Covariant derivative (or connection) of and along a curve












0












$begingroup$


Let $c:[a,b] rightarrow M$ be a curve parametrized by arc-length.



Then $c': [a,b] rightarrow TM$ such that $c'(t) in T_{c(t)}(M).$



So essentially, I want to understand how $nabla_{c'}c'$ is defined then in coordinates.



Cause normally, we are dealing with vector fields $X : M rightarrow TM$ when we talk about the covariant derivative or connection.



In this case, we have for $X,Y: M rightarrow TM$ that $X= sum_{i} eta^i partial_i$ and $Y = sum_{i} zeta^i partial_i$



In this case,



$$nabla_YX = sum_{i,j,k} zeta^i( eta^j Gamma_{i,j}^k partial_k + partial_i eta^j partial_j)$$



and everything is well-defined.



The problem occurs, cause $c' : [a,b] rightarrow TM$ so there is no way we can take a partial derivative in some coordinate direction of $c',$
as it just depends on time. Is there still a way to explain how $nabla_c'c'$ is for example defined, although $c'$ is not really a vector field?






real-analysis analysis differential-geometry differential-topology covariance







share|cite|improve this question











$endgroup$












  • $begingroup$
    Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
    $endgroup$
    – Braindead
    May 13 '15 at 14:05












  • $begingroup$
    You can also extend the vector field "trivially" within a coordinate patch.
    $endgroup$
    – Braindead
    May 13 '15 at 14:08
















0












$begingroup$


Let $c:[a,b] rightarrow M$ be a curve parametrized by arc-length.



Then $c': [a,b] rightarrow TM$ such that $c'(t) in T_{c(t)}(M).$



So essentially, I want to understand how $nabla_{c'}c'$ is defined then in coordinates.



Cause normally, we are dealing with vector fields $X : M rightarrow TM$ when we talk about the covariant derivative or connection.



In this case, we have for $X,Y: M rightarrow TM$ that $X= sum_{i} eta^i partial_i$ and $Y = sum_{i} zeta^i partial_i$



In this case,



$$nabla_YX = sum_{i,j,k} zeta^i( eta^j Gamma_{i,j}^k partial_k + partial_i eta^j partial_j)$$



and everything is well-defined.



The problem occurs, cause $c' : [a,b] rightarrow TM$ so there is no way we can take a partial derivative in some coordinate direction of $c',$
as it just depends on time. Is there still a way to explain how $nabla_c'c'$ is for example defined, although $c'$ is not really a vector field?






real-analysis analysis differential-geometry differential-topology covariance







share|cite|improve this question











$endgroup$












  • $begingroup$
    Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
    $endgroup$
    – Braindead
    May 13 '15 at 14:05












  • $begingroup$
    You can also extend the vector field "trivially" within a coordinate patch.
    $endgroup$
    – Braindead
    May 13 '15 at 14:08














0












0








0





$begingroup$


Let $c:[a,b] rightarrow M$ be a curve parametrized by arc-length.



Then $c': [a,b] rightarrow TM$ such that $c'(t) in T_{c(t)}(M).$



So essentially, I want to understand how $nabla_{c'}c'$ is defined then in coordinates.



Cause normally, we are dealing with vector fields $X : M rightarrow TM$ when we talk about the covariant derivative or connection.



In this case, we have for $X,Y: M rightarrow TM$ that $X= sum_{i} eta^i partial_i$ and $Y = sum_{i} zeta^i partial_i$



In this case,



$$nabla_YX = sum_{i,j,k} zeta^i( eta^j Gamma_{i,j}^k partial_k + partial_i eta^j partial_j)$$



and everything is well-defined.



The problem occurs, cause $c' : [a,b] rightarrow TM$ so there is no way we can take a partial derivative in some coordinate direction of $c',$
as it just depends on time. Is there still a way to explain how $nabla_c'c'$ is for example defined, although $c'$ is not really a vector field?






real-analysis analysis differential-geometry differential-topology covariance







share|cite|improve this question











$endgroup$




Let $c:[a,b] rightarrow M$ be a curve parametrized by arc-length.



Then $c': [a,b] rightarrow TM$ such that $c'(t) in T_{c(t)}(M).$



So essentially, I want to understand how $nabla_{c'}c'$ is defined then in coordinates.



Cause normally, we are dealing with vector fields $X : M rightarrow TM$ when we talk about the covariant derivative or connection.



