Isometries of $\mathbb C^2$

I have studied isometries of the Euclidean plane $\mathbb R^2$ but am trying to understand what happens when we study $\mathbb C^2$ (over $\mathbb C$) instead.

In particular, I would like to see if a certain way of decomposing (uniquely) an isometry of the plane has some nice analogue in $\mathbb C^2$. To this end, I decided to generalize the notions of translation, rotation (about $\vec 0$), and reflection about the $x$-axis to get the following conjecture.

Given an isometry $\mathcal I$ of $\mathbb C^2$ (under the metric induced by the norm $\|(z,w)\|^2=z\bar{z}+w\bar{w}$), we have that $\mathcal I= T_{\vec v}\circ O\circ R$ for some translation $T$ by $\vec v$, some orthogonal transformation $O$, and some map $R$ which denotes either the reflection $F:(z,w)\mapsto (z, \bar{w})$ or the identity $I$.

My initial thought-process is as follows. Set $\vec x =\mathcal I(\vec 0)$. Then the translation in the desired decomposition would be $T_{\vec x}$. Moreover, notice that the map $A:=T_{-\vec x}\circ \mathcal I$ is linear. If $\det A>0$, then set $O=A$. Otherwise, we need $O$ to be something else. An intuitive choice is to set $O=AF$ in this case. But $F$ is not linear considered as map on $\mathbb C^2$ over $\mathbb C$. However, $F$ is linear as a map on the four-dimensional space $\mathbb R^4$--that is, as the map $(a,b,c,d)\mapsto (a,b,-c,-d)$. And this should be fine, as I think that we can adjust the other maps easily to maps on $\mathbb R^4$.

Now, here are my questions.

(1) Since I want $O$ to be orthogonal, is there a relatively easy way to show that $A$ or $AF$ (depending on the case) is orthogonal? I am unsure how to write down the matrix representations of these maps.

(2) And of course, I haven't shown the details of the actual proof of my conjecture. I just have been able to outline what I think should be the case. But I see no way of getting from the fact that $\mathcal I$ is distance-preserving to the fact that it can be decomposed in the way described. I have vaguely tried to play around with the eigenvalues for $O$, which I know should be complex conjugate pairs with norm $1$. But is it problematic if none of these is real, considering that we're technically using $\mathbb R^4$ and no longer $\mathbb C^2$? And to be honest, I am not sure anyway how studying the eigenvalues would help me with the proof ultimately.

• It is not clear to me. Let $P$ be the set of $4 \times 4$ orthogonal matrices with only $1,-1,0$. Then the $4 \times 4$ orthogonal matrices are generated by $P$ and the rotations in $\mathbb{R}^2$. As those rotations are $\mathbb{C}$-linear, it reduces to finding which matrices of $P$ are $\mathbb{C}$-linear. – reuns Jul 26 '17 at 7:36

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$When you speak without qualification of an isometry with respect to the standard Hermitian norm, you're speaking of an isometry of Euclidean $4$-space. The story of Euclidean isometries is more-or-less uniform independently of dimension, so let's speak of Euclidean $n$-space with $n$ an arbitrary positive integer. Here are some useful factlets:

• If an isometry $F$ fixes the origin and the standard basis vectors, then $F$ is the identity.

Sketch of proof: This is clear if $n = 1$. To argue inductively on the dimension, decompose a point of $\Reals^{m+1}$ as $(x, x_{m}) = ((x_{1}, \dots, x_{m}), x_{m+1})$, and show $F$ preserves both $\|x\|$ (so $F(x) = x$ by the inductive hypothesis) and the $(m + 1)$th coordinate $x_{m+1}$.

• If an isometry $F$ fixes the origin, there exists a unique $n \times n$ (real) orthogonal matrix $A$ such that $F(x) = Ax$ for all $x$ in $\Reals^{n}$.

