Covariant derivative induced by Levi-Civita connection and compatibility with Lie brackets The...

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Covariant derivative induced by Levi-Civita connection and compatibility with Lie brackets



The Next CEO of Stack OverflowLie bracket of canonical vectors on tangent space to a point on a manifold is zero.Expression for Levi-Civita ConnectionShape Operators and Symmetric Linear TransformationsParallel translation via $e$-connectionDerivations on a ManifoldLee's Riemannian Manifold, Zero curvature implies flatnesslocal expression for affine connectionsAbout a Morse function on the euclidean $n$-sphere.Compute $nabla_{alpha '} V$ in local coordinateVolume form is parallel with respect to Levi-Civita connectionAre Christoffel symbols structure coefficients?












2












$begingroup$


Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
$$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.




  1. $ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $


  2. $D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$



  3. $D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$


  4. $D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.


where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $



I want to prove
$$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$



Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
$$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
$$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
$$= X(V^j) partial_j - Y(U^i) partial_i$$



The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
$$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
$$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
$$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
$$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$



Where is my mistake?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
    $$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.




    1. $ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $


    2. $D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$



    3. $D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$


    4. $D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.


    where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $



    I want to prove
    $$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$



    Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
    $$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
    where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
    $$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
    $$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
    $$= X(V^j) partial_j - Y(U^i) partial_i$$



    The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
    $$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
    $$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
    $$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
    $$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$



    Where is my mistake?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
      $$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.




      1. $ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $


      2. $D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$



      3. $D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$


      4. $D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.


      where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $



      I want to prove
      $$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$



      Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
      $$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
      where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
      $$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
      $$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
      $$= X(V^j) partial_j - Y(U^i) partial_i$$



      The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
      $$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
      $$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
      $$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
      $$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$



      Where is my mistake?










      share|cite|improve this question











      $endgroup$




      Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
      $$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.




      1. $ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $


      2. $D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$



      3. $D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$


      4. $D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.


      where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $



      I want to prove
      $$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$



      Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
      $$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
      where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
      $$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
      $$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
      $$= X(V^j) partial_j - Y(U^i) partial_i$$



      The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
      $$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
      $$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
      $$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
      $$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$



      Where is my mistake?







      differential-geometry riemannian-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 4 at 15:12









      J. W. Tanner

      4,2361320




      4,2361320










      asked Feb 4 at 14:49









      Bremen000Bremen000

      517310




      517310






















          1 Answer
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          $begingroup$

          There is no mistake. Actually the remaining term is zero:



          We have



          begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
          =V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
          =V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}



          Similar



          $$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$



          so the remaining term is given by



          $$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$



          since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.






          share|cite|improve this answer









          $endgroup$














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            $begingroup$

            There is no mistake. Actually the remaining term is zero:



            We have



            begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
            =V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
            =V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}



            Similar



            $$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$



            so the remaining term is given by



            $$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$



            since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              There is no mistake. Actually the remaining term is zero:



              We have



              begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
              =V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
              =V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}



              Similar



              $$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$



              so the remaining term is given by



              $$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$



              since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                There is no mistake. Actually the remaining term is zero:



                We have



                begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
                =V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
                =V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}



                Similar



                $$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$



                so the remaining term is given by



                $$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$



                since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.






                share|cite|improve this answer









                $endgroup$



                There is no mistake. Actually the remaining term is zero:



                We have



                begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
                =V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
                =V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}



                Similar



                $$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$



                so the remaining term is given by



                $$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$



                since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 18 at 2:04









                triitrii

                81817




                81817






























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