Pull back from $dx \wedge dy \wedge dz$ to $dx \wedge dy$

Question

What is the pull-back of $dx \wedge dy \wedge dz$ under $\varphi : \mathbb{R}^2 \rightarrow \mathbb{R}^3 : (a,b) \rightarrow (0,a,b)$? I would guess that $$ \varphi^{\ast}(dx \wedge dy \wedge dz) = \pm da \wedge db, $$ where I am not sure about the sign, but why is this the case?

Answer 1

If I let $(a,b)$ be standard coordinates on $\mathbb{R}^2$ and $(x,y,z)$ be standard coordinates on $\mathbb{R}^3$, then the transformation is given by the equations

$$ x = 0 \qquad \qquad y = a \qquad \qquad z = b $$

and thus the pullback is

$$ \varphi^*(\mathrm{d} x \wedge \mathrm{d} y \wedge \mathrm{d} z) = \mathrm{d} 0 \wedge \mathrm{d} a \wedge \mathrm{d} b = 0 $$

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Of course, there's a simpler shortcut here; the pullback of a 3-form is a 3-form, and there are no nonzero 3-forms on a 2-dimensional manifold.