How does following formula of the Lie bracket of two vector fields arise?












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Let X be a smooth manifold and $v, w in Gamma(TX)$ be two vector fields. Under the natural correspondence of vector fields and derivations on $C^infty(X)$ let $delta, epsilon: C^infty(X) rightarrow C^infty(X)$ be the corresponding derivations.



Then the commutator $[delta, epsilon] = delta circ epsilon - epsilon circ delta$ is another derivation. Define the Lie bracket $[v,w]$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $a in C^infty(X)$ as
$$delta(a) = sum v_i frac{partial a}{partial x_i} text{ and } epsilon(a) = sum w_i frac{partial a}{partial x_i}$$



Using this how can $[v,w]$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.



$$[v,w] = sum_{i,j} bigg(v_i frac{partial w_j}{partial x_i}- w_i frac{partial v_j}{partial x_i}bigg) frac{partial}{partial x_j}$$






calculus geometry differential-geometry manifolds







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    $begingroup$


    Let X be a smooth manifold and $v, w in Gamma(TX)$ be two vector fields. Under the natural correspondence of vector fields and derivations on $C^infty(X)$ let $delta, epsilon: C^infty(X) rightarrow C^infty(X)$ be the corresponding derivations.



    Then the commutator $[delta, epsilon] = delta circ epsilon - epsilon circ delta$ is another derivation. Define the Lie bracket $[v,w]$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $a in C^infty(X)$ as
    $$delta(a) = sum v_i frac{partial a}{partial x_i} text{ and } epsilon(a) = sum w_i frac{partial a}{partial x_i}$$



    Using this how can $[v,w]$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.



    $$[v,w] = sum_{i,j} bigg(v_i frac{partial w_j}{partial x_i}- w_i frac{partial v_j}{partial x_i}bigg) frac{partial}{partial x_j}$$






    calculus geometry differential-geometry manifolds







    share|cite|improve this question









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      0





      $begingroup$


      Let X be a smooth manifold and $v, w in Gamma(TX)$ be two vector fields. Under the natural correspondence of vector fields and derivations on $C^infty(X)$ let $delta, epsilon: C^infty(X) rightarrow C^infty(X)$ be the corresponding derivations.



      Then the commutator $[delta, epsilon] = delta circ epsilon - epsilon circ delta$ is another derivation. Define the Lie bracket $[v,w]$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $a in C^infty(X)$ as
      $$delta(a) = sum v_i frac{partial a}{partial x_i} text{ and } epsilon(a) = sum w_i frac{partial a}{partial x_i}$$



      Using this how can $[v,w]$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.



      $$[v,w] = sum_{i,j} bigg(v_i frac{partial w_j}{partial x_i}- w_i frac{partial v_j}{partial x_i}bigg) frac{partial}{partial x_j}$$






      calculus geometry differential-geometry manifolds







      share|cite|improve this question









      $endgroup$




      Let X be a smooth manifold and $v, w in Gamma(TX)$ be two vector fields. Under the natural correspondence of vector fields and derivations on $C^infty(X)$ let $delta, epsilon: C^infty(X) rightarrow C^infty(X)$ be the corresponding derivations.



      Then the commutator $[delta, epsilon] = delta circ epsilon - epsilon circ delta$ is another derivation. Define the Lie bracket $[v,w]$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $a in C^infty(X)$ as
      $$delta(a) = sum v_i frac{partial a}{partial x_i} text{ and } epsilon(a) = sum w_i frac{partial a}{partial x_i}$$



      Using this how can $[v,w]$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.



      $$[v,w] = sum_{i,j} bigg(v_i frac{partial w_j}{partial x_i}- w_i frac{partial v_j}{partial x_i}bigg) frac{partial}{partial x_j}$$






      calculus geometry differential-geometry manifolds




      calculus geometry differential-geometry manifolds






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      asked Dec 16 '18 at 21:12









      gengen

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          Your suspicion is correct. In the same local coordinates you have
          begin{equation*}
          begin{aligned}
          [v,w](a)&=deltacirc epsilon(a)-epsiloncirc delta(a)\
          &=Big(sum_i v_ifrac{partial}{partial x_i}Big(sum_j w_jfrac{partial a}{partial x_j}Big)Big)-Big(sum_j w_jfrac{partial}{partial x_j}Big(sum_i v_ifrac{partial a}{partial x_i}Big)Big)\
          &=sum_{i,j} Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}+v_iw_jfrac{partial^2a}{partial x_ix_j}Big)-sum_{j,i} Big(w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}+w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
          &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}-w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}Big)+sum_{i,j} Big(v_iw_jfrac{partial^2a}{partial x_ix_j}-w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
          &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial a}{partial x_j},
          end{aligned}
          end{equation*}
          where we smoothness of $a$ to cancel $frac{partial^2a}{partial x_ix_j}-frac{partial^2a}{partial x_jx_i}=0.$ Therefore
          $$[v,w]=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial}{partial x_j}.$$






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            $begingroup$

            Associating the vector field $v_i$ with the differential operator $v:=v_ipartial^i$ gives $vwf=v_ipartial^i(w_jpartial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,,w]f=Rf$, where $R$ is the expression for $[v,,w]$ you desire. The rest is an exercise.






