Riesz representation theorem, there is an element $*\lambda\in \Lambda^{n-k}(V)$ such that $f_{\lambda}(\theta)=(\theta,*\lambda)$ for all $\theta$.

My question is, how does this give us what we need? I've checked that this is indeed $\Bbb R$-linear, but all the construction immediately gives us is $\lambda\wedge\theta=f_{\lambda}(\theta)\omega=(\theta,*\lambda)\omega$ for any $\lambda\in \Lambda^k(V)$, $\theta\in \Lambda^{n-k}(V)$.

How can I see that the desired relation holds for $\alpha,\beta\in \Lambda^k(V)$?

## 1 Answer

Let's not take my answer too seriously because I am not a specialist.

Due to switching between scalar products of each $\Lambda^k(V)$ and in order to not lose the sight on the compatibility between them, I prefer to construct $\star$, first, on orthonormal basis of $\Lambda^k(V)$ - by the definition of the scalar products on $\Lambda^k(V)$ inherited from the scalar product of $V$ - naturally associated to the orthonormal basis $\{e_1,\cdots,e_n\}$ of $\Lambda(V)=V$. Let us denote this orthonormal basis of $\Lambda^k(V)$ by $e_1^k,\cdots,e_{C_n^k}^k$.

Provided that $V$ is oriented, $\star$ is the only linear operator such that $e_i^k\wedge \star e_i^k = \omega$ for all $k=1,\cdots,n$ and $i=1,\cdots,C_n^k$ (the unicity is derived from the same argument you have used). Using the orthonormality and linearity we get that, $\alpha \wedge \beta = \{\alpha,\beta\}\omega$, where $\alpha = \sum \alpha_i e_i^k$, $\beta = \sum \beta_i e_i^k$ and $\{\alpha,\beta\} = \sum \alpha_i \beta_i$. Note that this definition is independent from the choice of the orthonormal basis

I think, If you breakdown your formalism into orthonormal basis, it will be easy to find your wanted formula. By the way, $\omega$ is chosen to be the volume form, i.e $(\omega,\omega)=1$. this is another reason why I think, using the decomposition in the orthonormal basis is the only way to show the formula.