13
$\begingroup$

Let $f: M\rightarrow N$ be a smooth map between smooth closed oriented connected manifolds of same dimension.

Question: is it true that $f$ is smoothly homotopic to some smooth map $g: M\rightarrow N$ such that

  • The determinant of the differential $\det (dg(m))\geq 0$ for all $m\in M$
  • or $\det(dg(m))\leq 0$ for all $m\in M$.
$\endgroup$
5
  • $\begingroup$ This has the flavor of the "cancellation" result that if an oriented vector bundle has Euler class $0$, then it has a nowhere-vanishing section. Since you may as well assume that the degree of $f$ is nonnegative, I am now wondering about the case that the degree of $f$ is $0$. Then I believe the only way you can obtain your desired conclusion is for $g$ to be constant. So can we find an example of a degree zero map $f\colon M\to N$ that is not nullhomotopic? $\endgroup$ Nov 24, 2023 at 20:58
  • $\begingroup$ Nope, that is wrong. Every map $S^3\to S^2\times S^1$ has degree $0$ (e.g., by cohomological reasoning). What about taking $f=(\pi,c)$ where $\pi$ is the Hopf map and $c$ is a constant map? Here $f$ is singular at every point ($\det(df_p)=0$ for all $p$) but it certainly is not constant. $\endgroup$ Nov 24, 2023 at 21:09
  • 1
    $\begingroup$ A map $f:M \to N$ of degree $0$ which is "essentially surjective", in the sense that every map homotopic to $f$ is surjective, would provide a counterexample to the question. However, according to this MO post such maps do not exist! $\endgroup$ Nov 25, 2023 at 5:01
  • $\begingroup$ Origin/motivation of the problem? $\endgroup$
    – C.F.G
    Nov 29, 2023 at 6:40
  • $\begingroup$ To @TedShifrin's first comment: $g$ doesn't have to be constant. Consider the map from the 2-sphere to itself that sends $(\phi, \theta)$ to $(\phi, 0)$ (i.e., everything ends up on the prime meridian. The differential has rank 1 everywhere except the poles, where the rank is zero, so the det is zero everywhere. $\endgroup$ Dec 4, 2023 at 14:37

3 Answers 3

1
$\begingroup$

I don't think so.

Consider a three-holed torus, which looks like this (roughly) when seen lying on the XY-plane, seen from along the Z-axis:

()()()

with the center of the middle hole being at $(x,y,z) = (0,0,0)$, and the line containing the centers of the three holes being the $x$-axis. I hope that's clear.

Now I'm going to edit that slightly, by inserting a pair of tubes to widen the middle hole, so that the schematic picture now looks like this:

   __
()(__)()

To make that concrete, I'll say the cylindrical part extends from $x = -1$ to $x = 1$. I assume that you'll have no trouble seeing that this can be given the structure of a smooth manifold.

Now I'm going to define a map from this object -- let's call it $T$ -- to its right half. Let $$ F(x, y, z) = \begin{cases} (|x|, y, z) & |x| \ge 2 \ (k(x), y, z) & |x| \le 2 \end{cases} where $k$ is a smooth even function from $[-2, 2] \to [0, 2]$ with the property that $k(\pm 2) = 2$, $k'(2) = 1$, and $k'', k''', \ldots$ at $2$ are all zero.

Then $F$ is smooth, and has $\det(DF(m)) = 1$ for points where $x > 2$, but has $\det(DF(m)) = -1$ for points where $x < -2$. But $F$, restricted to either a meridian or longitude on the left handle of $T$ is orientation reversing, while restricted to a meridian or longitude on the right 'handle' of $T$ is the identity.

I believe that's enough to show that you can't get rid of the positive/negative determinant.

My first draft was applying this idea to a single torus: send $(x, y, z)$ to a smoothed version of $(|x|, y, z)$. The problem there is that the entire image of the map can be deformed into a single circle in the $xz$-plane, where the determinant is zero everywhere. I added the two extra handles to prevent that kind of 'collapse'.

What about a proof? I suspect (but it's been a long time) that some relationship between the degree of the map as a smooth map of manifolds and what the map does to (co?)homology classes makes it obvious that you can't get the differential to be zero everywhere is enough to make the case, but I can't claim to have a real proof.

$\endgroup$
3
  • $\begingroup$ Thanks for your answer, I do believe that there should be a counterexample to my question, but still I do not see how your answer could provide a proof. For instance, I do not understand why your function is not smoothly homotopic to some smooth function satisfying the property asked in my question. Probably, I will need to read more carefully your example. $\endgroup$
    – GSM
    Nov 23, 2023 at 16:00
  • $\begingroup$ My suspicion is that it can't be smoothly homotopic to such a function because of cohomology/degree reasons. I'll think about this tonight and see whether I can recall enough algebraic/differential topology to make it into a proof. $\endgroup$ Nov 23, 2023 at 23:23
  • $\begingroup$ No ... on reflection, I think you're right. The right-hand half of my example is topologically a twice-punctured torus. If we build this from the usual cut-and-paste gluing, with one of the punctures at the center of the disk, there's an obvious retraction to the boundary. That provides a homotopy from $F$ to a map $G$ whose image lies in the 1-skeleton, so the rank of the differential is always at most one, hence the det is zero. I'll ponder some more. $\endgroup$ Nov 24, 2023 at 2:59
0
$\begingroup$

A proposal for a counterexample, hopefully to be continued by someone else.

The idea is to use as $M=N$ a torus "strangled" on two opposite points.
Similarly, one could use two spheres, glue the north pole of first sphere to the south pole of second sphere, then (after stretching one sphere) glue the north pole of second sphere to the south pole of first sphere.

Usual torus has parametric equations, with $a$ and $b$ being the two circle radii, $a > b > 0$:
$\quad \begin{cases} x = (a + b \cos v) \cos u\\ y = (a + b \cos v) \sin u\\ z = b \sin v \end{cases}$

Strangling it on two opposite points $u=0$ and $u=\pi$ by replacing radius $b$ by $b \,(1 - \cos (2u))$, this gives $M$:
$\quad \begin{cases} x = (a + b \,(1 - \cos (2u)) \cos v) \cos u\\ y = (a + b \,(1 - \cos (2u)) \cos v) \sin u\\ z = b \,(1 - \cos (2u)) \sin v \end{cases}$

Then define $f: M \to M$ as:
$\quad \begin{cases} u \in [0, \pi): f(x,y,z) = (x,y,z)\\ u \in [\pi, 2\pi): f(x,y,z) = (x,y,-z) \end{cases}$

On $u \in [0, \pi)$, $f$ is identity so its Jacobian is $1$.
On $u \in [\pi, 2\pi)$, $f$ is a symetry by plane $z=0$, so it inverts orientation, its Jacobian is $-1$.

Now it seems $f$ is not smoothly homotopic to a smooth $g$ whose Jacobian is always $\ge 0$ or always $\le 0$, but I don't know how to prove it.

And a doubt: is strangling the surface to a point allowed on a smooth manifold?

$\endgroup$
0
$\begingroup$

This is true in two dimensions by a result of Edmonds: every map is homotopic to the composition of a pinch and a branched cover. In turn, pinches and branched covers have the desired property.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .