# geometry of null homotopies

Given a homotopy class of map $f$ between unit spheres $S^n \to S^m, n>m$, let "stretch" be its "stretch factor" ( = inf over the homotopy class of the sup norm on the ( operator) norm on the first derivative). Suppose, as is usually the case, that some $k$th power of $f$ is null homotopic. For a fixed representative $f$, $$\textrm{stretch} (f^k) \leq k\cdot \textrm{stretch} (f).$$ Does anyone know an example where the stretching of the null homotopy $h$ for $f^k$ ( as a map from the $n+1$-disk, $h:D^{n+1}\to S^m$, $h_{|\partial D^{n+1}}=f^k$) must be "enormously" larger (or even just larger) than $k\cdot \textrm{stretch}(f)$? Since simply connected homotopy theory is decidable, the minimum stretch of the null homotopy $h$ is bounded by a recursive function of the stretch of the boundary condition, stretch($f^k$) - but this still allows scope for potentially extravagant examples. Are any known?

-
What exactly do you mean by f^k? Are you iterating? smashing? Does it matter? –  Tom Goodwillie Aug 19 '10 at 19:44
I think he must mean the group operation $f^k = f \cdot f \cdot \cdots \cdot f$. –  Ryan Budney Aug 19 '10 at 21:10
The group being $\pi_n(S^m)$... But the $k+1$-disk must then be $n+1$. –  BS. Aug 19 '10 at 23:03
My initial inclination is that the interesting examples (probably there are some) would appear in the unstable range. I'm thinking about this from the framed cobordism sense -- in the stable range all you need is any manifold that bounds the representing manifold, so choose an "efficient" one. But in the unstable range, not all bounding manifolds neccessarily embed as the co-dimension is too small. So you might have to find "elaborate" bounding manifolds, resulting in lots of stretching. Your question has a Thurston norm type feel to it. –  Ryan Budney Aug 20 '10 at 6:28
incidentally, Pete Storm and Schmuel Weinberger thought about Lipschitz constants of maps between spheres. I think they showed that the Hopf map has minimal Lipschitz constant: atlas-conferences.com/c/a/w/j/13.htm –  Ian Agol Aug 20 '10 at 15:39

## 1 Answer

Your question should follow from: given a homotopically trivial map $g:S^n\to S^m$, is there a homotopy $h: D^{n+1} \to S^m, h_{| \partial D^{n+1}} = g$ such that $\textrm{stretch}(h) \leq \textrm{stretch}(g)$? Or maybe there exists a reasonable function $L_n:[0,\infty) \to [0,\infty)$ such that $\textrm{stretch}(h) \leq L_n(\textrm{stretch}(g))$. I think you're claiming that there exists $L_n$ which is computable. It seems to me that since $\textrm{stretch}(f^k)\leq k\cdot\textrm{stretch}(f)$, the fact that $g=f^k$ doesn't really matter in the problem.

Interpreting the problem this way, one approach is to try to find for $z\in S^m$ a factorization $g': S^n \to T_z S^m \cong \mathbb{R}^m$, such that $g=exp_z\circ g'$ using the fact that $g$ is homotopically trivial. If one could control the diameter of the image of $g'$, then one ought to be able to bound the stretch of the trivial filling by coning to the origin in $T_z S^m$ (this amounts to filling in $g$ by coning by geodesics to a point). I don't know when such a lift is possible though, since obviously $exp_z: \mathbb{R}^m \to S^m$ is far from being a fibration at concentric spheres about the origin.

Here's a possible approach: Take a map $h: D^{n+1} \to S^m, h_{| \partial D^{n+1}} = g$ where $h(0)=p$. Think of this a map $H: S^{n} \to [([0,1],0), (S^m,p)]$, that is a map from $S^{n}$ to the space of intervals at p. One may show that we may homotope $h$, keeping $h_{| S^{n}=\partial D^{n+1}}$ fixed, to a map sending each interval $[0,1]x, x\in S^{n}$ to a piecewise linear path. This is the process described in Theorem 17.1 of Milnor's book. The idea then would be to try to homotope all of the intervals to be piecewise geodesics of bounded length (above I suggested homotoping all of the intervals to geodesics, which is absurd), and then try to show that the resulting map has bounded stretch. Like the fundamental theorem of Morse theory, I would expect to be able to homotope things down onto $n+1$-cells which correspond to index $n+1$ geodesics, and therefore have bounded length, by something like cellular approximation. But I don't quite know how to complete the argument yet.

-