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I am puzzled by the concept of scalar fields that arise in conformal field theory in curved backgrounds. In general relativity, so far as I understand it, a scalar field is basically a function defined on the spacetime manifold, meaning that it is invariant under any smooth coordinate transformation $ f(x)=x' $. In special relativity, these objects are defined as things that remain invariant under the Lorentz transformations. So far, everything seems to be ok.

Now imagine a lorentz scalar under the scale transformations: $x'=\lambda x$. We know that a Lorentz scalar field transforms like $\phi'(x')=\lambda^{-\Delta}\phi(x)$ as a scale transformation is not a Lorentz transformation, so the field $\phi(x)$ does not remain invariant under it.

The problem for me arises when people try to write conformal field theories in curved manifolds in $d$ dimensions. For example, they write the following action for a conformal field theory with scalars:

$$ S=\frac{1}{2}\int d^dx\sqrt{g}(g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi+\xi R \phi^2).$$

Then they say that because the field $\phi(x)$ transforms like $\phi'(x')=\lambda^{-\Delta}\phi(x)$ by setting $\xi=(d-2)/4(d-1)$ the action would be invariant under conformal transformations.

However, I do not understand this setup. If $\phi(x)$ is a scalar on the manifold, then it should be invariant under any smooth transformation of coordinates, such as scale transformations. Thus there should not be a $\phi'(x')=\lambda^{-\Delta}\phi(x)$ behavior to begin with. If $\phi(x)$ is not a scalar on the manifold, what mathematical object allows for this behavior?

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    $\begingroup$ "They" are talking about Weyl transformations. $\endgroup$ Oct 26, 2023 at 8:09
  • $\begingroup$ @PeterKravchuk Thanks for the comment. By weyl transformation you mean $g'=\Omega(x)g$? If so shouldn't that only affect the objects that depend on the metric? $\endgroup$ Oct 26, 2023 at 8:23
  • $\begingroup$ If so, your action wouldn't be invariant. In formal language, $\phi$ is not a scalar but rather a section of a weighted version of the trivial line bundle. In other words, it has a non-zero Weyl weight. $\endgroup$ Oct 26, 2023 at 8:30
  • $\begingroup$ @PeterKravchuk Can you point me to a source that expands on that? $\endgroup$ Oct 26, 2023 at 8:35
  • $\begingroup$ As it is often the case in (theoretical) physics, one often assumes that the reader will be able to 'guess' the context; in this case scalar with respect to what? For instance, Lorentz transforms? Yes. Scalar under $U(1)$? We don't know, maybe it's charged, i.e. $\phi \mapsto e^{i\alpha(x)} \phi$. Scalar under conformal transformations? No, as you pointed out it does not transform as a scalar. We often mean by scalar that it's a Poincare scalar (roughly meaning it's spin is 0), but that can (as you see) lead to ambiguity, if one is not careful about interpreting the meaning of 'scalar'. $\endgroup$
    – mb28025
    Oct 26, 2023 at 10:17

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