Prove that the integral of a differential form is zero












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Let $mathbb{X}$ and $mathbb{Y}$ be vector fields on $mathbb{R}^n$ such that $[mathbb{X},mathbb{Y}]=0$, and let $Phi_t$ and $Psi_s$ denote their respective flows. Let $c:[0,1]times [0,1]to mathbb{R}^n$ denote the singular $2$-cube given by
$$c(s,t)=Psi_s(Phi_t(0)),$$
where $0$ denotes the origin of $mathbb{R}^n$. Let $omega=omega_idx^i$ be a $1$-form on $mathbb{R}^n$ and suppose that $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$. Show that
$$int_c domega =0.$$




Firstly, as the Jacobi bracket vanishes, the flows of the vector fields commutes. Then,
$$frac{partial c}{partial t}(s,t)=frac{partial}{partial t}(Psi_s(Phi_t(0)))=frac{partial}{partial t}(Phi_t(Psi_s(0)))=mathbb{X}(Phi_t(Psi_s(0)))=mathbb{X}(c(s,t))$$
$$frac{partial c}{partial s}(s,t)=frac{partial}{partial s}(Psi_s(Phi_t(0)))=mathbb{Y}(Psi_s(Phi_t(0)))=mathbb{Y}(c(s,t))$$
If we denote as $c_{(j,alpha)}$ the $(j,alpha)$-th face of $c$ and $c_{(j,alpha)}^*$ its pullback, then
$$c_{(1,0)}^*omega(t)=omega_i(c(0,t))mathbb{X}^i(c(0,t))dt$$
$$c_{(1,1)}^*omega(t)=omega_i(c(1,t))mathbb{X}^i(c(1,t))dt$$
$$c_{(2,0)}^*omega(s)=omega_i(c(s,0))mathbb{Y}^i(c(s,0))ds$$
$$c_{(2,1)}^*omega(s)=omega_i(c(s,1))mathbb{Y}^i(c(s,1))ds$$
Using the Stokes theorem,
$$int_c domega=int_{partial c}omega = sum_{j=1}^2sum_{alpha=0}^1 (-1)^{j+alpha}int_0^1 c_{(j,alpha)}^*omega$$



However, I don't know how to conclude the question. I think that the hypothesis of $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$, but I'm not sure where to use it. Any help would be appreciate.






differential-forms vector-fields stokes-theorem







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    1















    Let $mathbb{X}$ and $mathbb{Y}$ be vector fields on $mathbb{R}^n$ such that $[mathbb{X},mathbb{Y}]=0$, and let $Phi_t$ and $Psi_s$ denote their respective flows. Let $c:[0,1]times [0,1]to mathbb{R}^n$ denote the singular $2$-cube given by
    $$c(s,t)=Psi_s(Phi_t(0)),$$
    where $0$ denotes the origin of $mathbb{R}^n$. Let $omega=omega_idx^i$ be a $1$-form on $mathbb{R}^n$ and suppose that $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$. Show that
    $$int_c domega =0.$$




    Firstly, as the Jacobi bracket vanishes, the flows of the vector fields commutes. Then,
    $$frac{partial c}{partial t}(s,t)=frac{partial}{partial t}(Psi_s(Phi_t(0)))=frac{partial}{partial t}(Phi_t(Psi_s(0)))=mathbb{X}(Phi_t(Psi_s(0)))=mathbb{X}(c(s,t))$$
    $$frac{partial c}{partial s}(s,t)=frac{partial}{partial s}(Psi_s(Phi_t(0)))=mathbb{Y}(Psi_s(Phi_t(0)))=mathbb{Y}(c(s,t))$$
    If we denote as $c_{(j,alpha)}$ the $(j,alpha)$-th face of $c$ and $c_{(j,alpha)}^*$ its pullback, then
    $$c_{(1,0)}^*omega(t)=omega_i(c(0,t))mathbb{X}^i(c(0,t))dt$$
    $$c_{(1,1)}^*omega(t)=omega_i(c(1,t))mathbb{X}^i(c(1,t))dt$$
    $$c_{(2,0)}^*omega(s)=omega_i(c(s,0))mathbb{Y}^i(c(s,0))ds$$
    $$c_{(2,1)}^*omega(s)=omega_i(c(s,1))mathbb{Y}^i(c(s,1))ds$$
    Using the Stokes theorem,
    $$int_c domega=int_{partial c}omega = sum_{j=1}^2sum_{alpha=0}^1 (-1)^{j+alpha}int_0^1 c_{(j,alpha)}^*omega$$



    However, I don't know how to conclude the question. I think that the hypothesis of $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$, but I'm not sure where to use it. Any help would be appreciate.






    differential-forms vector-fields stokes-theorem







    share|cite|improve this question

























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      Let $mathbb{X}$ and $mathbb{Y}$ be vector fields on $mathbb{R}^n$ such that $[mathbb{X},mathbb{Y}]=0$, and let $Phi_t$ and $Psi_s$ denote their respective flows. Let $c:[0,1]times [0,1]to mathbb{R}^n$ denote the singular $2$-cube given by
      $$c(s,t)=Psi_s(Phi_t(0)),$$
      where $0$ denotes the origin of $mathbb{R}^n$. Let $omega=omega_idx^i$ be a $1$-form on $mathbb{R}^n$ and suppose that $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$. Show that
      $$int_c domega =0.$$




