0

I am working on J.Bochi and Avila's article "Uniformly hyperbolic finite-valued $SL(2,\mathbb{R})$-cocycles".As you may know,there is classification of $SL(2,\mathbb{R})$ which depends on the Trace of these matrices.for example if $M$ is our intended matrix, if $\vert Tr M\vert>2$ it is hyperbolic,if$\vert Tr M\vert<2$ then it is elliptic and if $\vert Tr M\vert=2$ then it is parabolic.i want to know why we give such as this names"hyperbolic,elliptic and parabolic" just by traces and why the value 2 is so important to determine the kind of classification.What is the role of trace in this classification ?if you know the answer i would be very happy to give me information or if there exist some free eBook to get some information please let me know.

Thank you in advance.

pershina olad
  • 383
  • 1
    Geometrically, you can get at this classification via the action of $SL(2,\mathbb{R})$ on the hyperbolic plane. If the action has a fixed point in the interior, it is elliptic, if it has one on the boundary, it is parabolic, and if it has two on the boundary, it is hyperbolic. If you solve the fixed point equation for a Mobius transformation you will see that the trace determines the kind of fixed points. – Max Apr 21 '20 at 00:13
  • @Max It helped me thank for your useful comment,but Is the any eBook to study more about what you said?why we use trace if it works with fixed points? – pershina olad Apr 21 '20 at 10:35
  • There are many books on hyperbolic geometry and I'm not sure what is best suited to your interests. I've read parts of Casson and Bleiler and Hubbard's book on Teichmuller theory. – Max Apr 21 '20 at 16:12
  • See https://math.stackexchange.com/questions/275079/books-for-hyperbolic-geometry – Max Apr 21 '20 at 16:14
  • 1
    It's convenient to use the trace: it's something very easy to compute just by looking at the matrix and of course the classification works out nicely. What's ``really going on" is a relationship between eigenvalues and the characteristic polynomial. – Max Apr 21 '20 at 16:15
  • @Max I get it,thanks a lot .But i know almost it doesn't work with hyperbolic geometry and its more about lie groups. – pershina olad Apr 22 '20 at 09:12

0 Answers0