# Computing $d\omega$ and $g^\ast \omega$ when $(x,y)=g(s,t)=(st,e^t)$ and $\omega= xdy$

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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Your $d\omega$ cannot be correct: differentiating a 1-form you get a 2-form but $t ds + e^t dt$ is a 1-form. –  Giuseppe Negro Dec 30 '11 at 15:41
@GiuseppeNegro You're right. There's a typo, it should be $tds \wedge e^tdt$. I've corrected it. –  user20353 Dec 30 '11 at 15:47

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 $g^*$.)