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.
thanks for your helps.
Why did you put parentheses on "local"? It is true with or without, but it is much easier with the parentheses, and you would have been given a hint to do the global version, so I guess what you need is the local version and that's what I'll be writing about.
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.