Derivation of the infinitesimal Special Conformal Transformation

This question is somewhat connected to my last question on special conformal transformations. I'm considering the derivation of a special conformal transformation. Namely, the quadratic translation

$$\delta x^\mu=c_{\mu\nu \rho}x^\nu x^\rho$$

From the conformal Killing equation, I know that

$$c_{\mu\nu\rho}=g_{\mu\rho}b_\nu +g_{\mu\nu}b_\rho-g_{\nu\rho}b_\mu$$

Plugging this into the above, I have

\begin{align}c_{\mu\nu\rho}x^\nu x^\rho&=(g_{\mu\rho}b_\nu +g_{\mu\nu}b_\rho-g_{\nu\rho}b_\mu)x^\nu x^\rho\notag\\ &=b_\nu g_{\mu\rho}x^\nu x^\rho+b_\rho g_{\mu\nu}x^\nu x^\rho-b_\mu g_{\nu\rho}x^\nu x^\rho\notag\\ &=b_\nu x^\nu x_\mu+b_\rho x_\mu x^\rho-b_\mu x^\nu x_\nu\notag\\ &=2(b\cdot x)x_\mu-b_\mu (x\cdot x) \end{align}

However, on page 11 of Alday's notes on CFTs, the final answer should have the $x$ and $b$ on the right hand side of the above in covariant form--i.e.,

$$c_{\mu\nu\rho}x^\nu x^\rho=2(b\cdot x)x^\mu-b^\mu(x\cdot x)$$

I don't see what exactly I did wrong in my calculation. I want to make sure I understand how to derive this equation, and I can't find an explanation anywhere online or in my textbooks.

• I believe I figured it out--in Alday's notes, what he calls $\epsilon^\mu$ in equation 2.11 should be $\epsilon_\mu$ if we want the left hand sides and the right hand sides to agree. Then everything works out as it short. So, technically, in my first equation, $\delta x_\mu$ should replace $\delta x^\mu$. This will assure agreement of the covariant and contravariant forms of the special conformal transformation differential equation that results from this exercise. Feb 11, 2018 at 5:12
• Hint: It seems that some of your equations are not covariant in the sense that sub and super indices on the two sides don't match, and that you just have to raise and lower the indices with the metric in appropriate places so that they do match. Feb 11, 2018 at 6:08