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Fractional Linear Transformation Question

  1. Jul 5, 2011 #1
    I am a graduate assistant and was asked a question about FLTs (Mobius Transformations). The student was asked to prove that any FLT can be written as an FLT with determinant 1.

    However, I can't make sense of that. If I look at the possible Jordan Canonical forms of 2-by-2's, it would seem that the matrix
    [x 0]
    [0 y]
    where x and y are distinct eigenvalues cannot be represented as an FLT with determinant 1 (since it would require finding a complex number that when multiplied with both x and y gives 1 which violates the uniqueness of multiplicative inverse).

    Am I thinking about this the wrong way, or was there a typo in the problem?
     
  2. jcsd
  3. Jul 5, 2011 #2
    To fix notation: a Mobius map is a map of the form
    [tex]z\mapsto \frac{az+b}{cz+d}[/tex]
    for some complex constants [itex]a,b,c,d[/itex] satisfying [itex]ad-bc\neq0[/itex]. It is this number [itex]ad-bc[/itex] that is meant by the 'determinant' in the problem (though that name is not justified quite yet, see later).

    The constants are not, however, unique for any one map: if you multiply all of them by a nonzero number, [itex]\lambda[/itex], you get the same map. The 'determinant' is then multiplied by [itex]\lambda^2[/itex], so by choice of [itex]\lambda[/itex] we can pick it to be one.

    The name determinant comes from the fact that there is an isomorphism between Mobius maps and the projective linear group [itex]SL(2,\mathbb{C})/\{1,-1\}[/itex] where a,b,c,d become the entries of the matrix (the freedom to still pick [itex]\lambda=\pm1[/itex] is why we have to take the quotient).

     
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