Equations 2-7 on page 21 of these notes, http://www.math.ias.edu/QFT/fall/NewGaw.ps seems to give a fairly compact definition of what a CFT is.

But I have two questions,

  • This definition is specific to 2 dimensions. Is there an analogue of this definition for higher dimensions?

  • I want to know if there is an equivalent definition of a CFT in terms of the conformal group in the specific dimension.

Like I would like to know if I can make precise a statement like this (for any dimension), "CFT is a QFT such that its Hilbert space splits into Verma modules and the correlation function of its primary fields is invariant under the conformal group in that dimension."

We do have odd-dimensional CFTs and there I don't know what is a "conformal group"!

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    $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Mar 1, 2014 at 8:28

1 Answer 1


The conformal group is defined for any spacetime you want. The conformal group of d-dimensional Euclidean space, which has isometry group SO(d), is SO(d+1,1). The conformal group of d+1 dimensional Minkowski space, whose isometry group is SO(d,1), is SO(d+1,2). The defining property of the conformal group is that its the set of transformations that leave the metrix $g_{\mu\nu}$ invariant up to a scaling factor $e^{\omega(x)}$.

A more general definition for CFTs in d>2 dimensions is that a CFT is a QFT whose Hilbert space breaks up into representations of the conformal group and whose correlation functions are invariant under any conformal transformation (I don't believe we have to restrict to primary operators).

The special case of two dimensions is that the conformal algebra is infinite dimensional. The group of globally defined conformal transformations is still finite, but there are an infinite number of local conformal transformations. So 2d CFTs are a lot more restricted then higher dimensional CFTs.

For a good introduction look at https://sites.google.com/site/slavarychkov/

  • $\begingroup$ Here are my 2 questions rephrased better, (1) Isn't it true that the conformal group in $(d>1)+1$ Minkowski manifold is only locally isomorphic to $SO(d+1,2)$ and not globally? (2) Is there a proof for $(d>1)+1$ Minskowskian space-time that demand of the classical Lagrangian being invariant under Weyl rescalings of the metric implies that the Hilbert space of the corresponding QFT will split into $SO(d+1,2)$ representations? [...AFAI have seen this crucial link between the classical and the quantum picture is provable only on Riemann surfaces with a fixed metric...] $\endgroup$
    – user6818
    Mar 3, 2014 at 5:47
  • $\begingroup$ Also is it true that in a CFT Hilbert space all operators will be eigenoperators under commutation with the dilatation operator? (...because when one argues what is the highest-weight sate in a conformal module one looks for that operator among only such operators which commutes with the $K$ - and that is the primary...) $\endgroup$
    – user6818
    Mar 3, 2014 at 5:55
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    $\begingroup$ To answer your first question, yes the isomorphism is only local (typically). The conformal group of 3+1 dimensional Minkowski space is doubly covered by SO(4,2), which in turn is doubly covered by SU(2,2). 2) I think Weyl rescalings are in general a different beast then conformal transformations (rescaling of metric vs coordinate transformations) and there is no guarantee a conformally invariant lagrangian gives rise to a CFT. In general a clasically conformally invariant lagrangian gives rise to a CFT only if its a fixed point of the RG flow (or typically that the beta function vanishes) $\endgroup$
    – David M
    Mar 3, 2014 at 23:43
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    $\begingroup$ And yes the Hilbert space of a CFT all operators will be eigenoperators of the dilatation operator. Basically thats how the Hilbert space is constructed, you start with the primary operators annihilated by K and construct all the other states in the conformal module via P. $\endgroup$
    – David M
    Mar 3, 2014 at 23:45
  • $\begingroup$ Thanks for the reply! (1) So if SO(d+1,2) is only a local model for whatever one wants to call as the "conformal group" (the group of all conformal diffeomorphisms - diffeomorphisms which keep the metric invariant upto a Weyl rescaling) then why should the Hlbert space of the CFT split under its representations? QFTs are supposed to see the full space-time and not just a local symmetry group - right? SO we are saying that no matter how complicated is the space-time topology a CFT on it will have its Hilbert space split under $SO(2,d+1)$! $\endgroup$
    – user6818
    Mar 4, 2014 at 3:32

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