an isometry $\phi$ of a Riemannian manifold $(M, g)$ is called a Clifford translation if the functionp $p→ d(p, φ(p)),p\in M$ is constant, where d is the riemannian distance and is defined as:
$d(p,q)=inf\{L[c] |\quad c:[a,b]\to M, \text{ c piece wise smooth} , c(a)=p,c(b)=q\}$
1) . Show that the Clifford translations on $\mathbb R^n$ with the Euclidean metric are the ordinary translations.
First I convinced myself that for translation $d$ is really const, by drawing a picture of two arbitrary point $p_1,p_2$ and shift them by a const $v$. The result is parallelogram.
To show analytically that $\phi$ must be a translation $\phi(p)=p+v$, i used the fact that the riemannian distance between two points in $R^n$ (eucl. metric) is just given by the straight line
$d(p,q) = L[[0,1]\ni t \to c(t)=p+(q-p)t]=\int_0^1 |\dot{c}(t)|dt= |q-p|=\left(\sum_i \left(q_i -p_i\right)^2 \right)^{1/2} \qquad $
Let $\phi$ be some arbitrary isometrie for which
$d(p_1,\phi(p)) = d(p_2,\phi(p_2))$
holds true. Using the result from above i get
$|\phi(p_1)-p_1|=|\phi(p_2)-p_2| \Leftrightarrow \left(\sum_i \left(\phi(p_1)_i -p_{1,i}\right)^2 \right)^{1/2} = \left(\sum_i \left(\phi(p_2)_i -p_{2,i}\right)^2 \right)^{1/2} $
$\Rightarrow \sum_i \phi(p_1)_i -p_{1,i}=\sum_i \phi(p_2)_i -p_{2,i} $
How can i show, that this equality only holds if $\phi$ is of the form $\phi(p)_i=p_i +v_i$?