In this case, we have for $X,Y: M rightarrow TM$ that $X= sum_{i} eta^i partial_i$ and $Y = sum_{i} zeta^i partial_i$



In this case,



$$nabla_YX = sum_{i,j,k} zeta^i( eta^j Gamma_{i,j}^k partial_k + partial_i eta^j partial_j)$$



and everything is well-defined.



The problem occurs, cause $c' : [a,b] rightarrow TM$ so there is no way we can take a partial derivative in some coordinate direction of $c',$
as it just depends on time. Is there still a way to explain how $nabla_c'c'$ is for example defined, although $c'$ is not really a vector field?






real-analysis analysis differential-geometry differential-topology covariance




real-analysis analysis differential-geometry differential-topology covariance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 13 '15 at 13:49







BanachSteinhaus

















asked May 13 '15 at 13:31









BanachSteinhausBanachSteinhaus

193




193












  • $begingroup$
    Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
    $endgroup$
    – Braindead
    May 13 '15 at 14:05












  • $begingroup$
    You can also extend the vector field "trivially" within a coordinate patch.
    $endgroup$
    – Braindead
    May 13 '15 at 14:08


















  • $begingroup$
    Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
    $endgroup$
    – Braindead
    May 13 '15 at 14:05












  • $begingroup$
    You can also extend the vector field "trivially" within a coordinate patch.
    $endgroup$
    – Braindead
    May 13 '15 at 14:08
















$begingroup$
Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
$endgroup$
– Braindead
May 13 '15 at 14:05






$begingroup$
Here is one way: think of $c'$ as a section of the pull-back bundle $c^*TM$ over $[a,b]$. Then $nabla_c' c'$ can be thought of as the derivative of $c'$ in the time direction: $(c^*nabla)_{d/dt} c'$.
$endgroup$
– Braindead
May 13 '15 at 14:05














$begingroup$
You can also extend the vector field "trivially" within a coordinate patch.
$endgroup$
– Braindead
May 13 '15 at 14:08




$begingroup$
You can also extend the vector field "trivially" within a coordinate patch.
$endgroup$
– Braindead
May 13 '15 at 14:08










2 Answers
2






active

oldest

votes


















0












$begingroup$

HINT: Write $c'(t) = sumlimits_i a^i(t)partial_ibig|_{c(t)}$. Then recall that for any vector field $X$ we can compute $nabla_{c'}X = dfrac D{dt}(Xcirc c)(t)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
    $endgroup$
    – RealAnalysis
    May 13 '15 at 17:47












  • $begingroup$
    I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
    $endgroup$
    – RealAnalysis
    May 14 '15 at 16:17












  • $begingroup$
    @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
    $endgroup$
    – Ted Shifrin
    May 14 '15 at 16:43





















0












$begingroup$

Actually a very easy but nice way to see this is to assume first that $c'$ is extended smoothly to a vector field $X$. Now like you already wrote locally $X = sum_i X^i partial_i$ and$$nabla_X X = sum_k (X(X^k) + sum_{i,j}X^{i}X^j Gamma^k_{ij}), partial_k$$
To get $nabla_{c'}c'$ we only need to evaluate $nabla_X X$ along $c$, which is straight forward except for $X(X^k)$. Therefore we calculate (using a chart $x = (x^1,dots,x^n))$: $$X_{c(t)}(X^k) = frac{d}{d tau}bigg{rvert}_{t}( X^k circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} (X(x^k) circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} X_{c(tau)}(x^k) \ = frac{d}{d tau}bigg{rvert}_{t} frac{d}{d s}bigg{rvert}_tau (x^k circ c)(s) = {(x^k circ c)}''(t)$$
Here I simply used that $X$ is tangent along $c$. Moreover this calculation shows that we only need to know $X$ along $c$, that is $c'$. Therefore $nabla_{c'} c' $ is well defined along $c$ and by the above formula (writing $c_j = x^j circ c$):



$$nabla_{c'} c' = sum_k (c_j'' + sum_{i,j}c_i'c_j' Gamma^k_{ij}), partial_k $$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    HINT: Write $c'(t) = sumlimits_i a^i(t)partial_ibig|_{c(t)}$. Then recall that for any vector field $X$ we can compute $nabla_{c'}X = dfrac D{dt}(Xcirc c)(t)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
      $endgroup$
      – RealAnalysis
      May 13 '15 at 17:47












