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Let $A$ be a $2\times2$ real matrix, then define $f:S^1\to S^1$ by $f(\phi)=\arctan(A\cdot(\cos \phi,\sin \phi))$.

This can be viewed as a discrete dynamical system on $\mathbb{S}^1$ and I am trying to figure out its behaviour. Are there any results about it? Does it have a name?

EDIT: What actually $f$ does, is the following. Takes a normalized vector, applies the matrix on the vector and renormalizes it. Then it can be considered as a map $S^1\to S^1$.

Then obviously an eigenvector corresponds to a fixed point.

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This doesn't quite make sense. – Ted Shifrin May 9 '13 at 21:42
Why? What's its problem? – tst May 9 '13 at 21:45
@tst:I don't understand why you multiply/divide angles by $2\pi$. Why can't you define the function as $\arctan(A\cdot [\cos \phi, \sin \phi])$? – TCL May 9 '13 at 23:39
Oh, it's just to normalize the the angle to $[0,1)$ and not to $[0,2\pi)$. It is not important at all. Maybe I should remove it. – tst May 9 '13 at 23:42
up vote 1 down vote accepted

You are essentially describing the action of elements of the projective general linear group $\mathit{PGL}(2,\mathbb{R})$ on the projective line. Instead of using the angle $\phi$, it is more common to describe the state space as the space of all lines through the origin. (Of course, it is slightly different because there are two angles $\phi$ for every line, but it makes sense to identify antipodal points on the circle since they remain antipodal under any linear transformation.)

The dynamics of these elements are completely understood. Assuming $A$ is non-singular and has positive determinant, the action of $A$ can be classified as elliptic, parabolic, or hyperbolic (see this Wikipedia article). All elliptic elements are topologically conjugate to a rotation of the circle. Parabolic elements have a single fixed point, and are all topologically conjugate to one another. Hyperbolic elements have two fixed points (one attracting and one repelling) and are again all topologically conjugate to one another.

By the way, this action is usually thought of somewhat differently: it is the action of the isometry group of the hyperbolic plane on the circle at infinity. In this case, elliptic transformations correspond to rotations, parabolic elements are "limit rotations", and hyperbolic elements are translations along hyperbolic geodesics.

The dynamics is much more interesting if you consider the group generated by several such transformations. Such a group is known as a Fuchsian group, and can be used to define a fractal subset of the circle known as a limit set. For Fuchsian groups the limit set is always either finite, or the entire circle, or a Cantor set, but for the closely related Kleinian groups (which act on a sphere instead of a circle) the limit set can be much more interesting.

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Thank you very much. Do you have any introductory book to suggest on the subject. Also as I wrote to my other question, a similar map can be defined between (hyper)spheres instead of projective spaces. Is this also a studied subject? – tst May 14 '13 at 1:23

If you are just considering the map on angles induced by matrix transformations on the circle, note that the matrix must be a rotation or reflection.

In the first case, your map is just addition of a constant $\theta$ modulo $2\pi$. This is a simple, well known map.

In the second case, you have a period 2 map $\phi\to\theta-\phi$ modulo $2\pi$. This is a very simple map too!

Things are slightly more interesting if you rescale the angle first, if you want to have a play!

Okay, if you rescale, then first note that $f_A\circ f_A= f_{A^2}$ etc. by linearity.

Hence the question is really about the possible forms which $A^n$ can take.

Check out the end of for a discussion of the map from $\lambda=x_1/x_2=\cot\phi$. The pictures are easy enough to draw. Think about the possible eigenvalues of $A$ etc. to guide you, then think about the cases with fewer eigenvalues.

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Well, I am describing something more general. I do not assume that the circle is fixed by the matrix and $f$ is not linear in general. – tst May 9 '13 at 23:19
If the circle isn't fixed then how is this a map from $S^1\to S^1$...? $A$ is linear, it's a matrix, right? – Sharkos May 9 '13 at 23:34
The idea is the following. Take a normalized vector, apply the matrix and then normalize the vector again. So it is a map from $S^1\to S^1$ but the matrix does not fix the circle. – tst May 9 '13 at 23:40
Oh, well I didn't write it but I've figured out that the Jordan form of $A$ determines a lot for $f$. For example, if $A$ has 2 eigenvector, then the one with the largest eigenvalue is an attracting fixed point and the other is repelling. If it has only one, then it attracts from one direction and repels to the other. Finally when the eigevalues are complex, it is a non-linear rotation. I wanted to see if there are any known results because I need to complexify it. So instead of taking a normalised vector in $\mathbb{R}^2$, i take it in $\mathbb{C}^2$ and do the same. – tst May 10 '13 at 0:59
Then $f$ is a map from the unit "circle" in $\mathbb{C}^2$ (which is $S^3$) to the unit "circle" in $\mathbb{C}^2$ and I don't know what to do with it. – tst May 10 '13 at 1:01

By linearity of $A$, $f^n(\phi)=\arctan(A^n\cdot [\cos\phi, \sin\phi])$, so the behavior of $f^n$ reduces to the behavior of $A^n$.

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