# What are the exact holomorphic Lagrangians in complex 2-space?

In an exact symplectic manifold, i.e. where the symplectic form can be written $\omega = d \lambda$, it's natural to look for exact Lagrangians, i.e. $L$ on which $\lambda_L = df$. One reason is that, any disk ending on such a Lagrangian has vanishing area:

$$\int_D \omega = \int_{\partial D} \lambda = \int_{\partial \partial D} f = 0$$

On the other hand, the search for such manifold-without-boundary Lagrangians in cotangent bundles of manifolds-without-boundary is expected to be fruitless -- Arnol'd conjectured that all such are homotopic to the zero section, and Abouzaid has proven that they're at least homotopy equivalent to the zero section.

I am interested in understanding how hard such things are to come by in the cotangent bundle of an open manifold. For cotangent bundles of one dimensional manifolds, the question is trivial -- a Lagrangian encloses area if it's a closed curve, and otherwise has no topology.

So the first case is $T^* \mathbb{C}$. What I really want to know is what the holomorphic exact Lagrangian manifolds are; this is motivated in part by the article by Xin Jin which explains that such exact holomorphic Lagrangians give rise to perverse sheaves under the Nadler-Zaslow correspondence.

All this is the motivation for the following entirely elementary question:

Characterize all algebraic curves $C \subset \mathbb{C}^2$ on which the one-form $y dx$ is exact.

• When we define exact Lagrangian, are we fixing a choice of $\lambda$ once and for all? – S. Carnahan Sep 14 '14 at 10:43
• To add to Robert Bryant's point: are you asking that the one-form be exact in the category of holomorphic functions, or in the category of polynomial functions? If you use the category of holomorphic functions, there are tons of examples because of the Poincar'e lemma, e.g., all algebraic embeddings of $\mathbb{C}$ in $\mathbb{C}^2$. – Jason Starr Sep 14 '14 at 12:04
• @JasonStarr, sure, but those are all pretty similar to the zero section. – Vivek Shende Sep 14 '14 at 19:02
• @VivekShende: Can you explain what you mean by 'pretty similar'? I don't see how such a curve as Jason suggests is at all similar to the zero section. Any rational curve parametrized in the form $(x,y) = \bigl(p(t),q(t)\bigr)$ where $p$ and $q$ are polynomials in one variable will have $y\ \mathrm{d}x$ be exact (in fact, it will the the differential of another polynomial in $t$), and, if $x$ has degree greater than $1$, this will usually be nowhere near a section, which would essentially require that $y$ be expressible as a (single valued) function of $x$. For example: $(x,y) = (t^2,t^3-t)$. – Robert Bryant Sep 15 '14 at 16:16

## 1 Answer

I'm not sure what you mean by 'characterize'. For example, here is a trivial characterization: Let $C$ be any Riemann surface, with meromorphic functions $a$ and $b$ with $da\not=0$. Then let $(x,y) = (a, db/da)$. The image of $C$ (minus a polar divisor) in $\mathbb{C}^2$ will be an algebraic curve on which $y\ dx = db$. Thus, there are tons of these.