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Given a riemannian Mainifold $(M,g)$, the Hodge-$\star$-Operator, the codifferential as $$\delta:\Omega^k(M)\rightarrow\Omega^{k-1}(M):\omega\mapsto(-1)^{n(k-1)-1}\star d\star \omega$$ and the Laplacian as $$\Delta:\Omega^k(M)\rightarrow\Omega^k(M):\omega\mapsto d\delta \omega+\delta d \omega.$$ I want to proof that the Laplacian in local coordinates is given by $$\Delta f=-\frac{1}{\sqrt{\det g}}\sum_{ij}\frac{\partial}{\partial x_i}(\sqrt{\det g}\cdot g^{ij}\frac{\partial f}{\partial x_j}).$$ I only know about the Hodge-$\star$-Operator that $\star^2=(-1)^{k(n-k)}$ and $\alpha\wedge\star\beta=<\alpha,\beta>dVol$.

I started with \begin{align} \Delta f=&d(\delta f) + \delta d f\\ =&\delta d f\\ =& (-1)^{n-1}\star d \star d f\\ =&(-1)^{n-1}\star d \star \sum \frac{\partial f}{\partial x_i} dx_i\\ =&(-1)^{n-1}\star d \sum \frac{\partial f}{\partial x_i} \star dx_i \end{align} Here i got stuck with the identities of the Hodge-$\star$-Operator. I don't see where i could get that $g^{ij}$ from. I have seen definitions of the $\star$-Operator which involve these but i couldn't show any equivalence.

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    $\begingroup$ If you are using $\alpha \wedge \star \beta = \langle \alpha,\beta \rangle ~d\text{Vol}$, then on the right side $\langle~,~\rangle_g$ and $d\text{Vol}_g$ both depend on the metric. Do you know what the dependence is? You can use this to figure out how $\star$ should act on say $dx_i$. $\endgroup$
    – Keshav
    May 10, 2020 at 12:47
  • $\begingroup$ $dVol_g= \sqrt{\det g} dx_1\dots dx_n$ depends on $g$, but in don't see it for $<.,.>$ if i remember correcltly it was defined via an ONB $\{v_i\}$ of $T_pM$ and $<\alpha,\beta>=\sum\alpha (v_I)\cdot\beta (v_I)$. So i use $g$ only there to find the orthonormal basis. $\endgroup$
    – Timmathy
    May 10, 2020 at 13:04
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    $\begingroup$ $\langle~,~\rangle$ is exactly what the matrix $g_{ij}$ is for (in particular for non-orthonormal bases). If $\alpha = \sum_i \alpha_i dx^i$, etc. then $\langle \alpha,\beta\rangle = \sum_{ij}g^{ij}\alpha_i \beta_j$. $\endgroup$
    – Keshav
    May 10, 2020 at 14:11
  • $\begingroup$ Oh, so it was hidden in plain sight... Thank you very much! $\endgroup$
    – Timmathy
    May 10, 2020 at 14:18

1 Answer 1

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Given an orientable Riemannian manifold $(M,g)$, the Laplacian and the Hodge-star-operator as before.

To show:$$\Delta f=-\frac{1}{\sqrt{\det g}}\sum_{ij}\frac{\partial}{\partial x_i}(\sqrt{\det g}\cdot g^{ij}\frac{\partial f}{\partial x_j}).$$ We know that $\star^2=(-1)^{k(n-k)}$ and $\alpha\wedge\star\beta=\langle\alpha,\beta\rangle dVol$.

Using the hints of @Keshav. We can display the product of the basis of differential forms as \begin{align} \langle dx_i,dx_j\rangle = \sum g^{kl}\delta_{ik}\delta_{jl}=g^{ij}. \end{align}

Also looking at the base of n-1-forms we can display $\star dx_i$ as $\star dx_i=\sum_{l=1}^n P_l \hat{dx}_l\wedge dx_1\wedge \dots \wedge dx_n$ for some $P_l:M\rightarrow \mathbb{R}$.

Let's combine these two ideas \begin{align} g^{ij}dVol&=\langle dx_i,dx_j\rangle dVol\\ &=dx_i\wedge\star dx_j\\ &=\sum_{l=1}^n P_l dx_l\wedge \hat{dx_l}\wedge dx_i\wedge\dots\wedge dx_n\\ &\text{the base of the n-1-forms and the 1 forms cancels out if }i\neq j\\ &=P_i(-1)^i dx_1\wedge\dots\wedge dx_n \end{align} So we can finish the proof with \begin{align} \Delta f=&d(\delta f) + \delta d f\\ =&\delta d f\\ =& (-1)^{n-1}\star d \star d f\\ =&(-1)^{n-1}\star d \star \sum \frac{\partial f}{\partial x_i} dx_i\\ =&(-1)^{n-1}\star d (\sum \frac{\partial f}{\partial x_i} \star dx_i)\\ =&(-1)^{n-1}\star d (\sum \frac{\partial f}{\partial x_i}(\sum (-1)^ig^{ji} \sqrt{|g|}\hat{dx_j}\wedge dx_1\wedge\dots\wedge dx_n))\\ &\text{putting things in order and using linearity of the exterior derivative we get}\\ =& (-1)\frac{1}{\sqrt{|g|}}\sum\frac{\partial}{\partial x_i}(\frac{\partial f}{\partial x_i} g^{ij}\sqrt{|g|})dx_1\wedge\dots\wedge dx_n \end{align}

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