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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.

soap
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    I asked a related question here. In general, notation like $\text{blah} \to \text{blah}'$ is inherently extremely vague and should best be ignored. – knzhou Apr 27 '18 at 14:10
  • @knzhou Yes, but the problem is: I want to understand the physics literature, and it is riddled with this. In particular, I can't make sense of this definition in blumenhagen's book – soap Apr 27 '18 at 16:14

2 Answers2

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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.

knzhou
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  • So are you saying that $\phi'=f^*\circ \phi$? But then $\phi'(z,\bar z)=\phi(f(z),\bar f(\bar z)) = \phi(\lambda z, \bar \lambda \bar z)$, right? or am I confused? – soap Apr 27 '18 at 18:05
  • @Soap No, $\phi$ is a quantum field and not a function. Classically all scaling is trivial! We're really talking about how $\phi$ behaves when, e.g. it's under a path integral. – knzhou Apr 27 '18 at 18:42
  • @Soap I don't really know any conformal field theory, so I can't say much more, sorry. – knzhou Apr 27 '18 at 18:42
  • Oh, so do you happen to know how is the "what happens to fields upon pullback by this map" defined in this case (in which $\phi$ is a quantum field)? – soap Apr 27 '18 at 19:34
  • @Soap I imagine that you would then integrate out modes of the field in the path integral to restore the cutoff to its original value, and rescale the field to give it a canonical kinetic term. But I’m just spitballing here, I really don’t know any CFT specifics. – knzhou Apr 27 '18 at 19:37
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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.

Oktay Doğangün
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