Let $M$ be a $n$-dimensional manifold and let

$$\star : \Lambda^p(M) \longrightarrow \Lambda^{n-p} : \beta \longmapsto \star \beta$$

be the Hodge star operator, such that

$$ ( \alpha, \beta ) = \int_M \alpha \wedge \star \beta$$

for all $\alpha \in \Lambda^p$. Here, $\big( \boldsymbol{\cdot}, \boldsymbol{\cdot} \big)$ denotes the inner product, defined as

$$ (\alpha, \beta) = \frac{1}{p!} \int_M \alpha_{i_1 \ldots i_p} \; \beta_{j_1 \ldots j_p} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; dV$$

Show that

$$\star \big( dx^{i_1} \wedge \ldots \wedge dx^{i_p} \big) = \frac{\sqrt{\det g}}{(n-p)!} \cdot g^{i_1 j_1} \cdots g^{i_p j_p} \cdot \varepsilon_{j_1 \ldots j_n} \cdot dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n}$$

Now, I suppose this is a pretty standard problem, when dealing with differential geometry -- however, I neither managed to solve it myself nor did I find a complete proof of that proposition. Most of the proofs I found were either missing some vital step, or did not bother to expand the proof as detailed as I would need it to understand.

Can anyone help me by providing a source for a proof or can contribute a proof itself? I would really appreciate it.

P.S.: I found the answer to How to derive the coordinate expression of the Hodge dual? useful, but I don't understand some of the steps. I copied the proof and marked the steps I don't understand with a question mark (?)

Since both sides are linear in $\omega$, it suffices to check on a basis. Let $\omega := \mathrm{d}x^{1}\wedge...\wedge \mathrm{d}x^{k}$. Then, one immediately notices that $\eta \overset{?}{=} \varepsilon^{i_1 ... i_k} \omega$. Observe that \begin{align*} ⟨\omega,\eta⟩ \ vol \enspace &= \enspace \varepsilon^{i_1 ... i_k} \ ⟨\omega,\omega⟩ \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &\overset{?}{=} \enspace \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ g^{1 j_1} ... g^{k j_k} \ \varepsilon_{j_1 ... j_k} \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \enspace \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \enspace \sqrt{g} \ \Big( \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \Big) \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &\overset{?}{=} \enspace \mathrm{d} x^1 \wedge ... \wedge \mathrm{d} x^k \wedge \Big( \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \Big) \ \mathrm{d} x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\\ &= \enspace \omega \wedge \widetilde{\eta} \end{align*}


1 Answer 1


I now figured out something, but I am not sure whether my calculations are correct, so please tell me if my approach is acceptable.

Let $\alpha \in \Lambda^p$ such that $\alpha = \frac{1}{p!} \; \alpha_{k_1 \ldots k_p} \; dx^{k_1} \wedge \ldots dx^{k_p}$ and let $\beta = dx^{j_1} \wedge \ldots \wedge dx^{j_p}$. Then by assumption

\begin{align} \alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha \wedge\Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big) \\ &= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha_{k_1 \ldots k_p} \; \Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge \Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big) \end{align}

By renaming the indices, one obtains

\begin{align} \alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \underbrace{\Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge\Big(dx^{k_{p+1}} \wedge \ldots \wedge dx^{k_n} \Big)}_{= \; \varepsilon^{k_1 \ldots k_n} \; dx^1 \wedge \ldots \wedge dx^n} \\ &= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \; \varepsilon^{k_1 \ldots k_n} \; \Big( dx^1 \wedge \ldots \wedge dx^n \Big) \end{align}

Using $dV =\sqrt{\det g} \cdot dx^1 \wedge \ldots \wedge dx^n$, one gets

\begin{align} \alpha \wedge \star \beta \enspace &= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \varepsilon^{k_1 \ldots k_n} \; \alpha_{k_1 \ldots k_p}\\ &= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; (n-p)! \; \delta^{k_1 \ldots k_p}_{j_1 \ldots j_p} \; \alpha_{k_1 \ldots k_p} \\ &= \enspace \frac{dV}{p!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; p! \; \alpha_{[j_1 \ldots j_p]} \\ &= \enspace dV \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \alpha_{j_1 \ldots j_p} \end{align}

which coincides with the inner product. Here, the antiymmetry of $\alpha$ was used. Note that the factor of $\tfrac{1}{p!}$ within the inner product arises from a $p$-form $\gamma = \tfrac{1}{p!} \, \gamma_{m_1 \ldots m_p} \, dx^{m_1} \wedge \ldots \wedge dx^{m_p}$ being inserted into the inner product, i.e. $(\alpha, \gamma)$.

Please tell me what you think.


You must log in to answer this question.

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