4
$\begingroup$

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 ?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\mathcal{M/D}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.