By a Banach space $X$ I mean, a complete normed vector space and by a Clifford isometry I mean a surjective isometry $\gamma$ of $X$ such that the distance $d(\gamma x, x)$ is constant on $X$. Inherently, $\gamma$ acts as a "translation" so Clifford isometries are sometimes called Clifford translations. As an example, in Euclidean space any translation "is" a Clifford isometry.

My question is: are there examples of such an $X$ where the set of Clifford isometries consists only of the identity?

  • 2
    $\begingroup$ I don't understand: every translation in a normed space is a Clifford isometry, as $\|(x+v)-x\| = \|v\|$. Are you asking whether there are Clifford isometries that aren't translations? $\endgroup$ – t.b. May 15 '12 at 21:44
  • $\begingroup$ Sure, just consider the vector space with one element $\{ 0 \}$. =) $\endgroup$ – Patrick Da Silva May 15 '12 at 21:47

By Mazur-Ulam theorem every surjective isometry is an affine map, i.e. $$ \gamma(x)=x_0+T(x) $$ for some fixed vector $x_0\in X$ and necessary isometric isomorphism $T\in\mathcal{B}(X)$.

Assume that surjective isometry $\gamma$ is a Clifford isometry, then we have $C\geq 0$ such that for all $x\in X$ we have $$ \Vert x_0+T(x)-x\Vert=C $$ I think it's quite obvious that in this case $T$ is neccessary the identity map on $X$. Hence, every Clifford isometry is a translation on some vector $x_0\in X$.

  • $\begingroup$ Just a clarification. Having taken care of translation, we are left with a Clifford isometry that fixes $0$. It must be a linear operator, by Mazur-Ulam. For all $x$ we have $2\|Tx-x\|=\|T(2x)-2x\|=\|Tx-x\|$, hence $Tx=x$. $\endgroup$ – user31373 May 16 '12 at 0:14
  • $\begingroup$ @Leonid, thanks! It is a bit easier. $\endgroup$ – Norbert May 16 '12 at 8:56

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