I came across a function $f(\mathbf{z})$ that maps a vector of $\mathbb{C}^3$, such as $\mathbf{z}=(a+ib,c+id,e+if)$, as follows:
$$ f(\mathbf{z})=\mathbf{z}^*\mathbf{z}+(\Re[\mathbf{z}] \times \Im [\mathbf{z}])^*(\Re[\mathbf{z}] \times \Im [\mathbf{z}]) $$
where $\times$ denotes the cross-product.
I am looking for the invariance group of this function. Essentially, the function achieves two things:
- The term $\mathbf{z}^*\mathbf{z}$ constrains the transformation to be invariant with respect to $U^*U=I$, or in this case $U(3)$.
- It adds the additional constraints provided by the secondary (right-most) term.
My intuition is that the second part constricts $U(3)$ to $SU(3)$ but I have had trouble verifying my intuition. Here is what I have done so far:
Injecting a transformation $M$ into $f(\mathbf{z})$ I get:
$$ f(M\mathbf{z})=(M\mathbf{z}^*)M\mathbf{z}+(\Re[M\mathbf{z}] \times \Im [M\mathbf{z}])^*(\Re[M\mathbf{z}] \times \Im [M\mathbf{z}]) $$
It is easy to see that the term $(M\mathbf{z}^*)M\mathbf{z}$ eliminates as $M^*M=I$.
However, for the other term, I am stuck.
The identity $Ma\times M b=(\det M)(M^{-1})^T (a\times b)$ might be useful if there is a way to "pull" $M$ out of $\Re[M\mathbf{z}]$ and $\Im[M\mathbf{z}]$.
To "pull" $M$ out, I am trying to use the equivalences:
$$ \Re[z]=\frac{z+z^*}{2}\\ \Im[z]=\frac{z-z^*}{2i} $$
edit: (already here I am confused because if I replace $z$ (a scalar) by $\mathbf{z}$ a vector, I just realized I am adding a $3\times1$ with a $1\times 3$ matrix).
So how can I find the invariance group of $f(M\mathbf{z})=f(\mathbf{z})$?
edit2: The goal is to pull "M" out the of cross-product and of the functions $\Re[\mathbf{z}]$ and $\Im[\mathbf{z}]$ so that the matrix is eventually eliminated via the relation $((\det M)(M^{-1})^T)^*(M^{-1})^T(\det M)$, leading to $(\det M)^2=1$ because $M^*M=I\implies (M^{-1})^T)^*(M^{-1})^T=I$.
Let us define the vectorial form of $\Re[z]$ and $\Im[z]$:
$$ \vec{\Re}[\mathbf{z}]=\frac{\mathbf{z}+\overline{\mathbf{z}}}{2}\\ \vec{\Im}[\mathbf{z}]=\frac{\mathbf{z}-\overline{\mathbf{z}}}{2i} $$
where $\overline{\mathbf{z}}$ is a component-by-component complex conjugate.
$$ \begin{align} &||\Re[M \mathbf{z}] \times \Im [M \mathbf{z}]||^2 \\ =& \frac{1}{16} ||(M\mathbf{z}+\overline{M\mathbf{z}}) \times (M\mathbf{z}-\overline{M\mathbf{z}})||^2\\ =& \frac{1}{16}||(M\mathbf{z}+\overline{M\mathbf{z}}) \times (M\mathbf{z}-\overline{M\mathbf{z}})||^2\\ =& \frac{1}{16}||M\mathbf{z}\times M\mathbf{z}+ M\mathbf{z}\times -\overline{M\mathbf{z}} + \overline{M\mathbf{z}} \times M\mathbf{z} + \overline{M\mathbf{z}} \times - \overline{M\mathbf{z}} ||^2\\ =& \frac{1}{16}||M\mathbf{z}\times M\mathbf{z}+ 2M\mathbf{z}\times -\overline{M\mathbf{z}} + \overline{M\mathbf{z}} \times M\mathbf{z} ||^2 \end{align} $$
I just realized that $\overline{M\mathbf{z}} \times - \overline{M\mathbf{z}}=0$ and $M\mathbf{z}\times M\mathbf{z}=0$. We thus get:
$$ \begin{align} &\frac{1}{4}|| M\mathbf{z}\times -\overline{M\mathbf{z}}||^2\\ =&\frac{1}{4}|| M\mathbf{z}\times \overline{M\mathbf{z}}||^2 \end{align} $$