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Theorem 5.1 on page 80 of this paper says that

Assuming that the matter fields satisfy their equations of motion, the matter field action is locally Weyl invariant if and only if the corresponding energy-momentum tensor is traceless.

If I have understood this right, Weyl invariance reduces to scale invariance in flat spacetime, which is basically what I am up to at the moment.

My question now is:

If one has Weyl or scale invariance in accordance with this theorem, does this mean that global Weyl or scale invariance (such that the scaling parameter does not depend on position) is guaranteed too? If not why not?

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Dear Dilaton, the Weyl invariance is scaling invariance but the coefficient of scaling must be allowed to depend on the position, otherwise it's not Weyl. The global scaling is always a subgroup of the Weyl symmetry if the latter exists, so the scaling is guaranteed if Weyl holds. So Weyl is always local. Above 2 spacetime dimensions, it's somewhat hard or unnatural to find Weyl-invariant theories. In 2 dimensions, the 2-derivative kinetic terms are naturally Weyl-symmetric because the scaling of $g^{\mu\nu}$ contracting $\partial_\mu$ and $\partial_\mu$ cancels the transformation of $\sqrt g$ –  Luboš Motl Nov 27 '12 at 14:07
    
Thanks @LubošMotl for this helpful and enlightening comment. Is the vanishing of the trace of the energy-momentum tensor a neccessary and sufficient contition for global scale invariance to hold then? –  Dilaton Nov 28 '12 at 9:58
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Actually, from the proof in that paper it would appear that what you propose is true. They use an arbitrary conformal scaling of the metric so it should be true globally. –  levitopher Nov 28 '12 at 15:42
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I do not think that local Weyl (conformal) invariance implies global invariance. The fields are only defined over local open sets $U_i$, so if the action is invariant under a constant conformal transformation (which does not depend on $x\in U_i$)

$\phi\to \Omega_i^2 \phi$

this does not imply anything about the global properties of $\Omega:M\to \mathbb{R}$, only that $\Omega|U_i$ is constant. There is something called a "conformal connection" which I don't know very much about, but I think the short answer is that if you restrict the fields to satisfy some specific symmetry group $G$ (i.e. they are part of a $G$-bundle and $G$ is some group of conformal transformations) then the fields will always be conformally invariant, locally and globally.

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