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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?

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    If $x=uv$, then $dx =u dv +v du$. Workout what $dy$ has to be using the same rules, and then wedge them together and expand using the linearity of the wedge product. – 2017-02-05
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    $y=u^2$ so $dy=2udu+0dv=2udu$ I do not know what you mean exactly. – 2017-02-05
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    I'll turn it into an answer. – 2017-02-05

2 Answers 2

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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, $du \wedge du = 0$.

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I think -

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

One other mistake -

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

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    you are right. I edited it, but my calculation is still wrong – 2017-02-05
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    What is the original answer? – 2017-02-05
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    I do not know, but Mark B said there is smth wrong as well. – 2017-02-05
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    I don't think so. It seems to correct for me. – 2017-02-05
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    If it should be correct, then is for $$\omega_1=xydx+2zdy-ydz$$ $$ d(\phi ^* \omega_1)=\phi ^*(d \omega_1)$$, but it is not true. So where is the mistake? While $ d\omega_1= \omega $ – 2017-02-05
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    I think I got it. – 2017-02-05
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53079/discussion-between-kanwaljit-singh-and-john-doe). – 2017-02-05