# Scale transformation of scalars in curved backgrounds

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?

• "They" are talking about Weyl transformations. Commented Oct 26, 2023 at 8:09
• @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? Commented Oct 26, 2023 at 8:23
• 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. Commented Oct 26, 2023 at 8:30
• @PeterKravchuk Can you point me to a source that expands on that? Commented Oct 26, 2023 at 8:35
• 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'. Commented Oct 26, 2023 at 10:17