# Determinant of a special conformal transformation

1. Jun 17, 2012

### maverick280857

Hi,

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

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

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.

2. Jun 17, 2012

### fzero

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.

3. Jun 17, 2012

### 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.

4. Jun 18, 2012

### maverick280857

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.

5. Jun 18, 2012

### 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: $$(x^a + b^a x^2)/x^2 \rightarrow \frac{x^2 (x^a + b^a x^2)}{(x + b x^2)^2} = \frac{x^a + b^a x^2}{1 + 2 bx + x^2}$$.

6. Jun 18, 2012

### maverick280857

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.