compute Derivative exterior of 1-form

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Let
$X=xfracpartial partial x+2xyfracpartial partial y$



$Y=yfracpartial partial y$ vector fields on $mathbbR^2$ with 1-form $w=(x^2+2y)dx+(x+y^2)dy$



Show that $dw(X,Y)=X(w(Y))-Y(w(X))-w([X,Y])$



I have this



$X(w(Y))=X((x^2+2y)dx+(x+y^2)dy)Y=
X[(x^2+2y)Y(x)+(x+y^2)Y(y)]
=X[(x+y^2)Y(y)]=X[(x+y^2)y]=X(x)y+xX(y)+6xy^3$



therefore $X(w(Y))=X(x)y+xX(y)+6xy^3$



$Y(w(X))=Y((x^2+2y)dx+(x+y^2)dy)X=Y[(x^2+2y)X(x)+(x+y^2)X(y)]=
Y[(x^2+2y)x+(x+y^2)2xy]=y2x+(x^2+2y)Y(x)+y2y2xy+(x+y^2)y2x=2xy+6xy^3+2x^2y$



therefore $Y(w(X))=2xy+6xy^3+2x^2y$ $quad$ ( $Y(x)=0$)



Now
$w([X,Y])=((x^2+2y)dx+(x+y^2)dy)(XY-YX)=
(x^2+2y)dx(XY)+(x+y^2)dy(XY)-(x^2+2y)dx(YX)-(x+y^2)dy(YX)=
(x^2+2y)XY(x)+(x+y^2)XY(y)-(x^2+2y)YX(x)-(x+y^2)YX(y)=
(x^2+2y)XY(x)+(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)Y(2xy)=(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)y2x=(x+y^2)X(y)-2x^2y-2x^2y^3$



therefore $w([X,Y])=(x+y^2)X(y)-2x^2y-2x^2y^3 $



then I have the following



$X(w(Y))-Y(w(X))-w([X,Y])=-xy-4x^2y-2xy^3-2x^2y^3$



but $dw(X,Y)=-xy $



Is it a numerical error or something conceptual?



Thanks




differential-geometry




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  • I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
    – md2perpe
    Jul 23 at 19:47










  • I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
    – md2perpe
    Jul 23 at 19:53














up vote
0
down vote

favorite












Let
$X=xfracpartial partial x+2xyfracpartial partial y$



$Y=yfracpartial partial y$ vector fields on $mathbbR^2$ with 1-form $w=(x^2+2y)dx+(x+y^2)dy$



Show that $dw(X,Y)=X(w(Y))-Y(w(X))-w([X,Y])$



I have this



$X(w(Y))=X((x^2+2y)dx+(x+y^2)dy)Y=
X[(x^2+2y)Y(x)+(x+y^2)Y(y)]
=X[(x+y^2)Y(y)]=X[(x+y^2)y]=X(x)y+xX(y)+6xy^3$



therefore $X(w(Y))=X(x)y+xX(y)+6xy^3$



$Y(w(X))=Y((x^2+2y)dx+(x+y^2)dy)X=Y[(x^2+2y)X(x)+(x+y^2)X(y)]=
Y[(x^2+2y)x+(x+y^2)2xy]=y2x+(x^2+2y)Y(x)+y2y2xy+(x+y^2)y2x=2xy+6xy^3+2x^2y$



therefore $Y(w(X))=2xy+6xy^3+2x^2y$ $quad$ ( $Y(x)=0$)



Now
$w([X,Y])=((x^2+2y)dx+(x+y^2)dy)(XY-YX)=
(x^2+2y)dx(XY)+(x+y^2)dy(XY)-(x^2+2y)dx(YX)-(x+y^2)dy(YX)=
(x^2+2y)XY(x)+(x+y^2)XY(y)-(x^2+2y)YX(x)-(x+y^2)YX(y)=
(x^2+2y)XY(x)+(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)Y(2xy)=(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)y2x=(x+y^2)X(y)-2x^2y-2x^2y^3$



therefore $w([X,Y])=(x+y^2)X(y)-2x^2y-2x^2y^3 $



then I have the following



$X(w(Y))-Y(w(X))-w([X,Y])=-xy-4x^2y-2xy^3-2x^2y^3$



but $dw(X,Y)=-xy $



Is it a numerical error or something conceptual?



