We call the transformation $x^i=x^i(z^1,...,z^n)$ an isometry

if $g'_{ij}(z^1,...,z^n)=g_{ij}(x^1(z),...,x^n(z))$, where $g'_{ij}=\frac{\partial x^k}{\partial z^i}g_{kl}\frac{\partial x^l}{\partial z^j}$ (use the Einstein notation)

Want to obtain: Isometry of Euclidean space ( by the Riemannian metric ) is affine. (No need to use manifold language here.)

There are many discussions follow from other definitions, but I want to know how I can solve this problem from the definition above.

Here is my idea:

Consider coordinate $z$ be the usual Euclidean coordinate.

Notice that the matrix $(g'_{ij})$and$(g_{kl})$are identity matrix.

Then $1=(\frac{\partial x^k}{\partial z^i})^2$ for all $i$ .

and $0=\frac{\partial x^k}{\partial z^i}\frac{\partial x^k}{\partial z^j}$ for all $i\neq j$. (use Einstein notation on $k$)

I want to show that $\frac{\partial^2 x^a}{\partial z^b\partial z^c}=0$ for all $a,b,c$ , and hence $\frac{\partial x^a}{\partial z^b}=const.$

Then the result follows.

But I have difficulty with the last step, do I need more procedure before ?

  • $\begingroup$ Notations for those who are not used to: a vector $\vec{x}$ has components $(x^1,\cdots, x^n)$ (index could also go from 0 to n-1) and the scalar product of two vectors is written $\langle \vec{x},\vec{y}\rangle = g_{ij} x^i y^j$. Second comment, the proof below works for an isometry from a Euclidean affine space $X$ to different one $Y$ of the same dimension. $\endgroup$
    – Noix07
    Commented Mar 25, 2017 at 18:32
  • $\begingroup$ here is a variant of the proof, showing that it still holds if the "scalar product" is not definite positive, but still non-degenerate. $\endgroup$
    – Noix07
    Commented Mar 26, 2017 at 14:11

1 Answer 1


As you pointed out the coordinate notations and the partial derivatives are useless and just make things complicated, and please justify a minimum what you have done (why $g_{ij}= \mathrm{id}, (\frac{\partial ...}{\partial ...})= ...$) if you want people to be interested in/understand your question and take the time to write an answer.

One can find the answer in Geometry I (Springer, Universitext 1987) by Marcel Berger, Proposition 9.1.3 p.202: (reformulated with words)

Let $(X,V)$ be a Euclidean affine space (finite dimensional) and $f: X \to X$ a map.

$f$ is an isometry iff it is affine, invertible and its associated linear map $\vec{f}$ is orthogonal (i.e. $\vec{f}$ rotation, possibly composed with a reflection).


  1. By definition of an affine space, for any two points $x,y \in X$, there exists a unique vector that can be denoted $\vec{xy}\in V$ such that $y= x + \vec{xy}$ (where + should rigorously be understood as the action of $V$ on X). It is also natural to write $y-x = \vec{xy} \in V$.
  2. By definition, an affine map $f: X \to Y$ between two affine spaces is of the form $f(x)= p+ \vec{f}(x-x_0)\ $ (evaluating at $x_0$ yields $p= f(x_0)$) with $\vec{f}$ a linear map on the underlying vector spaces of the respective affine spaces.
  3. An isometry $f$ is a map that preserves the norm associated to the scalar product, i.e. with the previous notation $\Vert f(y)-f(x)\Vert = \Vert y-x \Vert$ (or first notation $\Vert \overset{\longrightarrow}{f(x)f(y)}\Vert = \Vert \vec{xy} \Vert$ or if $f$ is affine $\Vert \vec{f}\big(\vec{xy}\big)\Vert = \Vert \vec{xy} \Vert$). Your definition of isometry naturally implies this weaker form: if $\Big\langle f(y)-f(x), f(y')-f(x')\big)\Big\rangle = \langle y-x, y'-x'\rangle$ for all $\ x,y,x',y' \in X$ then in particular $$ \Big\langle f(y)-f(x), f(y)-f(x)\Big\rangle = \langle y-x, y-x\rangle \quad \Longrightarrow\quad \Vert f(y)-f(x) \Vert = \Vert y-x\Vert$$ The two definitions coincide if $f$ is affine.

