Problem with $Pullback$ Calculation

Question

Let $$\omega=-xdx \wedge dy-3dy\wedge dz$$ and $$\phi:\mathbb R^2 \to \mathbb R^3, (u,v) \to(uv,u^2,3u+v)$$ I tried to compute the pullback $\phi^*(\omega)$, but was not able to solve it. $$\phi^*(\omega)=(-uv)(vdu+udv)-3(2udu)\wedge(3du+dv)=-uv^2du-u^2vdv-6udu\wedge(3du+dv)=uv^2du-u^2vdv-6udu\wedge3du-6udu\wedge dv$$ Could someone tell me, what I am doing wrong?

Answer 1

We first note that since $$x = uv$$$$y=u^2$$ $$z=3u +v$$ we have $$dx = u\,dv + v\,du$$ $$dy = 2u\,du$$ $$dz = 3\,du + dv$$Then we simply substitute these values into the value of $\omega = -x \,dx \wedge dy - 3 dy \wedge dz$. We get:

$$\phi^{\ast}\omega = -(uv)(u\,dv + v\,du) \wedge (2u\,du) - 3(2u\,du)\wedge(3\,du+dv)$$ $$=(-u^2v\,dv - uv^2 \, du) \wedge (2u \, du) - 6u \, du \wedge (3 \, du + dv)$$ We then expand the brackets using linearity of the wedge product: $$= - 2u^3v \, dv \wedge du - 2u^2 v^2 \, du \wedge du - 18u \, du \wedge du - 6u\, du \wedge dv$$ $$=2u^3v \, du \wedge dv - 6u \, du \wedge dv$$ $$=(2u^3v - 6u)\, du \wedge dv$$

Notice that in the second last line we have used the fact that the wedge product is antisymmetric, which means $dv \wedge du = - du \wedge du$, and in particular, $du \wedge du = 0$.

Answer 2

I think -

$-6udu\wedge(3du+dv)=-6udu\wedge3du-6udu\wedge dv$

One other mistake -

$\phi^*(\omega)=(-uv)(vdu+udv)\wedge(2udu)-3(2udu)\wedge(3du+dv)$