What is the "prime operator"?

Question

When studying GR, for example, it is common to find sentences like:

The metric, being a $2$-tensor, transforms as $g_{\mu\nu}\rightarrow g'_{\mu\nu}=g_{\alpha\beta}\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^{\nu} }$ under a coordinate transformation $x^\mu\rightarrow x'^\mu$.

So it seems that there is a "priming" map of the type $(-)':2tensors\rightarrow2tensors$ that takes a metric $g$ to a primed metric $g'$. As I understand it, this is just an illusion: in fact the metric is always the sames, what we are doing is changing the charts and thus changing the basis $\{dx^\mu\otimes dx^\nu\}$ in which we are writing our $2$-tensor. So we have, for $p\in M$ and $(x^\mu)^{-1}(x)=(x'^\mu)^{-1}(x')=p$ (with $(x^\mu)$ and $(x'^\mu)$ two charts on $M$)

\begin{align} g(p)&=g(p)\\ g_{\mu\nu}(x) dx^\mu\otimes dx^\nu &= g'_{\mu\nu}(x') dx'^\mu\otimes dx'^\nu \\ g_{\mu\nu}(x) \frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^{\nu} }dx'^\mu\otimes dx'^\nu &= g'_{\mu\nu}(x') dx'^\mu\otimes dx'^\nu \end{align} and this explains the result above.

There are other contexts in which this "priming operator" seems to mean something else, but I don't know what. For example, what does $\phi'$ mean (mathematically) in this definition from "An introduction to conformal field theory" (Blumenhagen, Plauschinn):

Definition 4. If a field $\phi(z,\bar z)$ transforms under scalings $z \mapsto \lambda z$ according to

$$\phi(z, \bar z) \mapsto \phi'(z, \bar z) = \lambda^h \bar{\lambda}^{\bar{h}} \phi(\lambda z, \bar \lambda \bar z)$$

It is said to have conformal dimension $(h, \bar h)$.

Note: I am expecting an answer of the type: $\phi'=\phi\circ f$ where $f$ is... If something is unclear, please ask.

Also, does this generalize to what one sees in other branches of physics, like QFT, or each case has its own interpretation?

Edit: I should add this: I asked my teacher, and he said that we could see $\phi'$ as $\phi$ under a (left) action of the conformal group. This starts to be more satisfactory to me, but still too non-concrete.

Answer 1

In your first example, we are performing a passive coordinate transformation, i.e. merely relabeling the points. Hence the metric tensor $g(p)$ is not changing at all. The confusion is because we write the "new" metric with a prime. Really, we should write $$g_{\mu\nu}(p) \to g_{\mu' \nu'}(p)$$ indicating that $g$ stays the same, and we're just changing the coordinate system from unprimed to primed, which changes the component functions. But this isn't done because it's annoying to write primes on half of the indices. In any case, here the "$\rightarrow$" map is a map on component functions.

In your second example, we are considering an active transformation. Consider the map on spacetime which, in coordinates, maps $x$ to $\lambda x$. Then we are considering what happens to fields upon pullback by this map.

Answer 2

Prime only means "the transformed quantity", it does not have any context about the transformation itself. It is only a labeling convention to indicate the element you arrive after applying the transformation in subject.