Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we have a sequence, { f_n }, of symplectic diffeomorphisms of R^{2n} converging to a function f. By f_n converging to f I mean: f_n converges to f in C^k(B_r) for every r > 0, where B_r is the ball centered at the origin of radius r. Suppose k > 0.

Question: Is f is diffeomorphism? Certainly, f is symplectic and hence a local diffeomorphism, and so my question is really: is f invertible?

Interestingly enough, this is not true if f_n is not symplectic (e.g. take f_n = x/n, which converges to 0 in C^k(B_r) for every r > 0).


share|improve this question
Why doesn't your exactly same example work in the simplectic case? The map $f_n:x\in\mathbb R^{2n}\mapsto x/n\in\mathbb R^{2n}$ is quite symplectic. –  Mariano Suárez-Alvarez Jul 21 '10 at 16:50
The fact that $\operatorname{SP}_{2n}(\mathbb{R})$ is closed in $M_{2n}(\mathbb{R})$ but $\operatorname{GL}_{2n}(\mathbb{R})$ is not seems relevant here. –  Pete L. Clark Jul 21 '10 at 16:50
@Mariano: it is only a symplectomorphism if you take the source and target with differently scaled symplectic structure, no? So it is not really a symplectic automorphism, which is presumeably what the OP wants. –  Willie Wong Jul 21 '10 at 17:43
Willie is correct in saying that I'd like each f_n to be a symplectic automorphism. –  user7807 Jul 21 '10 at 17:57
add comment

2 Answers

up vote 8 down vote accepted

No, $f$ doesn't have to be invertible. It has to be injective, even in the volume-preserving category rather than the symplectic category. None of the singular values of $Df$ can go to 0, because some other singular value would go to $\infty$. This shows that it is locally injective, and then globally it is not possible to make the locally injective pieces overlap.

But even when $n=1$, it does not have to be surjective. In this case the symplectic condition says just that $f_n$ is area-preserving. You can reshape larger and larger circles centered at $0$ to non-concentric circles with the same area that lie in the upper half plane. So the image of $f$ could be the upper half plane. Actually, I think that the image of $f$ can be any open topological disk with infinite area.

share|improve this answer
Thanks for the reply. I thought about whether the image of f can be the upper half plane, but wasn't sure. Are you 100% sure that this example works? I tried to construct f_n somewhat more explicitly, but had some trouble. In the examples I try to come up with the f_n's do not converge (e.g. my first naive attempt to explicitly carry out your example was to take f_n(x,y) = (x, y + n)). Do you know any conditions that one can give on f_n to guarantee that f is invertible? –  user7807 Jul 21 '10 at 17:55
It's a bit tricky to write explicit formulas, but yeah, I'm fairly sure. You can even make $f_n = f$ on the disk $B_n(0)$. Maybe the trick is to make $f$ first, so that it takes circles at the origin to circles in the upper half plane, and then change it outside of a disk of radius $n$ to make $f_n$. –  Greg Kuperberg Jul 21 '10 at 18:23
Thanks, that seems to work. –  user7807 Jul 21 '10 at 23:18
add comment

To find explicit examples of what Greg Kuperberg suggests, maybe one can try the following. Consider $\mathbb{R}^{2n}$ as the cotangent bundle of $\mathbb{R}^n$. Then diffeomorphisms between opens subsets of $\mathbb{R}^n$ lift to symplectomorphisms of the "cylinders" over these open subsets. So you can find a sequence of diffeomorphisms of $\mathbb{R}^n$ converging to a diffeomorphism of $\mathbb{R}^n$ onto a proper open subset of $\mathbb{R}^n$. When $n =1$ you can take a sequence converging to $f(x) = e^x$ (for instance $f_n(x) = e^x + \frac{x}{n}$). The symplectomorphisms obtained as the lift of these maps should give you an example of what Greg Kuperberg suggested.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.