exterior derivative of the flow












2












$begingroup$


Let $E(x, y, z) = (x, y, z)$ be a vector field on $mathbb{R}^3$.



$α(x, y, z) = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy$ is a 2-form.



Find $phi^{t*}_Ealpha$.



The flow of X is $phi^t = (x_0e^t,y_0e^t,z_0e^t)$.



I need to compute $dphi_t^{*}$ as $phi_t^{*}alpha = alpha_{phi_t}(d phi_{t})$.



1) Is the pullbak of a k-form always a k-form?



2) If we consider $phi_{t}$ as a 1-form, $dphi_{t}$ is a 2-form, how to compute it?



Thank you for your help!






differential-geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 21:30












  • $begingroup$
    Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
    $endgroup$
    – PerelMan
    Dec 24 '18 at 23:28






  • 1




    $begingroup$
    Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 23:33
















2












$begingroup$


Let $E(x, y, z) = (x, y, z)$ be a vector field on $mathbb{R}^3$.



$α(x, y, z) = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy$ is a 2-form.



Find $phi^{t*}_Ealpha$.



The flow of X is $phi^t = (x_0e^t,y_0e^t,z_0e^t)$.



I need to compute $dphi_t^{*}$ as $phi_t^{*}alpha = alpha_{phi_t}(d phi_{t})$.



1) Is the pullbak of a k-form always a k-form?



2) If we consider $phi_{t}$ as a 1-form, $dphi_{t}$ is a 2-form, how to compute it?



Thank you for your help!






differential-geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 21:30












  • $begingroup$
    Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
    $endgroup$
    – PerelMan
    Dec 24 '18 at 23:28






  • 1




    $begingroup$
    Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 23:33














2












2








2





$begingroup$


Let $E(x, y, z) = (x, y, z)$ be a vector field on $mathbb{R}^3$.



$α(x, y, z) = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy$ is a 2-form.



Find $phi^{t*}_Ealpha$.



The flow of X is $phi^t = (x_0e^t,y_0e^t,z_0e^t)$.



I need to compute $dphi_t^{*}$ as $phi_t^{*}alpha = alpha_{phi_t}(d phi_{t})$.



1) Is the pullbak of a k-form always a k-form?



2) If we consider $phi_{t}$ as a 1-form, $dphi_{t}$ is a 2-form, how to compute it?



Thank you for your help!






differential-geometry







share|cite|improve this question









$endgroup$




Let $E(x, y, z) = (x, y, z)$ be a vector field on $mathbb{R}^3$.



$α(x, y, z) = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy$ is a 2-form.



Find $phi^{t*}_Ealpha$.



The flow of X is $phi^t = (x_0e^t,y_0e^t,z_0e^t)$.



I need to compute $dphi_t^{*}$ as $phi_t^{*}alpha = alpha_{phi_t}(d phi_{t})$.



1) Is the pullbak of a k-form always a k-form?



2) If we consider $phi_{t}$ as a 1-form, $dphi_{t}$ is a 2-form, how to compute it?



Thank you for your help!






differential-geometry




differential-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 17:24









PerelManPerelMan

538311




538311








  • 1




    $begingroup$
    How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 21:30












  • $begingroup$
    Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
    $endgroup$
    – PerelMan
    Dec 24 '18 at 23:28






  • 1




    $begingroup$
    Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 23:33














  • 1




    $begingroup$
    How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 21:30












  • $begingroup$
    Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
    $endgroup$
    – PerelMan
    Dec 24 '18 at 23:28






  • 1




    $begingroup$
    Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
    $endgroup$
    – Ted Shifrin
    Dec 24 '18 at 23:33








1




1




$begingroup$
How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
$endgroup$
– Ted Shifrin
Dec 24 '18 at 21:30






$begingroup$
How are you arriving at thinking of $phi_t$ (which is a diffeomorphism of $Bbb R^3$) as a $1$-form? That's not remotely correct. There's a difference between a function and a vector field (which is naturally dual to a $1$-form). You compute pullback the way you compute pullback by any function.
$endgroup$
– Ted Shifrin
Dec 24 '18 at 21:30














$begingroup$
Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
$endgroup$
– PerelMan
Dec 24 '18 at 23:28




$begingroup$
Indeed! $phi_t$ is just a diffeomorphism. However, we can still view $dphi_t$ as a 1-form right?
$endgroup$
– PerelMan
Dec 24 '18 at 23:28




1




1




$begingroup$
Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 24 '18 at 23:33




$begingroup$
Because we are living on $Bbb R^3$ and not a general manifold, yes, it’s a vector-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 24 '18 at 23:33










1 Answer
1






active

oldest

votes


















0












$begingroup$

I will partially answer my question.



For $2)$ :



Let $v,w$ be two vectors of $mathbb{R}^3$. Let u be a point in $mathbb{R}^3$.



The usual expression of the pullback gives:



$phi_t^* alpha(u)(v,w) = alpha(phi_t(u))(d phi_t(u).v, d phi_t(u).w)$



Which gives:



$phi_t^* alpha(u)(v,w) = alpha(e^tu)(ve^t, we^t) = e^{3t}alpha(u)(v,w)$.






share|cite|improve this answer









$endgroup$













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    active

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    0












    $begingroup$

    I will partially answer my question.



    For $2)$ :



    Let $v,w$ be two vectors of $mathbb{R}^3$. Let u be a point in $mathbb{R}^3$.



    The usual expression of the pullback gives:



    $phi_t^* alpha(u)(v,w) = alpha(phi_t(u))(d phi_t(u).v, d phi_t(u).w)$



    Which gives:



    $phi_t^* alpha(u)(v,w) = alpha(e^tu)(ve^t, we^t) = e^{3t}alpha(u)(v,w)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I will partially answer my question.



      For $2)$ :



      Let $v,w$ be two vectors of $mathbb{R}^3$. Let u be a point in $mathbb{R}^3$.



      The usual expression of the pullback gives:



      $phi_t^* alpha(u)(v,w) = alpha(phi_t(u))(d phi_t(u).v, d phi_t(u).w)$



      Which gives:



      $phi_t^* alpha(u)(v,w) = alpha(e^tu)(ve^t, we^t) = e^{3t}alpha(u)(v,w)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I will partially answer my question.



        For $2)$ :



        Let $v,w$ be two vectors of $mathbb{R}^3$. Let u be a point in $mathbb{R}^3$.



        The usual expression of the pullback gives:



        $phi_t^* alpha(u)(v,w) = alpha(phi_t(u))(d phi_t(u).v, d phi_t(u).w)$



        Which gives:



        $phi_t^* alpha(u)(v,w) = alpha(e^tu)(ve^t, we^t) = e^{3t}alpha(u)(v,w)$.






        share|cite|improve this answer









        $endgroup$



        I will partially answer my question.



        For $2)$ :



        Let $v,w$ be two vectors of $mathbb{R}^3$. Let u be a point in $mathbb{R}^3$.



        The usual expression of the pullback gives:



        $phi_t^* alpha(u)(v,w) = alpha(phi_t(u))(d phi_t(u).v, d phi_t(u).w)$



        Which gives:



        $phi_t^* alpha(u)(v,w) = alpha(e^tu)(ve^t, we^t) = e^{3t}alpha(u)(v,w)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 23:41









        PerelManPerelMan

        538311




        538311






























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