Action of 3x3 invertible matrices on $\mathbb{C}$?

Question

One of the many beautiful properties of mobius transformations includes the fact they form a natural map of equivalence classes (related by scalar multiplication of $2 \times 2$ complex matrices

$$\begin{pmatrix} a & b\\ c&d \end{pmatrix} \ \text{~} \ \frac{ax+b}{cx+d}$$

This had got me wondering.. the space of meromorphic functions is pretty massive. Does there exist then a natural action of equivalence classes $3 \times 3$ complex matrices (up to multiplication by scalar) into complex functions?

I tried to explore and find some, outside of the mobius transformations and came across the group $z^{\theta}$ where $\theta \in \mathbb{C}, |\theta| = 1$ as an action of the circle group into the complex plane which generalizes naturally to the group $e^{\frac{a\log z +b}{c \log z + d}}$ (where $c=b=0$, $|a/d| =1$) corresponds to the circle group. But the general form suggests this isn't anything "new" from the mobius transformations we have already seen.

I suspect it may be possible there is no "elementary" set of functions (as in involving algebraic, and exponents), but then the problem gets a lot more complex since the space of meromorphic functions has a lot of exotic and complicated infinite series.

Answer 1

Möbius transformations are the only meromorphic functions on the Riemann sphere $\hat{\mathbb{C}}$ that are invertible with respect to composition. So no, there is not any natural meromorphic group action on $\hat{\mathbb{C}}$ other than groups that have a natural homomorphism to the group $PGL_2(\mathbb{C})$ of Möbius transformations.

In particular, for the group $PGL_3(\mathbb{C})$ of complex $3\times 3$ invertible matrices modulo scalars, there does not exist any nontrivial continuous homomorphism $PGL_3(\mathbb{C})\to PGL_2(\mathbb{C})$ at all. This follows from the fact that $PGL_3(\mathbb{C})$ is a simple group, so any nontrivial homomorphism would be injective, but there cannot exist any continuous injection $PGL_3(\mathbb{C})\to PGL_2(\mathbb{C})$ since $PGL_3(\mathbb{C})$ has higher dimension than $PGL_2(\mathbb{C})$. (In fact, I suspect the assumption of continuity can be dropped, though I don't see a way to prove it at the moment. In any case, certainly any "natural" action would be continuous.)

The natural action of $PGL_3(\mathbb{C})$ is instead on the complex projective plane $\mathbb{CP}^2$: $GL_3(\mathbb{C})$ acts on $\mathbb{C}^3$ linearly, and thus also acts on the space $\mathbb{CP}^2$ of $1$-dimensional linear subspaces of $\mathbb{C}^3$, and this action descends to $PGL_3(\mathbb{C})$. This is a direct generalization of the action of $PGL_2(\mathbb{C})$ on $\hat{\mathbb{C}}$, as $\hat{\mathbb{C}}$ can be identified with $\mathbb{CP}^1$ and the action again just comes from the action of $GL_2(\mathbb{C})$ on $\mathbb{C}^2$.

Answer 2

This remains a conjecture but I have good reason to believe the following will work:

Lets call this group of functions the $M_3$ transforms. We define them as the set of complex functions which send cubic plane curves to cubic plane curves.

To be concrete on what this means: we characterize our functions $M_3: \mathbb{C} \rightarrow \mathbb{C}$ instead as $M_3: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Then in the cartesian plane given a cubic plane curve on $\mathbb{R}^2$ of the form

$$ Ax^3 + Bx^2y + Cxy^2 + Dy^3 + Ex^2 + Fxy + Gy^2 + Hx + Iy + J = 0$$

Where $A... J \in \mathbb{R}$ the image of this set under an element of $M_3$ will be:

$$ A_fx^3 + B_fx^2y + C_fxy^2 + D_fy^3 + E_fx^2 + F_fxy + G_fy^2 + H_fx + I_fy + J_f = 0$$

Where $A_f ... J_f \in \mathbb{R}$.

I claim that with a certain "niceness" condition that is yet to be found [ex: continuous] that this set of complex functions represents a natural action of $3\times 3$ matrices upon the complex plane.

