find a vector field in $mathbb{R^3}$ with specific properties












0












$begingroup$


Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$



Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$



What I did:



I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$



We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)



and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.



but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$



and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$



and $ dx wedge dy (bpartial_y) = 0$



which means finally
$ ady(.) = bdx(.) $ (II)



Could we find a,b and c from the two results (I) and (II)?



Thank you!






differential-geometry vector-fields exterior-derivative







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$endgroup$








  • 1




    $begingroup$
    (II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
    $endgroup$
    – Dog_69
    Dec 9 '18 at 23:32










  • $begingroup$
    Indeed, $alpha(Z) in mathbb{R}$. Thank you!
    $endgroup$
    – PerelMan
    Dec 9 '18 at 23:41








  • 1




    $begingroup$
    You're welcome :)
    $endgroup$
    – Dog_69
    Dec 10 '18 at 9:05
















0












$begingroup$


Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$



Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$



What I did:



I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$



We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)



and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.



but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$



and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$



and $ dx wedge dy (bpartial_y) = 0$



which means finally
$ ady(.) = bdx(.) $ (II)



Could we find a,b and c from the two results (I) and (II)?



Thank you!






differential-geometry vector-fields exterior-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    (II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
    $endgroup$
    – Dog_69
    Dec 9 '18 at 23:32










  • $begingroup$
    Indeed, $alpha(Z) in mathbb{R}$. Thank you!
    $endgroup$
    – PerelMan
    Dec 9 '18 at 23:41








  • 1




    $begingroup$
    You're welcome :)
    $endgroup$
    – Dog_69
    Dec 10 '18 at 9:05














0












0








0





$begingroup$


Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$



Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$



What I did:



I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$



We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)



and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.



but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$



and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$



and $ dx wedge dy (bpartial_y) = 0$



which means finally
$ ady(.) = bdx(.) $ (II)



Could we find a,b and c from the two results (I) and (II)?



Thank you!






differential-geometry vector-fields exterior-derivative







share|cite|improve this question











$endgroup$




Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$



Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$



What I did:



I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$



We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)



and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.



but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$



and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$



and $ dx wedge dy (bpartial_y) = 0$



which means finally
$ ady(.) = bdx(.) $ (II)



Could we find a,b and c from the two results (I) and (II)?



Thank you!






differential-geometry vector-fields exterior-derivative




differential-geometry vector-fields exterior-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 23:23







PerelMan

















asked Dec 9 '18 at 23:04









PerelManPerelMan

679313




679313








  • 1




    $begingroup$
    (II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
    $endgroup$
    – Dog_69
    Dec 9 '18 at 23:32










  • $begingroup$
    Indeed, $alpha(Z) in mathbb{R}$. Thank you!
    $endgroup$
    – PerelMan
    Dec 9 '18 at 23:41








  • 1




    $begingroup$
    You're welcome :)
    $endgroup$
    – Dog_69
    Dec 10 '18 at 9:05














  • 1




    $begingroup$
    (II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
    $endgroup$
    – Dog_69
    Dec 9 '18 at 23:32










  • $begingroup$
    Indeed, $alpha(Z) in mathbb{R}$. Thank you!
    $endgroup$
    – PerelMan
    Dec 9 '18 at 23:41








  • 1




    $begingroup$
    You're welcome :)
    $endgroup$
    – Dog_69
    Dec 10 '18 at 9:05








1




1




$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32




$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32












$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41






$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41






1




1




$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05




$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05










1 Answer
1






active

oldest

votes


















1












$begingroup$

As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
    $endgroup$
    – PerelMan
    Dec 10 '18 at 13:40






  • 1




    $begingroup$
    It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
    $endgroup$
    – Andreas Cap
    Dec 11 '18 at 8:01













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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

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1












$begingroup$

As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
    $endgroup$
    – PerelMan
    Dec 10 '18 at 13:40






  • 1




    $begingroup$
    It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
    $endgroup$
    – Andreas Cap
    Dec 11 '18 at 8:01


















1












$begingroup$

As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
    $endgroup$
    – PerelMan
    Dec 10 '18 at 13:40






  • 1




    $begingroup$
    It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
    $endgroup$
    – Andreas Cap
    Dec 11 '18 at 8:01
















1












1








1





$begingroup$

As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.






share|cite|improve this answer









$endgroup$



As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 11:30









Andreas CapAndreas Cap

11.3k923




11.3k923












  • $begingroup$
    How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
    $endgroup$
    – PerelMan
    Dec 10 '18 at 13:40






  • 1




    $begingroup$
    It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
    $endgroup$
    – Andreas Cap
    Dec 11 '18 at 8:01




















  • $begingroup$
    How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
    $endgroup$
    – PerelMan
    Dec 10 '18 at 13:40






  • 1




    $begingroup$
    It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
    $endgroup$
    – Andreas Cap
    Dec 11 '18 at 8:01


















$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40




$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40




1




1




$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01






$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01




















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