# Weyl + Diffeomorphism invariance = Conformal symmety only in flat space?

Let $$(M,g)$$ be a Riemannian manifold and consider a classical field theory $$\phi: M \to \mathbb{R}$$ given by some action functional $$S$$ satisfying Weyl & diffeomorphism invariance $$S(\phi, e^\omega g) = S(\phi,g),$$ $$S(\psi^* \phi, \psi^* g) = S(\phi,g).$$ Now, if we further assume that $$\psi$$ is a conformal map on the manifold $$(M,g)$$, then $$\psi^* g = e^\omega g$$ for some $$\omega$$. Then $$S(\phi \circ \psi^{-1},g) = S(\phi, \psi^* g) = S(\phi, e^\omega g) = S(\phi,g)\,,$$ i.e. the action describes a conformally symmetric field. I guess this is the reason why Weyl+Diffeo invariance is often the definition of a conformal field theory.

My question is that why do some textbooks and other sources say that "Weyl+Diffeo = Conformal Symmetry in flat background" (for example Erbin's recent String Field Theory book)? Is taking $$g$$ to be flat necessary?

Why I find this weird is that many Riemann surfaces don't admit a flat metric. (Of course Weyl+Diffeo invariance makes sense in any surface, so maybe that's the reason it is a better definition of conformal symmetry)

I wonder if in my computation above something is wrong with saying that $$\psi^*g = e^\omega g$$. Certainly this is true for Euclidean metric $$g=\delta$$ and conformal $$\psi$$, but for arbitrary $$g$$ the pullback changes the point where we evaluate $$g$$ $$\psi^* g = (D\psi)^t (g\circ \psi) D \psi\,,$$ but it seems that one could still take $$e^\omega = |\psi'(z)|^2 (g_{z \bar z}\circ \psi)/ g_{z \bar z}$$ in conformal coordinates.

Consider the fact that the conformal group in $$d$$ Euclidean dimensions is $$SO(d + 1, 1)$$ which means it has a finite number of generators. Diffeomorphism groups contain these generators along with infinitely many other generators not in the conformal group. So saying that Weyl plus diffeomorphism invariance is ever "equal" to conformal invariance can't be right.
Conformal invariance is only a useful concept when the background is fixed. In practice, flat space is the background discussed the most. But it can also be another manifold which admits conformal Killing vectors. Diffeomorphism invariance, on the other hand, is something you can reasonably expect to find if you allow yourself to perform active transformations on the metric. The interesting question is how much of this symmetry will survive after you change viewpoints (or gauge fix) to make transformations passive on the manifold and only active on the other fields. Whether this residual symmetry includes not just $$SO(d)$$ but the conformal group as well depends on whether the original theory with a dynamical metric respected Weyl invariance.
• Now that I think of it, I should have formulated the question as does Weyl+Diffeo imply that $S(\phi \circ \psi,g) = S(\phi,g)$ for conformal $\psi$, since to me the property $S(\phi \circ \psi,g) = S(\phi,g)$ is the simplest possible way to formulate something that would be "conformal symmetry". Of course in many cases this is too naive, since you might not have global conformal maps on your manifold. Sep 2, 2021 at 12:43
• Of course the generators of diffeos contain generators of conformal maps, but my point is that the diffeo invariance (or covariance?) is formulated as $S(\psi^* \phi, \psi^* g) = S(\phi,g)$, i.e. the value of the action does not change if you change both the field and the metric, but in "conformal symmetry" I would expect $S(\psi^* \phi,g) = S(\phi,g)$ for $\psi$ conformal. I don't see why flat background would be special here (except maybe because it is just a common choice). Sep 2, 2021 at 12:45
• Ah yes. Upon re-reading your question it looks like you were already aware of the main point I was making! Consider this a vote for the "$g$ does not have to be flat" scenario. Sep 2, 2021 at 13:48