# Does there exist a leaf of this holomorphic foliation with non trivial holonomy?

Let's $\mathcal{F}$ be the holomorphic foliation of $\mathbb{C}^2$ tangent to the kernel of $\alpha=(sin x) dx -(cos x)dy$.

Are all leaves of $\mathcal{F}$ simply connected? If the answer is no, does there exist a leaf with non trivial holonomy?

• No and no. The leaves are the level sets of the function $e^ycosx$. – Tom Goodwillie Jun 15 '18 at 13:18
• @TomGoodwillie Thank you. Is it obvious that the connected components of level sets correspond to non zero values $c\neq 0$ are diffeomorphic to $\mathbb{C}$? – Ali Taghavi Jun 15 '18 at 13:39
• On the contrary, for each complex number $a$ such that $cos a=0$ there is the leaf $x=a$, but each other leaf is the graph of a branch of the function $y=-log(cos x)$, and is therefore an infinite cyclic cover of the set of all $x\in\mathbb C$ such that $cos x\neq 0$. – Tom Goodwillie Jun 15 '18 at 14:35
• @TomGoodwillie I am sorry if I change my question on this system as follows: let's consider the same $1$-form but as a real form on $\mathbb{R}^2$. Is there a Riemannian metric on $\mathbb{R}^2$ such that the foliation would be foliation by geodesic? – Ali Taghavi Jun 16 '18 at 10:45
• This seems like an unrelated question, now, but yes: let $u=e^ycosx$, let $v=e^ysinx$, let the metric be $ds^2=du^2+dv^2$. – Tom Goodwillie Jun 16 '18 at 13:57