Determinant of a special conformal transformation

Hi,

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

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

where |.| denotes the determinant. I know that

$$x'^{\mu} = \frac{x^\mu - b^\mu x^2}{1-2 b\cdot x + b^2 x^2}$$

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

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 Quote by maverick280857 Hi, I am working through Chapter 4 of Francesco, Mathieu and Senechal's CFT book (http://www.amazon.com/Conformal-Theo.../dp/038794785X). Equation 4.52 states that for a special conformal transformation $$\left|\frac{\partial\textbf{x'}}{\partial\textbf{x}}\right| = \frac{1}{(1-2(\textbf{b}\cdot\textbf{x})+b^2 x^2)^{d}}$$ where |.| denotes the determinant. I know that $$x'^{\mu} = \frac{x^\mu - b^\mu x^2}{1-2 b\cdot x + b^2 x^2}$$ How does this give the determinant above? I would appreciate a hint. Thanks in advance!
Note that

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

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.
 Blog Entries: 2 Recognitions: Science Advisor 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.

Determinant of a special conformal transformation

 Quote by fzero 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.
 Quote by Bill_K 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.
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.
 Recognitions: Homework Help Science Advisor 1. Invert: $x^a \rightarrow x^a/x^2$. 2. Add $b$: $x^a/x^2 \rightarrow x^a/x^2 + b^a = (x^a + b^a x^2)/x^2$. 3. Invert again: $$(x^a + b^a x^2)/x^2 \rightarrow \frac{x^2 (x^a + b^a x^2)}{(x + b x^2)^2} = \frac{x^a + b^a x^2}{1 + 2 bx + x^2}$$.

 Quote by Physics Monkey 1. Invert: $x^a \rightarrow x^a/x^2$. 2. Add $b$: $x^a/x^2 \rightarrow x^a/x^2 + b^a = (x^a + b^a x^2)/x^2$. 3. Invert again: $(x^a + b^a x^2)/x^2 \rightarrow x^2 (x^a + b^a x^2)/((x + b x^2)^2 = (x^a + b^a x^2)/(1 + 2 bx + x^2)$.
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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