Evolution equation of Christoffel symbols under Mean Curvature Flow












2












$begingroup$


I'm reading a master thesis about Mean Curvature Flow and I'm trying understand how was developed the following equation:




$$frac{partial Gamma^i_{jk}}{partial t} = nabla A ast A.$$




It's how the equation above was developed:



enter image description here



I didn't understand how the terms highlighted appears. I know that $A ast nabla A$ means that I'm considering the linear combination of the contraction of $A$ and $nabla A$ with respect to the metric $g$ (as you can read on the beginning of the section $13$ on page $40$ of Hamilton's article), but I can't see why the line above $(3)$ it's a linear combination of contractions of $nabla H, A, H$ and $nabla A$ and why $nabla H ast A + H ast nabla A = nabla A ast A$. I think if I have more details about the operator $ast$, then I will be able to understand $(3)$.



Thanks in advance!



$textbf{EDIT:}$



I understood how obtain $(1)$ and (2) and I have an idea for answer my doubt
about $(3)$, but I couldn't understand how obtain $(3)$ yet.



$(1)$: using the fact that $frac{partial g^{il}}{partial t} = 2H h^{il}$, we observe that



$$frac{partial g^{il}}{partial t} = frac{partial (g^{is}g_{sz}g^{zl})}{partial t} = frac{partial (g^{is}g^{zl})}{partial t} g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



$$ = left( frac{partial g^{is}}{partial t} g^{zl} + g^{is} frac{partial g^{zl}}{partial t} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



$$ = left( 2H h^{is} g^{zl} + g^{is} 2H h^{zl} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



$$= 2H h^{is} g^l_s + g^i_z 2H h^{zl} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



$$= 2H h^{il} + 2H h^{il} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



$$=2 (2H h^{il}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



$$=2 (frac{partial g^{il}}{partial t}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



$$Longrightarrow frac{partial g^{il}}{partial t} = - g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



$(2)$: it's just observe that



$$Gamma^z_{jk} = g^{zl} left( frac{partial g_{kl}}{partial x_j} + frac{partial g_{jl}}{partial x_k} - frac{partial g_{jk}}{partial x_l} right)$$



$(3)$: Computing in normal coordinates, we observe that



$(circ) - H(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk}) = - g^{ij}h_{ij}(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk})$



$(circ circ) -h^i_k nabla_j H -h^i_j nabla_k H + h_{jk} nabla^i H = -h^i_k nabla_j (g^{rs}h_{rs}) -h^i_j nabla_k (g^{rs}h_{rs}) + h_{jk} nabla^i (g^{rs}h_{rs}) = -h^i_k left( g^{rs} nabla_j h_{rs} right) -h^i_j left( g^{rs} nabla_k h_{rs} right) + h_{jk} left( g^{rs} nabla^i h_{rs} right)$,



I don't be sure if I'm using the definition of $ast$ correctly, but if I'm not wrong, then $(circ)$ it's a linear combination which terms involve the trace of the tensor $ A$ and the components of the tensor $nabla A$ (it's the components of the tensor $nabla A$ if we rewrite $nabla_j h^i_k = g^{is} nabla_j h_{sk}$ and $nabla_k h^i_j = g^{is} nabla_k h_{sj}$), while $(circ circ)$ it's a linear combination which terms involve the trace of the tensor $nabla A$ and the components of the tensor $A$ (again, it's the components of the tensor $A$ if we rewrite $h^i_k = g^{is}h_{sk}$ and $h^i_j = g^{is}h_{sj}$) nad this justify why $(circ) + (circ circ) = nabla A ast A$, but this not answer why
$(circ) + (circ circ) = nabla H ast A + H ast nabla A$, which lead us to my question:



let be $T$ and $S$ $(0,2)$-tensors (just for simplicity) with components $T_{ij}$ and $S_{kl}$ and denote by $text{tr}_g T$ the trace of the tensor $T$ with respect to the metric $g$. Is it $T ast S$ a linear combination with terms like $text{tr}_g T S_{ij}$, $T_{ij} text{tr}_g S$, $text{tr}_g T text{tr}_g S$ or it's a linear combination with other kind of terms?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm reading a master thesis about Mean Curvature Flow and I'm trying understand how was developed the following equation:




    $$frac{partial Gamma^i_{jk}}{partial t} = nabla A ast A.$$




    It's how the equation above was developed:



    enter image description here



    I didn't understand how the terms highlighted appears. I know that $A ast nabla A$ means that I'm considering the linear combination of the contraction of $A$ and $nabla A$ with respect to the metric $g$ (as you can read on the beginning of the section $13$ on page $40$ of Hamilton's article), but I can't see why the line above $(3)$ it's a linear combination of contractions of $nabla H, A, H$ and $nabla A$ and why $nabla H ast A + H ast nabla A = nabla A ast A$. I think if I have more details about the operator $ast$, then I will be able to understand $(3)$.



