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Confusion about tensor fields taking values in vector fields rather than functions



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A question about Killing vector and Riemann curvature tensorA question about a definition of Ricci curvatureShow isometry of flow on a compact Riemannian manifold where the vector field is KillingUsing index notation to write $d^2=0$ in terms of a torsion free connection.connection laplacian on general vector bundlesBasic question: Riemannian Curvature is nondegenerateHow to derive the Riemann Curvature Tensor?Curvature tensor for a manifold with torsionWhat comes after the curvature tensor in “higher derivatives”?Relation between the Curvature of a Connection and the Curvature of the Induced Connection on the Frame Bundle










0












$begingroup$


I'm beginning with tensorial calculus and I have some questions. Let $(M,g)$ a riemannian manifold with $nabla$ his Levi Civita connection. The curvature tensor $R$ is defined as



beginalign*
R : mathfrakX(M) times mathfrakX(M) times mathfrakX(M) &to mathfrakX(M)\
(X,Y,Z) &mapsto R(X,Y)Z
endalign*



where



$$R(X,Y)Z = nabla_X nabla_Y Z - nabla_Y nabla_X Z - nabla_[X,Y]Z.$$



In all books of differential geometry, they said that $R$ is a $(1,3)$ tensor but I don't know why, because a $(1,3)$ tensor is a multilinear map



$$Omega^1(M) times mathfrakX(M) times mathfrakX(M)timesmathfrakX(M) to C^infty(M).$$



I think that I don't understand any concept or because for example, let $X in mathfrakX(M)$, then $nabla X : TM to TM$, $(nabla X)(v_p) = nabla_v_pX$, is a $(1,1)$ tensor field and I don't know why.



$textbfRemark$: $Omega^1(M)$ is the set of all 1-forms, $alpha : M to TM^*$.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I'm beginning with tensorial calculus and I have some questions. Let $(M,g)$ a riemannian manifold with $nabla$ his Levi Civita connection. The curvature tensor $R$ is defined as



    beginalign*
    R : mathfrakX(M) times mathfrakX(M) times mathfrakX(M) &to mathfrakX(M)\
    (X,Y,Z) &mapsto R(X,Y)Z
    endalign*



    where



    $$R(X,Y)Z = nabla_X nabla_Y Z - nabla_Y nabla_X Z - nabla_[X,Y]Z.$$



    In all books of differential geometry, they said that $R$ is a $(1,3)$ tensor but I don't know why, because a $(1,3)$ tensor is a multilinear map



    $$Omega^1(M) times mathfrakX(M) times mathfrakX(M)timesmathfrakX(M) to C^infty(M).$$



    I think that I don't understand any concept or because for example, let $X in mathfrakX(M)$, then $nabla X : TM to TM$, $(nabla X)(v_p) = nabla_v_pX$, is a $(1,1)$ tensor field and I don't know why.



    $textbfRemark$: $Omega^1(M)$ is the set of all 1-forms, $alpha : M to TM^*$.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I'm beginning with tensorial calculus and I have some questions. Let $(M,g)$ a riemannian manifold with $nabla$ his Levi Civita connection. The curvature tensor $R$ is defined as



      beginalign*
      R : mathfrakX(M) times mathfrakX(M) times mathfrakX(M) &to mathfrakX(M)\
      (X,Y,Z) &mapsto R(X,Y)Z
      endalign*



      where



      $$R(X,Y)Z = nabla_X nabla_Y Z - nabla_Y nabla_X Z - nabla_[X,Y]Z.$$



      In all books of differential geometry, they said that $R$ is a $(1,3)$ tensor but I don't know why, because a $(1,3)$ tensor is a multilinear map



      $$Omega^1(M) times mathfrakX(M) times mathfrakX(M)timesmathfrakX(M) to C^infty(M).$$



      I think that I don't understand any concept or because for example, let $X in mathfrakX(M)$, then $nabla X : TM to TM$, $(nabla X)(v_p) = nabla_v_pX$, is a $(1,1)$ tensor field and I don't know why.



      $textbfRemark$: $Omega^1(M)$ is the set of all 1-forms, $alpha : M to TM^*$.










      share|cite|improve this question











      $endgroup$




      I'm beginning with tensorial calculus and I have some questions. Let $(M,g)$ a riemannian manifold with $nabla$ his Levi Civita connection. The curvature tensor $R$ is defined as



      beginalign*
      R : mathfrakX(M) times mathfrakX(M) times mathfrakX(M) &to mathfrakX(M)\
      (X,Y,Z) &mapsto R(X,Y)Z
      endalign*



      where



      $$R(X,Y)Z = nabla_X nabla_Y Z - nabla_Y nabla_X Z - nabla_[X,Y]Z.$$



      In all books of differential geometry, they said that $R$ is a $(1,3)$ tensor but I don't know why, because a $(1,3)$ tensor is a multilinear map



      $$Omega^1(M) times mathfrakX(M) times mathfrakX(M)timesmathfrakX(M) to C^infty(M).$$



      I think that I don't understand any concept or because for example, let $X in mathfrakX(M)$, then $nabla X : TM to TM$, $(nabla X)(v_p) = nabla_v_pX$, is a $(1,1)$ tensor field and I don't know why.



