# All the isometries of $\mathbb{C}^n$ into itself are made like these

This is again a request for references. I'd appreciate a pointer to any published proof of the following:

Proposition. Given $n \in \mathbb{N}^+$, let $\Phi$ be a function $\mathbb{C}^n > \to \mathbb{C}^n$. Then $\Phi$ is an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself if and only if there exist a unitary $U \in > \mathbb{C}^{n,n}$ and an orthogonal $O \in \mathbb{R}^{n,n}$ such that $\Phi(z) - \Phi(0) = U\; (\Re(z) + i\; O \; > \Im(z))$ for all $z \in > \mathbb{C}^n$.

Here $\|\cdot\|$ stands for the usual norm $\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{R}: (z_1, \ldots, z_n) \mapsto \left(\sum_{k=1}^n |z_k|^2\right)$, and an isometry of $(\mathbb{C}^n,\|\cdot\|)$ into itself is a function $f: \mathbb{C}^n \to \mathbb{C}^n$ such that $\|f(z)-f(w)\| = \|z-w\|$ for all $z,w \in \mathbb{C}^n$.

Edit. I almost forgot! Thanks in advance for any feedback.

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The proposition of the posting can't be true in general, I'm afraid: e.g., the isometry group of $\mathbb{C}^2$ has (real) dimension $4+6=10$, while $\dim U(2)+\dim O(2)+4=9$. However, the isometries of $\mathbb{R}^n$ (and in particular, of $\mathbb{C}^n$) are easy to describe.

Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a map such that $d(x,y)=d(f(x,f(y))$ for all $x,y\in\mathbb{R}^n$ where $d$ is the metric induced by a positive definite quadratic form $(\cdot,\cdot)$.

Suppose $f$ preserves the origin. Then for all $x,y\in\mathbb{R}^n$ of unit length such that $(x,y)=0$ we have $(f(x),f(y))=0$, since $f(x)$ and $f(y))$ are of unit length and the distance between them is $\sqrt{2}$.

Now choose an orthogonal frame $e_1,\ldots,e_n$ of $\mathbb{R}^n$ and take a composition $h=g\circ f$ such that $g$ is of the form $x\mapsto Ax+b$ for some $A\in O(n),b\in\mathbb{R}^n$ and $h$ preserves the origin and any $e_i$. If $x=\sum a_i x_i\in\mathbb{R}^n$ and we know $d=d(x,0)$ and all $d_i=d(x,e_i)$, then we can recover $a_i$ as $a_i=\frac{1}{2}(d^2-d_i^2+1)$. So the map $h$ is in fact the identity, which means that $f$ is a composition of a translation and an orthogonal map.

The question of the posting is the case when $\mathbb{R}^n=\mathbb{C}^{n/2}$ and $(\cdot,\cdot)$ is the real part of the standard hermitian form.

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Yes, I see it now (the conclusions concerning $\mathbb{R}^n$ follow in a more general form from the Mazur-Ulam theorem, though this is a bit overkilling here). Thanks! P.S.: you may want to edit a minor error in your LaTeX (I think a symbol = is missing where you define the $d_i$'s). – Salvo Tringali Oct 30 '11 at 23:34
Salvo -- welcome! I've fixed the typo. – algori Oct 31 '11 at 1:36

As algori mentions, since the distance on $\mathbb{C}^n$ only depends on the underlying real affine space, the question is actually about isometries of the Euclidean space $\mathbb{R}^{2n}$.

Now observe that any isometry of Euclidean space $\mathbb{R}^n$ must be affine. Reason: you have a metric characterization of segments ($[a,b]$ is the set of points $c$ for which equality holds in the triangle inequality $d(a,b)\leq d(a,c)+d(c,b)$), so isometries map segment to segment, i.e. are affine. From this you easily get (I think this was first proved by Darboux) that every isometry of Euclidean space is uniquely given as the product of a linear isometry with a translation.

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