What does the torsion-free condition for a connection mean in terms of its horizontal bundle? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A geometric interpretation of the Levi-Civita connection?Are bundle gerbes bundles of algebras? Terminology of “covariant derivative” and various “connections”How do we use an Ehresmann connection to define a semispray?Torsion-free $G$-StructuresPrincipal bundles and Subriemannian GeometryGeometric interpretation of horizontal and vertical lift of vector fieldLevi-Civita connections from metrics on the orthogonal frame bundleProjectively flat Weyl connection on closed higher genus surfaceIs a symmetric, parallel (0,2)-tensor a metric?

What does the torsion-free condition for a connection mean in terms of its horizontal bundle?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A geometric interpretation of the Levi-Civita connection?Are bundle gerbes bundles of algebras? Terminology of “covariant derivative” and various “connections”How do we use an Ehresmann connection to define a semispray?Torsion-free $G$-StructuresPrincipal bundles and Subriemannian GeometryGeometric interpretation of horizontal and vertical lift of vector fieldLevi-Civita connections from metrics on the orthogonal frame bundleProjectively flat Weyl connection on closed higher genus surfaceIs a symmetric, parallel (0,2)-tensor a metric?










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$begingroup$


I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.



The aim of this question is to try to finally put this uncomfortable condition to rest.



Ehresmann Connections



Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $Erightarrow M$ is just a choice of a complementary subbundle to $ker(TE rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.



If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)



I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $mathbbR$-linear map $Gamma(E)rightarrowGamma(Eotimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.



Torsion-Freeness



A Levi-Civita connection is a connection that:
1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.)
2. It is torsion-free.
Torsion free means $nabla_XY - nabla_YX = [X,Y]$.



This definition very heavily uses the less intuitive notion of connection.



So:



Questions



  1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)

  2. I realized that I don't actually have handy an example of a connection on $mathbbR^2$ that preserves the canonical Riemannian metric on $mathbbR^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.









share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.



    The aim of this question is to try to finally put this uncomfortable condition to rest.



    Ehresmann Connections



    Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $Erightarrow M$ is just a choice of a complementary subbundle to $ker(TE rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.



    If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)



    I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $mathbbR$-linear map $Gamma(E)rightarrowGamma(Eotimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.



    Torsion-Freeness



    A Levi-Civita connection is a connection that:
    1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.)
    2. It is torsion-free.
    Torsion free means $nabla_XY - nabla_YX = [X,Y]$.



    This definition very heavily uses the less intuitive notion of connection.



    So:



    Questions



    1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)

    2. I realized that I don't actually have handy an example of a connection on $mathbbR^2$ that preserves the canonical Riemannian metric on $mathbbR^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.









    share|cite|improve this question









    $endgroup$














      3












      3








      3


      1



      $begingroup$


      I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.



      The aim of this question is to try to finally put this uncomfortable condition to rest.



      Ehresmann Connections



      Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $Erightarrow M$ is just a choice of a complementary subbundle to $ker(TE rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.



      If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)



      I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $mathbbR$-linear map $Gamma(E)rightarrowGamma(Eotimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.



      Torsion-Freeness



      A Levi-Civita connection is a connection that:
      1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.)
      2. It is torsion-free.
      Torsion free means $nabla_XY - nabla_YX = [X,Y]$.



      This definition very heavily uses the less intuitive notion of connection.



      So:



      Questions



      1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)

      2. I realized that I don't actually have handy an example of a connection on $mathbbR^2$ that preserves the canonical Riemannian metric on $mathbbR^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.









      share|cite|improve this question









      $endgroup$




      I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.



      The aim of this question is to try to finally put this uncomfortable condition to rest.



      Ehresmann Connections



      Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $Erightarrow M$ is just a choice of a complementary subbundle to $ker(TE rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.



      If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)



      I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $mathbbR$-linear map $Gamma(E)rightarrowGamma(Eotimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.



      Torsion-Freeness



      A Levi-Civita connection is a connection that:
      1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.)
      2. It is torsion-free.
      Torsion free means $nabla_XY - nabla_YX = [X,Y]$.



      This definition very heavily uses the less intuitive notion of connection.



      So:



      Questions



      1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)

      2. I realized that I don't actually have handy an example of a connection on $mathbbR^2$ that preserves the canonical Riemannian metric on $mathbbR^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.






      dg.differential-geometry intuition






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      asked 5 hours ago









      Andrew NCAndrew NC

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          $begingroup$

          You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.



          An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of




          S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963







          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:




            1. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.



            Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              2












              $begingroup$

              You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.



              An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of




              S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963







              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.



                An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of




                S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963







                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.



                  An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of




                  S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963







                  share|cite|improve this answer









                  $endgroup$



                  You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.



                  An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of




                  S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Liviu NicolaescuLiviu Nicolaescu

                  25.9k260111




                  25.9k260111





















                      0












                      $begingroup$

                      Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:




                      1. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.



                      Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$






                      share|cite|improve this answer











                      $endgroup$

















                        0












                        $begingroup$

                        Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:




                        1. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.



                        Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$






                        share|cite|improve this answer











                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:




                          1. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.



                          Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$






                          share|cite|improve this answer











                          $endgroup$



                          Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:




                          1. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.



                          Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$







                          share|cite|improve this answer














                          share|cite|improve this answer



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                          edited 13 mins ago

























                          answered 18 mins ago









                          Ivo TerekIvo Terek

                          442211




                          442211



























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