In QFT , The action should be invariant under poincare symmetry $g_{\mu\rho}(x)=g_{\mu\rho}^{\prime}(x^\prime)$ . We can generalize this by considering theories invariant under conformal symmetry $g_{\mu\rho}^{\prime}(x^\prime)=A(x)g_{\mu\rho}(x)$ . If we consider more general variations ( Diffeomorphisms on spacetime manifolds ) What will happen ? If we gauge fix the redundant variables , Can we remove the metric completely from the action? The theory .Will the theory be background independent ? Can we classify topological invariant properities using correlation functions from that theory ?
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2 Answers
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First of all, the term "conformal symmetry" means something else than indicated in the question. A conformal symmetry is purely a transformation of coordinates – no additional change of the fields than the transformation derived from the change of coordinates is allowed – that preserves the metric tensor up to a scaling (equivalently that preserves the angles at each point).
On the other hand, an explicit additional change of the metric by the multiplicative factor $A(x)$ – something that is added on top of the transformation induced by the change of coordinates – is called "Weyl symmetry". Similar transformations are talked about in both but one must carefully think about the logic because conformal symmetries and Weyl-times-diffeomorphism symmetries are completely different.
In particular, the conformal group in $D$ dimensions is finite-dimensional, namely $SO(D+1,1)$ or $SO(D,2)$ where the latter holds in the Lorentzian signature. The Weyl group, the diffeomorphism group, and their union (semidirect product) are always infinite-dimensional because one may simply choose new coordinates and a new $A(x)$ at each point. A conformal symmetry is essentially an "isometry up to a local rescaling of the metric". Isometries are rare. That's very different from diffeomorphisms and Weyl symmetries that can always be written down.
The only exception is the case of 2 dimensions where the conformal group is infinite-dimensional, too (at least locally). In the Euclidean signature, every holomorphic (or, if we allow poles, meromorphic) function of a complex variable is conformal. In the Lorentzian signature, any separate redefinition $x^{\prime+ } = f(x^+)$, $x^{\prime-}=f(x^-)$, also preserves the angles. This special enhanced group in 2 dimensions is the reason why string theory based on 2-dimensional world sheet is the only case of consistent extended objects that may be defined by a straightforward gravitational action in the world volume: all the gravitational excitations along with all the usual ultraviolet problems that they typically imply become unphysical or decoupled, at least locally. Two parameters per point that define a diffeomorphism in $D=2$ plus one Weyl scaling parameter per point gives you three parameters, exactly enough to set the three components of the 2-dimensional metric to any conventional predetermined non-singular values, at least locally. The number of parameters in diff-Weyl is totally insufficient to eliminate the whole metric above $D=2$.
In no other dimension, there is enough freedom to eliminate the dynamical degrees of freedom from the theory completely. The theories in $D=3$ would deserve a special treatment, however, because the Ricci flatness implies the Riemann flatness in $D=3$. But even in $D=3$, it's fair to say that the usual problems with the quantization of gravity remain there (although it's a bit subtle to see what the problems are in $D=3$).
Every theory with a diffeomorphism symmetry (which includes theories with both diffeomorphism and Weyl symmetry) should be viewed as a theory of quantum gravity, not an ordinary quantum field theory, and as I mentioned, only in $D=2$, such a theory may be consistent if defined by the usual Lagrangian methods.
However, every theory that respects the diffeomorphism symmetry – which must be a gauge symmetry, for consistency – is "background independent" when it comes to the metric. Whether it's background-independent concerning other fields that define the background depends on the particular theory. Also, even if such a theory is background-independent, it may be hard to see this fact; the fact may refuse to be manifest. Whether it's manifest or not depends on the particular formalism we find to describe such a theory.
answered Jun 18, 2013 at 7:05
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This doesn't happen often, but I have to disagree with Luboš's answer. The OP seems to be asking if a diffeomorphism invariant is "background independent", in the sense that it is independent of the background metric, or topological, as the last part of his question seems to indicate.
No, a diffeomorphism invariant theory is not necessarily metric-independent.
The group of diffeomorphisms $\mathcal{D}$ acts on the space of metrics $\mathcal{M}$ on the space on which the theory is defined. This action is not transitive, which means that not any two metrics can be related by a diffeomorphism. The action of $\mathcal{D}$ on $\mathcal{M}$ therefore decomposes into a set of orbits $\mathcal{M/D}$.
The statement that the theory is diffeomorphism invariant is that the partition function of the theory and its correlation functions, a priori defined as functions over $\mathcal{M}$, are invariant along the orbits of the diffeomorphism group. They can therefore be seen as functions over $\mathcal{M/D}$. But they still can have a very non-trivial dependence on the metric.
The statement that the theory is topological or metric-independent is that the partition function is constant over $\mathcal{M}$ (hence also over $\mathcal{M/D}$).
