# Some questions about calculation central charge in a CFT in $d$ spacetime dimensions

This is based on this paper, http://arxiv.org/abs/hep-th/0212138

• For a CFT on a $S^d$ spacetime (of radius R) it seems to be claimed that the central charge is given by, $c = \langle \int_{S^d_R} d^dx \sqrt{g} T_\mu ^\mu \rangle$

• Their equation 23 (on page 6) seems to indicate that if $W = -log Z$ is the free energy of the theory then it further follows that, $c = \frac{1}{d}R \frac{\partial }{\partial R} W$ (...I believe that the derivative is being evaluated at the value of the radius of the sphere..)

• Just below equation 26 it is claimed that, "...the central charge can be read off from the coefficient of log R in an expansion of W[R]..."

I would like to know the proof/derivation of three methods that have been spelt out to calculate the central charge of a CFT.

• "Method" $3$ is an obvious consequence of Method $2$ ($c = \frac{1}{d}R \frac{\partial }{\partial R} W$). If you get $W = a Log R +b$, you will have $c=\frac{a}{d}$ – Trimok Oct 28 '13 at 13:06
• @Trimok Isn't your argument just one-way? What if it were $W = aR^2 + bR$? Then $c = (R/d)(2aR + b)$. I don't get what you are saying. Firstly one can't have a term like "log(R)" in $W$ since $R$ has dimensions and one can't take log of something with dimensions. So even if such a log term arises it will come compensated with some scale to make the argument dimensionless. So it can still be something like, $W = alog(\Lambda R) + bR^2 + cR$ say and then this method-3 will not work!. And also why should such a "log(R)" term be natural? – user6818 Oct 28 '13 at 23:22
• @Trimok One way the things can match up is if in the method-3 there is an implicit limit of $R \rightarrow 0$..but why should that be? – user6818 Oct 28 '13 at 23:26
• If you look at expressions $(36), (40), (48),(50)$, you see that $V_1$ and $V_2$ ($W_f \sim V_1+V_2$), respect the expression $(a \log R+b)$. – Trimok Oct 29 '13 at 4:21
• In the large $R$ limit (see text between equations $26$ and $27$), $fR^{d-2\Delta} \to \infty$ , that is $fg_l \to \infty$, we may replace, in equation $25$, $\log (1+fg_l)$ by $\log (fg_l)$, and, in equation $24$, we have $g_l \sim R^{d-2\Delta}$, so we naturally get $\log R$ terms in the large $R$ limit. – Trimok Oct 29 '13 at 5:57