An explicit description of all conjugations on $\mathbb{C}$

Question

The "standard" conjugation for a complex number $a+bi$ is $\overline{(a+bi)}=a-bi$. If we see $\mathbb{C}$ as a one dimensional abelian Lie algebra, we can associate to this conjugation a real form, as the fixed point set of numers on the real line (i.e. with $0$ imaginary part), which is isomorphic to $R$ as a vector space.

However, if I take the antilinear map $\phi: \mathbb{C} \to \mathbb{C}$ that sends $a+bi \to -a+bi$, this satisfies all the properties of a conjugation, as:

1. $\phi^2=id_v$
2. $\phi(zv)=\overline{z}\phi(v)$ for all $z,v \in \mathbb{C}$ as, if $z=\zeta_r+\zeta_i i$ and $v=\gamma_r+\gamma_i i$, I have that $\phi(zv)=\phi((\zeta_r+\zeta_i i)(\gamma_r+\gamma_i i))=\phi(\zeta_r\gamma_r-\zeta_i\gamma_i+(\zeta_r\gamma_i+\zeta_i\gamma_r)i)=\zeta_i\gamma_i-\zeta_r\gamma_r+(\zeta_r\gamma_i+\zeta_i\gamma_r)i=(\zeta_r-\zeta_i i)(-\gamma_r+\gamma_i i)=\overline{z}\phi(v)$
3. $\phi(v_1+v_2)=\phi(v_1)+\phi(v_2)$

In this case the real form corresponding to this conjugation is the imaginary axis, which is again isomorphic to $\mathbb{R}$.

What if I want to find all the possible real structures on $\mathbb{C}$?

This is how I think I may proceed. I take the realification $\mathbb{R}^2$ of $\mathbb{C}$ and I consider all the linear (3) involutions (1) on $\mathbb{R}^2$ which I can represent as a matrix of the form $\begin{bmatrix}a & b\\ c & -a\end{bmatrix}$ with $a^2 + bc =1$. I then apply also condition (2), by taking the standard linear complex structure $J=\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}$ that allows me to represent the multiplication by any complex scalar $e+fi$ in $\mathbb{C}$ as the linear transformation in $\mathbb{R}^2$ given by $\begin{bmatrix}e & -f\\ f & e\end{bmatrix}$ and I finally impose the condition for each $e+fi$:

$$\begin{bmatrix}a & b\\ c & -a\end{bmatrix}\begin{bmatrix}e & -f\\ f & e\end{bmatrix}=\begin{bmatrix}e & f\\ -f & e\end{bmatrix}\begin{bmatrix}a & b\\ c & -a\end{bmatrix}$$

From this condition I get $b=c$ so that my conjugations are represented by linear matrices

$$\begin{bmatrix}a & \sqrt{1-a^2}\\ \sqrt{1-a^2} & -a \end{bmatrix} or \begin{bmatrix}a & -\sqrt{1-a^2}\\ -\sqrt{1-a^2} & -a \end{bmatrix}$$

I see that $a=1$ is the "standard" conjugation while $a=-1$ is the one that fixes the imaginary axis. What can be said about the others? As for each $-1<a<1$ the two matrixes have eigenvalues $1,-1$ and orthogonal eigenvalues (one corresponding to the eigenvalue 1 and another corresponding to the eigenvalue -1), we can say that all the possible conjugations in $\mathbb{C}$ are given by the reflections across all the axes that pass from the origin, and the axis of each reflection will be the corresponding real form for that conjugation.

Is this correct?

Answer 1

This looks correct to me, but here is an alternative way to view these things:

I claim all "conjugation maps" as you define them, on the $1$-dimensional $\mathbb C$-vector space $\mathbb C$, are given by fixing one basis vector $v$ and then setting the map $\phi_v : z v \mapsto \bar z v$. Compare some arguments in https://math.stackexchange.com/a/4399985/96384.

Note that once we view this chosen basis vector $v$ as element of $\mathbb C^\times$, the map $\phi_v$ can also be described, on a general complex input $x$, as $x \mapsto v \cdot \overline{x \cdot v^{-1}}$. I.e. your general "conjugation maps" are the standard complex conjugation, "semi-linear-matrix-conjugated" with some nonzero complex number $v$. Your example $a+bi \mapsto -a+bi$ is the case $\phi_i$, i.e. naturally viewed as $x \mapsto i \cdot \overline{x \cdot (-i)} = (-1)\cdot \bar x$.

More generally, writing $v$ as $r e^{i\varphi}$, one sees the $r$ cancels out of the equation and we have $\phi_v (x) = e^{2i\varphi} \cdot \bar x$, and the matrix this corresponds to on the two-dimensional real vector space with basis $1, i$ is

$$\pmatrix{\cos(2\varphi) &\sin(2\varphi)\\\sin(2\varphi) &-\cos(2\varphi)}$$

and you cover all possibilities by just $\varphi \in [0, \pi/2]$. Each one corresponds to fixing the "(real) line" through $v=e^{i\varphi}$, with the "standard" conjugation at $\varphi=0 \Leftrightarrow v=1$, and your other extreme example at $\varphi=\pi/2 \Leftrightarrow v=i$.

You'll see this is is a somewhat more natural parametrization of the matrices you found, as the $\sin$ terms naturally switch sign at $\varphi=\pi/4$.