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$\newcommand{\SO}[1]{\text{SO}(#1)}$ $\newcommand{\Hom}[1]{\text{Hom}(#1)}$ $\newcommand{\R}{\mathbb{R}}$

Let $U \subseteq \mathbb{S}^n$ be an open subset of the round sphere.

There exist an $\epsilon >0$ such that there is no smooth map $f:U \to \mathbb{R}^n$ satisfying $\text{dist}(df,\text{SO}) \le \epsilon$ everywhere on $U$.

Is there an elementary argument for showing this?

Here is how I define $\text{dist}(df,\text{SO})$. Let $p \in U$, then $df_p \in \Hom{T_pU,\mathbb{R}^n} $. The the inner products on $T_pU,\mathbb{R}^n$ induce an an inner product on the Hom-space; (This is just the tensor product on $T_p^*U \otimes \R^n$, or less abstractly, this product reduces to the Euclidean (Frobenius) product on matrices, when we represent linear maps w.r.t orthonormal bases).

Now, using the induced distance on the Hom-space, we have a notion of $ \text{dist}(df,\text{SO})(p)= \text{dist}\big(df_p,\SO{T_pU,\R^n}\big)$ .

The only proof I know for the claim above is hard; it is a proof for a stronger claim:

With the notations as above, consider the functional $E(f)=\int_U \text{dist}^2(df_n,\text{SO}) $. (The integration is w.r.t the Riemannian volume form of the sphere).

Suppose that there exist a sequence of maps $f_n:U \to \R^n$ satisfying $\text{dist}(df_n,\text{SO}) \le \frac{1}{n}$. Then $E(f_n) \to 0$. This implies, according to theorem 1.4 here, that $f_n$ converges (strongly) in the Sobolev space $W^{1,2}(U,\R^n)$ to a smooth isometric immersion. A contradiction.

Of course, requiring a small uniform bound is much stronger than requiring
a small integral bound, so I expect there should be an easier proof. Also, the theorem cited above holds only for the distance from $\text{SO}(n)$, while I expect one cannot find maps with arbitrarily small uniform distance from $\text{O}(n)$. (i.e. the claim in this question should hold if we replace $\text{SO}(n)$ with $\text{O}(n)$).

Edit:

I tried showing that an "approximate isometry" $f$ (with very small $\text{dist}(df,SO)$) maps geodesics to paths with small geodesic curvature, and then use the fact the rates of spread of geodesics in the sphere and the Euclidean space are different. (governed by the curvature).

However, this does not seem to be true in general; The condition on $\text{dist}(df,SO)$ is a condition on first derivatives, while the geodesic curvature of the mapped paths is affected also by the second derivative of $f$. Of course, for an exact isometry, $\nabla df=0$, but it seems false that small $\text{dist}(df,SO)$ implies small $|\nabla df|$. Indeed, think of the one-dimensional case: a map $\mathbb{R} \to \mathbb{R}$ can have $f'$ uniformly close to $1$, and crazy second derivative.

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  • $\begingroup$ The Frobenius distance $d(df,O)$ is attained at the orthogonal factor $\Theta$ of the polar decomposition $df = S \Theta,$ i.e. $d(df,O) = |S-I|.$ One idea is to interpret this $\Theta$ as an orthonormal coframe for $U$ and try to bound the curvature in terms of $|S-I|$ along with derivatives of $S.$ Seems to get messy, though. $\endgroup$ – Anthony Carapetis Feb 23 '18 at 11:52
  • $\begingroup$ Thanks, yeah I know about the polar factor. My intuition is also related to using the curvature as an obstruction. I tried using Jacobi fields; the rate of spread of geodesics is governed by the curvature, and an "approximate isometry" should map geodesics to "approximate geodesics". This might give a proof by contradiction. However, the details seem messy indeed. $\endgroup$ – Asaf Shachar Feb 23 '18 at 20:04
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$U$ can be taken to be a metric ball of radius $\delta$ around $s\in U.$ I claim:

  1. The image of $f$ includes at least a ball of radius $\delta c_1(\epsilon)$ around $f(s)$ where $c_1(\epsilon)\to 1$ as $\epsilon\to 0.$

    When $\text{dist}(df,\text{SO})$ is small, path lengths are approximately preserved; in particular there is some inequality of the form $|\gamma'|\leq c_1(\epsilon)^{-1}|(f\circ \gamma)'|$ for paths $\gamma:[0,1]\to U,$ with $c_1(\epsilon)\to 1$ as $\epsilon\to 0.$ Given a straight-line path of length at most $\delta c_1(\epsilon)$ starting at $f(s),$ we can lift it to a path of length at most $\delta$ in $U$ starting at $s.$

    To perform this lift we just need the solution of the ODE $df_{\gamma(t)}\circ \gamma'(t)=\text{const}.$ This exists by Picard–Lindelöf, shrinking $\delta$ if necessary to ensure that the second derivative of $f$ is bounded.

  2. The volume of the image of $f$ is at most $c_2(\epsilon)\operatorname{vol}(U)$ where $c_2(\epsilon)\to 1$ as $\epsilon\to 0.$

    I mean the usual Riemannian volume, $\sqrt{|g|}dx_1\wedge \dots \wedge dx_n$ in co-ordinates. This definition (or whatever other definition you prefer) shows the volume form is controlled by $g.$

But 1 and 2 are in contradiction for small $\epsilon$ because $\operatorname{vol}(U)$ is smaller than the volume of a Euclidean $\delta$-ball.

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  • $\begingroup$ Thanks, this looks like a nice solution! Regarding the volume estimate of the image of $f$, here is my way of seeing this: The Hausdorff measure and the Riemannian measures coincide, and the Hausdorff measure is controlled by the Lipschitz constant. Was that your thinking? $\endgroup$ – Asaf Shachar Feb 27 '18 at 7:16
  • $\begingroup$ @AsafShachar: hmm, I guess I am thinking of the "change of variables formula" - as formulated as an inequality here math.stackexchange.com/q/59115 $\endgroup$ – Dap Feb 27 '18 at 8:06
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Here is a simple answer for a simpler question: For every domain $V\subset R^n$, there exist an $\epsilon>0$ such that there is no diffeomorphism $f:U→V$ satisfying $\text{dist}(df,\text{SO})< \epsilon$ everywhere on $U$.

Proof: assume otherwise, then you have a sequence $f_k:U\to V$ such that $\text{dist}(df_k,\text{SO})< 1/k$. This implies that the local Lipschitz constant of $f_k$ is $L(f_k)< 1+ C/k$ (for some $C$ depending only on the dimension). By Arcela-Ascoli you get that $f_k$ converges uniformly to $f:U\to V$ and that $L(f)\le 1$. A simple topological argument gives you that $f$ is surjective. The same arguments work for $f_k^{-1}$, converging to a surjective $g:V\to U$ with $L(g)\le 1$. This implies that $U$ and $V$ are isometric (see Theorem 1.6.15 in Burago-Burago-Ivanov).

Maybe one can improve this proof into what you want...

Note that condition $\text{dist}(df,\text{SO})< \epsilon$ is equivalent to the purely metric conditions (i) $L(f) < 1+\epsilon$, (ii) $f$ is locally invertible and (iii) $L(f^{-1})< 1+ \epsilon$, where $L$ is the local Lipschitz constant. The argument above therefore works for any two non-isometric compact metric spaces $U,V$.

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