Why I think this might work:

Observe:

Mobius Group of Transformations is isomorphic $PGL(2,\mathbb{C})$

The elements of $PGL(2,\mathbb{C})$ can be represented by $2\times 2$ matrices and are determined entirely by their image of 2 vectors of size 2, however, this results in 4 equations (for each coordinate of the 2 complex vectors) for 3 unknowns (the topological dimension of $PGL(2,\mathbb{C})$ is 3). So really each element of $PGL(2,\mathbb{C})$ can be uniquely identified with its action on 3 points on the complex plane [naturally then Mobius transformations are entirely determined by their action on 3 points in the complex plane, or riemann sphere since we now consider infinity to be a point].

Now it is not a coincidence that mobius transformations map generalized circles to generalized circles. A generalized circle (that is object of constant curvature) is ENTIRELY determined by 3 points.

So the idea is then that mobius transformation should be entirely determined by their actions on 3 points, and those 3 points have a unique generalized circle passing through them, and thus the mobius transformation should be entirely determined by their action on a generalized circle $C_1$, that is given a generalized circle $C_1$ what generalized circle $C_2$ does it get mapped to? That determines a mobius transformation uniquely.

We now have all the ingredients to lift this idea into the space of $3\times 3$ complex matrices. In this case i'm consider $GL(3,\mathbb{C})$ NOT $PSL(3, \mathbb{C})$ so these matrices are determined by their action on 3 vectors of size 3, or in other words 9 points in the complex plane.

So now I have a set of functions that need to be entirely determined by their action on 9 points in the complex plane. Through every 9 points there happens to pass a unique cubic curve. It is therefore natural for me to consider those functions which exclusively map cubic curves to cubic curves, since they will naturally form a group under composition.

What I need to now add is a condition [ex: continuity or meromorphicity etc... ] (but likely not meromorphicity due to @Eric Wofsey's findings) so that I restrict my space of all $f$'s to a set of functions that isn't so pathological that they are not UNIQUELY determined by their action on 9 points [I believe simply continuous except at isolated points might be sufficient]

Answer 3

After spending more time thinking about this we have two of the following results:

There is a natural action of $3 \times 3$ real matrices on the euclidean plane $\mathbb{R}^2$ given by:

$$ \begin{bmatrix} x \\ y \end{bmatrix} \rightarrow \frac{1}{gx+hy+i} \begin{bmatrix} ax+by+c \\ dx+ey+f \end{bmatrix} $$

(It's a lot of work algebraically but one can verify that composing such transformations respects matrix multiplication). This implies by considering a real part and imaginary part function, denoted as $\text{Re}(z)$ and $\text{Im}(z)$ then an action of $3\times 3$ real matrices can be realized as a mobius transformations is given as

$$ z \rightarrow \frac{ (a\text{Re}(z)+b\text{Im}(z)+c) + i (d\text{Re}(z)+e\text{Im}(z)+f)}{g\text{Re}(z)+h\text{Im}(z)+j} $$

For your choice of 9 real numbers $a,b,c,d,e,f,g,h,j$ (note that $i$ is also used for the imaginary constant so we skip it in the previous list).

But the introduction of real parts and imaginary parts this way is a little less than ideal, and this doesn't let us consider complex valued matrices.

Developing a functional equation trick:

The mobius transformations can identified as a class of smooth solutions to the following functional equation

$$ f: \mathbb{C}^5 \rightarrow \mathbb{C} $$

$$ f(a,b,c,d,f(e,f,g,h,z)) = f(ae+bg, af+bh, ce+dg, cf+dh,z)$$

(I conjecture solution to this all are of the form $\Omega^{-1} \left( \frac{a \Omega(z)+b}{c\Omega(z)+d} \right) $, for a choice of bijection $\Omega: \mathbb{C} \rightarrow \mathbb{C}$. The case of $\Omega$ being the identity are the usual mobius transforms we encounter).

This strategy should be able to easily scale to let us consider

$$ f: \mathbb{C}^{10} \rightarrow \mathbb{C} $$

$$ f(A, f(B,z)) = f(AB, z) $$

where $A,B$ are $3 \times 3$ complex matrices that represent our 9 other arguments to $f$.

Developing strategies to extract solutions from these functional equations currently remains a mystery to me.