Fractional Linear Transformation Question

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SUMMARY

The discussion centers on the properties of Fractional Linear Transformations (FLTs), specifically Mobius Transformations, and their representation with a determinant of 1. It is established that any FLT can indeed be expressed in this form by appropriately choosing a scaling factor, λ, which allows for the determinant to be adjusted. The confusion arises from the Jordan Canonical forms of 2-by-2 matrices, particularly when distinct eigenvalues are involved. The determinant in this context refers to ad-bc, which is crucial for understanding the isomorphism with the projective linear group SL(2,ℂ)/{1,-1}.

PREREQUISITES
  • Understanding of Fractional Linear Transformations (Mobius Transformations)
  • Familiarity with determinants in linear algebra
  • Knowledge of Jordan Canonical forms for 2-by-2 matrices
  • Basic concepts of projective linear groups, specifically SL(2,ℂ)
NEXT STEPS
  • Study the properties of Mobius Transformations in detail
  • Learn about the Jordan Canonical form and its implications for linear transformations
  • Explore the projective linear group SL(2,ℂ) and its significance in complex analysis
  • Investigate the role of scaling factors in transforming determinants of matrices
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Graduate students in mathematics, particularly those studying complex analysis and linear algebra, as well as educators and researchers interested in the properties of Mobius Transformations and their applications.

joeblow
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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?
 
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To fix notation: a Mobius map is a map of the form
z\mapsto \frac{az+b}{cz+d}
for some complex constants a,b,c,d satisfying ad-bc\neq0. It is this number ad-bc 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 anyone map: if you multiply all of them by a nonzero number, \lambda, you get the same map. The 'determinant' is then multiplied by \lambda^2, so by choice of \lambda 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 SL(2,\mathbb{C})/\{1,-1\} where a,b,c,d become the entries of the matrix (the freedom to still pick \lambda=\pm1 is why we have to take the quotient).

joeblow said:
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?
 

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