Differential map of velocity vector












1












$begingroup$


This is a very basic differential geometry question (please be patient, I am learning)
I am given the definition of the differential map of $phi:M to N$ as



$$dphi_p(v)(g)=v(gcircphi)$$



where $vin T_pM$ is some vector and $g$ is a smooth function on $N$ and I also know about the chain rule for these. (I suppose, I need to use it but cannot see how just yet)



Now, we have defined tangent vectors as derivations and only later I learned that a velocity vector to a curve at a particular point of the manifold lies in the tangent space to that point. Let the curve be $alpha(t)$ and its tangent vector $alpha '(t)$.
Now, its just stated that when pushing forward the velocity vector, I should do it like this



$$d phi (alpha'(t))=(phicircalpha)'(t)$$



But I cannot derive, why this is the right way and in particular I am also buzzed by the fact that I do not have to care about a particular point anymore. This pushforward seems to work fine for every $tin mathbb{R}$ whereas above I had to specify the point $p$ and my reference says explicitly:
The differential map of $phi: M to N$ moves individual tangent vectors from
$M$ to $N$ , but in general provides no way to move vector fields from $M$ to $N$
(or the reverse).



So why does it work for the entire tangents to the curve?






differential-geometry curves tangent-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    This is a very basic differential geometry question (please be patient, I am learning)
    I am given the definition of the differential map of $phi:M to N$ as



    $$dphi_p(v)(g)=v(gcircphi)$$



    where $vin T_pM$ is some vector and $g$ is a smooth function on $N$ and I also know about the chain rule for these. (I suppose, I need to use it but cannot see how just yet)



    Now, we have defined tangent vectors as derivations and only later I learned that a velocity vector to a curve at a particular point of the manifold lies in the tangent space to that point. Let the curve be $alpha(t)$ and its tangent vector $alpha '(t)$.
    Now, its just stated that when pushing forward the velocity vector, I should do it like this



    $$d phi (alpha'(t))=(phicircalpha)'(t)$$



    But I cannot derive, why this is the right way and in particular I am also buzzed by the fact that I do not have to care about a particular point anymore. This pushforward seems to work fine for every $tin mathbb{R}$ whereas above I had to specify the point $p$ and my reference says explicitly:
    The differential map of $phi: M to N$ moves individual tangent vectors from
    $M$ to $N$ , but in general provides no way to move vector fields from $M$ to $N$
    (or the reverse).



    So why does it work for the entire tangents to the curve?






    differential-geometry curves tangent-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This is a very basic differential geometry question (please be patient, I am learning)
      I am given the definition of the differential map of $phi:M to N$ as



      $$dphi_p(v)(g)=v(gcircphi)$$



      where $vin T_pM$ is some vector and $g$ is a smooth function on $N$ and I also know about the chain rule for these. (I suppose, I need to use it but cannot see how just yet)



      Now, we have defined tangent vectors as derivations and only later I learned that a velocity vector to a curve at a particular point of the manifold lies in the tangent space to that point. Let the curve be $alpha(t)$ and its tangent vector $alpha '(t)$.
      Now, its just stated that when pushing forward the velocity vector, I should do it like this



      $$d phi (alpha'(t))=(phicircalpha)'(t)$$



      But I cannot derive, why this is the right way and in particular I am also buzzed by the fact that I do not have to care about a particular point anymore. This pushforward seems to work fine for every $tin mathbb{R}$ whereas above I had to specify the point $p$ and my reference says explicitly:
      The differential map of $phi: M to N$ moves individual tangent vectors from
      $M$ to $N$ , but in general provides no way to move vector fields from $M$ to $N$
      (or the reverse).



