Let S be the open unit square in R^2: the set of points (x,y) with 0 < x < 1 and 0 < y < 1. Consider an area-preserving smooth map S --> R^2, that is, a map whose Jacobian has determinant 1 at every point. What can the image of S look like?

Can the image of the square have a smooth boundary? I think you can smooth out the two corners on top with a transformation of the form (x,y) --> (x,y+f(x)), but this makes the other two corners sharper.

(I had added, and now removed, some nonsense about Gromov's nonsqueezing theorem, which does not hold in dimension 2, and doesn't say what I thought it said besides.)

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    $\begingroup$ 1) Gromov nonsqueezing theorem is a statment about symplectic geometry of dimension 4 and higher. In dimension 2 a symplectic map is just a volume preserving map. Notice that (x,y)-->(x/2,2y) is a volume preserving map. So what is written in 1 is just not correct. Look in en.wikipedia.org/wiki/Nonsqueezing_theorem 2) This depends on the class of maps that you consider. Since you have a "manifold" with a conner you should specify what happen on the conner. If you don't put any restriction, you can do whatever you want $\endgroup$ – Dmitri Panov Nov 12 '09 at 9:20

Yes, you can smooth all corners. Say, in the class of maps $(x,y)\mapsto (f(x),g(x,y))$.

Moreover, one can map open unit square to any domain bounded by convex smooth curve by a map of the above type.

I'm sure that any simply connected domain of the same area can be also obtained as an image...

  • $\begingroup$ Can you say more about how to choose f and g? For (f(x), g(x,y)) to be area-preserving, g has to be linear in y, which feels like too strong a condition. I should say what I know about Gromov's nonsqueezing theorem in the question. What I know isn't much, but it contradicts your intuition about being able to get anything as the image. $\endgroup$ – David Treumann Nov 12 '09 at 5:48
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    $\begingroup$ Yes $g(x,y)=a(x)+b(x)y$ and the functions $f$, $a$ and $b$ depend on the shape you want to get (any smooth closed curve which bounds a domain of area 1 will give unique $f$, $a$ and $b>0$. $\endgroup$ – Anton Petrunin Nov 12 '09 at 15:32

Your statement of Gromov's theorem is wrong. Of course you can map a square to any rectangle of the same area! Just by diagonal linear matrix with eigenvalues (k, 1/k) (i.e. shrink in one direction, stretch in another).

Gromov nonsqueezing says $B^{2n}(1)$ can not be put inside $B(r)x\mathbb{C}^{n-1}$ for $B(r)$ - real 2-d ball in $\mathbb{C}$ with $r<1$ by a symplectomorphism. This means by a map that preserve symplectic form, which in turn can be viewed as measuring the "sum of the areas of projections of a 2-d object on the planes x1-y1, x2-y2 ....xn-yn". The non-squeezing in some sense is a statement about not being able to trade these "complex plane" directions for each other. It says nothing about squeezing inside those planes.

I believe by a general volume preserving map you can take any contractible domain to any other of the same volume...

  • $\begingroup$ This answer is more helpful and embarrassing for me, but I think it means that Anton is dead right. $\endgroup$ – David Treumann Nov 12 '09 at 15:35
  • $\begingroup$ For volume prserving embeddings, see "Embedding problems in symplectic geometry" by Felix Schlenk, Appendix B: volume is indeed the only constraint. $\endgroup$ – Max M Nov 12 '09 at 17:03

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