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Math Help - Matrix trace relations

  1. #1
    Member alunw's Avatar
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    Matrix trace relations

    The trace of a matrix is the sum of the entries on the leading diagonal. So for
    a 2*2 matrix a11 a12
    a21 a22
    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
    = x*x+y*y+w*w-x*y*w-2
    = x*x+y*y-z*w-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.
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  2. #2
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    The book suggests that for \frac{\mathop{tr}(A^nBA^{-n}B^{-1})-2}{(\mathop{tr}A^n)^2-4} we should write A in standard form. For a 2\times2 matrix A of determinant 1 with (\mathop{tr}A)^2\neq4 this means a matrix of the form \begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}. Since \mathop{tr} CD=\mathop{tr}(U^{-1}CUU^{-1}DU) for any invertible matrix U, we may assume that A is standard and B is any \textrm{SL}(2,\mathbb C) matrix.

    Thus A^nBA^{-n}B^{-1}=\begin{pmatrix}t^n&0\\0&t^{-n}\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix  }\begin{pmatrix}t^{-n}&0\\0&t^n\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \ =\begin{pmatrix}ad-t^{2n}bc&\dots\\\dots&ad-t^{-2n}bc\end{pmatrix}.

    Thus \mathop{tr}(A^nBA^{-n}B^{-1})-2=2ad-bc(t^{2n}+t^{-2n})-2=-bc(t^{2n}+t^{-2n}-2) using ad-bc=1.

    Also (\mathop{tr}A^n)^2-4=(t^n+t^{-n})^2-4=t^{2n}+t^{-2n}-2. Therefore \frac{\mathop{tr}(A^nBA^{-n}B^{-1})-2}{(\mathop{tr}A^n)^2-4}=-bc, a constant. (Of course this assumes t^{2n}\neq 1.)
    Last edited by halbard; July 6th 2009 at 09:58 AM. Reason: Forgot something!
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  3. #3
    Member alunw's Avatar
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    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.
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