The trace of a matrix is the sum of the entries on the leading diagonal. So for
a 2*2 matrix a11 a12
the trace is a11+a22.
Let a and b be matrices in SL(2,C) i.e. 2*2 matrices with entries that are
complex numbers and with determinant 1. Let us call the inverse matrices A,B
Preliminaries: show that trace is invariant under conjugation and inversion i.e. that trace(B*a*b) = trace(a)=trace(A)
Show that trace(a*b) + trace(a*B) = trace(a)*trace(b)
Show that if a*a = -I if and only if a has trace 0 (where I is the identity matrix).
Show that a^3 = I if and only if a has trace -1 and a^6=I if and only if
a has trace 1.
Let x = trace(a),y = trace(b),z = trace(a*b),w = trace(a*B)
Show that trace(a*b*A*B) = x*x+y*y+z*z-x*y*z-2
(The above relations, especially the first, are very important for geometry and in the theory of Kleinian Groups)
Now the hard part:
Suppose x^2 <> 4. Show that for n>=1 (trace(a^n*b*A^N*B)-2)/(trace^2(a^n)-4) is constant.
I got this result from the book "Outer Circles" which says that you can prove it for n=2 and then use induction. But I could not do it by induction.
Instead I found a general expression for calculating trace(a^n*w) given knowledge of trace(a), trace(w) and trace(a*w) by using a difference equation, and then proved it by brute force.
July 6th 2009, 08:42 AM
The book suggests that for we should write in standard form. For a matrix of determinant with this means a matrix of the form . Since for any invertible matrix , we may assume that is standard and is any matrix.
Thus using .
Also . Therefore , a constant. (Of course this assumes .)
July 6th 2009, 09:52 AM
That's much shorter than my proof and will teach me not to assume avoiding multiplying matrices and using trace identities instead is always is a good plan. I notice you haven't proved it by induction either. Your remark shows the result is not actually true if A has finite order.