3
$\begingroup$

This question is in some sense a continuation of this question, though more focused in its scope.

Let $(M,g)$ be an $n$-dimensional, connected, compact Riemannian manifold with boundary. Assume we are given an immersion $f:M \to \mathbb{R}^n$. (i.e $df$ is invertible at every point $p \in M$, note that I assume $n$ is the dimension of $M$).

Let $\omega:(M,g) \to (\mathbb{R}^n,e)$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $\omega|_{\partial M}=f|_{\partial M}$

Is it true that $\omega$ must be an immersion? Does anything change if we assume $f$ is (globally) injective?


Note that since $f$ is an immersion, it is locally injective, hence $f|_{\partial M}$ is locally injective. (In particular naive "counter-examples" like taking $\,f|_{\partial M}$ to be constant so $\omega$ is constant do not work). In fact we have $\text{rank}(d\omega_p)\ge \text{rank}\big(d(\omega|_{\partial M})_p\big)= \text{rank}\big(d(f|_{\partial M})_p\big)=n-1$ for every $p \in \partial M$

$\endgroup$
1
  • $\begingroup$ I deleted them because it was wrong, as Anthony Carapetis (who has now written a nice answer) pointed out to me: I gave examples where there are immersions that do not bound an immersion from $M$. But you have assumed already that this is possible. $\endgroup$ – user98602 Apr 13 '16 at 17:24
4
+150
$\begingroup$

Here's a counterexample on the unit disc. Let $f : \mathbb D^2 \to \mathbb R^2$ be defined by

$$ f(x,y) = (x-2y^2,y). $$

This has differential $$\left(\begin{matrix}1 & -4y \\ 0 & 1\end{matrix}\right)$$

and thus is an immersion. It's also injective.

The solution to the Dirichlet problem in this case is $$\omega(x,y) = (x^2 - y^2 + x - 1,y);$$ you can check that this is harmonic and agrees with $f$ on the boundary. However, its differential $$\left(\begin{matrix}1+2x & -2y \\ 0 & 1\end{matrix}\right)$$ is degenerate along the line $x = -1/2$, so $\omega$ is not an immersion.

It seems that you at least need to assume that $f(M)$ is convex - in the two-dimensional case this is enough (see https://eudml.org/doc/152349), but I'm not sure whether such a result holds in higher dimensions.

$\endgroup$
8
  • $\begingroup$ Thanks! Two remarks: 1) In your example $d\omega$ was non-invertible only on a set of measure zero. My interest is actually about integral quantities and hence the behaviour on sets of measure zero might not have any effect. I wonder if you have an example in mind (or can construct one) where the set of points where $d\omega$ is non-invertible has a positive measure. $\endgroup$ – Asaf Shachar Apr 13 '16 at 19:58
  • $\begingroup$ 2) I skimmed through the paper you linked to, and it seems their condition is a little different than what you have stated, i.e they assume the boundary map is a homeomorphism, not just injective, and I think only $f(\partial M)$ has to be convex, not all $f(M)$, but it's quite possible I misunderstood something. Thanks anyway. $\endgroup$ – Asaf Shachar Apr 13 '16 at 19:58
  • $\begingroup$ @AsafShachar: (2) See invariance of domain - since $f$ is between domains of equal dimension, injective immersion implies homeomorphism. The assumption is geodesic convexity of $f(\partial M)$, which is just convexity of $f(M)$ in this case since $f(\partial M) = \partial f(M)$ and $f(M)$ is a contractible domain in Euclidean space. $\endgroup$ – Anthony Carapetis Apr 14 '16 at 1:13
  • $\begingroup$ (1) If $M$ is analytic then $\omega$ will be analytic, in which case the set of critical points is measure-zero. I think without this assumption you're probably still out of luck - maybe see if you can adapt the example in link.springer.com/article/10.1007/BF02571406 to meet your requirements. $\endgroup$ – Anthony Carapetis Apr 14 '16 at 1:32
  • $\begingroup$ correction to my previous comment: I meant "convex in the sense of geodesic curvature" rather than "geodesically convex". This theorem also only applies in the case where $M$ is simply connected. $\endgroup$ – Anthony Carapetis Apr 14 '16 at 1:39

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.