# Infinitesimal conformal transformations

From Paul Ginsparg's lecture note on Applied Conformal Field Theory, I am given the followings.

In $$\mathbb{R}^d$$, the infinitesimal transformation is given by $$x \rightarrow x+\epsilon$$ which leaves the metric to transform as $$g_{\mu\nu} \rightarrow g_{\mu\nu} +\partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu + O(\epsilon^2).$$Therefore, infinitesimal conformal transformation must satisfy $$\partial_\mu\epsilon_\nu + \partial_\nu \epsilon_\mu = \frac{2}{d}(\partial \cdot \epsilon) g_{\mu\nu} \tag{1}$$ where the proportionality constant can be checked from the trace.

From here, I am supposed to get $$(g_{\mu\nu} \Box+(d-2)\partial_\mu\partial_\nu )(\partial \cdot \epsilon)=0. \tag{2}$$ However, it seems to me that this is false. The reason is as follows.

From (1), $$g_{\mu\nu} \Box (\partial \cdot \epsilon) = \frac{d}{2} \Box(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu).$$ If (2) is true, then $$\frac{d}{2}(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)+(d-2)\partial_\mu\partial_\nu(\partial \cdot \epsilon)=0.$$ Tracing, I get $$d\Box(\partial \cdot \epsilon)+ (d-2)\Box (\partial \cdot \epsilon) = 2(d-1)\Box (\partial \cdot \epsilon) =0.$$ However, this is only generally true if $$d=1$$, which seems to be unintuitive.

My question is as follows. 1. Is my approach correct? 2. If so, what should the alternative for (2) be? It is used to argue that the third derivatives of $$\epsilon$$ must vanish. 3. Are there any remarks to be made here?

• OP's eq. (2) is explained in eq. (6) in my Phys.SE answer here. Commented May 13, 2020 at 14:13
• After my reading in your previous Phys.SE answer, as part of my question, although this might be trivial, does this result imply that at $d=2$, $\epsilon$ can have arbitrary dimension?
– Nugi
Commented May 13, 2020 at 14:39
• In $d=2$ the full conformal symmetry is infinite dimensional, if that's what you're asking. In fact, that's more or less the point of the linked question. Commented May 13, 2020 at 17:50