Determinant of a special conformal transformation

[maverick280857]

Hi,

I am working through Chapter 4 of Francesco, Mathieu and Senechal's CFT book (https://www.amazon.com/dp/038794785X/?tag=pfamazon01-20). 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!

 

[fzero]

Hi,

I am working through Chapter 4 of Francesco, Mathieu and Senechal's CFT book (https://www.amazon.com/dp/038794785X/?tag=pfamazon01-20). 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.

 

[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.

 

[maverick280857]

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.

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.

 

[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}.

 

[maverick280857]

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.