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I'm trying to prove that if $f\colon\mathbb{R}^n \to \mathbb{R}^n$ is a $\mathcal{C}^1$ mapping such that $f'(x)$ is a (linear) isometry for every $x \in \mathbb{R}^n$, then $f$ is an isometry. By an application of inverse mapping theorem and mean value theorem, we have that $|f(x) - f(y)| = |x-y|$ as long as $x$ and $y$ are sufficiently close. How to extend this to the whole space?

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I'm not really sure what you intend with $f'$ here, but do you really mean $f'(x)$ to be a map? Presumably evaluating a derivative at a point would result in an element of the field, but not a map. Please take some time to clarify these two points :) – rschwieb May 28 '12 at 1:55
@rschwieb: For maps $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$, the derivative at a point is defined to be a linear map satisfying certain properties. When viewed this way, for a function $f:\mathbb{R}\rightarrow\mathbb{R}$, the "usual" derivative at a point $x_0$ should be thought of as "multiplication by $f'(x_0)$." – Jason DeVito May 28 '12 at 1:59
I suspect he means that $f'(x)$ is the differential of $f$ at the point $x$. For example, in one variable the functions equal to their own derivatives in this sense are really the linear functions $f(x) = mx$, not $f(x) = e^x$ (whose derivative at $x_0$ is the map $x \mapsto e^{x_0}x$ on the tangent space at $x_0$). Maybe one can use some kind of connectedness argument to show a local formula for $f$ as an orthogonal map composed with a translation has to work globally. – KCd May 28 '12 at 2:00
His formulation is correct. The derivative is a linear transformation. This is the approach taken in Spivak's Calculus on Manifolds. In this case $f': \mathbb{R}^n \rightarrow L(\mathbb{R}^n)$. – ncmathsadist May 28 '12 at 2:02
OK just checking :) I wasn't able to guess all of this. Thanks! – rschwieb May 28 '12 at 2:07
up vote 3 down vote accepted


Here's a much simpler argument. Let $X\subseteq\mathbb{R}^n\times\mathbb{R}^n$ with $$X = \{(x,y)\in\mathbb{R}^n\times \mathbb{R}^n|d(x,y) = d(f(x),f(y)) \}.$$ Note that $(x,x)\in X$ for any $x$, so $X\neq \emptyset$. Your observation is equivalent to the statement that $X$ is open. To see $X$ is closed, note that if $g:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}$ with $g(x,y) = d(x,y) - d(f(x),g(y))$, then $g$ is continuous since $d$ and $f$ are, and $X = g^{-1}(0)$. Hence, $X$ is a nonempty clopen subset of $\mathbb{R}^n\times\mathbb{R}^n$, so $X = \mathbb{R}^n\times\mathbb{R}^n$, so $f$ is an isometry.

(End edit)

Here's one approach, borrowed from Riemannian geometry.

Let $\gamma:\mathbb{R}\rightarrow\mathbb{R}^n$ be any straight line paramaterized by arclength, meaning that $\|\gamma(t)-\gamma(s)\| = |t-s|$ for any $t$ and $s$. We will show that $f\circ \gamma$ is also a straight line parametrized by arclength.

Believing this for a second, for $x$ and $y$ in $\mathbb{R}^n$, if $\gamma$ is chosen to be the line going through $x$ and $y$ with $\gamma(t) = x$ and $\gamma(s) = y$, then we have \begin{align*} d(x,y) &= \|\gamma(t)-\gamma(s)\|\\\ &= |t-s|\\\ &= \|f(\gamma(t))-f(\gamma(s))\| \\\ &= d(f(x),f(y)), \end{align*} establishing what we want.

So, why is $f\circ \gamma$ a straight line parameterized by arclength? This follows from your observation that $\|f(x)-f(y)\| = \|x-y\|$ for $x$ and $y$ close together. More specifically, looking at the point $\gamma(t)$, we know that for $s$ close to $t$ (and therefore $\gamma(s)$ close to $\gamma(t)$), that $\|f(\gamma(t)) - f(\gamma(s))\| = \|\gamma(t)-\gamma(s)\|$. This implies that for the line segment of points near $\gamma(t)$, that $f($segment$)$ is another line segment, parametrized by arclength.

Since $\mathbb{R}$ is connected, so is $f(\gamma(\mathbb{R}^n))$. This implies that $f(\gamma(\mathbb{R}^n))$ is a union of line segments end point to end point where each segment is parameterized by arclength. This is a straight line iff there are no corners. But if $f(\gamma(t_0))$ is a corner, then $f$ brings points near $\gamma(t_0)$, but on either side of it, closer together, contradicting your earlier observation that locally $f$ preserves distance.

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Alternatively it is easy to see that $f$ is injective for every open $U$, then $f$ is a diffeomorphism of $U$ over the open $f(U)$. Showing that $f(\mathbb R^m)$ is a closed set solves the problem, since $\mathbb R^m$ is connected. – Aaron Maroja Feb 25 at 19:58

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