Zamolodchikov metric and coupling-dependent field rescalings this is my first post/question, so I apologize in advance for any mistakes in phrasing or format or anything else.
Consider a conformal field theory (CFT) with exactly marginal deformations described by local operators $O_i$ with associated local coordinates $t^i$ for the conformal manifold, described by an action of the form
$$S = S_0 + \sum_i \, t^i \int d^d x \, O_i(x) \, .$$
The ground-state correlators
$$\langle O_i(x) O_j(y)\rangle_t = \frac{G_{ij}(t)}{|x-y|^{2d}}$$
define the Zamolodchikov metric $G$ on the conformal manifold [1]. As an example, writing the classical action of $\mathcal{N}=4$ Super Yang-Mills according to
$$S_{\text{YM}} = \frac{1}{2g^2} \int \text{Tr} \, F \wedge \star F + \dots$$
the gauge coupling $g$ is a marginal deformation (actually the full complexified coupling $\tau$ is). Varying $g \to g + \delta g$, one finds
$$G_{gg}(g) \, dg \otimes dg \propto \frac{1}{g^2} \, dg \otimes dg$$
so that $g=0$ is at infinite distance with respect to a generic point. However, rescaling the fields such that the kinetic terms be canonically normalized, $S_\text{YM} = \frac{1}{2} \int \text{Tr} \, F \wedge \star F + \dots \,$, varying $g \to g + \delta g$ only leaves the vertices, not the full Lagrangian, as the operator of which to compute the correlator, and it does not seem to be a marginal operator (nor the resulting metric seems to be the same). Alternatively, one can still consider the full Lagrangian as a marginal operator, but then one would need to introduce an additional modulus in front of the action, since the coupling now multiplies the vertices only.
My questions are the following:

*

*Is the Zamolodchikov metric invariant under coupling-dependent field rescalings? If so, where is the mistake in the above argument? The metric is invariant under coupling redefinitions, but that seems to be a different type of redefinition.


*If the metric is not invariant, how is one to determine the "correct" one? Namely, the above metric for $\mathcal{N} = 4$ places $g=0$ at infinite distance, but this does not seem to be the case after the canonical field rescaling according to the above argument.


*More generally, if the "origin" $\{t^i = 0\}$ of the conformal manifold corresponds to a non-singular, perhaps free, CFT described by $S = S_0$, can this point be located at infinite distance? According to the CFT Distance Conjecture [2,3], this should be true (at least in dimensions higher than two), but it seems to me that the (asymptotic) Zamolodchikov metric would be given by the free correlator $\langle O_i(x) O_j(y)\rangle_0$ with no negative powers of the $t^i$ leading to infinite distance. In $\mathcal{N} = 4$ SYM with the usual convention, this does not happen because the action becomes singular at $g=0$, and the same happens for the free-boson CFT (more generally, for Narain moduli spaces corresponding to toroidal compactifications).
References:
[1] https://inspirehep.net/literature/240292
[2] https://arxiv.org/abs/2011.10040
[3] https://arxiv.org/abs/2011.03583
 A: I'll give this a shot even though it's easy to make a mistake when discussing these singular limits.
As I'm sure you know, the reason for $g \to g + t$ is because the marginal operators are tangent vectors as far as the Zamolodchikov metric is concerned. So in the neighbourhood of the theory at coupling $g$, I would actually write it as $\frac{1}{g^2} dt \otimes dt$. In this case, the first answer is that $G_{ij}$ is not invariant. It transforms the way any metric transforms when you rescale the co-ordinates $t$. I wouldn't say there's any "correct" metric since you can always do something like a Penrose map to bring $\infty$ to a finite point. However, the fact that the free theory is an infinite proper distance away from a generic point is physical and no co-ordinate transformation will be able to undo that.
So how then did you apparently end up with finite distances after rescaling the fields? This must have to do with the fact that you did something more drastic than just reparameterizing $t$... you rescaled the action $S_0$ that $t$ was perturbing around! It should be a red flag when a calculation depends crucially on doing something like this. The normalization of operators we use in a CFT is up to us and should have nothing to do with how the action is expressed. Also, in this case it is worth pointing out that $\mathrm{Tr} F \wedge \star F$ has an especially useful normalization since it's part of the same supermultiplet as the stress tensor. So a rescaling which makes it look like there's no perturbation by $t \int \mathrm{Tr} F \wedge \star F$ must be wrong.
This would probably be easier to see in the compact free boson since there the marginal operator is the entire Lagrangian density. In your rescaling, the $t$ dependence was shoved onto "other parts of the Lagrangian" which do not constitute marginal operators in the CFT. At some level, the reason why these show up in $\mathcal{N} = 4$ SYM is an artifact of our choice to regard it as a gauge theory in order to be able to use the action formalism. So going outside the scope of where the Zamolodchikov metric can be defined is interpreted as having gauge variant things like $t \bar{\psi} \gamma^\mu A_\mu \psi$ seemingly appear as perturbations.
In short, I would be very careful about using a kinetic term that is finite as $g \to 0$.
