I have a $2$-form given by $\omega = dx \wedge dp + dy \wedge dq$ and a map $i : (u,v) \mapsto (u,v,f_u,-f_v)$ for a general smooth map $f : (u,v) \mapsto f(u,v)$. I want to calculate the pullback of this map, i.e. $i^*\omega$. I understand that in this case,
\begin{align}
i^*\omega &= i^*(dx \wedge dp + dy \wedge dq) \\
&= d(x \circ i)\wedge d(p \circ i) + d(y \circ i)\wedge d(q \circ i).
\end{align}
Now, calculating each terms gives
\begin{align}
d(x \circ i) &= d(u) = du, \\
d(y \circ i) &= d(v) = dv, \\
d(p \circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \\
d(q \circ i) &= d(-f_v) = -f_{vu}du – f_{vv}dv.
\end{align}
Then, the pullback is given by
\begin{align}
i^*\omega &= du \wedge (f_{uu}du + f_{uv}dv) – dv \wedge (f_{vu}du + f_{vv}dv) \\
&= du \wedge (f_{uu}du) + du \wedge (f_{uv}dv) – dv \wedge (f_{vu}du) – dv \wedge (f_{vv}dv).
\end{align}
Could someone verify that it is correct? Can it be further simplified? I tried to follow the steps described in this question.
Note. This is not a homework question, I am just trying to teach myself some differential geometry.
Best Answer
Yes, your calculations are correct.
Two things to remember: the wedge product is $C^\infty(M)$-bilinear rather than just $\Bbb R$-bilinear, and $du \wedge dv = -dv\wedge du$ (or, more generally, if $\omega$ is a k-form and $\mu$ an $l$-form, then $\omega \wedge \mu = (-1)^{kl}\mu \wedge \omega$). Using these, you should be able to simplify your last equation to $$i^*\omega =2f_{uv} du \wedge dv.$$