Thanks




differential-geometry




share|cite|improve this question



















  • I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
    – md2perpe
    Jul 23 at 19:47










  • I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
    – md2perpe
    Jul 23 at 19:53












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let
$X=xfracpartial partial x+2xyfracpartial partial y$



$Y=yfracpartial partial y$ vector fields on $mathbbR^2$ with 1-form $w=(x^2+2y)dx+(x+y^2)dy$



Show that $dw(X,Y)=X(w(Y))-Y(w(X))-w([X,Y])$



I have this



$X(w(Y))=X((x^2+2y)dx+(x+y^2)dy)Y=
X[(x^2+2y)Y(x)+(x+y^2)Y(y)]
=X[(x+y^2)Y(y)]=X[(x+y^2)y]=X(x)y+xX(y)+6xy^3$



therefore $X(w(Y))=X(x)y+xX(y)+6xy^3$



$Y(w(X))=Y((x^2+2y)dx+(x+y^2)dy)X=Y[(x^2+2y)X(x)+(x+y^2)X(y)]=
Y[(x^2+2y)x+(x+y^2)2xy]=y2x+(x^2+2y)Y(x)+y2y2xy+(x+y^2)y2x=2xy+6xy^3+2x^2y$



therefore $Y(w(X))=2xy+6xy^3+2x^2y$ $quad$ ( $Y(x)=0$)



Now
$w([X,Y])=((x^2+2y)dx+(x+y^2)dy)(XY-YX)=
(x^2+2y)dx(XY)+(x+y^2)dy(XY)-(x^2+2y)dx(YX)-(x+y^2)dy(YX)=
(x^2+2y)XY(x)+(x+y^2)XY(y)-(x^2+2y)YX(x)-(x+y^2)YX(y)=
(x^2+2y)XY(x)+(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)Y(2xy)=(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)y2x=(x+y^2)X(y)-2x^2y-2x^2y^3$



therefore $w([X,Y])=(x+y^2)X(y)-2x^2y-2x^2y^3 $



then I have the following



$X(w(Y))-Y(w(X))-w([X,Y])=-xy-4x^2y-2xy^3-2x^2y^3$



but $dw(X,Y)=-xy $



Is it a numerical error or something conceptual?



Thanks




differential-geometry




share|cite|improve this question











Let
$X=xfracpartial partial x+2xyfracpartial partial y$



$Y=yfracpartial partial y$ vector fields on $mathbbR^2$ with 1-form $w=(x^2+2y)dx+(x+y^2)dy$



Show that $dw(X,Y)=X(w(Y))-Y(w(X))-w([X,Y])$



I have this



$X(w(Y))=X((x^2+2y)dx+(x+y^2)dy)Y=
X[(x^2+2y)Y(x)+(x+y^2)Y(y)]
=X[(x+y^2)Y(y)]=X[(x+y^2)y]=X(x)y+xX(y)+6xy^3$



therefore $X(w(Y))=X(x)y+xX(y)+6xy^3$



$Y(w(X))=Y((x^2+2y)dx+(x+y^2)dy)X=Y[(x^2+2y)X(x)+(x+y^2)X(y)]=
Y[(x^2+2y)x+(x+y^2)2xy]=y2x+(x^2+2y)Y(x)+y2y2xy+(x+y^2)y2x=2xy+6xy^3+2x^2y$



therefore $Y(w(X))=2xy+6xy^3+2x^2y$ $quad$ ( $Y(x)=0$)



Now
$w([X,Y])=((x^2+2y)dx+(x+y^2)dy)(XY-YX)=
(x^2+2y)dx(XY)+(x+y^2)dy(XY)-(x^2+2y)dx(YX)-(x+y^2)dy(YX)=
(x^2+2y)XY(x)+(x+y^2)XY(y)-(x^2+2y)YX(x)-(x+y^2)YX(y)=
(x^2+2y)XY(x)+(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)Y(2xy)=(x+y^2)X(y)-(x^2+2y)Y(x)-(x+y^2)y2x=(x+y^2)X(y)-2x^2y-2x^2y^3$



therefore $w([X,Y])=(x+y^2)X(y)-2x^2y-2x^2y^3 $



then I have the following



$X(w(Y))-Y(w(X))-w([X,Y])=-xy-4x^2y-2xy^3-2x^2y^3$



but $dw(X,Y)=-xy $



Is it a numerical error or something conceptual?



Thanks





differential-geometry





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share|cite|improve this question




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asked Jul 23 at 17:17









eraldcoil

386




386











  • I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
    – md2perpe
    Jul 23 at 19:47










  • I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
    – md2perpe
    Jul 23 at 19:53
















  • I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
    – md2perpe
    Jul 23 at 19:47










  • I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
    – md2perpe
    Jul 23 at 19:53















I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
– md2perpe
Jul 23 at 19:47




I get the same as you on $Y(w(X))$. What do you get after expanding $X(x)$ and $X(y)$ in $X(w(Y))=X(x)y+xX(y)+6xy^3$?
– md2perpe
Jul 23 at 19:47












I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
– md2perpe
Jul 23 at 19:53




I get $XY = YX$, so $[X, Y] = 0$ and $w([X, Y]) = 0.$
– md2perpe
Jul 23 at 19:53










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I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.






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    up vote
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    I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.






    share|cite|improve this answer

























      up vote
      0
      down vote













      I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.






      share|cite|improve this answer























        up vote
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        down vote










        up vote
        0
        down vote









        I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.






        share|cite|improve this answer













        I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 23 at 19:53









        Semsem

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