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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeMay 6th 2020

    stated the definition Δf=ddf\Delta f = \star d \star d f and spelled out how this gives the usual component formula:

    ddf =d( jf)dx j =d(1(D1)!|det((g ij))|g ij( jf)ε ik 2k Ddx k 2dx k D) = k 1(1(D1)!|det((g ij))|g ij( jf)ε ik 2k Ddx k 1dx k 2dx k D) =|det((g ij))|1D!(D1)!ε l 1l 2l Dg l 1k 1g l 2k 2g l Dk Dε ik 2k D=det((g ij) 1)δ i k 1 k 1(|det((g ij))|g ij( jf)) =1|det((g ij))|δ i k 1 k 1(|det((g ij))|g ij( jf)) =1|det((g ij))| i(|det((g ij))|g ij( jf)) \begin{aligned} \star d \star d f & = \star d \star (\partial_j f) d x^j \\ & = \star d \left( \tfrac{1}{ \color{green} (D-1)! } \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } \, g^{ i j} (\partial_j f) \, \epsilon_{ i {\color{green} k_2 \cdots k_{D} } } d x^{ \color{green} k_2 } \wedge \cdots \wedge d x^{ \color{green} k_{D} } \right) \\ & = \star \partial_{ \color{magenta} k_1} \left( \tfrac{1}{ \color{green} (D-1)! } \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } \, g^{i j} (\partial_j f) \, \epsilon_{ i {\color{green} k_2 \cdots k_{D} } } d x^{ \color{magenta} k_1 } \wedge d x^{ \color{green} k_2 } \wedge \cdots \wedge d x^{ \color{green} k_{D} } \right) \\ & = \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } \underset{ = \det\big( (g_{i j})^{-1} \big) \delta^{ \color{magenta} k_1 }_i }{ \underbrace{ \tfrac{1}{ { \color{orange} D! } { \color{green} (D-1)! } } \epsilon_{ \color{orange} l_1 l_2 \cdots l_D } g^{ { \color{orange} l_1 } { \color{magenta} k_1 } } g^{ { \color{orange} l_2 } { \color{green} k_2 } } \cdots g^{ { \color{orange} l_D} { \color{green} k_D } } \epsilon_{ i {\color{green} k_2 \cdots k_{D} } } } } \, \partial_{ \color{magenta} k_1 } \left( \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } g^{i j} (\partial_j f) \right) \\ & = \frac{1}{ \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } } \delta^{ \color{magenta} k_1 }_i \partial_{ \color{magenta} k_1 } \left( \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } g^{i j} (\partial_j f) \right) \\ & = \frac{1}{ \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } } \partial_{i} \left( \sqrt{ \left\vert det\big( (g_{i j}) \big) \right\vert } g^{i j} (\partial_j f) \right) \end{aligned}

    diff, v7, current

    • CommentRowNumber2.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 6th 2020

    I always forget we have these colour effects:

    a{\bgcolor{red} a}

    b{\color{green} b}

    c{\bgcolor{green} \color{red} c}

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 6th 2020

    Ah, thanks for reminding me that there is the background color option, too: \bgcolor. Maybe that can be put to useā€¦