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 $p \in \partial M$

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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