6
$\begingroup$

A $n-1$ dimensional submanifold $N\subset \mathbb{R}^n$ is called a convex submanifold if for every $x\in N$ ,ther is a neighborhood $W$ of $x$ in $N$ such that $W$ entirly lies at one side of $T_x N$. A (local) diffeomorphism $\phi$ on $\mathbb{R}^n$ is called a convex diffeomorphism if $\phi$ and its inverse preserves the convexity of all codimension $1$ submanifolds. For example every affine linear isomorphism is a convex diffeomorphism but the diffeomorphism $\phi(x,y)=(x,y-x^2+x^3)$ is not a convex diffeomorphism because it maps the convex curve $y=x^2$ to non convex curve $y=x^3$.

1.In the above definition, is the word "its inverse", redundant?

  1. Is there a well known description of the group of all smooth convex diffeomorphisms of $\mathbb{R}^n$?

3.Does every manifold admit an atlas whose all transition maps are convex diffeomorphism?

If the answer to the later question is affirmative, then we can define the concept of convexity for any codimension $1$ submanifold of an arbitrary manifold $M$. In particular we can speak of "convex foliation" of a manifold $M$, a codimension $1$ foliation of $M$ whose all leaves are convex submanifold. In this case the next question would be the following:

4.Is there a manifold which admit a codimension $1$ foliation but does not admit any convex foliation?In particular does $S^3$ admit a convex foliation?

$\endgroup$
  • $\begingroup$ Shouldn't the first sentence read "for every $x \in \partial N$"? $\endgroup$ – Arnaud Mortier Dec 30 '17 at 11:57
  • 1
    $\begingroup$ I understand that convex hypersurfaces are more general than subsets of boundary of a convex domain, in that for instance a boundary of a non-convex polygon, with smoothed corners, does satisfy the definition. But I'd say all convex diffeos are affinities. Or is there a counterexample? $\endgroup$ – Pietro Majer Dec 30 '17 at 18:08
  • 1
    $\begingroup$ I guess that if you do not abandon the condition $\phi^{-1}$ preserves the convexity of all codimension 1 sub-manifold then you definition of convex diffeomorphism is nothing but affine map. $\endgroup$ – Hu xiyu Dec 31 '17 at 4:20
  • 1
    $\begingroup$ For the last question I think the existence of global convex foliation induce a strong topology restriction of the underlying manifold $M$, may be $M$ could just be a ball, at least I think the fundamental map vanish, just look at the induce foliation on loops space. $\endgroup$ – Hu xiyu Dec 31 '17 at 4:23
  • 3
    $\begingroup$ @PietroMajer: In addition to the affine maps, there are also projective maps, which, on their domain of definition, carry line segments to line segments, and, hence, preserve convexity. Thus, for example, on the right half plane (defined by $x>0$, the diffeomorphism $f(x,y) = (1/x, y/x)$ carries convex sets to convex sets and is not an affinity. $\endgroup$ – Robert Bryant Dec 31 '17 at 11:14
6
$\begingroup$

If $n>1$, and a smooth diffeomorphism $f:U\to V$ (where $U$ and $V$ are, say, convex, open subsets of $\mathbb{R}^n$) carries convex sets to convex sets, then it is easy to show that it must carry each intersection $U\cap H$, where $H\subset\mathbb{R}^n$ is a hyperplane, to an intersection $V\cap H'$, where $H'\subset\mathbb{R}^n$ is another hyperplane. Hence $f$ must carry intersections of hyperplanes to intersections of hyperplanes, and hence line segments to line segements. In particular $f$ must be the restriction of a projective transformation of $\mathbb{RP}^n$ to $U\subset\mathbb{R^n}\subset\mathbb{RP}^n$. Conversely any such restriction carries convex sets to convex sets, as does its inverse. This answers Questions 1 and 2.

For Question 3, you are asking whether every smooth $n$-manifold $M$ admits a flat projective structure. The answer is no, for then the associated developing map on the simply connected cover would have a smooth immersion into $S^n$. However, already this is impossible for $n=4$ if one starts with a simply-connected compact $4$-manifold other than the $4$-sphere.

However, note that $n=2$ does work, since, by uniformization, every two-dimensional surface carries a metric of constant Gauss curvature, and the underlying projective structure associated to such a metric is projectively flat.

Finally, $S^3$, which does admit a codimension $1$ foliation and does admit a flat projective structure (as the double cover of $\mathbb{RP}^3$), does not admit a convex foliation. The reason is that the flat projective structure is unique (since $S^3$ is simply-connected) and at least one of the leaves of the foliation would have to be a compact torus because every codimension $1$ foliation of the $3$-sphere must have a Reeb component. However, it is not possible to immerse the standard torus in the standard $S^3$ smoothly as a convex surface (with respect to the projective structure): Endow $S^3$ with its standard 'round' metric of constant sectional curvature $+1$; its underlying projective structure is the flat one. For any torus $T^2\subset S^3$, the Gauss curvature of the induced metric must vanish somewhere (since the integral is zero), and, at such a point, the product of the principal curvatures must be $-1$, by the Gauss equation. In particular, the two principal curvatures are nonzero, with opposite signs, and the torus is not convex on a neighborhood of such a point. Thus, this gives an example of the kind asked for in Question 4.

$\endgroup$
  • $\begingroup$ Dear Prof. Bryant. Thank you very much for your great answer. I try to understand its details. $\endgroup$ – Ali Taghavi Jan 1 '18 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.