      So why does it work for the entire tangents to the curve?






      differential-geometry curves tangent-spaces







      share|cite|improve this question









      $endgroup$




      This is a very basic differential geometry question (please be patient, I am learning)
      I am given the definition of the differential map of $phi:M to N$ as



      $$dphi_p(v)(g)=v(gcircphi)$$



      where $vin T_pM$ is some vector and $g$ is a smooth function on $N$ and I also know about the chain rule for these. (I suppose, I need to use it but cannot see how just yet)



      Now, we have defined tangent vectors as derivations and only later I learned that a velocity vector to a curve at a particular point of the manifold lies in the tangent space to that point. Let the curve be $alpha(t)$ and its tangent vector $alpha '(t)$.
      Now, its just stated that when pushing forward the velocity vector, I should do it like this



      $$d phi (alpha'(t))=(phicircalpha)'(t)$$



      But I cannot derive, why this is the right way and in particular I am also buzzed by the fact that I do not have to care about a particular point anymore. This pushforward seems to work fine for every $tin mathbb{R}$ whereas above I had to specify the point $p$ and my reference says explicitly:
      The differential map of $phi: M to N$ moves individual tangent vectors from
      $M$ to $N$ , but in general provides no way to move vector fields from $M$ to $N$
      (or the reverse).



      So why does it work for the entire tangents to the curve?






      differential-geometry curves tangent-spaces




      differential-geometry curves tangent-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 25 at 23:01









      MarslMarsl

      716




      716






















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          $begingroup$

          If $alpha:I to M$ is a curve defined in an open interval, then $I$ itself is a smooth manifold with a global chart $tcolon I to Bbb R$, and so by definition we have $$alpha'(t) doteq {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) in T_{alpha(t)}M.$$Here, under the identification $T_tI cong Bbb R$, the coordinate vector $partial/partial t|_t$ corresponds to the number $1$. This being understood, if $phicolon M to N$ is smooth, we have that $phicircalphacolon I to N$ is a curve, and the above definition applied this time for $phicirc alpha$ gives $$(phicircalpha)'(t) = {rm d}(phicircalpha)_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)} circ {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)}(alpha'(t))in T_{phi(alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $alpha'(t)$ again. If what bothers you is writing ${rm d}phi(alpha'(t))$ without indicating the base point $alpha'(t)$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $alpha'(t)in T_{alpha(t)}M$ and that distinct tangent spaces are disjoint.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly what I was looking for, sharp answer. Thx
            $endgroup$
            – Marsl
            Jan 26 at 0:21










          • $begingroup$
            Glad to help! $$phantom{}$$
            $endgroup$
            – Ivo Terek
            Jan 26 at 0:58













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          $begingroup$

          If $alpha:I to M$ is a curve defined in an open interval, then $I$ itself is a smooth manifold with a global chart $tcolon I to Bbb R$, and so by definition we have $$alpha'(t) doteq {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) in T_{alpha(t)}M.$$Here, under the identification $T_tI cong Bbb R$, the coordinate vector $partial/partial t|_t$ corresponds to the number $1$. This being understood, if $phicolon M to N$ is smooth, we have that $phicircalphacolon I to N$ is a curve, and the above definition applied this time for $phicirc alpha$ gives $$(phicircalpha)'(t) = {rm d}(phicircalpha)_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)} circ {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)}(alpha'(t))in T_{phi(alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $alpha'(t)$ again. If what bothers you is writing ${rm d}phi(alpha'(t))$ without indicating the base point $alpha'(t)$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $alpha'(t)in T_{alpha(t)}M$ and that distinct tangent spaces are disjoint.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly what I was looking for, sharp answer. Thx
            $endgroup$
            – Marsl
            Jan 26 at 0:21










          • $begingroup$
            Glad to help! $$phantom{}$$
            $endgroup$
            – Ivo Terek
            Jan 26 at 0:58


















