Derivative of metric along curve












2












$begingroup$


Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
$$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).



I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
$$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
but
$$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$



I do not see how these two expressions are equal.



Is there something wrong? Is there something I am missing?



Thanks!






differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry







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    2












    $begingroup$


    Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
    $$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
    (Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).



    I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
    $$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
    but
    $$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$



    I do not see how these two expressions are equal.



    Is there something wrong? Is there something I am missing?



    Thanks!






    differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry







    share|cite|improve this question







    New contributor




    Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
      $$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
      (Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).



      I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
      $$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
      but
      $$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$



      I do not see how these two expressions are equal.



      Is there something wrong? Is there something I am missing?



      Thanks!






      differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry







      share|cite|improve this question







      New contributor




      Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
      $$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
      (Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).



      I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
      $$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
      but
      $$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$



      I do not see how these two expressions are equal.



      Is there something wrong? Is there something I am missing?



      Thanks!






      differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry




      differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry






      share|cite|improve this question







      New contributor




      Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






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      asked Jan 7 at 2:08









      Quim LlorensQuim Llorens

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      132




      New contributor




      Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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          1 Answer
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          $begingroup$

          From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as



          $$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$



          Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,



          $$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.






          share|cite|improve this answer









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          • $begingroup$
            Lovely, thank you!!!! :)
            $endgroup$
            – Quim Llorens
            Jan 7 at 12:38











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          1 Answer
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          1 Answer
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          active

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          active

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          0












          $begingroup$

          From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as



          $$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$



          Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,



          $$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Lovely, thank you!!!! :)
            $endgroup$
            – Quim Llorens
            Jan 7 at 12:38
















          0












          $begingroup$

          From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as



          $$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$



          Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,



          $$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Lovely, thank you!!!! :)
            $endgroup$
            – Quim Llorens
            Jan 7 at 12:38














          0












          0








          0





          $begingroup$

          From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as



          $$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$



          Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,



          $$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.






          share|cite|improve this answer









          $endgroup$



          From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as



          $$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$



          Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,



          $$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 7:24









          kobekobe

          34.8k22247




          34.8k22247












          • $begingroup$
            Lovely, thank you!!!! :)
            $endgroup$
            – Quim Llorens
            Jan 7 at 12:38


















          • $begingroup$
            Lovely, thank you!!!! :)
            $endgroup$
            – Quim Llorens
            Jan 7 at 12:38
















          $begingroup$
          Lovely, thank you!!!! :)
          $endgroup$
          – Quim Llorens
          Jan 7 at 12:38




          $begingroup$
          Lovely, thank you!!!! :)
          $endgroup$
          – Quim Llorens
          Jan 7 at 12:38










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