How do you find the conformal generators from their infinitesimal form? For instance, how does the infinitesimal special conformal transformation $$x'^{\mu} = x^\mu + 2(x\cdot b)x^\mu - x^2 b^\mu,\tag{2.22}$$

tell us that $$K_\mu = -i(2x_\mu x^\nu \partial_\nu -x^2 \partial_\mu)~?\tag{2.23}$$ The similarity is apparent but I still do not know the procedure.

I'm following http://arxiv.org/abs/1511.04074 (and using its notation, p.25) and it seems that this step should be very easy.I know that supposedly the (general) procedure is explained in the page above but I do not fully understand it.

P.S.: I would like the answer to not use the commutations relations between conformal quantities since these notes derive these relations from the form of the explicit form of the generators and we run into the danger of circular reasoning.


2 Answers 2


The generator gives the action of an infinitesimal transformation on a field $\phi$. For instance, for the easy case of a constant translation we get

$$ \phi(x+\epsilon) = (1+i \epsilon^\mu P_\mu )\phi(x) \tag{1}$$

where $P_\mu$ is found to be $-i\partial_\mu$ by a simple taylor expansion. Here we chose to keep a factor of i outside of $P^\mu$ as this is conventional (this is really just a choice whether you want your generator to be hermitian or anti-hermitian). Now for a special conformal transformation we can find the generator by the exact same method. Take

$$ \phi(x+\epsilon) = (1+i b^\mu K_\mu )\phi(x) \tag{2}$$

insert $\epsilon =2(x\cdot b)x^\mu - x^2 b^\mu$ and do a taylor expansion to first order in $b$. We then get

$$\begin{aligned} \phi(x^\mu) + (2(x\cdot b) x^\mu -x^2 b^\mu ) \partial_\mu \phi = (1+b^\mu (2x_\mu x^\rho \partial_\rho - x^2 \partial_\mu) ) \phi(x^\mu) \end{aligned} $$

Thus we see that it can be written in the form of equation (1) with $K_\mu = -i(2x_\mu x^\nu \partial_\nu -x^2 \partial_\mu)$.


For any $b^{\mu}$ the special conformal transformation is given by \begin{equation} x^{'\mu} = \frac{x^{\mu}-b^{\mu}x^{2}}{1-2b\cdot x+b^{2}x^{2}} \tag{1} \end{equation} To compute the generators $K_{\mu}$ of these special conformal transformations, we have to find $\frac{\partial x^{'\mu}}{\partial b^{\nu}}$. And here is the result:

\begin{equation} \frac{\partial x^{'\mu}}{\partial b^{\nu}} = \frac{({1-2b\cdot x+b^{2}x^{2}})(-\delta_{\mu}^{\nu}x^{2})-(x^{\mu}-b^{\mu}x^{2})(-2x_{\nu}+2b_{\nu}x^{2})}{({1-2b\cdot x+b^{2}x^{2}})^{2}} \tag{2} \end{equation}

Evaluating (2) at $b=0$ gives $$-\delta_{\mu}^{\nu}x^{2}+2x^{\mu}x_{\nu}$$ so the generator is $$K_{\nu}=-i(-\delta_{\mu}^{\nu}x^{2}+2x^{\mu}x_{\nu})\partial_{\mu} = -i\left(2x_{\nu}(x\cdot\partial)-x^{2}\partial_{\nu}\right)$$

  • 1
    $\begingroup$ Thanks, but now why is this the procedure to find the generator? Why take this derivative? Why evaluate at $b=0$? And you lose an index in the thirds step. $\endgroup$
    – Kvothe
    Sep 14, 2016 at 19:24
  • $\begingroup$ If you find the answer correct then please upvote it. $\endgroup$
    – Mass
    Sep 14, 2016 at 19:34
  • $\begingroup$ Let $\mathcal{G}$ is a Lie group depending on some parameters $\alpha_{i}(i=1,2,\cdots n)$, with corresponding generators say $L_{i}$. The $\mathcal{G}(\alpha_{1},\alpha_{2}..\alpha_{i})=e^{\sum_{i}}\alpha_{i}L_{i}$. Then $L_{i}=\left(\frac{\partial\mathcal{G}}{\partial\alpha_{i}}\right)_{\alpha_{i}}(\alpha_{i}=0)$. $\endgroup$
    – Mass
    Sep 14, 2016 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.