    • $begingroup$
      I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
      $endgroup$
      – RealAnalysis
      May 14 '15 at 16:17












    • $begingroup$
      @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
      $endgroup$
      – Ted Shifrin
      May 14 '15 at 16:43


















    0












    $begingroup$

    HINT: Write $c'(t) = sumlimits_i a^i(t)partial_ibig|_{c(t)}$. Then recall that for any vector field $X$ we can compute $nabla_{c'}X = dfrac D{dt}(Xcirc c)(t)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
      $endgroup$
      – RealAnalysis
      May 13 '15 at 17:47












    • $begingroup$
      I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
      $endgroup$
      – RealAnalysis
      May 14 '15 at 16:17












    • $begingroup$
      @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
      $endgroup$
      – Ted Shifrin
      May 14 '15 at 16:43
















    0












    0








    0





    $begingroup$

    HINT: Write $c'(t) = sumlimits_i a^i(t)partial_ibig|_{c(t)}$. Then recall that for any vector field $X$ we can compute $nabla_{c'}X = dfrac D{dt}(Xcirc c)(t)$.






    share|cite|improve this answer











    $endgroup$



    HINT: Write $c'(t) = sumlimits_i a^i(t)partial_ibig|_{c(t)}$. Then recall that for any vector field $X$ we can compute $nabla_{c'}X = dfrac D{dt}(Xcirc c)(t)$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 13 '15 at 18:49

























    answered May 13 '15 at 14:38









    Ted ShifrinTed Shifrin

    64.9k44792




    64.9k44792












    • $begingroup$
      but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
      $endgroup$
      – RealAnalysis
      May 13 '15 at 17:47












    • $begingroup$
      I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
      $endgroup$
      – RealAnalysis
      May 14 '15 at 16:17












    • $begingroup$
      @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
      $endgroup$
      – Ted Shifrin
      May 14 '15 at 16:43




















    • $begingroup$
      but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
      $endgroup$
      – RealAnalysis
      May 13 '15 at 17:47












    • $begingroup$
      I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
      $endgroup$
      – RealAnalysis
      May 14 '15 at 16:17












    • $begingroup$
      @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
      $endgroup$
      – Ted Shifrin
      May 14 '15 at 16:43


















    $begingroup$
    but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
    $endgroup$
    – RealAnalysis
    May 13 '15 at 17:47






    $begingroup$
    but isn't this exactly the question of the TES, that the derivative $partial_j a^i$ (which would appear in the coordinate representation of the connection) is not defined?
    $endgroup$
    – RealAnalysis
    May 13 '15 at 17:47














    $begingroup$
    I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
    $endgroup$
    – RealAnalysis
    May 14 '15 at 16:17






    $begingroup$
    I thought about this hint and don't understand it, actually. If you say that $nabla_{c'}X(t) := (X circ c)'(t)$ then the problem seems to be that the right-hand side is not defined on a general manifold, as we don't know how the time-derivative is defined on the manifold or am I missunderstand your notation?
    $endgroup$
    – RealAnalysis
    May 14 '15 at 16:17














    $begingroup$
    @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
    $endgroup$
    – Ted Shifrin
    May 14 '15 at 16:43






    $begingroup$
    @RealAnalysis: You don't misunderstand the notation, but you need to think about the chain rule on a manifold.
    $endgroup$
    – Ted Shifrin
    May 14 '15 at 16:43













    0












    $begingroup$

    Actually a very easy but nice way to see this is to assume first that $c'$ is extended smoothly to a vector field $X$. Now like you already wrote locally $X = sum_i X^i partial_i$ and$$nabla_X X = sum_k (X(X^k) + sum_{i,j}X^{i}X^j Gamma^k_{ij}), partial_k$$
    To get $nabla_{c'}c'$ we only need to evaluate $nabla_X X$ along $c$, which is straight forward except for $X(X^k)$. Therefore we calculate (using a chart $x = (x^1,dots,x^n))$: $$X_{c(t)}(X^k) = frac{d}{d tau}bigg{rvert}_{t}( X^k circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} (X(x^k) circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} X_{c(tau)}(x^k) \ = frac{d}{d tau}bigg{rvert}_{t} frac{d}{d s}bigg{rvert}_tau (x^k circ c)(s) = {(x^k circ c)}''(t)$$
    Here I simply used that $X$ is tangent along $c$. Moreover this calculation shows that we only need to know $X$ along $c$, that is $c'$. Therefore $nabla_{c'} c' $ is well defined along $c$ and by the above formula (writing $c_j = x^j circ c$):



