Nice parameterization of linear complex structures on the real plane?

Question

A linear complex structure on a real vector space $V$ is an endomorphism $J$ such that $J \circ J=-\mathrm{id}$. What do all the linear complex structures on $\mathbb{R}^2$ look like? If we let

$$ J = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right],\ I = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] $$

then we can grind out the defining relationship $J^2=-I$ and discover that $J$ must have the form

$$ J = \left[ \begin{array}{cc} a & b \\ \frac{-1-a^2}{b} & -a \end{array} \right], $$

i.e., the space of linear complex structures on $\mathbb{R}^2$ on the plane is parameterized by two values $a,b \in \mathbb{R}$. However, this parameterization is not very nice -- the lower-left and upper-right entries look very different (probably for no good reason), and more importantly there is a singularity at $b=0$. So my first question is:

Question 1: what's a better parameterization of the linear complex structures on $\mathbb{R}^2$?

Geometric intuition may help here. In other words, what degrees of freedom do we have geometrically? If we let $a=0$, for instance, then $J$ looks like

$$ \left[ \begin{array}{cc} 0 & b \\ \frac{1}{b} & 0 \end{array} \right]\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right], $$

i.e., we rotate by a quarter turn, then apply a scaling by $b$ and its reciprocal. If we apply this same transformation again, then we basically rotate again and "undo" the scaling. So one degree of freedom is "how much stretching."

Question 2: what is our other degree of freedom (geometrically)?

Answer 1

A linear complex structure on $\mathbb{R}^2$ is characterized uniquely by where it sends a single nonzero element $v$, and it can send it to any element not in the span of $v$.

More precisely, let $0\neq v\in \mathbb{R}^2$. Then, if $Jv = \lambda v$ with $\lambda \in \mathbb{R}$, $J^2v = \lambda^2 v \neq -v$ since $\lambda^2 \geq 0$. This implies that $Jv$ cannot be a multiple of $v$.

Since $Jv$ is not a multiple of $v$, it must be the case that $\{v, Jv\}$ is a basis of $\mathbb{R}^2$. Further, we must have $J(Jv) = J^2 v = -v$, so $J$ is determined on this basis. Since $J$ is linear, this determines $J$ on all vectors in $\mathbb{R}^2$ and it's easy to see this this linear map satisfies $J^2 x = -x$ for all $x$ simply because it does so for each basis element.

So, one way to parameterize all $J$s is by picking a particular $v$, such as $v = (1,0)$. Then the $J$s are parameterized by points in $\mathbb{R}^2$ which are NOT contained on the $x$ axis.

Alternatively, $G=GL(2,\mathbb{R})$ acts on the collection of all $J$s by $A*J = AJA^{-1}$. Note that $(AJA^{-1})^2 = AJA^{-1} AJA^{-1} = AJ^2 A^{-1} = A(-I) A^{-1} = -I$. I claim that the action is transitive.

To see this, I'll actually use the description above. Note that it is enough to see that $G$ can move the "standard" complex structure $$J_0 = \begin{bmatrix} 0& -1 \\ 1&0\end{bmatrix}$$ (which maps $x$ to $y$ and $y$ to $-x$) to any other complex structure. To do this, given any complex structure $J$ with nonzero vector $v$, let $A^{-1}$ be the unique matrix which takes $x$ to $v$ and $y$ to $Jv$.

That is, $A^{-1} x = v$ and $A^{-1} y = Jv$. I claim that $AJA^{-1} = J_0$. To see this, it's enough to check it on a basis. I'll just check that $AJA^{-1} y = J_0 y$, leaving the other calculation to you.

We have \begin{align*} AJA^{-1} y &= AJ(Jv) \\ &= AJ^2v\\ &= A(-v) &= -Av\\ &= -x\\&= J_0 y.\end{align*}

Great, so the action is transitive. It remains to compute the stabilizer at $J_0$. Algebriacally, we're asking for all matrices which commute with $J_0$. Writing this out and solving shows that the stabilizer is matrices of the form $$\begin{bmatrix} a & b \\ -b & a\end{bmatrix}$$ with $a^2 + b^2\neq 0$.

Geometrically, since $J_0$ is a rotation matrix, it commutes with all rotations and also with all scalings and the stabilizer subgroup is simply the subgroup generated by rotations and (nonzero) scalings. Note that rotating 180 degrees around is the same as scaling by $-1$ (and no rotations is the same as no scaling), but otherwise rotations and scalings are distinct.

It follows that the collection of linear complex structures is canonically parameterized the the coset space $G/(SO(2)\cdot \mathbb{R}^\ast)$ where $SO(2)\cdot \mathbb{R}^\ast \cong [SO(2)\times \mathbb{R}^\ast]/\{ \pm (I, 1)\}.$

Answer 2

You can also use $$J=\begin{pmatrix} \sinh \alpha & -e^\beta \cosh \alpha \\ e^{-\beta} \cosh \alpha & -\sinh\alpha \end{pmatrix} $$ where $\alpha$ and $\beta$ are real numbers.