Closed form on any submanifold closed? Let $M$ be a manifold and $\omega$ a closed differential form, so i.e. $d \omega =0.$ If I now consider a submanifold $N$ of $M$. Does this mean that $\omega$ is still closed on $N$?
This statement might be generalizable to the question whether $(d\omega)|N = d( \omega|_N)$ holds?
 A: "Restriction of $\omega$ to $N$" is the same thing as the pullback form $i^*\omega$, where $i:N \to M$ is the inclusion. Then all you need to know is that for any smooth map $f$, $df^*\omega = f^*d\omega$. You can prove this in coordinates, for instance. 
A: Mike Miller said you that $\omega|_N$ is the same thing as $i^*\omega.$ Why is it so?
(I assume that $\omega$ is a $k$-form) See first that $\omega|_N,$ by definition of restriction, is just a function and for any $x\in N$ $$\omega|_N(x)\in\bigwedge^kT_x^*\color{red}{M}$$
So it is not what we expected to get. Hence by definition
$$\omega|_N:=i^*\omega.$$
In my opinion notation is $\omega|_N$ is fine cause it captures the idea, but it does not fit pefectly to set theoretic realm.
Again as Mike Miller said, to proof that $(d_M\omega)|_N=d_N(\omega|_N)$ you just need to know that for any smooth function $f:N\rightarrow M$ and any $\omega\in\Omega^*(M).$
$$d_Nf^*\omega=f^*d_M\omega$$
Hence in your case it is
$$(d_M\omega)|_N=i^*d_M\omega=d_Ni^*\omega=d_N(\omega|_N).$$
