How to calculate Dr. Curt McMullen's expanding eigenvalues for totally degenerate groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:06:13Z http://mathoverflow.net/feeds/question/74270 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74270/how-to-calculate-dr-curt-mcmullens-expanding-eigenvalues-for-totally-degenerate How to calculate Dr. Curt McMullen's expanding eigenvalues for totally degenerate groups? steve 2011-09-01T16:30:38Z 2011-09-05T18:47:28Z <p>What is required in order to derive the expanding eigenvalues of Dr. Curt McMullen's torus orbifold bundles over the circle and the corresponding totally degenerate groups, as presented in Section 3.7 of his book, Renormalization and 3-Manifolds which Fiber over the Circle? </p> <p>He provides this value for the orbifold bundle obtained from the torus with a singular point of order 2, but I cannot reproduce his method. I am eager to understand and, further, to derive the expanding eigenvalue for the orbifold bundle obtained from the torus with a singular point of order 3.</p> <p>My inability to calculate these values is disconcerting because I am able to derive the values for the fixed points and the lengths of the singular closed geodesics analytically. My ambition is to have a coherent and comprehensive understanding of Dr. McMullen's work on the totally degenerate group which is isomorphic to the fundamental group of a 2-dimensional orbifold of genus 1 with a single cone point of order 3. </p> <p>I believe that my fundamental problem is that I do not grasp the mathematical motivation for Dr. McMullen's reference for his work: </p> <blockquote> <p>T. Jørgensen, "Compact 3-manifolds of constant negative curvature ﬁbering over the circle", Ann. of Math. (2) 106: 61–72 (1977).</p> </blockquote> <p>Thank you in advance for your assistance.</p> http://mathoverflow.net/questions/74270/how-to-calculate-dr-curt-mcmullens-expanding-eigenvalues-for-totally-degenerate/74607#74607 Answer by Ian Agol for How to calculate Dr. Curt McMullen's expanding eigenvalues for totally degenerate groups? Ian Agol 2011-09-05T18:47:28Z 2011-09-05T18:47:28Z <p>Let me first make a comment on Jorgensen's work. His Annals paper that you refer to was based on computations he first made in the cusped case, but did not appear in <a href="http://www.ams.org/mathscinet-getitem?mr=2044551" rel="nofollow">print until recently</a>. He then did "orbifold <a href="http://en.wikipedia.org/wiki/Dehn_filling" rel="nofollow">Dehn filling</a>" to get a closed example, which he was then able to publish explicitly, without revealing the origin of his techniques. I believe Thurston was inspired by Jorgensen's examples to prove his geometrization theorem for fibered manifolds. One may see pictures of the Ford domains and limit sets for the cusped case using the program <a href="http://delta.math.sci.osaka-u.ac.jp/OPTi/index.html" rel="nofollow">Opti</a>. Unfortunately, though, you can't draw the types of "snowflake curves" with this program, unless maybe you know the coordinates of group precisely. </p> <p>Given a mapping class of a punctured torus, you get a corresponding map of its representation variety. If you want the commutator to lie in a specific conjugacy class (like an elliptic of order 3), then you get a very specific equation for the traces of elements in the group related to the Markoff equation, which is given in Equation 3.3 in McMullen's book (for derivation of these types of relations in $SL(2,C)$, you could refer to Chapter 1 of <a href="http://www.math.umn.edu/~am/book/outercircles.pdf" rel="nofollow">Marden's book</a> on Kleinian groups). </p> <p>So now you get an action of the mapping class on the traces, which McMullen computes explicitly for the transformation $LR$, and will be the same algebraic transformation regardless of the value of $tr[A,B]$. </p> <p>McMullen computes (see Marden) $$L(\alpha, \beta,\gamma)=(\gamma, \beta, \beta\gamma-\alpha),$$ $$R(\alpha,\beta,\gamma)=(\alpha,\gamma,\alpha\gamma-\beta).$$</p> <p>These preserve the Markoff equation $\alpha^2+\beta^2+\gamma^2=\alpha\beta\gamma+3$ which holds in your case of an elliptic commutator of order 3. Solving for the fixed point $LR(\alpha,\beta,\gamma)=(\alpha\gamma-\beta,\gamma,\gamma^2\alpha-\beta\gamma-\alpha)=(\alpha,\beta,\gamma)$, we obtain the fixed point $(\alpha,\alpha/(\alpha-1),\alpha/(\alpha-1))$, subject to the Markov equation. Plugging in, we see that $\alpha$ is a root of the equation $\alpha^4-3\alpha^3+6\alpha-3=0$. This equation has two real roots, and two complex conjugate roots, either of which gives the desired fixed point (up to complex conjugation) giving the traces of a totally degenerate group. </p> <p>Now, to compute the eigenvalue of the fixed point, we compute the characteristic polynomial of the derivative of $LR$ (which is a 3x3 matrix). I won't go through the computation, but I get the polynomial $\lambda^3-(2\alpha^2/(\alpha-1))\lambda^2 +2\alpha^2/(\alpha-1)\lambda-1$ after substituting for the fixed point. Dividing by the obvious factor $\lambda-1$ (which I think comes from the fact that the mapping class group has a conserved quantity $tr[A,B]$ on the character variety), we have your desired eigenvalue is a root of the polynomial $\lambda^2 - (2\alpha^2/(\alpha-1) +1)\lambda +1$. </p>