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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.

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I assume you want also some connectivity conditions. I don't know precisely which version of "manifold" you use, but presumably I can draw a half-space parallel to the first $n$ axes, through every point with totally rational coordinates. This is an immersion of countably many disjoint copies of $\mathbb R^n$. But I'm sure this is not what you want. –  Theo Johnson-Freyd Jan 17 '12 at 5:06
    
@S. Carnahan In the original post of Richard Montgomery there should be a link to a paper of Panov. Because for some reason it is missing and I have not enough reputation to edit it, would you make it a real link. Thank you. It is: www2.imperial.ac.uk/~dpanov/TORUS.PDF –  Giuseppe Tortorella Jan 17 '12 at 8:46
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Why not take the product of $n$ 1-dimensional examples in $\mathbb R^{2n}=(\mathbb R^2)^n$ ? –  BS. Jan 17 '12 at 13:12
    
@Giuseppe, Richard Montgomery: I fixed the link. It seems one must include the "http://" for links to work properly. –  j.c. Jan 17 '12 at 17:16
    
@Giuseppe. thanks. –  Richard Montgomery Jan 18 '12 at 14:33
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2 Answers

up vote 17 down vote accepted

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.

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@ Robert You can take direct products of your example to produce higher dimensional examples, $n$ even. –  Liviu Nicolaescu Jan 17 '12 at 15:01
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@Liviu: Yes, you are right, the $n$ even case is clear for this reason, but, somehow, it seems like cheating. Moreover, one could get examples for all $n$ by starting with a dense (analytic, if you like) curve in $\mathbb{R^2}$ and taking the $n$-fold product of it in $\mathbb{R}^{2n}$ (as BS pointed out in the comments to the question). It seems that there should be some construction that doesn't depend on complex analysis or products. –  Robert Bryant Jan 17 '12 at 15:50
    
@ Robert: Very nice examples. –  Liviu Nicolaescu Jan 17 '12 at 18:22
    
You the man, Robert! Very clear. And the concrete example of the dense curve is a big help. -thank you. –  Richard Montgomery Jan 18 '12 at 14:33
    
@Richard: Thanks. I have a question about the $n=1$ example you cited. I looked at the paper and got a general idea of how the construction goes, but it wasn't clear to me (since I haven't had time to study the details) whether this example of a dense, connected $1$-dimensional submanifold in $\mathbb{R}^2$ is real-analytic. Is it? The dense real-analytic curves that I describe in my answer are not submanifolds, of course. –  Robert Bryant Jan 19 '12 at 12:34
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This is more a remark than an answer.

The typical solution of the typical polynomial ODE is uniformized by the Poincaré disc not by the complex line.

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.

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