      $textbfRemark$: $Omega^1(M)$ is the set of all 1-forms, $alpha : M to TM^*$.







      differential-geometry tensors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 25 at 19:19









      Michael Albanese

      64.8k1599315




      64.8k1599315










      asked Mar 25 at 18:58









      hal97hal97

      907




      907




















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^infty(M)$-multilinear map $T : Omega^1(M)^ptimesmathfrakX(M)^q to C^infty(M)$.



          Given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to mathfrakX(M)$, there is an associated $(p + 1, q)$-tensor field $T : Omega^1(M)^p+1timesmathfrakX(M)^q to C^infty(M)$ defined by



          $$T(beta, alpha^1, dots, alpha^p, X_1, dots, X_q) := beta(S(alpha^1, dots, alpha^p, X_1, dots, X_q)).$$



          Likewise, given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : Omega^1(M)^ptimesmathfrakX(M)^q+1 to C^infty(M)$ defined by



          $$T(alpha^1, dots, alpha^p, Y, X_1, dots, X_q) := (S(alpha^1, dots, alpha^p, X_1, dots, X_q))(Y).$$






          share|cite|improve this answer









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            $begingroup$

            A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^infty(M)$-multilinear map $T : Omega^1(M)^ptimesmathfrakX(M)^q to C^infty(M)$.



            Given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to mathfrakX(M)$, there is an associated $(p + 1, q)$-tensor field $T : Omega^1(M)^p+1timesmathfrakX(M)^q to C^infty(M)$ defined by



            $$T(beta, alpha^1, dots, alpha^p, X_1, dots, X_q) := beta(S(alpha^1, dots, alpha^p, X_1, dots, X_q)).$$



            Likewise, given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : Omega^1(M)^ptimesmathfrakX(M)^q+1 to C^infty(M)$ defined by



            $$T(alpha^1, dots, alpha^p, Y, X_1, dots, X_q) := (S(alpha^1, dots, alpha^p, X_1, dots, X_q))(Y).$$






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^infty(M)$-multilinear map $T : Omega^1(M)^ptimesmathfrakX(M)^q to C^infty(M)$.



              Given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to mathfrakX(M)$, there is an associated $(p + 1, q)$-tensor field $T : Omega^1(M)^p+1timesmathfrakX(M)^q to C^infty(M)$ defined by



              $$T(beta, alpha^1, dots, alpha^p, X_1, dots, X_q) := beta(S(alpha^1, dots, alpha^p, X_1, dots, X_q)).$$



              Likewise, given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : Omega^1(M)^ptimesmathfrakX(M)^q+1 to C^infty(M)$ defined by



              $$T(alpha^1, dots, alpha^p, Y, X_1, dots, X_q) := (S(alpha^1, dots, alpha^p, X_1, dots, X_q))(Y).$$






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^infty(M)$-multilinear map $T : Omega^1(M)^ptimesmathfrakX(M)^q to C^infty(M)$.



                Given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to mathfrakX(M)$, there is an associated $(p + 1, q)$-tensor field $T : Omega^1(M)^p+1timesmathfrakX(M)^q to C^infty(M)$ defined by



                $$T(beta, alpha^1, dots, alpha^p, X_1, dots, X_q) := beta(S(alpha^1, dots, alpha^p, X_1, dots, X_q)).$$



                Likewise, given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : Omega^1(M)^ptimesmathfrakX(M)^q+1 to C^infty(M)$ defined by



                $$T(alpha^1, dots, alpha^p, Y, X_1, dots, X_q) := (S(alpha^1, dots, alpha^p, X_1, dots, X_q))(Y).$$






                share|cite|improve this answer









                $endgroup$



                A $(p, q)$-tensor field on a smooth manifold $M$ is a $C^infty(M)$-multilinear map $T : Omega^1(M)^ptimesmathfrakX(M)^q to C^infty(M)$.



                Given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to mathfrakX(M)$, there is an associated $(p + 1, q)$-tensor field $T : Omega^1(M)^p+1timesmathfrakX(M)^q to C^infty(M)$ defined by



                $$T(beta, alpha^1, dots, alpha^p, X_1, dots, X_q) := beta(S(alpha^1, dots, alpha^p, X_1, dots, X_q)).$$



                Likewise, given a $C^infty(M)$-multilinear map $S : Omega^1(M)^ptimesmathfrakX(M)^q to Omega^1(M)$, there is an associated $(p, q + 1)$-tensor field $T : Omega^1(M)^ptimesmathfrakX(M)^q+1 to C^infty(M)$ defined by



                $$T(alpha^1, dots, alpha^p, Y, X_1, dots, X_q) := (S(alpha^1, dots, alpha^p, X_1, dots, X_q))(Y).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 25 at 19:19









                Michael AlbaneseMichael Albanese

                64.8k1599315




                64.8k1599315



























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