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              1












              $begingroup$

              Your suspicion is correct. In the same local coordinates you have
              begin{equation*}
              begin{aligned}
              [v,w](a)&=deltacirc epsilon(a)-epsiloncirc delta(a)\
              &=Big(sum_i v_ifrac{partial}{partial x_i}Big(sum_j w_jfrac{partial a}{partial x_j}Big)Big)-Big(sum_j w_jfrac{partial}{partial x_j}Big(sum_i v_ifrac{partial a}{partial x_i}Big)Big)\
              &=sum_{i,j} Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}+v_iw_jfrac{partial^2a}{partial x_ix_j}Big)-sum_{j,i} Big(w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}+w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
              &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}-w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}Big)+sum_{i,j} Big(v_iw_jfrac{partial^2a}{partial x_ix_j}-w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
              &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial a}{partial x_j},
              end{aligned}
              end{equation*}
              where we smoothness of $a$ to cancel $frac{partial^2a}{partial x_ix_j}-frac{partial^2a}{partial x_jx_i}=0.$ Therefore
              $$[v,w]=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial}{partial x_j}.$$






              share|cite|improve this answer









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                1












                $begingroup$

                Your suspicion is correct. In the same local coordinates you have
                begin{equation*}
                begin{aligned}
                [v,w](a)&=deltacirc epsilon(a)-epsiloncirc delta(a)\
                &=Big(sum_i v_ifrac{partial}{partial x_i}Big(sum_j w_jfrac{partial a}{partial x_j}Big)Big)-Big(sum_j w_jfrac{partial}{partial x_j}Big(sum_i v_ifrac{partial a}{partial x_i}Big)Big)\
                &=sum_{i,j} Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}+v_iw_jfrac{partial^2a}{partial x_ix_j}Big)-sum_{j,i} Big(w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}+w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}-w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}Big)+sum_{i,j} Big(v_iw_jfrac{partial^2a}{partial x_ix_j}-w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial a}{partial x_j},
                end{aligned}
                end{equation*}
                where we smoothness of $a$ to cancel $frac{partial^2a}{partial x_ix_j}-frac{partial^2a}{partial x_jx_i}=0.$ Therefore
                $$[v,w]=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial}{partial x_j}.$$






                share|cite|improve this answer









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                  1












                  1








                  1





                  $begingroup$

                  Your suspicion is correct. In the same local coordinates you have
                  begin{equation*}
                  begin{aligned}
                  [v,w](a)&=deltacirc epsilon(a)-epsiloncirc delta(a)\
                  &=Big(sum_i v_ifrac{partial}{partial x_i}Big(sum_j w_jfrac{partial a}{partial x_j}Big)Big)-Big(sum_j w_jfrac{partial}{partial x_j}Big(sum_i v_ifrac{partial a}{partial x_i}Big)Big)\
                  &=sum_{i,j} Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}+v_iw_jfrac{partial^2a}{partial x_ix_j}Big)-sum_{j,i} Big(w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}+w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                  &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}-w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}Big)+sum_{i,j} Big(v_iw_jfrac{partial^2a}{partial x_ix_j}-w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                  &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial a}{partial x_j},
                  end{aligned}
                  end{equation*}
                  where we smoothness of $a$ to cancel $frac{partial^2a}{partial x_ix_j}-frac{partial^2a}{partial x_jx_i}=0.$ Therefore
                  $$[v,w]=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial}{partial x_j}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Your suspicion is correct. In the same local coordinates you have
                  begin{equation*}
                  begin{aligned}
                  [v,w](a)&=deltacirc epsilon(a)-epsiloncirc delta(a)\
                  &=Big(sum_i v_ifrac{partial}{partial x_i}Big(sum_j w_jfrac{partial a}{partial x_j}Big)Big)-Big(sum_j w_jfrac{partial}{partial x_j}Big(sum_i v_ifrac{partial a}{partial x_i}Big)Big)\
                  &=sum_{i,j} Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}+v_iw_jfrac{partial^2a}{partial x_ix_j}Big)-sum_{j,i} Big(w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}+w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                  &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}frac{partial a}{partial x_j}-w_jfrac{partial v_i}{partial x_j}frac{partial a}{partial x_i}Big)+sum_{i,j} Big(v_iw_jfrac{partial^2a}{partial x_ix_j}-w_jv_ifrac{partial^2a}{partial x_jx_i}Big)\
                  &=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial a}{partial x_j},
                  end{aligned}
                  end{equation*}
                  where we smoothness of $a$ to cancel $frac{partial^2a}{partial x_ix_j}-frac{partial^2a}{partial x_jx_i}=0.$ Therefore
                  $$[v,w]=sum_{i,j}Big(v_ifrac{partial w_j}{partial x_i}-w_ifrac{partial v_j}{partial x_i}Big)frac{partial}{partial x_j}.$$







                  share|cite|improve this answer












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                  answered Dec 19 '18 at 21:15









                  positrón0802positrón0802

                  4,353520




                  4,353520























                      0












                      $begingroup$

                      Associating the vector field $v_i$ with the differential operator $v:=v_ipartial^i$ gives $vwf=v_ipartial^i(w_jpartial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,,w]f=Rf$, where $R$ is the expression for $[v,,w]$ you desire. The rest is an exercise.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Associating the vector field $v_i$ with the differential operator $v:=v_ipartial^i$ gives $vwf=v_ipartial^i(w_jpartial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,,w]f=Rf$, where $R$ is the expression for $[v,,w]$ you desire. The rest is an exercise.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Associating the vector field $v_i$ with the differential operator $v:=v_ipartial^i$ gives $vwf=v_ipartial^i(w_jpartial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,,w]f=Rf$, where $R$ is the expression for $[v,,w]$ you desire. The rest is an exercise.






                          share|cite|improve this answer









                          $endgroup$



                          Associating the vector field $v_i$ with the differential operator $v:=v_ipartial^i$ gives $vwf=v_ipartial^i(w_jpartial^jf)$ for a sufficiently nice function $f$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $[v,,w]f=Rf$, where $R$ is the expression for $[v,,w]$ you desire. The rest is an exercise.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 16 '18 at 21:16









                          J.G.J.G.

                          27.1k22843




                          27.1k22843






























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