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