Euclidean Connection and Constant Vector fields












1












$begingroup$


Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$



Then the Euclidean connection is defined as:



$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$



I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.



What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$






differential-geometry manifolds smooth-manifolds







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
    $endgroup$
    – Aweygan
    Dec 6 '18 at 2:01










  • $begingroup$
    You're totally right, thanks. That was silly on my part
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:02










  • $begingroup$
    @Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:05


















1












$begingroup$


Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$



Then the Euclidean connection is defined as:



$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$



I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.



What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$






differential-geometry manifolds smooth-manifolds







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
    $endgroup$
    – Aweygan
    Dec 6 '18 at 2:01










  • $begingroup$
    You're totally right, thanks. That was silly on my part
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:02










  • $begingroup$
    @Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:05
















1












1








1





$begingroup$


Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$



Then the Euclidean connection is defined as:



$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$



I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.



What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$






differential-geometry manifolds smooth-manifolds







share|cite|improve this question









$endgroup$




Let $nabla$ be the Euclidean connection on a manifold $M = mathbb{R}^n$. The definition I'm following is if $X, Y$ is are smooth vector fields on $M$ with $Y$ given by:
$$
Y = sum_{i}^{n} Y^i frac{partial}{partial x_i}
$$



Then the Euclidean connection is defined as:



$$
nabla_Y X = sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}
$$



I want to show that given a curve $gamma: I rightarrow M $, a vector field whose euclidean connection with $gamma'$ is $0$ has to be constant.



What I don't understand is how we can act on $gamma^i$ with $X$ when $gamma ^i$ is a map from $I subset mathbb{R}$ into $M$, unlike $Y^i$ which is $C^infty(M)$






differential-geometry manifolds smooth-manifolds




differential-geometry manifolds smooth-manifolds






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 1:55









David FengDavid Feng

929




929








  • 1




    $begingroup$
    When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
    $endgroup$
    – Aweygan
    Dec 6 '18 at 2:01










  • $begingroup$
    You're totally right, thanks. That was silly on my part
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:02










  • $begingroup$
    @Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:05
















  • 1




    $begingroup$
    When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
    $endgroup$
    – Aweygan
    Dec 6 '18 at 2:01










  • $begingroup$
    You're totally right, thanks. That was silly on my part
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:02










  • $begingroup$
    @Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
    $endgroup$
    – David Feng
    Dec 6 '18 at 2:05










1




1




$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01




$begingroup$
When you define $nabla_YX$, shouldn't that be $nabla_XY$? Usually a connection is $C^infty(M)$-linear in the subscript. This would also eliminate the issue your having.
$endgroup$
– Aweygan
Dec 6 '18 at 2:01












$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02




$begingroup$
You're totally right, thanks. That was silly on my part
$endgroup$
– David Feng
Dec 6 '18 at 2:02












$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05






$begingroup$
@Aweygan If you'd like to post your comment I could mark it as the answer, I think I've figured out the rest :)
$endgroup$
– David Feng
Dec 6 '18 at 2:05












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$begingroup$

The definition of the Euclidean connection should be



$$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$



(This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.






share|cite|improve this answer









$endgroup$













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    $begingroup$

    The definition of the Euclidean connection should be



    $$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$



    (This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The definition of the Euclidean connection should be



      $$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$



      (This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The definition of the Euclidean connection should be



        $$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$



        (This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.






        share|cite|improve this answer









        $endgroup$



        The definition of the Euclidean connection should be



        $$nabla_XY=sum_{i}^{n} X(Y^i)frac{partial}{partial x_i}.$$



        (This makes $nabla$ satisfy the standard definition of a connection.) Using this, it should be easy to define $nabla_{gamma'}Y$ for a curve $gamma$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 2:08









        AweyganAweygan

        14.1k21441




        14.1k21441






























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