Sketch: By the isometry condition and the polarization identity $u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})$, the points $F(e_{i})$ constitute an orthonormal set in $\Reals^{n}$. Let $A$ be the unique orthogonal matrix satisfying $F(e_{i}) = Ae_{i}$ for each $i$, and apply the first item to $A^{-1} \circ F$. (This answers your first question affirmatively.)

• If $F$ is an isometry, there exists a unique $n \times n$ orthogonal matrix and a unique $b$ in $\Reals^{n}$ such that $$F(x) = Ax + b\quad\text{for all x in \Reals^{n}.}$$ (You already have this proof. :)

• If $p$ is a point of $\Reals^{n}$ and $u$ is a unit vector, then reflection in the hyperplane through $p$ orthogonal to $u$ is the mapping $$R(x) = x - 2[(x - p) \cdot u] u,$$ namely, $x$ minus "twice the $u$-component of $x - p$". For example, if $u = e_{n}$ and $p$ is the origin, then $$R(x_{1}, \dots, x_{n-1}, x_{n}) = (x_{1}, \dots, x_{n-1}, x_{n}) - 2x_{n} e_{n} = (x_{1}, \dots, x_{n-1}, -x_{n}).$$ The Euclidean group is generated by the set of reflections. In fact, if $F$ is a Euclidean isometry, there exist at most $(n + 1)$ reflections (not uniquely-defined!) whose composition is $F$.

A non-trivial translation is a composition of two reflections. (Why?)

A rotation is the composition of at most $n$ reflections, and is always the composition of an even number of reflections (because a reflection reverses orientation while a rotation preserves orientation). Thus, for example, every rotation of Euclidean $3$-space is a composition of at most two reflections, while a rotation of Euclidean $4$- or $5$-space is a composition of at most four reflections. (To prove this inductively, note that if $u$ and $v$ are arbitrary unit vectors, there exists a reflection exchanging $u$ and $v$.)

If, on the other hand, you want to speak only of Euclidean isometries of $\Cpx^{m} \simeq \Reals^{2m}$ for which the "rotation part" is complex-linear, you have the much smaller "holomorphic isometry group" corresponding to the unitary group $U(m) \subset SO(2m)$.

Thinking of $\Cpx^{2}$, the set of complex lines (a special class of real $2$-plane) is the well-known complex projective line, topologically a $2$-sphere, while the set of (oriented) real $2$-planes is topologically a product $S^{2} \times S^{2}$. This shows geometrically how small $U(2)$ is compared to $SO(4)$,

• I know this is late, but how did you get that the unitary matrices are the elements of $SO(4)$ that are complex linear? – CuriousKid7 Aug 9 '17 at 16:22
• Equip $\Reals^{2m}$ with the Euclidean inner product $g$, and the complex structure $J$ defined by $J(x, y) = (-y, x)$ for $x$, $y$ in $\Reals^{m}$. If we identify $(x, y)$ with $x + iy$ in $\Cpx^{m}$, then $g$ is precisely the Hermitian structure $$\langle(x, y), (u, v)\rangle = x \cdot u + y \cdot v.$$ A linear transformation $T$ is orthogonal if and only if $g(Tx, Ty) = g(x, y)$ for all $x$, $y$, is complex-linear if and only if $T(Jx) = JT(x)$ for all $x$, and (since the Euclidean and Hermitian structures agree) is unitary if and only if $T$ is both orthogonal and complex-linear. – Andrew D. Hwang Aug 9 '17 at 16:37
• The only thing I don't see is that $T$ is complex linear iff $T(Jx)=JT(x)$. Could you please briefly say why it's true? – CuriousKid7 Aug 10 '17 at 3:10
• Because $i(x + iy) = -y + ix$, the transformation $J(x, y) = (-y, x)$ "implements" multiplication by $i$. To say $TJ = JT$ is to say "complex scalars factor out of $T$"; formally, $T(ix) = iT(x)$. – Andrew D. Hwang Aug 10 '17 at 10:42