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









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1280286%2fcovariant-derivative-or-connection-of-and-along-a-curve%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1280286%2fcovariant-derivative-or-connection-of-and-along-a-curve%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            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 ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            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:"...
            Read more