Differential 1-form of line












0












$begingroup$


This problem is from V.I Arnold's book Mathematics of Classical Mechanics.
Q) Show that every differential 1-form on line is differential of some function




  1. Relevant equations
    The differential of any function is
    $$df_{x}(psi): TM_{x} rightarrow R$$

  2. The attempt at a solution


The tangent to line is line itself. The differential 1-form is $dy-dx=0$. Here I am struct. I don't know how to find out the differential. Can anyone help?enter image description here






differential-geometry differential-forms







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$endgroup$












  • $begingroup$
    If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
    $endgroup$
    – Gabriele Cassese
    Feb 1 at 15:57












  • $begingroup$
    $dy-dx$ is not a differential form on the line, it is a differential form on the plane
    $endgroup$
    – Gabriele Cassese
    Feb 3 at 15:15
















0












$begingroup$


This problem is from V.I Arnold's book Mathematics of Classical Mechanics.
Q) Show that every differential 1-form on line is differential of some function




  1. Relevant equations
    The differential of any function is
    $$df_{x}(psi): TM_{x} rightarrow R$$

  2. The attempt at a solution


The tangent to line is line itself. The differential 1-form is $dy-dx=0$. Here I am struct. I don't know how to find out the differential. Can anyone help?enter image description here






differential-geometry differential-forms







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
    $endgroup$
    – Gabriele Cassese
    Feb 1 at 15:57












  • $begingroup$
    $dy-dx$ is not a differential form on the line, it is a differential form on the plane
    $endgroup$
    – Gabriele Cassese
    Feb 3 at 15:15














0












0








0


0



$begingroup$


This problem is from V.I Arnold's book Mathematics of Classical Mechanics.
Q) Show that every differential 1-form on line is differential of some function




  1. Relevant equations
    The differential of any function is
    $$df_{x}(psi): TM_{x} rightarrow R$$

  2. The attempt at a solution


The tangent to line is line itself. The differential 1-form is $dy-dx=0$. Here I am struct. I don't know how to find out the differential. Can anyone help?enter image description here






differential-geometry differential-forms







share|cite|improve this question









$endgroup$




This problem is from V.I Arnold's book Mathematics of Classical Mechanics.
Q) Show that every differential 1-form on line is differential of some function




  1. Relevant equations
    The differential of any function is
    $$df_{x}(psi): TM_{x} rightarrow R$$

  2. The attempt at a solution


The tangent to line is line itself. The differential 1-form is $dy-dx=0$. Here I am struct. I don't know how to find out the differential. Can anyone help?enter image description here






differential-geometry differential-forms




differential-geometry differential-forms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 1 at 15:41









Abhi7731756Abhi7731756

94




94












  • $begingroup$
    If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
    $endgroup$
    – Gabriele Cassese
    Feb 1 at 15:57












  • $begingroup$
    $dy-dx$ is not a differential form on the line, it is a differential form on the plane
    $endgroup$
    – Gabriele Cassese
    Feb 3 at 15:15


















  • $begingroup$
    If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
    $endgroup$
    – Gabriele Cassese
    Feb 1 at 15:57












  • $begingroup$
    $dy-dx$ is not a differential form on the line, it is a differential form on the plane
    $endgroup$
    – Gabriele Cassese
    Feb 3 at 15:15
















$begingroup$
If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
$endgroup$
– Gabriele Cassese
Feb 1 at 15:57






$begingroup$
If you consider $omega(x)dx$ as the differential form, what can you say about $f(x):=int_0^x omega(t)dt$?
$endgroup$
– Gabriele Cassese
Feb 1 at 15:57














$begingroup$
$dy-dx$ is not a differential form on the line, it is a differential form on the plane
$endgroup$
– Gabriele Cassese
Feb 3 at 15:15




$begingroup$
$dy-dx$ is not a differential form on the line, it is a differential form on the plane
$endgroup$
– Gabriele Cassese
Feb 3 at 15:15










1 Answer
1






active

oldest

votes


















0












$begingroup$

$dy-dx$ is not a differential form on $mathbb{R}$, since $mathbb{R}$ has only one coordinate for every chart you choose.
It is instead a differential form on $mathbb{R}^2$, where is the differential of
$f(x,y)=y-x$.



