Consider the subset $S=\{(x,y,z):x^2-y^2-z^2+1=0\}$ of $\Bbb R^3$. Defining a function $f:\Bbb R^3\to \Bbb R$ by $f(x,y,z)=x^2-y^2-z^2+1$, it is easily seen that $0$ is a regular value of $f$, so it follows that $S=f^{-1}(0)$ is an embedded submanifold of $\Bbb R^3$ of codimension $1$. I am trying to find a closed $1$-form on $S$ which is not exact.
It is easy to find a closed $2$-form on $S$ which is not exact: Since $S$ is an embedded submanifold of codimension $1$ of the orientable manifold $\Bbb R^3$, it follows that $S$ is also orientable, so we can take an orientation form of $S$. But how can I find a non-exact closed $1$-form on $S$? Using wolframalpha.com, I saw how $S$ looks like; it clearly seems that $S$ is homotopy equivalent to the circle $S^1$, so $H^1(S)\neq 0$. Thus $S$ must indeed have a non-exact closed $1$-form but I have no idea to find it. Any hints?
You can use the fact that $S$ contains a circle, since $S\cap\{x=0\}=\{(0,y,z):y^{2}+z^{2}=1\}$.
On $\mathbb{R}^{3}\setminus\{y=z=0\}$, we have the following "winding form" that is closed but not exact, given by $$ \omega=\frac{ydz-zdy}{y^{2}+z^{2}}. $$ Since $S\subset\mathbb{R}^{3}\setminus\{y=z=0\}$, we can pull back this one-form to $S$. The pullback is still closed, but not exact: if it were exact, then the integral around the loop $S\cap\{x=0\}=\{(0,y,z):y^{2}+z^{2}=1\}$ would be zero by Stokes' theorem. This is not the case, since the integral is $2\pi$.