I've read at least two conventions for the sign of the codifferential. The first of them \begin{align} \delta &= (-1)^{k}\star^{-1}\operatorname{d}\star\\ &= (-1)^{kn+n+1}\operatorname{sgn}(g)\star\operatorname{d}\star\tag{1} \end{align} comes from

Let $M$ be a smooth orientable $n$-dimensional manifold with exterior derivative $\operatorname{d}:\Omega^{k}(M)\to\Omega^{k+1}(M)$, metric $g\in\Gamma(T^{0,2}M)$ and volume form $\mu\in\Omega^{n}(M)$.

One extends the metric $g$ to a metric $g_{(k)}$ on each $\Omega^{k}(M)$ requiring that for every $\alpha,\beta\in\Omega^{k}(M)$ we have $$g_{(k)}(\alpha,\beta) := \frac{1}{k!} g^{a_1b_1}\cdots g^{a_kb_k}\alpha_{a_1\dots a_k}\beta_{b_1\dots b_k}$$ and define the sign of the metric by $$\operatorname{sgn}(g) := g_{(n)}(\mu,\mu) = \pm 1$$ Then define the Hodge dual to be the operator $\Omega^{k}(M)\to\Omega^{n-k}(M)$ such that for every $\alpha,\beta\in\Omega^{k}(M)$ we have $$\alpha\wedge\star\beta = g_{(k)}(\alpha,\beta)\mu$$ which leads to $$\star\star = (-1)^{k(n-k)}\operatorname{sgn}(g)$$ and this implies $$\star^{-1} = (-1)^{k(n-k)}\operatorname{sgn}(g)\star$$

Now if we take the definition $$\delta\omega = (-1)^{k}\star^{-1}\operatorname{d}\star\omega$$ And we calculate the action of $\star^{-1}$ over $\operatorname{d}\star\omega$: If $\omega$ is of grade $k$, $\star\omega$ is of grade $n-k$ and $\operatorname{d}\star\omega$ is of grade $n-k+1$. Hence we have \begin{align} \star^{-1}\operatorname{d}\star\omega &= (-1)^{(n-k+1)(n-(n-k+1))}\star\operatorname{d}\star\omega\\ &= (-1)^{nk+n+k+1}\star\operatorname{d}\star\omega \end{align} Now, multiplying by $(-1)^{k}$ we get $$\delta\omega = (-1)^{kn+n+1}\operatorname{sgn}(g)\star\operatorname{d}\star\omega$$ Which is formula $(1)$. However, I've also encountered the formula \begin{align} \delta &= (-1)^{kn}\operatorname{sgn}(g)\star\operatorname{d}\star\tag{2} \end{align} Where does $(2)$ come from?

  • $\begingroup$ I'm aware that some people define the Hodge dual such that $$\alpha\wedge\star\beta = \operatorname{sgn}(g)g_{(k)}(\alpha,\beta)\mu$$ but that doesn't change $\star\star$ nor $\star^{-1}$ $\endgroup$ – Jackozee Hakkiuz Jun 23 '18 at 3:32

Okay. I've done it. I'll post it here. I think it may be of help to others.

The whole purpose of the codifferential is to be the adjoint of the exterior derivative with respect to the Hodge inner product. I.e. $$(\delta\omega,\eta) = (\omega,\operatorname{d}\eta)$$ Then, if you define the inner product by $$(\alpha,\beta) = \int_{M}\alpha\wedge\star\beta$$ you end up with the requirement that $$\delta=(-1)^{k}\star^{-1}\operatorname{d}\star$$ However, some people define the Hodge inner product to be $$(\alpha,\beta) = \int_{M}\star\alpha\wedge\beta$$ and this leads us to set the codifferential to $$\delta=(-1)^{n+k+1}\star^{-1}\operatorname{d}\star$$ which yields formula $(2)$ in the original post.

EDIT I'll leave the calculations here, for the curious ones. I'll denote the grade of $\omega$ by $|\omega|$, and set $|\eta|=|\omega|-1$. We also assume $\partial M = \emptyset$.

Case 1: $(\alpha,\beta) := \int_{M}\alpha\wedge\star\beta$

We have $$(\operatorname{d}\eta,\omega) = \int_{M}\operatorname{d}\eta\wedge\star\omega$$

Using $\operatorname{d}(\eta\wedge\star\omega) = \operatorname{d}\eta\wedge\star\omega + (-1)^{|\eta|} \eta\wedge\operatorname{d}\star\omega$ and $|\eta| = |\omega|-1$ $$=\int_{M}\operatorname{d}(\eta\wedge\star\omega) - (-1)^{|\omega|-1}\int_{M}\eta\wedge\operatorname{d}\star\omega$$ since $\partial M = \emptyset$, the first summand vanishes by Stokes' theorem, and we get \begin{align} &=(-1)^{|\omega|}\int_{M}\eta\wedge\operatorname{d}\star\omega\\ &=(-1)^{|\omega|}\int_{M}\eta\wedge\star\star^{-1}\operatorname{d}\star\omega\\ &=(-1)^{|\omega|}(\eta,\star^{-1}\operatorname{d}\star\omega) \end{align} and we define $\delta\omega = (-1)^{|\omega|}\star^{-1}\operatorname{d}\star\omega$ so that this equals $(\eta,\delta\omega)$

Case 2: $(\alpha,\beta) := \int_{M}\star\alpha\wedge\beta$

We have $$(\omega,\operatorname{d}\eta) = \int_{M}\star\omega\wedge\operatorname{d}\eta$$

Using $\operatorname{d}(\star\omega\wedge\eta) = \operatorname{d}\star\omega\wedge\eta + (-1)^{|\star\omega|} \star\omega\wedge\operatorname{d}\eta$ and $|\star\omega| = n-|\omega|$ $$=(-1)^{n-|\omega|}\left(\int_{M}\operatorname{d}(\star\omega\wedge\eta) - \int_{M}\operatorname{d}\star\omega\wedge\eta\right)$$ the first summand vanishes by Stokes' theorem (again since $\partial M = \emptyset$), and we get \begin{align} &=(-1)^{n-|\omega|+1}\int_{M}\operatorname{d}\star\omega\wedge\eta\\ &=(-1)^{n-|\omega|+1}\int_{M}\star\star^{-1}\operatorname{d}\star\omega\wedge\eta\\ &=(-1)^{n-|\omega|+1}(\star^{-1}\operatorname{d}\star\omega,\eta) \end{align} If we want this to equal $(\delta\omega,\eta)$, we need to define $\delta\omega = (-1)^{n+|\omega|+1}\star^{-1}\operatorname{d}\star\omega$ (since $(-1)^{n-|\omega|+1} = (-1)^{n+|\omega|+1}$).

And there you go.

  • 1
    $\begingroup$ I guess I'll accept my own answer. If someone has something to add, please do. I'd love to listen. Especially if there is a mistake. $\endgroup$ – Jackozee Hakkiuz Jun 27 '18 at 6:14

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