    Thanks in advance!



    $textbf{EDIT:}$



    I understood how obtain $(1)$ and (2) and I have an idea for answer my doubt
    about $(3)$, but I couldn't understand how obtain $(3)$ yet.



    $(1)$: using the fact that $frac{partial g^{il}}{partial t} = 2H h^{il}$, we observe that



    $$frac{partial g^{il}}{partial t} = frac{partial (g^{is}g_{sz}g^{zl})}{partial t} = frac{partial (g^{is}g^{zl})}{partial t} g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



    $$ = left( frac{partial g^{is}}{partial t} g^{zl} + g^{is} frac{partial g^{zl}}{partial t} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



    $$ = left( 2H h^{is} g^{zl} + g^{is} 2H h^{zl} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



    $$= 2H h^{is} g^l_s + g^i_z 2H h^{zl} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



    $$= 2H h^{il} + 2H h^{il} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



    $$=2 (2H h^{il}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



    $$=2 (frac{partial g^{il}}{partial t}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



    $$Longrightarrow frac{partial g^{il}}{partial t} = - g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



    $(2)$: it's just observe that



    $$Gamma^z_{jk} = g^{zl} left( frac{partial g_{kl}}{partial x_j} + frac{partial g_{jl}}{partial x_k} - frac{partial g_{jk}}{partial x_l} right)$$



    $(3)$: Computing in normal coordinates, we observe that



    $(circ) - H(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk}) = - g^{ij}h_{ij}(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk})$



    $(circ circ) -h^i_k nabla_j H -h^i_j nabla_k H + h_{jk} nabla^i H = -h^i_k nabla_j (g^{rs}h_{rs}) -h^i_j nabla_k (g^{rs}h_{rs}) + h_{jk} nabla^i (g^{rs}h_{rs}) = -h^i_k left( g^{rs} nabla_j h_{rs} right) -h^i_j left( g^{rs} nabla_k h_{rs} right) + h_{jk} left( g^{rs} nabla^i h_{rs} right)$,



    I don't be sure if I'm using the definition of $ast$ correctly, but if I'm not wrong, then $(circ)$ it's a linear combination which terms involve the trace of the tensor $ A$ and the components of the tensor $nabla A$ (it's the components of the tensor $nabla A$ if we rewrite $nabla_j h^i_k = g^{is} nabla_j h_{sk}$ and $nabla_k h^i_j = g^{is} nabla_k h_{sj}$), while $(circ circ)$ it's a linear combination which terms involve the trace of the tensor $nabla A$ and the components of the tensor $A$ (again, it's the components of the tensor $A$ if we rewrite $h^i_k = g^{is}h_{sk}$ and $h^i_j = g^{is}h_{sj}$) nad this justify why $(circ) + (circ circ) = nabla A ast A$, but this not answer why
    $(circ) + (circ circ) = nabla H ast A + H ast nabla A$, which lead us to my question:



    let be $T$ and $S$ $(0,2)$-tensors (just for simplicity) with components $T_{ij}$ and $S_{kl}$ and denote by $text{tr}_g T$ the trace of the tensor $T$ with respect to the metric $g$. Is it $T ast S$ a linear combination with terms like $text{tr}_g T S_{ij}$, $T_{ij} text{tr}_g S$, $text{tr}_g T text{tr}_g S$ or it's a linear combination with other kind of terms?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm reading a master thesis about Mean Curvature Flow and I'm trying understand how was developed the following equation:




      $$frac{partial Gamma^i_{jk}}{partial t} = nabla A ast A.$$




      It's how the equation above was developed:



      enter image description here



      I didn't understand how the terms highlighted appears. I know that $A ast nabla A$ means that I'm considering the linear combination of the contraction of $A$ and $nabla A$ with respect to the metric $g$ (as you can read on the beginning of the section $13$ on page $40$ of Hamilton's article), but I can't see why the line above $(3)$ it's a linear combination of contractions of $nabla H, A, H$ and $nabla A$ and why $nabla H ast A + H ast nabla A = nabla A ast A$. I think if I have more details about the operator $ast$, then I will be able to understand $(3)$.