          0












          $begingroup$

          If $alpha:I to M$ is a curve defined in an open interval, then $I$ itself is a smooth manifold with a global chart $tcolon I to Bbb R$, and so by definition we have $$alpha'(t) doteq {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) in T_{alpha(t)}M.$$Here, under the identification $T_tI cong Bbb R$, the coordinate vector $partial/partial t|_t$ corresponds to the number $1$. This being understood, if $phicolon M to N$ is smooth, we have that $phicircalphacolon I to N$ is a curve, and the above definition applied this time for $phicirc alpha$ gives $$(phicircalpha)'(t) = {rm d}(phicircalpha)_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)} circ {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)}(alpha'(t))in T_{phi(alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $alpha'(t)$ again. If what bothers you is writing ${rm d}phi(alpha'(t))$ without indicating the base point $alpha'(t)$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $alpha'(t)in T_{alpha(t)}M$ and that distinct tangent spaces are disjoint.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Exactly what I was looking for, sharp answer. Thx
            $endgroup$
            – Marsl
            Jan 26 at 0:21










          • $begingroup$
            Glad to help! $$phantom{}$$
            $endgroup$
            – Ivo Terek
            Jan 26 at 0:58
















          0












          0








          0





          $begingroup$

          If $alpha:I to M$ is a curve defined in an open interval, then $I$ itself is a smooth manifold with a global chart $tcolon I to Bbb R$, and so by definition we have $$alpha'(t) doteq {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) in T_{alpha(t)}M.$$Here, under the identification $T_tI cong Bbb R$, the coordinate vector $partial/partial t|_t$ corresponds to the number $1$. This being understood, if $phicolon M to N$ is smooth, we have that $phicircalphacolon I to N$ is a curve, and the above definition applied this time for $phicirc alpha$ gives $$(phicircalpha)'(t) = {rm d}(phicircalpha)_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)} circ {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)}(alpha'(t))in T_{phi(alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $alpha'(t)$ again. If what bothers you is writing ${rm d}phi(alpha'(t))$ without indicating the base point $alpha'(t)$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $alpha'(t)in T_{alpha(t)}M$ and that distinct tangent spaces are disjoint.






          share|cite|improve this answer









          $endgroup$



          If $alpha:I to M$ is a curve defined in an open interval, then $I$ itself is a smooth manifold with a global chart $tcolon I to Bbb R$, and so by definition we have $$alpha'(t) doteq {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) in T_{alpha(t)}M.$$Here, under the identification $T_tI cong Bbb R$, the coordinate vector $partial/partial t|_t$ corresponds to the number $1$. This being understood, if $phicolon M to N$ is smooth, we have that $phicircalphacolon I to N$ is a curve, and the above definition applied this time for $phicirc alpha$ gives $$(phicircalpha)'(t) = {rm d}(phicircalpha)_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)} circ {rm d}alpha_tleft(frac{partial}{partial t}bigg|_tright) = {rm d}phi_{alpha(t)}(alpha'(t))in T_{phi(alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $alpha'(t)$ again. If what bothers you is writing ${rm d}phi(alpha'(t))$ without indicating the base point $alpha'(t)$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $alpha'(t)in T_{alpha(t)}M$ and that distinct tangent spaces are disjoint.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 23:53









          Ivo TerekIvo Terek

          46.5k954143




          46.5k954143












          • $begingroup$
            Exactly what I was looking for, sharp answer. Thx
            $endgroup$
            – Marsl
            Jan 26 at 0:21










          • $begingroup$
            Glad to help! $$phantom{}$$
            $endgroup$
            – Ivo Terek
            Jan 26 at 0:58




















          • $begingroup$
            Exactly what I was looking for, sharp answer. Thx
            $endgroup$
            – Marsl
            Jan 26 at 0:21










          • $begingroup$
            Glad to help! $$phantom{}$$
            $endgroup$
            – Ivo Terek
            Jan 26 at 0:58


















          $begingroup$
          Exactly what I was looking for, sharp answer. Thx
          $endgroup$
          – Marsl
          Jan 26 at 0:21




          $begingroup$
          Exactly what I was looking for, sharp answer. Thx
          $endgroup$
          – Marsl
          Jan 26 at 0:21












          $begingroup$
          Glad to help! $$phantom{}$$
          $endgroup$
          – Ivo Terek
          Jan 26 at 0:58






          $begingroup$
          Glad to help! $$phantom{}$$
          $endgroup$
          – Ivo Terek
          Jan 26 at 0:58




















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