Closed differential form. In the paper, page 5, Line 5, why the form $\eta_j$ is closed? We have to show that $d \eta_j=0$. Here $\eta_j = f(t)dt$ for some function $f$. Is it always true that $d\eta_j$ for any $f$? Thank you very much.
 A: This is just a definition. Any differential form $\eta $ such that $d\eta = 0$ is called a closed form. So in your case if $\eta _j=f(t)dt$ and you are able to show that $d\eta _j=0$, then it follows that $\eta _j$ is a closed form. 
Since, $\eta _j=f(t)dt$ is a 1-form. By definition of the exterior differential operator you get $d\eta _j=f'(t)dt\wedge dt=0$, since $dt\wedge dt=0$. 
This holds for the case where $f(t)$ is real valued. 
A: Your notation, though taken from the paper, does not make it clear that we are dealing with a complex paramter $t$ and complex valued $f$. In that setting, the form $f(t) dt$ is closed iff $f$ is holomorphic. To see this write $fdt = fdx+i fdy $ and apply $d$ to see that this is closed iff $\frac{\partial}{\partial y} f=  i\frac{\partial}{\partial x} f $. In other words iff
$$ 0 = \left(\frac{\partial}{\partial y} - i\frac{\partial}{\partial x}\right)f = -i\left( \frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)f = -2i\frac{\partial}{\partial \overline{t}}f $$
(with $t=x+iy$, which is kind of against conventions). Now just check whether your $\eta_i$ is holomorphic.
(In general you will get a $dt \wedge  d\overline{t}$ term).
