In a nutshell: If we start from an immersion, and look at its harmonic decomposition, are the components also immersions?


Let $(M,g)$ be an $n$-dimensional, compact Riemannian manifold with boundary. Assume $f:M \to \mathbb{R}^n$ is an immersion which is not harmonic*. (i.e $df$ is invertible at every point $p \in M$, note I assume $n$ is the dimension of $M$). Let $(\omega,z)$ be the harmonic decomposition of $f$, i.e: $\omega,z:M \to \mathbb{R}^n$ satisfying:

(1) $f=\omega+z$

(2) $\omega$ is harmonic ($\Delta_B\omega^i=0,\text{where } \, \omega^i \, \, \text{ are the components of } \, \,\omega$ and $\Delta_B$ is the Laplace-Beltrami operator)

(3) $\omega|_{\partial M}=f|_{\partial M}$

Is it true $dw$,$dz$ must be invertible?

*Note: If $f$ is harmonic ($f=\omega,z=0$ then of course $dz$ is non-invertible).

Update: As showed by Anthony Carapetis $dz$ can be in general non-invertible on an open subset of $M$. The question whether $d\omega$ can be non-invertible is still open.


1 Answer 1


Let's take $M$ to be a connected domain in $\mathbb R^n$ and the boundary condition to be the standard immersion $\partial M \to \mathbb R^n$ (so that $\omega$ is the inclusion map $M \to \mathbb R^n$).

Then let $\phi$ be a function $M \to \mathbb R$ vanishing on the boundary satisfying $|\nabla \phi| < 1$ and define the deformation $f(x) = x + \phi(x)e_1$. This has differential $Df(v) = v + D_v\phi\ e_1$, so in particular $|Df(v)| \ge |v|(1 - |D_{v/|v|} \phi|) > 0$; i.e. $f$ is an immersion. Additionally we know $\Delta f = \Delta \phi\ e_1$, so we can even arrange for $f$ to be non-harmonic everywhere on the interior (take $\phi$ to be a small multiple of the first Dirichlet eigenfunction of $\Delta$).

Since $\omega(x) = x$, we get $z(x) = \phi(x) e_1$, which has 1-dimensional image and is clearly not an immersion.

I imagine you can do a similar "bad" deformation starting from any harmonic local diffeomorphism.

  • $\begingroup$ Thanks, but I meant for the domain and the codomain to be of the same dimension. (i.e $n$ is the same). I will stress this point better in the phrasing of the question. $\endgroup$ Commented Apr 4, 2016 at 10:04
  • $\begingroup$ Actually I now see that I wrote $df$ should be invertible,... so this was implicitly stated in the question $\endgroup$ Commented Apr 4, 2016 at 10:07
  • $\begingroup$ @AsafShachar: I believe the mistake was mine - you did state $M$ was $n$-dimensional. Anyway, I've got a new counterexample for you. $\endgroup$ Commented Apr 4, 2016 at 10:30
  • $\begingroup$ Thanks (I upvoted:). In your example $\omega$ was an immersion. I am still interested to know whether it's possible for $d \omega$ to be non-invertible (on a set of positive measure). Perhaps you can think of an example where this situation occurs? Thanks again. $\endgroup$ Commented Apr 4, 2016 at 16:18
  • $\begingroup$ @AsafShachar: I think $\omega$ will always be an immersion. Try looking for some literature on the Dirichlet problem for harmonic diffeomorphisms. $\endgroup$ Commented Apr 4, 2016 at 22:49

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