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Determinant of a special conformal transformation

  1. Jun 17, 2012 #1

    I am working through Chapter 4 of Francesco, Mathieu and Senechal's CFT book (https://www.amazon.com/Conformal-Theory-Graduate-Contemporary-Physics/dp/038794785X). Equation 4.52 states that for a special conformal transformation

    [tex]\left|\frac{\partial\textbf{x'}}{\partial\textbf{x}}\right| = \frac{1}{(1-2(\textbf{b}\cdot\textbf{x})+b^2 x^2)^{d}}[/tex]

    where |.| denotes the determinant. I know that

    [tex]x'^{\mu} = \frac{x^\mu - b^\mu x^2}{1-2 b\cdot x + b^2 x^2}[/tex]

    How does this give the determinant above? I would appreciate a hint.

    Thanks in advance!
  2. jcsd
  3. Jun 17, 2012 #2


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    Note that

    [tex]\frac{\partial x'^{\mu}}{\partial x^\nu} = \frac{\delta^\mu_\nu}{1-2 b\cdot x + b^2 x^2} + f_\nu(x) b^\mu + g^\mu(x) b_\nu, [/tex]

    where ##f,g## can be easily determined. If you go ahead and express the determinant in your favorite way (using epsilon symbols is most straightforward), you'll find

    $$\left| \frac{\partial x'^{\mu}}{\partial x^\nu} \right| = \frac{1}{(1-2 b\cdot x + b^2 x^2)^d} + \epsilon_{\mu_1\mu_2\cdots} b^{\mu_1} b^{\mu_2} \cdots + \cdots.$$

    I haven't specified all of the extra terms, but you can see that they always involve antisymmetric combinations of ##b^{\mu_1} b^{\mu_2}## (and similar products with factors of ##f,g##). But all of these terms vanish because products like ##b^{\mu_1} b^{\mu_2}## are actually symmetric.
  4. Jun 17, 2012 #3


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    Make use of the fact that conformal transformations form a group. Write your special transformation as a product of translations and inversions. It's determinant will then be the product of the determinants of the individual transformations.
  5. Jun 18, 2012 #4
    Thank you fzero and Bill_K. I figured it out using the idea suggested by fzero. But Bill_K, what is the role of b in the translation*inversion product? I remember reading that a special conformal transformation can be decomposed this way, but I didn't quite understand it in the first place. Could you please elaborate.
  6. Jun 18, 2012 #5

    Physics Monkey

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    1. Invert: [itex] x^a \rightarrow x^a/x^2 [/itex].
    2. Add [itex] b [/itex]: [itex] x^a/x^2 \rightarrow x^a/x^2 + b^a = (x^a + b^a x^2)/x^2 [/itex].
    3. Invert again: [tex] (x^a + b^a x^2)/x^2 \rightarrow \frac{x^2 (x^a + b^a x^2)}{(x + b x^2)^2} = \frac{x^a + b^a x^2}{1 + 2 bx + x^2} [/tex].
  7. Jun 18, 2012 #6
    Thanks. I'm so stupid -- I didn't think of breaking it down this way, and it totally slipped my mind that operation 1 is the inversion step.
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