    $$nabla_{c'} c' = sum_k (c_j'' + sum_{i,j}c_i'c_j' Gamma^k_{ij}), partial_k $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Actually a very easy but nice way to see this is to assume first that $c'$ is extended smoothly to a vector field $X$. Now like you already wrote locally $X = sum_i X^i partial_i$ and$$nabla_X X = sum_k (X(X^k) + sum_{i,j}X^{i}X^j Gamma^k_{ij}), partial_k$$
      To get $nabla_{c'}c'$ we only need to evaluate $nabla_X X$ along $c$, which is straight forward except for $X(X^k)$. Therefore we calculate (using a chart $x = (x^1,dots,x^n))$: $$X_{c(t)}(X^k) = frac{d}{d tau}bigg{rvert}_{t}( X^k circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} (X(x^k) circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} X_{c(tau)}(x^k) \ = frac{d}{d tau}bigg{rvert}_{t} frac{d}{d s}bigg{rvert}_tau (x^k circ c)(s) = {(x^k circ c)}''(t)$$
      Here I simply used that $X$ is tangent along $c$. Moreover this calculation shows that we only need to know $X$ along $c$, that is $c'$. Therefore $nabla_{c'} c' $ is well defined along $c$ and by the above formula (writing $c_j = x^j circ c$):



      $$nabla_{c'} c' = sum_k (c_j'' + sum_{i,j}c_i'c_j' Gamma^k_{ij}), partial_k $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Actually a very easy but nice way to see this is to assume first that $c'$ is extended smoothly to a vector field $X$. Now like you already wrote locally $X = sum_i X^i partial_i$ and$$nabla_X X = sum_k (X(X^k) + sum_{i,j}X^{i}X^j Gamma^k_{ij}), partial_k$$
        To get $nabla_{c'}c'$ we only need to evaluate $nabla_X X$ along $c$, which is straight forward except for $X(X^k)$. Therefore we calculate (using a chart $x = (x^1,dots,x^n))$: $$X_{c(t)}(X^k) = frac{d}{d tau}bigg{rvert}_{t}( X^k circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} (X(x^k) circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} X_{c(tau)}(x^k) \ = frac{d}{d tau}bigg{rvert}_{t} frac{d}{d s}bigg{rvert}_tau (x^k circ c)(s) = {(x^k circ c)}''(t)$$
        Here I simply used that $X$ is tangent along $c$. Moreover this calculation shows that we only need to know $X$ along $c$, that is $c'$. Therefore $nabla_{c'} c' $ is well defined along $c$ and by the above formula (writing $c_j = x^j circ c$):



        $$nabla_{c'} c' = sum_k (c_j'' + sum_{i,j}c_i'c_j' Gamma^k_{ij}), partial_k $$






        share|cite|improve this answer









        $endgroup$



        Actually a very easy but nice way to see this is to assume first that $c'$ is extended smoothly to a vector field $X$. Now like you already wrote locally $X = sum_i X^i partial_i$ and$$nabla_X X = sum_k (X(X^k) + sum_{i,j}X^{i}X^j Gamma^k_{ij}), partial_k$$
        To get $nabla_{c'}c'$ we only need to evaluate $nabla_X X$ along $c$, which is straight forward except for $X(X^k)$. Therefore we calculate (using a chart $x = (x^1,dots,x^n))$: $$X_{c(t)}(X^k) = frac{d}{d tau}bigg{rvert}_{t}( X^k circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} (X(x^k) circ c)(tau) = frac{d}{d tau}bigg{rvert}_{t} X_{c(tau)}(x^k) \ = frac{d}{d tau}bigg{rvert}_{t} frac{d}{d s}bigg{rvert}_tau (x^k circ c)(s) = {(x^k circ c)}''(t)$$
        Here I simply used that $X$ is tangent along $c$. Moreover this calculation shows that we only need to know $X$ along $c$, that is $c'$. Therefore $nabla_{c'} c' $ is well defined along $c$ and by the above formula (writing $c_j = x^j circ c$):



        $$nabla_{c'} c' = sum_k (c_j'' + sum_{i,j}c_i'c_j' Gamma^k_{ij}), partial_k $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 18:55









        T'xT'x

        373212




        373212






























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            How to change which sound is reproduced for terminal bell?

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            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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