On diffeomorphisms that preserve the metric

Question

Suppose $\Omega\subset \mathbb R^2$ is a bounded domain with smooth boundary and suppose that $$ F: \Omega \to \Omega,$$ is a diffeomorphism that fixes $\partial \Omega$ (i.e $F|_{\partial \Omega}$ is equal to the identity map) and such that the pull back of the Euclidean metric under $F$, namely $F^\star e$ is again equal to the Euclidean metric $e$. Can we conclude that $F$ is the identity map?

Answer 1

The expression "fixes $\partial\Omega$" is ambiguous. Do you mean that $f(\partial\Omega)=\partial\Omega$ or that $f(z)=z$ for all $z\in\partial\Omega$?

For the first, weaker condition, all exceptional exceptional domains $\Omega$ and functions can be easily described.

Preservation of Euclidean metric implies that $f$ is conformal, therefore it is complex analytic, and $|f'(z)|=1$ everywhere. By Complex Analysis, this easily implies that $f(z)=cz+b,$ where $|c|=1$. Now it is easy to describe all domains $\Omega$ which can mapped homemoprphically onto themselves by such functions.

If $c$ is not a root of unity, they must be disks, including the whole plane, and any disk with center $b$ is mapped onto itself by $f(z)=cz+b-cb$. If $c$ is a root of unity but not $1$, then $\Omega$ must have a rotational symmetry of some finite order, and again all such domains possess such homeomorphisms. Finally if $c=1$, a domain must be periodic (with period $b$).

These are all exceptional domains.

So for the second condition (fixing the boundary pointwise) we have only one exceptional domain, namely the plane, and $f(z)=cz+b$ with arbitrary $c, |c|=1$.

Answer 2

This is true if $\Omega\neq\mathbb{R}^2$ (so that every path component of $\Omega$ has nonempty boundary). Firstly, $F^*e=e$ means that $F$ is a local isometry. We know that two local isometries $f,g:X\to Y$ between connected Riemannian manifolds that satisfy $f(p)=g(p)$ and $df_p=dg_p$ for some $p\in M$ are the same (see for example exercise 5.7 of Lee's $\textit{Riemannian Manifolds}$).

To prove that $F(x)=x$ for all $x$, let $y$ be a point of $\partial\Omega$ in the same path component as $x$. Then $Df_y=Id_{T_y\Omega}$, because it is the identity in $T_y\partial\Omega$ and fixes the unit vector perpendicular to $\partial\Omega$. So for any geodesic from $y$ to some point $z$ in the interior of $\Omega$, $F^*e=e$ implies that $F$ has to fix that geodesic so $F(z)=z$. This means that there is some open set in the path component of $x$ which is fixed by $F$, so $F$ is the identity in that path component of $X$.