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I am reading the paper on the $c$-theorem [1:Zamolodchikov 1986] and the entropic $c$-theorem [2:Casini, Huerta 2006].


Q1. What do you mean by "monotonically decreasing under RG flow"?

I am unclear on the specific dependencies of the c-function.

According to a review article [3:Nishioka 2018], Zamolodchikov's c-function satisfies the following properties:

  • It takes the same value as the central charge $c$ of the CFT corresponding to each fixed point of RG flows: $$ c(\tau, g)|_{g = g_\text{CFT}} = c. $$
  • It monotonically decreases along any RG flow: $$ \tau \frac{\mathrm{d} c(\tau, g)}{\mathrm{d}\tau} \leq 0. $$
  • It is stationary only at the fixed points: $$ \left. \frac{\partial c(\tau, g)}{\partial g} \right|_{g = g_{\text{CFT}}} = 0. $$

I don't understand the derivative on the left-hand side. I understand that the RG flow $g = g(\tau)$ is generated by $\beta$-function as $$ \tau \frac{\mathrm{d}g(\tau)}{\mathrm{d}\tau} = \beta(g(\tau)) \quad \text{(RG flow)}. $$ I think the left-hand side of the inequality $$ \tau \frac{\mathrm{d} c(\tau, g(\tau))}{\mathrm{d} \tau} = \tau \frac{\partial c(\tau, g(\tau))}{\partial\tau} + \tau \frac{\mathrm{d}g(\tau)}{\mathrm{d}\tau} \frac{\partial c(\tau, g(\tau))}{\partial g} $$ is always zero by definition since $c$ is a well-defined observable independent of the cutoff $\tau$ given by hand, or, in other words, by the renormalization equation $$ \tau \frac{\partial c(\tau, g(\tau))}{\partial\tau} = - \tau \frac{\mathrm{d}g(\tau)}{\mathrm{d}\tau} \frac{\partial c(\tau, g(\tau))}{\partial g} \quad (\text{RG for } c). $$ Is the derivative only the first partial term? If so, "the monotonicity of c-function under RG flow" means that the quantity above is non-positive, right?


Q2. Why can one say that the monotonicity of $C(R)$ leads to monotonicity under RG flow?

In [2], we have an inequality $C'(R) \leq 0$ for $C(R) := 3RS'(R)$. By dimensional analysis, $R$ in $C(R)$ always appears with $\tau$ as $\tau R$, which means $$ \tau\frac{\partial C(R, \tau, g(\tau))}{\partial \tau} = R\frac{\partial C(R, \tau, g(\tau))}{\partial R}. $$ So, we have the monotonicity under RG flow in the same sense as Q1, right?

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Regarding your first question:

In the 2d-case the c-function is a function on the space of string theories. Every point is represented by an operator $\mathcal O$ deforming the sigma model. The operator can be expanded in terms of couplings $\phi^i$ in the form $$\mathcal O=\phi^iO_i ,$$ which are the background fields of the sigma model. These fields are dynamical objects of the effective background theory $S[\phi^i]$. Depending on the dimension of the operator, three different kinds of flows in the space of couplings can be triggered. These flows are integral flow-lines of a gradient flow, where the gradient vector field is the $\beta$-function. The $\beta$-function can be obtained by the variation of the effective background theory $$\frac{\delta S[\phi^i]}{\delta\phi_i}=\beta^i(\phi)$$ and therefore its zeroes correspond to the equations of motion . On the other hand the $\beta$-function is the change of these couplings with the logarithmic energy scale $$\beta^i(\phi^i)=-\frac{\partial\phi^i}{\partial\tau}.$$ This implies that the zeros of the $\beta$-function correspond to the conformally invariant theories. So the derivative with respect to the scale of the c-function also vanishes, and hence it is constant at the fixed-points of the RG-flow. Due to the positivity of the beta function for such sigma models, we always flow from a UV fixed-point to a IR fixed-point. Finally, from the positivity of the $\beta$-function and $$\frac{\partial c(\phi)}{\partial \phi^i}\sim -\beta^jG_{ij}$$ $(G_{ij}$ is a metric on the space of couplings) it follows that the c-function always decreases if we follow a RG-trajectory in the space of couplings. Thus $c_{UV} \geq c_{IR}$.

Regarding question 2: Yes, indeed.

Edit: For the open string aka boundary CFTs there is an analogue of the c-theorem called the g-theorem. The g-function represents the boundary entropy at the fixed points and it is given by $g[\phi^i]:=(1+\beta)Z[\phi^i]$ with $Z$ the partition function.

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    $\begingroup$ Thank you for your comment! I think I understand. In your notation, the c-theorem is $$ \beta^i \frac{\partial c}{\partial \phi^i} \sim - \beta^i \beta^j G_{ij} \leq 0, $$ with the inequality coming from the positive definite metric! $\endgroup$
    – Nek
    Commented Sep 9 at 6:08
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    $\begingroup$ Yes the metric is positiv if and only if the CFT is unitary. $\endgroup$
    – Simp
    Commented Sep 9 at 23:46

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