Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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).

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook