Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:26:00Z http://mathoverflow.net/feeds/question/85881 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85881/can-a-lagrangian-submanifold-of-mathbb-r2n-be-dense-n1 Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)? Richard Montgomery 2012-01-17T04:51:22Z 2012-01-17T18:56:39Z <p>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 ${\mathbb R}^{2n}$?</p> <p>(For an $n =1$ example, one can use the leaves arising from <a href="http://www2.imperial.ac.uk/~dpanov/TORUS.PDF" rel="nofollow">this modification by Panov</a> of irrational flow on the two-torus.) </p> <p>Strong preference given to analytic immersions of ${\mathbb R}^n$.</p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/85881/can-a-lagrangian-submanifold-of-mathbb-r2n-be-dense-n1/85901#85901 Answer by Robert Bryant for Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)? Robert Bryant 2012-01-17T12:38:46Z 2012-01-17T18:56:39Z <p>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.</p> <p><strong>Added comment about injectivity:</strong> 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 <em>injective</em>). 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 <code>$\mathbb{C}\setminus\{x_1,\ldots,x_k\}$</code> 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.)</p> <p><strong>Dense analytic curves in $\mathbb{R}^2$:</strong> 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.</p> <p>I'll think about the case $n>2$. I don't see it yet either, but maybe it's not too hard.</p> http://mathoverflow.net/questions/85881/can-a-lagrangian-submanifold-of-mathbb-r2n-be-dense-n1/85902#85902 Answer by jvp for Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)? jvp 2012-01-17T12:53:03Z 2012-01-17T13:33:08Z <p>This is more a remark than an answer.</p> <p>The typical solution of the typical polynomial ODE is uniformized by the Poincaré disc not by the complex line. </p> <p>Indeed, after the work of McQuillan, it is known that the existence of a non-algebraic leaf uniformized by $\mathbb C$ imposes strong restrictions on the polynomial vector field. It turns out that there exits a projective surface birational to $\mathbb C^2$ where the foliation defined by the vector field has at worst canonical singularities and its cotangent sheaf has Kodaira dimension zero or one. </p>