About Arnold's exercise, I would insted prove it in this way:
Let $omega(x)dx $ be the general differential form on $mathbb{R}$, where $omega(x)$ is $C^{infty}[mathbb{R}]$. Then, let $f:=int_0^xomega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that:
$df=partial_{x}( int_0^x omega(t)dt)dx=omega(x)dx$
QED






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @Abhi7731756 Is there anything wrong or unclear with the answer?
    $endgroup$
    – Gabriele Cassese
    Feb 4 at 20:43












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

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votes









0












$begingroup$

$dy-dx$ is not a differential form on $mathbb{R}$, since $mathbb{R}$ has only one coordinate for every chart you choose.
It is instead a differential form on $mathbb{R}^2$, where is the differential of
$f(x,y)=y-x$.



About Arnold's exercise, I would insted prove it in this way:
Let $omega(x)dx $ be the general differential form on $mathbb{R}$, where $omega(x)$ is $C^{infty}[mathbb{R}]$. Then, let $f:=int_0^xomega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that:
$df=partial_{x}( int_0^x omega(t)dt)dx=omega(x)dx$
QED






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @Abhi7731756 Is there anything wrong or unclear with the answer?
    $endgroup$
    – Gabriele Cassese
    Feb 4 at 20:43
















0












$begingroup$

$dy-dx$ is not a differential form on $mathbb{R}$, since $mathbb{R}$ has only one coordinate for every chart you choose.
It is instead a differential form on $mathbb{R}^2$, where is the differential of
$f(x,y)=y-x$.



About Arnold's exercise, I would insted prove it in this way:
Let $omega(x)dx $ be the general differential form on $mathbb{R}$, where $omega(x)$ is $C^{infty}[mathbb{R}]$. Then, let $f:=int_0^xomega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that:
$df=partial_{x}( int_0^x omega(t)dt)dx=omega(x)dx$
QED






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @Abhi7731756 Is there anything wrong or unclear with the answer?
    $endgroup$
    – Gabriele Cassese
    Feb 4 at 20:43














0












0








0





$begingroup$

$dy-dx$ is not a differential form on $mathbb{R}$, since $mathbb{R}$ has only one coordinate for every chart you choose.
It is instead a differential form on $mathbb{R}^2$, where is the differential of
$f(x,y)=y-x$.



About Arnold's exercise, I would insted prove it in this way:
Let $omega(x)dx $ be the general differential form on $mathbb{R}$, where $omega(x)$ is $C^{infty}[mathbb{R}]$. Then, let $f:=int_0^xomega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that:
$df=partial_{x}( int_0^x omega(t)dt)dx=omega(x)dx$
QED






share|cite|improve this answer











$endgroup$



$dy-dx$ is not a differential form on $mathbb{R}$, since $mathbb{R}$ has only one coordinate for every chart you choose.
It is instead a differential form on $mathbb{R}^2$, where is the differential of
$f(x,y)=y-x$.



About Arnold's exercise, I would insted prove it in this way:
Let $omega(x)dx $ be the general differential form on $mathbb{R}$, where $omega(x)$ is $C^{infty}[mathbb{R}]$. Then, let $f:=int_0^xomega(t)dt$. By definition of differential and by the fundamental theorem of calculus, we obtain that:
$df=partial_{x}( int_0^x omega(t)dt)dx=omega(x)dx$
QED







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 3 at 15:34

























answered Feb 3 at 14:47









Gabriele CasseseGabriele Cassese

1,211316




1,211316












  • $begingroup$
    @Abhi7731756 Is there anything wrong or unclear with the answer?
    $endgroup$
    – Gabriele Cassese
    Feb 4 at 20:43


















  • $begingroup$
    @Abhi7731756 Is there anything wrong or unclear with the answer?
    $endgroup$
    – Gabriele Cassese
    Feb 4 at 20:43
















$begingroup$
@Abhi7731756 Is there anything wrong or unclear with the answer?
$endgroup$
– Gabriele Cassese
Feb 4 at 20:43




$begingroup$
@Abhi7731756 Is there anything wrong or unclear with the answer?
$endgroup$
– Gabriele Cassese
Feb 4 at 20:43


















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