      Thanks in advance!



      $textbf{EDIT:}$



      I understood how obtain $(1)$ and (2) and I have an idea for answer my doubt
      about $(3)$, but I couldn't understand how obtain $(3)$ yet.



      $(1)$: using the fact that $frac{partial g^{il}}{partial t} = 2H h^{il}$, we observe that



      $$frac{partial g^{il}}{partial t} = frac{partial (g^{is}g_{sz}g^{zl})}{partial t} = frac{partial (g^{is}g^{zl})}{partial t} g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$ = left( frac{partial g^{is}}{partial t} g^{zl} + g^{is} frac{partial g^{zl}}{partial t} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$ = left( 2H h^{is} g^{zl} + g^{is} 2H h^{zl} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$= 2H h^{is} g^l_s + g^i_z 2H h^{zl} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$= 2H h^{il} + 2H h^{il} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$=2 (2H h^{il}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $$=2 (frac{partial g^{il}}{partial t}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $$Longrightarrow frac{partial g^{il}}{partial t} = - g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $(2)$: it's just observe that



      $$Gamma^z_{jk} = g^{zl} left( frac{partial g_{kl}}{partial x_j} + frac{partial g_{jl}}{partial x_k} - frac{partial g_{jk}}{partial x_l} right)$$



      $(3)$: Computing in normal coordinates, we observe that



      $(circ) - H(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk}) = - g^{ij}h_{ij}(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk})$



      $(circ circ) -h^i_k nabla_j H -h^i_j nabla_k H + h_{jk} nabla^i H = -h^i_k nabla_j (g^{rs}h_{rs}) -h^i_j nabla_k (g^{rs}h_{rs}) + h_{jk} nabla^i (g^{rs}h_{rs}) = -h^i_k left( g^{rs} nabla_j h_{rs} right) -h^i_j left( g^{rs} nabla_k h_{rs} right) + h_{jk} left( g^{rs} nabla^i h_{rs} right)$,



      I don't be sure if I'm using the definition of $ast$ correctly, but if I'm not wrong, then $(circ)$ it's a linear combination which terms involve the trace of the tensor $ A$ and the components of the tensor $nabla A$ (it's the components of the tensor $nabla A$ if we rewrite $nabla_j h^i_k = g^{is} nabla_j h_{sk}$ and $nabla_k h^i_j = g^{is} nabla_k h_{sj}$), while $(circ circ)$ it's a linear combination which terms involve the trace of the tensor $nabla A$ and the components of the tensor $A$ (again, it's the components of the tensor $A$ if we rewrite $h^i_k = g^{is}h_{sk}$ and $h^i_j = g^{is}h_{sj}$) nad this justify why $(circ) + (circ circ) = nabla A ast A$, but this not answer why
      $(circ) + (circ circ) = nabla H ast A + H ast nabla A$, which lead us to my question:



      let be $T$ and $S$ $(0,2)$-tensors (just for simplicity) with components $T_{ij}$ and $S_{kl}$ and denote by $text{tr}_g T$ the trace of the tensor $T$ with respect to the metric $g$. Is it $T ast S$ a linear combination with terms like $text{tr}_g T S_{ij}$, $T_{ij} text{tr}_g S$, $text{tr}_g T text{tr}_g S$ or it's a linear combination with other kind of terms?










      share|cite|improve this question











      $endgroup$




      I'm reading a master thesis about Mean Curvature Flow and I'm trying understand how was developed the following equation:




      $$frac{partial Gamma^i_{jk}}{partial t} = nabla A ast A.$$




      It's how the equation above was developed:



      enter image description here



      I didn't understand how the terms highlighted appears. I know that $A ast nabla A$ means that I'm considering the linear combination of the contraction of $A$ and $nabla A$ with respect to the metric $g$ (as you can read on the beginning of the section $13$ on page $40$ of Hamilton's article), but I can't see why the line above $(3)$ it's a linear combination of contractions of $nabla H, A, H$ and $nabla A$ and why $nabla H ast A + H ast nabla A = nabla A ast A$. I think if I have more details about the operator $ast$, then I will be able to understand $(3)$.



