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Define $g:\mathbb{R^2} \rightarrow \mathbb{R^2}$ by $(x,y)=g(s,t)=(st,e^t)$ and let $\omega= xdy$. How can I compute $d\omega$ and $g^\ast \omega$?

Actually, I computed $d\omega =tds \wedge e^tdt$. (Is this true?) However I am confused about the pullback map.

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    @GiuseppeNegro You're right. There's a typo, it should be $tds \wedge e^tdt$. I've corrected it.2011-12-30

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Let's deal with $\mathrm{d} \omega$ first. The exterior derivative of a differential form is defined purely in terms of that differential form, so $g$ never enters the picture here. The Leibniz rule for exterior derivatives is $\mathrm{d} (\alpha \wedge \beta) = \mathrm{d} \alpha \wedge \beta + (-1)^p \alpha \wedge \mathrm{d} \beta$ where $\alpha$ is a $p$-form. Hence, $\mathrm{d} (x \, \mathrm{d} y) = \mathrm{d} x \wedge \mathrm{d} y - x \, \mathrm{d}^2 y = \mathrm{d} x \wedge \mathrm{d} y$ since $\mathrm{d}^2 y = 0$.

The pullback of a differential form by a smooth map is a little more confusing. Recall that a $1$-form $\alpha$ on a manifold $M$ is a smooth map $\alpha : M \to T^* M$ such that $\alpha |_p \in T^*_p M \text{ for each } p \text{ in } M$ where I have written $\alpha |_p$ to mean the value of $\alpha$ at $p$. But an element of $T^*_p M$ is a linear map $T_p M \to \mathbb{R}$, so given a smooth map of manifolds g : M' \to M, at each point p' of M', we have \alpha |_{g(p')} \circ \mathrm{D}g |_{p'} : T_{p'} M' \to \mathbb{R} where \mathrm{D}g |_{p'} : T_{p'} M' \to T_{g(p')} M is the Jacobian of $g$ at p'. Thus this yields a $1$-form on M' g^* \alpha : M' \to T^* M' defined by g^* \alpha |_{p'} = \alpha |_{g(p')} \circ \mathrm{D}g |_{p'}

Now, this only looks complicated, but it really isn't. Since $g(s, t) = (x, y) = (st, \exp t)$, we have $\begin{align} g^* \mathrm{d} x & = t \, \mathrm{d} s + s \, \mathrm{d} t \\ g^* \mathrm{d} y & = \exp(t) \, \mathrm{d} t \end{align}$ and $g^*$ is a homomorphism of exterior differential algebras, so $g^* (x \, \mathrm{d} y) = s t \exp(t) \, \mathrm{d} t$ which is exactly what you would expect from naïvely pushing around differential operators and substituting! (This is partly the reason why some people simply omit the notation $g^*$.)