TITLE: Can a  Lagrangian submanifold of  ${\mathbb R}^{2n}$ be dense ($n>1$)?
QUESTION [19 upvotes]: I'm betting `yes, sure!', but don't see it.  Could  someone please point me toward, 
or construct for me, a Lagrangian submanifold immersed in 
standard symplectic ${\mathbb R}^{2n}$ for $n > 1$,  whose closure is all of ${\mathbb R}^{2n}$?
(For an  $n =1$ example, one can use  the leaves arising from 
this modification by Panov of irrational flow on the two-torus.) 
Strong preference given to  analytic immersions of ${\mathbb R}^n$.
Holomorphically  immersed complex lines which are dense  in complex 2-space -- i.e. dense ${\mathbb C}$'s in ${\mathbb C}^2$ -- are well-known.  Ilyashenko in 1968 showed that the typical solution of the typical polynomial  ODE (in complex time) yields such a curve. Following his line
of thought, it  might be easier to construct an entire  singular Lagrangian foliation of ${\mathbb R}^{2n}$ whose typical leaf is dense, rather than the one submanifold. 
Motivation: I have a certain unstable manifold  related to a Hamiltonian system.
It is Lagrangian.  I would like to be ``as dense as can be'', so I'd like to know 
how dense can that be.

REPLY [22 votes]: Your question already has the answer in it for $n=2$.  Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$.  Then $L$ is Lagrangian for the real part of the holomorphic $2$-form $\Upsilon = dz^1\wedge dz^2$.  This real part of $\Upsilon$ is equivalent to the standard symplectic structure on $\mathbb{R}^4$ by a linear change of variables.
Added comment about injectivity:  Note, by the way, that one can easily arrange for such an $L$ to be a submanifold, not just the image of an immersion (i.e., the immersion is injective).  One simple explicit way to do this is to select constants $\lambda_1,\ldots,\lambda_k$ such that the subgroup in $\mathbb{C}^\times$ generated by the numbers $\mathrm{e}^{2\pi i\lambda_1},\ldots,\mathrm{e}^{2\pi i\lambda_k}$ is dense in $\mathbb{C}^\times$ and consider the linear differential equation
$$
\frac{dy}{dx} = \left(\frac{\lambda_1}{x-x_1}+\cdots + \frac{\lambda_k}{x-x_k}\right)\ y
$$
where $x_1,\ldots,x_k\in \mathbb{C}$ are distinct.  The graph of any nonzero multi-valued solution $y(x)$ over $\mathbb{C}\setminus\{x_1,\ldots,x_k\}$ will then be dense in $\mathbb{C}^2$. (Consider the holonomy around the punctures $x_j$.)  Of course, these graphs are the Riemann surfaces associated to the multivalued functions
$$
y = y_0 (x{-}x_1)^{\lambda_1}\cdots(x{-}x_k)^{\lambda_k}
$$
(when $y_0\not=0$).  These are obviously integral curves (leaves) of the polynomial $1$-form
$$
\omega = (x{-}x_1)\cdots(x{-}x_k)\ dy - q(x) y\ dx
$$
for some polynomial $q$ of degree at most $k{-}1$ in $x$.  Aside from the obvious closed leaves $x-x_j=0$ and $y=0$, the rest of the leaves are dense submanifolds.  (This just gives a simple, explicit example of the general theorem that Richard quoted.)
Dense analytic curves in $\mathbb{R}^2$:  It is not hard to construct dense, connected analytic curves in $\mathbb{R}^2$: There exist analytic metrics on the $2$-sphere that have geodesics that wander densely over the surface.  Now take such a geodesic and remove a point from $S^2$ through which the geodesic doesn't pass.  What's left is a dense analytic curve in $\mathbb{R}^2$.  If you are willing to use Finsler metrics, you can even do this with a rotationally invariant real analytic Finsler metric on the $2$-sphere (eg. Katok's examples), so that you can write down the dense analytic curve very explicitly.
I'll think about the case $n>2$.  I don't see it yet either, but maybe it's not too hard.