Proof: Let's start with the easy implication ($\Leftarrow$). $f$ affine means that there exists a linear map $\vec{f}$ such that $f(y)-f(x)= \vec{f}(y-x)$, moreover this $\vec{f}$ is orthogonal, i.e. $\Vert \vec{f}\big(\vec{xy}\big)\Vert = \Vert \vec{xy} \Vert$ hence $\Vert f(y)-f(x) \Vert = \Vert y-x\Vert$.

($\Rightarrow$) Conversely, assume that $f$ is an isometry. Choose an arbitrary point $p \in X$ and consider the isometry $T: X\longrightarrow X,\ x \longmapsto x + \overset{\longrightarrow}{f(p) p}\ $ designed so that $p$ is a fixed point of $T\circ f$, i.e. $T\circ f (p)=p$. (One can check that the composition $T\circ f$ is still an isometry).

The tricky part is to use the polarization identity (real case) which takes several forms (and only holds for norms that come from a scalar product).

$$\langle \mathbf{u},\ \mathbf{v} \rangle := \frac{1}{4}\Big(\Vert \mathbf{u}+\mathbf{v}\Vert^2 - \Vert\mathbf{u} -\mathbf{v}\Vert^2 \Big) = \frac{1}{2} \Big(\Vert\mathbf{u}+\mathbf{v}\Vert^2 - \Vert\mathbf{u}\Vert^2 - \Vert\mathbf{v}\Vert^2\Big) = \frac{1}{2} \Big( \Vert\mathbf{u}\Vert^2 + \Vert\mathbf{v}\Vert^2 - \Vert\mathbf{u}-\mathbf{v}\Vert^2\Big)$$

The last one is the most convenient in our case because only the $\mathbf{u}-\mathbf{v}$ appears and not $\mathbf{u}+\mathbf{v}$. (remember also $T\circ f (p)=p$) $$ \Big\langle T\circ f (x)-p ,\ T\circ f (y)-p \Big\rangle = \frac{1}{2} \Big( \Big\Vert T\circ f (x)-p \Big\Vert^2 + \Big\Vert T\circ f (y)-p \Big\Vert^2 - \Big\Vert T\circ f (x)- T\circ f (y)\Big\Vert^2\Big) = \frac{1}{2} \Big( \Vert\vec{px}\Vert^2 + \Vert\vec{py}\Vert^2 - \Vert\vec{xy}\Vert^2\Big) = \langle \vec{px}, \vec{py}\rangle$$

This is your stronger definition of isometry, which as explained above shows that if $T\circ f$ were affine, its associated linear map would be orthogonal.

Recall that $V$ is the underlying vector space, and define the map $$ \vec{Tf}: V \longrightarrow V,\ \vec{x} \longmapsto T\circ f \big(p+\vec{x}\big)-p$$

Showing that $T\circ f$ is affine is equivalent to $\vec{Tf}$ linear and actually follows from that same scalar product preserving property: pick $(x_i),\ i=1, \cdots , n$ such that $\big(\mathbf{e}_i:=\vec{px_i}\big)_{1\leq i\leq n}$ is an orthonormal basis of the underlying vector space. The scalar product equality then implies that $\big(\mathbf{f}_i:= \vec{Tf}(\mathbf{e}_i)\big)_{1\leq i\leq n}$ is another O.N.Basis. One has for any $\vec{x}\in V$

$$ \Big\langle \vec{Tf}(\vec{x}),\ \mathbf{f}_i \Big\rangle = \langle \vec{x}, \mathbf{e}_i\rangle =: x_i$$

This means that $\vec{Tf}(\vec{x})$ and $\vec{x}$ have the same expansion w.r.t. the two orthonormal basis, i.e. $ \vec{Tf}(\vec{x}) = \sum_{i=1}^n x_i\, \mathbf{f}_i $ and $\vec{x} = \sum_{i=1}^n x_i\, \mathbf{e}_i$ but that is exactly linearity $$\vec{Tf}(\vec{x})=\vec{Tf}\Big(\sum_{i=1}^n x_i\Big) = \sum_{i=1}^n x_i\, \mathbf{f}_i = \sum_{i=1}^n x_i\, \vec{Tf}(\mathbf{e}_i)$$


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