Let $X=x\frac{{\partial }}{{\partial x}}+2xy\frac{{\partial }}{{\partial y}}$
$Y=y\frac{{\partial }}{{\partial y}}$ vector fields on $\mathbb{R}^2$ 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
I think the first and second terms are correct whereas the third is incorrect. The Lie bracket should be zero.