---- > [!definition] Definition. ([[Hodge star]]) > Let $(M,g)$ be an [[orientation of a vector bundle|oriented]] [[Riemannian manifold|Riemannian]] [[smooth manifold|manifold]] of dimension $n$. Let $\omega=\omega_{g}$ be the [[Riemannian volume form]] induced by the orientation and metric. Let $x \in M$, $p \in [n]$, and recall [[inner product]] $\langle -,- \rangle_{g}$ on $\Lambda^{p}T^{*}_{x}M$ induced by $g$. > > The **Hodge star operator** on $(M,g,\omega)$ is the unique [[linear map]] $\star: \Lambda^{p}T_{x}^{*}M \to \Lambda^{n-p}T_{x}^{*}M$ satisfying $\alpha \wedge \star \beta= \langle \alpha, \beta \rangle_{g} \omega_{g} $for all $\alpha, \beta \in \Lambda^{p}T_{x}^{*}M$. It is uniquely determined by the identity $\star(e^{1} \wedge \dots \wedge e^{p})=e^{p+1} \wedge \dots \wedge e^{n},$ > where $(e^{i}) \subset \Gamma(T^{*}U)$ form the orthonormal coframe field giving rise to $\omega_{g}$ (i.e. locally $\omega_{g} |_{U}=e^{1} \wedge \dots \wedge e^{n}$.) > > The **Hodge codifferential**[^1] is the map $\delta:\Omega^{p}(M) \to \Omega^{p-1}(M)$ defined as the composition ($\Omega^{-1}(M):=0$) > $\Omega^{p}(M) \xrightarrow{ \star} \Omega^{n-p}(M) \xrightarrow{d} \Omega^{n-p+1}(M) \xrightarrow{ \star} \Omega^{p-1}(M) \xrightarrow{\cdot(-1)^{n(p+1)+1}} \Omega^{p-1}(M), $ > that is, $\delta=\begin{cases} > (-1)^{n(p+1)+1} \star d\star & p > 0; \\ > 0 & p=0. > \end{cases}$ > It is (formally) [[adjoint|adjoint]] to the [[exterior derivative|exterior differential]] $d:\Omega^{p}(M) \to \Omega^{p+1}(M)$, in the $L^{2}$-[[L2 inner product of differential forms|sense]] (see below). > [!basicexample] > With the help of $\star$, any $2$-form on a surface in $\mathbb{R}^{3}$ can be realized as a normal field. ^basic-example > [!basicproperties] > > - $\star1=\omega_{g}$ > - $\star \omega_{g}=1$ > - $\star \star |_{\Lambda^{p}T_{x}^{*}M}=(-1)^{p(n-p)} \id$ > - As a corollary of the [[geometer's integration by parts]], [[integration of a compactly supported volume form on an oriented smooth manifold|one obtains]] the (formal) adjointness relation $\begin{align} \langle \langle d \xi, \eta \rangle \rangle_{M,g}&= \langle \langle \xi, \delta \eta \rangle \rangle _{M, g}, \text{ that is, } \\ \int _{M} \langle d \xi, \eta \rangle _{g} \omega_{g} &= \int _{M} \langle \xi, \delta \eta \rangle _{g} \omega_{g}, \end{align} $ for all [[compact|compactly]] [[support|supported]] $\xi\in \Omega^{p-1}(M), \eta \in \Omega^{p}(M)$ Summary: 1. Start with $\langle \langle d \alpha, \beta \rangle \rangle=\int _{M} \overbrace{ \langle d\alpha, \beta \rangle \omega_{g} }^{ d\alpha \wedge \star \beta }$. 2. Then $\langle \langle \alpha, \delta \beta \rangle \rangle=\int _{M} \langle \alpha, \delta \beta \rangle \omega _g=\int _{M}\overbrace{ \langle \alpha, (-1)^{\text{smt}} \star d \star \beta \rangle \omega_{g} }^{ \alpha \wedge \star (\star d \star \beta) }$. Up to sign stuff, this is just $\int _{M} \alpha \wedge d \star \beta$. 3. Apply Stokes theorem to $d(\alpha \wedge \star \beta)$ and notice that the terms that come out are the two sides of the desired equality, up to sign. 4. Figure out the signs By [[Generalized Stokes' Theorem|Stokes' Theorem]], $\int _{M} d(\alpha \wedge \star \beta)=0$. By the defining property of [[exterior derivative]], $d(\alpha \wedge \star \beta)=d \alpha \wedge \star \beta + (-1)^{p-1}\alpha \wedge {d} \star{\beta}$. So [[geometer's integration by parts|we]] have $0= \int _{M} d(\alpha \wedge \star \beta)=\int _{M} \overbrace{ d \alpha \wedge \star \beta }^{ \langle d \alpha, \beta \rangle_{g} \omega _{g} } + \int _{M} (-1)^{p-1}\alpha \wedge {d} \star{\beta} ,$ hence $\langle \langle d \alpha, \beta \rangle \rangle_{M, g}=\int _{M} \langle d \alpha, \beta \rangle_{g} \omega _g=(-1)^{p} \alpha \wedge d \star \beta$. WTS the RHS equals $\langle \langle \alpha, \delta \beta \rangle \rangle$. $\begin{align} \langle \langle \alpha, \delta \beta \rangle \rangle &= \langle \langle \alpha, (-1)^{n(p+1)+1} \star d \star \beta\rangle \rangle \\ &=(-1)^{n(p+1)+1} \int _{M} \overbrace{ \langle \alpha, \star d \star \beta \rangle _{g} \omega _{g} }^{ = \alpha \wedge \star (\star d \star \beta), \text{ by def. of } \star }\\ &=(-1)^{n(p+1)+1} \int _{M} \alpha \wedge \overbrace{ \star \star }^{ (-1)^{(p-1)(n-p+1) \id} } d \star \beta \\ &= (-1)^{n(p+1)+1} (-1)^{(p-1)(n-p+1) } \int _{M} \alpha \wedge d \star \beta. \end{align}$ We are done if we can show the coefficient is in fact $(-1)^{p-1}$. This is true; details are left as an exercise. ---- #### [^1]: This terminology a bit [[(co)homology of a complex|(co)homologically]]-backward, but is easy to get used to. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```