# Thread: differentiable map and local isometry

1. ## differentiable map and local isometry

hi dear mathhelpform's citizens ı have a question,ı want to ask you;

prove;
if the differential of a differentiable map F preserves,orthonormal basis then F is a(local) isometry.

2. Originally Posted by sah_mat
hi dear mathhelpform's citizens ı have a question,ı want to ask you;

prove;
if the differential of a differentiable map F preserves,orthonormal basis then F is a(local) isometry.
I'll need an additional hypothesis: $F$ is continuously differentiable. Did you forget it?
Note that if $\varphi$ is a linear map, " $\varphi$ preserves orthornormal bases" implies $\|\varphi(x)\|=\|x\|$ for every $x$, (one says that $\varphi$ is orthogonal). This implies $\|\varphi\|=\max_{x\neq 0}\frac{\|\varphi(x)\|}{\|x\|}=1$.
Suppose $F$ is defined on an open convex set $U$. By the previous remark, we have $\|dF_x\|=1$ for every $x\in U$. As a consequence, for any $x,y\in U$, the mean-value theorem gives $\|F(x)-F(y)\|\leq \max_{z\in[x,y]}\|dF_z\|\times \|x-y\|=\|x-y\|$.
In order to get the reverse inequality, prove that, at any point, $F$ has locally an inverse function (by the inverse function theorem), notice that this inverse function satisfies the same hypothesis as $F$ and procede like above to find $\|F^{-1}(f(x))-F^{-1}(f(y))\|\leq \|f(x)-f(y)\|$ if $x,y$ are in a (possibly small) open convex set where $F$ is invertible. And this is it.