# Hodge double star operator

I want to prove that $**\omega=\left(-1\right)^{k\left(n-k\right)}\omega$, where $*$ is the Hodge star operator acting on differential $k$-forms $\omega$ on $\mathbb{R}^n$. Where can I find the proof of this?

• it's in Warner.. Jan 12, 2014 at 6:34
• In Warner (Foundations of differential manifolds and Lie groups) it is just given as an exercise 13 of chapter 2. The proof isn't there.
– Alem
Jan 12, 2014 at 17:14

The proof follows directly from applying the definition of the Hodge star twice. The most annoying thing is that you usually need some identity for the contraction of totally antisymmetric tensors. It's spelled out in Nakahara's "Geometry, Topology and Physics", page 291. He defines the dual as $$\star \omega = \frac{\sqrt{\vert g \vert}}{r!(m-r)!} \omega_{m_1 \dots m_r} \epsilon^{m_1 \dots m_r}_{\phantom{m_1\dots m_r} n_{r+1} \dots n_m} dx^{n_{r+1}} \wedge \dots \wedge dx^{n_m}.$$ Using this twice you'll need some identity for contracting the $\epsilon$'s, which you can calculate quickly using induction (or you can just guess it). If you don't want to get your hands dirty by shifting indices around, I found another version of the proof in Voisin's "Hodge Theory and Complex Algebraic Geometry, volume 1" around page 120. The key to that is that $\star$ preserves metrics, so $$\langle \alpha, \beta \rangle \text{vol} = \langle \star \alpha, \star \beta \rangle \text{vol}.$$

• If $\omega=\sum_{I}a_{I}dx_{I}$, where $I's$ are $k$-tuples increasingly ordered from $1,2,...n$, then it is sufficient to prove that $sign\left(I\right)+sign\left(I^{c}\right)$ is congruent to $k\left(n-k\right)$ modulo 2. Here $\left(I,I^{c}\right)$ is a permutation of $1,2,...,n$ and $I^{c}$ is increasingly ordered.But I don't know how to prove that.
– Alem
Jan 12, 2014 at 18:28
• This is sort of the same as the first method of proof I wrote down (which holds for manifolds). Restrict to $\mathbb{R}^n$ with Euclidean metric and your tuple $I$ corresponds to the indices $m_1, \dots, m_r$ while $I^c$ seems to be $n_{r+1}, \dots, n_m$. Since $\epsilon$ is totally antisymmetric with $\epsilon_{123\dots m} = +1$, you'll pick up some minus signs by bringing both tuples into that order. You need to do that twice for each time you're applying the above form. So it's in spirit the same as you're trying to do, and I think it's easier since it's more explicit.
– jws
Jan 12, 2014 at 19:17
• That is what I really did. I applied twice Hodge star and I saw what I need to prove in the language of the signs of permutations. But I just can't prove it.
– Alem
Jan 12, 2014 at 19:37
• Let's do an explicit example in $\mathbb{R}^2$. Take coordinates $x^1, x^2$. Then $\star dx^1 = \frac{1}{1} (dx^1)_m \epsilon^m_{\phantom{m}n} dx^n = \epsilon_{12} dx^2$, and $\star dx^2 = \epsilon_{21} dx^1 = -dx^1$. Thus $\star\star dx^m = -dx^m = (-1)^{1(2-1)} dx^m$. Do the same for $1, dx^1 \wedge dx^2$ in $\mathbb{R}^2$, if it's still not clear, try the same for $\mathbb{R}^3$.
– jws
Jan 12, 2014 at 20:12

Starting from \begin{align} &w_p = \frac{1}{p!} w_{m_1 m_2 \cdots m_p} dx^{m_1} \wedge dx^{m_2} \wedge \cdots \wedge dx^{m_p} \\ &*w_p = \frac{\sqrt{g}}{p!(n-p)!} w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n } dx^{m_{p+1}}\wedge \cdots \wedge dx^{m_n} \end{align} since $$*w_p$$ is $$n-p$$ forms we can write \begin{align} *w_p = \frac{1}{(n-p)!} \tilde{w}_{m_{p+1} \cdots m_n} dx^{m_{p+1}} \wedge \cdots \wedge dx^{m_n} \end{align} where $$\tilde{w}_{m_{p+1} \cdots m_n} = \frac{\sqrt{g}}{p!} w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n }$$ Then \begin{align} ** w_p &= \frac{\sqrt{g}}{p! (n-p)!} \tilde{w}_{m_{p+1} \cdots m_n} \epsilon^{m_{p+1} \cdots m_n }_{\phantom{m_{p+1} \cdots m_n } m_1 \cdots m_p} dx^{m_{1}} \wedge \cdots \wedge dx^{m_p} \\ & = \frac{\sqrt{g}}{p! (n-p)!} \frac{\sqrt{g}}{p!} w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n } \epsilon^{m_{p+1} \cdots m_n }_{\phantom{m_{p+1} \cdots m_n } m_1 \cdots m_p} dx^{m_{1}} \wedge \cdots \wedge dx^{m_p} \\ & = \frac{|g|}{p! (n-p)!} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n } \epsilon^{m_{p+1} \cdots m_n }_{\phantom{m_{p+1} \cdots m_n } m_1 \cdots m_p} w_p \\ & = \frac{1}{p! (n-p)!} \varepsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n } \varepsilon^{m_{p+1} \cdots m_n }_{\phantom{m_{p+1} \cdots m_n } m_1 \cdots m_p} w_p \\ & = (-1)^{p(n-p)} w_p \end{align} where we have been used contraction of the Levi-Civita tensor, denoted $$\varepsilon$$, while $$\epsilon$$ is the Levi-Civita symbol. The relation between the two is $$\varepsilon = \sqrt{g}\, \epsilon$$

• @phys_math Multiply repeated indices ($m_1,m_2,\ldots,m_p$) in the second line of $\ast \ast \omega_P$ calculation. Apr 10, 2022 at 11:43