This question about showing that an alternative construction of the Hodge dual operator satisfies to the universal property through which the Hodge dual is usually defined.

Let me give the construction. Let $V$ be an n-dimensional real vector space. Then, after choosing a volume form $\eta: \bigwedge{^n}V \rightarrow\mathbb{R}$, for all integer k such that $0\leq k \leq n$, the wedge product gives us the following linear map:\begin{align*} \wedge \colon \bigwedge{^k}V \otimes \bigwedge{^{n-k}}V &\to \mathbb{R} \\ s\otimes t &\mapsto \eta(s\wedge t) \end{align*} This allows us to define the isomorphism \begin{align*} \delta_k \colon \bigwedge{^k}V &\to (\bigwedge{^{n-k}}V)^\ast \\ s &\mapsto \eta(s\wedge -) \end{align*}

Now equip $V$ with an inner-product $\beta: V\otimes V \rightarrow \mathbb{R}$. This yields a canonical inner-product $<-,->_p: \bigwedge^p V\otimes \bigwedge^p V \rightarrow \mathbb{R}$ and hence, it also fixes a volume form $\tilde{\eta}:\bigwedge^n V\rightarrow \mathbb{R}$. From the canonical inner-product $<-,->_{n-k}$, we also get a canonical isomorphism $\phi_k: (\bigwedge{^{n-k}}V)^\ast \xrightarrow{\sim} \bigwedge{^{n-k}}V$.

My goal is to show, without introducing a basis, that the linear map $\phi_k \circ \delta_k : \bigwedge{^k}V \rightarrow \bigwedge{^{n-k}}V$ is precisely the Hodge dual operator.

Recall that on inner-product space $V$, the Hodge dual operator is defined as the unique linear isomorphism $\star : \bigwedge{^k} V \rightarrow \bigwedge{^{n-k}}V$ satisfying $ \tilde{\eta}(\star s \wedge t)= <s,t>_k$ for all $t\in \bigwedge{^k}V$.

When calculating this map with a basis on $V$, I do get that it equal the Hodge dual operator. Yet, I feel like this proof is unsatisfactory. Does anyone know of a way of proving this result without introducing a basis and calculating $\phi_k \circ \delta_k$ on basis elements. There must be a clever way of making use of universal properties for induced inner-products on $\bigwedge^p V$ and induced volume form $\tilde{\eta}$!

Thank you!


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