      Thanks in advance!



      $textbf{EDIT:}$



      I understood how obtain $(1)$ and (2) and I have an idea for answer my doubt
      about $(3)$, but I couldn't understand how obtain $(3)$ yet.



      $(1)$: using the fact that $frac{partial g^{il}}{partial t} = 2H h^{il}$, we observe that



      $$frac{partial g^{il}}{partial t} = frac{partial (g^{is}g_{sz}g^{zl})}{partial t} = frac{partial (g^{is}g^{zl})}{partial t} g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$ = left( frac{partial g^{is}}{partial t} g^{zl} + g^{is} frac{partial g^{zl}}{partial t} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$ = left( 2H h^{is} g^{zl} + g^{is} 2H h^{zl} right) g_{sz} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$= 2H h^{is} g^l_s + g^i_z 2H h^{zl} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$= 2H h^{il} + 2H h^{il} + (g^{is}g^{zl}) frac{partial g_{sz}}{partial t}$$



      $$=2 (2H h^{il}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $$=2 (frac{partial g^{il}}{partial t}) + g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $$Longrightarrow frac{partial g^{il}}{partial t} = - g^{is} frac{partial (g_{sz})}{partial t} g^{zl}$$



      $(2)$: it's just observe that



      $$Gamma^z_{jk} = g^{zl} left( frac{partial g_{kl}}{partial x_j} + frac{partial g_{jl}}{partial x_k} - frac{partial g_{jk}}{partial x_l} right)$$



      $(3)$: Computing in normal coordinates, we observe that



      $(circ) - H(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk}) = - g^{ij}h_{ij}(nabla_j h^i_k + nabla_k h^i_j - nabla^i h_{jk})$



      $(circ circ) -h^i_k nabla_j H -h^i_j nabla_k H + h_{jk} nabla^i H = -h^i_k nabla_j (g^{rs}h_{rs}) -h^i_j nabla_k (g^{rs}h_{rs}) + h_{jk} nabla^i (g^{rs}h_{rs}) = -h^i_k left( g^{rs} nabla_j h_{rs} right) -h^i_j left( g^{rs} nabla_k h_{rs} right) + h_{jk} left( g^{rs} nabla^i h_{rs} right)$,



      I don't be sure if I'm using the definition of $ast$ correctly, but if I'm not wrong, then $(circ)$ it's a linear combination which terms involve the trace of the tensor $ A$ and the components of the tensor $nabla A$ (it's the components of the tensor $nabla A$ if we rewrite $nabla_j h^i_k = g^{is} nabla_j h_{sk}$ and $nabla_k h^i_j = g^{is} nabla_k h_{sj}$), while $(circ circ)$ it's a linear combination which terms involve the trace of the tensor $nabla A$ and the components of the tensor $A$ (again, it's the components of the tensor $A$ if we rewrite $h^i_k = g^{is}h_{sk}$ and $h^i_j = g^{is}h_{sj}$) nad this justify why $(circ) + (circ circ) = nabla A ast A$, but this not answer why
      $(circ) + (circ circ) = nabla H ast A + H ast nabla A$, which lead us to my question:



      let be $T$ and $S$ $(0,2)$-tensors (just for simplicity) with components $T_{ij}$ and $S_{kl}$ and denote by $text{tr}_g T$ the trace of the tensor $T$ with respect to the metric $g$. Is it $T ast S$ a linear combination with terms like $text{tr}_g T S_{ij}$, $T_{ij} text{tr}_g S$, $text{tr}_g T text{tr}_g S$ or it's a linear combination with other kind of terms?







      riemannian-geometry mean-curvature-flows






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 1:10







      George

















      asked Nov 3 '18 at 18:59









      GeorgeGeorge

      801515




      801515






















          0






          active

          oldest

          votes











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2983271%2fevolution-equation-of-christoffel-symbols-under-mean-curvature-flow%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          0






          active

          oldest

          votes








          0






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2983271%2fevolution-equation-of-christoffel-symbols-under-mean-curvature-flow%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Mario Kart Wii

          What does “Dominus providebit” mean?

          File:Tiny Toon Adventures Wacky Sports JP Title.png