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Parsing notation for covariant derivative embedded in ambient euclidean space


Covariant derivative (or connection) of and along a curveBetter proof that vector fields on submanifolds extend globally iff submanifold is closedShow that the outward unit normal is smooth vector fieldCovariant derivative from an Ehresmann connection on a fibre bundleCovariant derivative of contracted tensor productCovariant derivative derived from embedding in higher-dimensional euclidean spaceCovariant derivative in Euclidean spaceComputing pullbackIntrinsic definition of Jacobian matrix on manifoldsCovariant Derivatives unaffected by change in basis













0












$begingroup$


$(triangledown_XY)_p=fracDdt|_p Y=pi(fracddt|_t=0(Y(gamma(t))=pi(dY_p(gamma'(0))$



Where $x_p = gamma'(0)$



So i'm still being thrown off sometimes by seeing vector fields as functions in $mathbbR^n$ and other times using them as derivative operators. My main question here is how exactly am I to understand $dY_p$?



Like this?:



$Y(p) = (y_1(p),..,y_n(p)) rightarrow dY_p=(dy_1(p),....,dy_n(p))$



so just taking the derivative of the components of $Y$?



And then from here, would we have $dY_p$ act on $gamma'(0)=x_p?$ I guess the obvious answer is just to write $dY_p$ as a jacobian matrix and $x_p$ as a vector and use matrix multiplication.



Is this all correct thinking?



Is this all there is to taking derivatives along vector fields when embedded in euclidean space? It's just one vector field you take the total derivative normally, and then apply that total derivative into whatever direction your other vector field at that point is pointing at and project to the tangent plane? Thanks










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    $(triangledown_XY)_p=fracDdt|_p Y=pi(fracddt|_t=0(Y(gamma(t))=pi(dY_p(gamma'(0))$



    Where $x_p = gamma'(0)$



    So i'm still being thrown off sometimes by seeing vector fields as functions in $mathbbR^n$ and other times using them as derivative operators. My main question here is how exactly am I to understand $dY_p$?



    Like this?:



    $Y(p) = (y_1(p),..,y_n(p)) rightarrow dY_p=(dy_1(p),....,dy_n(p))$



    so just taking the derivative of the components of $Y$?



    And then from here, would we have $dY_p$ act on $gamma'(0)=x_p?$ I guess the obvious answer is just to write $dY_p$ as a jacobian matrix and $x_p$ as a vector and use matrix multiplication.



    Is this all correct thinking?



    Is this all there is to taking derivatives along vector fields when embedded in euclidean space? It's just one vector field you take the total derivative normally, and then apply that total derivative into whatever direction your other vector field at that point is pointing at and project to the tangent plane? Thanks










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      $(triangledown_XY)_p=fracDdt|_p Y=pi(fracddt|_t=0(Y(gamma(t))=pi(dY_p(gamma'(0))$



      Where $x_p = gamma'(0)$



      So i'm still being thrown off sometimes by seeing vector fields as functions in $mathbbR^n$ and other times using them as derivative operators. My main question here is how exactly am I to understand $dY_p$?



      Like this?:



      $Y(p) = (y_1(p),..,y_n(p)) rightarrow dY_p=(dy_1(p),....,dy_n(p))$



      so just taking the derivative of the components of $Y$?



      And then from here, would we have $dY_p$ act on $gamma'(0)=x_p?$ I guess the obvious answer is just to write $dY_p$ as a jacobian matrix and $x_p$ as a vector and use matrix multiplication.



      Is this all correct thinking?



      Is this all there is to taking derivatives along vector fields when embedded in euclidean space? It's just one vector field you take the total derivative normally, and then apply that total derivative into whatever direction your other vector field at that point is pointing at and project to the tangent plane? Thanks










      share|cite|improve this question









      $endgroup$




      $(triangledown_XY)_p=fracDdt|_p Y=pi(fracddt|_t=0(Y(gamma(t))=pi(dY_p(gamma'(0))$



      Where $x_p = gamma'(0)$



      So i'm still being thrown off sometimes by seeing vector fields as functions in $mathbbR^n$ and other times using them as derivative operators. My main question here is how exactly am I to understand $dY_p$?



      Like this?:



      $Y(p) = (y_1(p),..,y_n(p)) rightarrow dY_p=(dy_1(p),....,dy_n(p))$



      so just taking the derivative of the components of $Y$?



      And then from here, would we have $dY_p$ act on $gamma'(0)=x_p?$ I guess the obvious answer is just to write $dY_p$ as a jacobian matrix and $x_p$ as a vector and use matrix multiplication.



      Is this all correct thinking?



      Is this all there is to taking derivatives along vector fields when embedded in euclidean space? It's just one vector field you take the total derivative normally, and then apply that total derivative into whatever direction your other vector field at that point is pointing at and project to the tangent plane? Thanks







      differential-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 16 at 23:06









      Mathematical MushroomMathematical Mushroom

      918




      918




















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