      Firstly, as the Jacobi bracket vanishes, the flows of the vector fields commutes. Then,
      $$frac{partial c}{partial t}(s,t)=frac{partial}{partial t}(Psi_s(Phi_t(0)))=frac{partial}{partial t}(Phi_t(Psi_s(0)))=mathbb{X}(Phi_t(Psi_s(0)))=mathbb{X}(c(s,t))$$
      $$frac{partial c}{partial s}(s,t)=frac{partial}{partial s}(Psi_s(Phi_t(0)))=mathbb{Y}(Psi_s(Phi_t(0)))=mathbb{Y}(c(s,t))$$
      If we denote as $c_{(j,alpha)}$ the $(j,alpha)$-th face of $c$ and $c_{(j,alpha)}^*$ its pullback, then
      $$c_{(1,0)}^*omega(t)=omega_i(c(0,t))mathbb{X}^i(c(0,t))dt$$
      $$c_{(1,1)}^*omega(t)=omega_i(c(1,t))mathbb{X}^i(c(1,t))dt$$
      $$c_{(2,0)}^*omega(s)=omega_i(c(s,0))mathbb{Y}^i(c(s,0))ds$$
      $$c_{(2,1)}^*omega(s)=omega_i(c(s,1))mathbb{Y}^i(c(s,1))ds$$
      Using the Stokes theorem,
      $$int_c domega=int_{partial c}omega = sum_{j=1}^2sum_{alpha=0}^1 (-1)^{j+alpha}int_0^1 c_{(j,alpha)}^*omega$$



      However, I don't know how to conclude the question. I think that the hypothesis of $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$, but I'm not sure where to use it. Any help would be appreciate.






      differential-forms vector-fields stokes-theorem







      share|cite|improve this question














      Let $mathbb{X}$ and $mathbb{Y}$ be vector fields on $mathbb{R}^n$ such that $[mathbb{X},mathbb{Y}]=0$, and let $Phi_t$ and $Psi_s$ denote their respective flows. Let $c:[0,1]times [0,1]to mathbb{R}^n$ denote the singular $2$-cube given by
      $$c(s,t)=Psi_s(Phi_t(0)),$$
      where $0$ denotes the origin of $mathbb{R}^n$. Let $omega=omega_idx^i$ be a $1$-form on $mathbb{R}^n$ and suppose that $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$. Show that
      $$int_c domega =0.$$




      Firstly, as the Jacobi bracket vanishes, the flows of the vector fields commutes. Then,
      $$frac{partial c}{partial t}(s,t)=frac{partial}{partial t}(Psi_s(Phi_t(0)))=frac{partial}{partial t}(Phi_t(Psi_s(0)))=mathbb{X}(Phi_t(Psi_s(0)))=mathbb{X}(c(s,t))$$
      $$frac{partial c}{partial s}(s,t)=frac{partial}{partial s}(Psi_s(Phi_t(0)))=mathbb{Y}(Psi_s(Phi_t(0)))=mathbb{Y}(c(s,t))$$
      If we denote as $c_{(j,alpha)}$ the $(j,alpha)$-th face of $c$ and $c_{(j,alpha)}^*$ its pullback, then
      $$c_{(1,0)}^*omega(t)=omega_i(c(0,t))mathbb{X}^i(c(0,t))dt$$
      $$c_{(1,1)}^*omega(t)=omega_i(c(1,t))mathbb{X}^i(c(1,t))dt$$
      $$c_{(2,0)}^*omega(s)=omega_i(c(s,0))mathbb{Y}^i(c(s,0))ds$$
      $$c_{(2,1)}^*omega(s)=omega_i(c(s,1))mathbb{Y}^i(c(s,1))ds$$
      Using the Stokes theorem,
      $$int_c domega=int_{partial c}omega = sum_{j=1}^2sum_{alpha=0}^1 (-1)^{j+alpha}int_0^1 c_{(j,alpha)}^*omega$$



      However, I don't know how to conclude the question. I think that the hypothesis of $i_{mathbb{X}}omega=i_{mathbb{Y}}omega=0$, but I'm not sure where to use it. Any help would be appreciate.






      differential-forms vector-fields stokes-theorem




      differential-forms vector-fields stokes-theorem






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      asked Jan 4 at 22:30









      user326159user326159

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          By definition, $iota_{X} omega = omega(X) = omega_i X^i$ and $iota_{Y} omega = omega(Y) = omega_i Y^i$



          So if $iota_{X}omega = iota_Yomega = 0$, then all four of your $c_{(j, alpha)}^star omega $'s are identically zero.






          share|cite|improve this answer























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            By definition, $iota_{X} omega = omega(X) = omega_i X^i$ and $iota_{Y} omega = omega(Y) = omega_i Y^i$



            So if $iota_{X}omega = iota_Yomega = 0$, then all four of your $c_{(j, alpha)}^star omega $'s are identically zero.






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              By definition, $iota_{X} omega = omega(X) = omega_i X^i$ and $iota_{Y} omega = omega(Y) = omega_i Y^i$



              So if $iota_{X}omega = iota_Yomega = 0$, then all four of your $c_{(j, alpha)}^star omega $'s are identically zero.






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                By definition, $iota_{X} omega = omega(X) = omega_i X^i$ and $iota_{Y} omega = omega(Y) = omega_i Y^i$



                So if $iota_{X}omega = iota_Yomega = 0$, then all four of your $c_{(j, alpha)}^star omega $'s are identically zero.






                share|cite|improve this answer














                By definition, $iota_{X} omega = omega(X) = omega_i X^i$ and $iota_{Y} omega = omega(Y) = omega_i Y^i$



                So if $iota_{X}omega = iota_Yomega = 0$, then all four of your $c_{(j, alpha)}^star omega $'s are identically zero.







                share|cite|improve this answer














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                edited Jan 4 at 22:53

























                answered Jan 4 at 22:43









                Kenny WongKenny Wong

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