This is sort of a silly question because I'm a total beginner, and I debated whether it was better to ask here or on Math.SE. I decided on here because it's about how physicists use terminology, even though the terminology is intrinsically mathematical and there's not actually any physical content in the question.

It's a well-known fact in theoretical physics that the conformal group of Minkowski space is $\operatorname{SO}(4,2)$ (or sometimes its double cover $\operatorname{SU}(2,2)$ is chosen instead). This is infinitesimally generated by the standard Poincare generators $M_{\mu \nu}$ and $P_\mu$ together with scale transformation $D$ and special conformal generators $K_\mu$. I've also seen conformal groups discussed for other manifolds (the case which lead me to this question is Witten's 2003 paper Perturbative Gauge Theory As A String Theory In Twistor Space, which includes the conformal group of $\mathbb{R}^4$ with all possible signatures.)

However, the above doesn't seem to agree with the definition of the term "conformal group" which I know from the minimal amount of conformal geometry I know. Specifically, the conformal group of a pseudo-Riemannian manifold $M$ is the group of conformal self-maps on $M$, that is, self-diffeomorphisms for which the pullback metric is conformally equivalent to the original metric. (Two metrics $g$, $h$ are conformally equivalent if there is a positive real-valued function $u$ such that at each point $x \in M$ we have $g_{x, \mu \nu}=u(x)h_{x, \mu \nu}$.)

The trouble is that the conformal group of Minkowski space isn't $\operatorname{SO}(4,2)$ with the above definition. The special conformal transformations, given by the expression (for some vector $b_\mu$) $$ x' ^\mu = \frac{x^\mu - x^2 b^\mu}{1-2b \cdot x +b^2 x^2}$$ aren't self-maps of Minkowski space at all. The denominator can become 0 (while the numerator remains finite), and at that particular $x^\mu$ the transformation given by $b^\mu$ is undefined. For an example, take $x=(1,0,0,0)$ and $b=(0,1,0,0)$. So in fact the special conformal transformations are only maps from a subset of Minkowski space back to the full Minkowski space, and shouldn't be included as part of the conformal group if we take the above definition. It's difficult to tell from this formula, but from the fact that $$ \frac{x'^\mu}{x'^2} = \frac{x^\mu}{x^2} - b^\mu$$ we can tell that the inverse transformation is just given by $-b^\mu$, another special conformal transformation, which means these maps aren't surjective either.

These problems stem from the fact that the conformal group is typically derived in physics by considering the algebra of vector fields (which we think of as infinitesimal coordinate transformations) which satisfy the conformal Killing equation. There's in general no reason to expect that the corresponding finite transformation to be well-behaved away from the origin.

How, exactly, are conformal groups defined in physics for arbitrary (in particular noncompact) pseudo-Riemannnian manifolds so that we get the expected result for Minkowski space?

To be perfectly clear, I don't want answers of the form "find the Lie algebra, then exponentiate everything". I know that this works, but I want a definition for the conformal group as a group, without referencing algebras. In my opinion, there ought to be a way to define it this way, and it seems a bit important to me since a priori I don't see any reason why the group of conformal self-maps of a manifold needs to be a (possibly infinite-dimensional) Lie group at all. There are probably some high-power theorems (perhaps in PDE) which would imply this trivially, but naturally we need to know an actual group structure to be able to apply them. In the case of the conformal group as I defined it above (so excluding special conformal transformations) I believe I know a way to prove this for the "mathematician" definition, but when we start throwing in maps that are not everywhere defined the approach I came up with fails. Hence, even if this is how most physicists would define it, I'm looking for an alternate global definition to get some better geometric intuition which produces the correct answers.

Some thoughts on this (feel free to ignore, especially if I'm on the wrong track):

It seems to me possible to remedy this in a fairly trivial way. We can consider all injective conformal maps from dense open sets of Minkowski space to (possibly distinct) dense open sets of Minkowski space. We also need to define compositions to be restricted to the subset of Minkowski space on which they are defined. Furthermore, we impose an equivalence relation that two maps which restrict to the same map on the intersection of their domains are equal. Using these definitions, we can at least get a group which contains all of the conformal transformations it should. It also contains some other less desirable sorts of transformations (for instance, take two rotations and patch them together on the interior and exterior respectively of $\mathbb{R} \times S^2$; incidentally, this is essentially why the proof I have could potentially fail for the "physicist's" conformal groups), but by imposing some conditions on the behavior of the maps in a neighborhood of the boundary of the region on which they are defined it should be possible to get back the get back the ordinary conformal group. (I have not checked this, nor do I know exactly what conditions should be applied, though I could figure it out in the case of Minkowski space based on an alternate model below for the conformal group.)

That approach, while easy to justify based on our above assumptions, is not very intuitive. It requires a lot of technical work to get everything to work out properly. We're implicitly using the fact that finite intersections of dense open sets are dense and open. The conditions required as the functions approach the boundary of their domains to get everything to work are also not obvious. More importantly, the whole thing seems to be suggesting an alternate approach which is much more geometrical; that we should view Minkowski space as contained within some slightly larger space on which the conformal group acts more nicely, where we view the complement of Minkowski space as points at $\infty$.

Specifically (and at this point I'm just making conjectures because they seem helpful) for any psuedo-Riemannian manifold $M$, there ought to be some unique space $\bar M$ which has the properties that $M \subset \bar M$ is dense (this is based on the above observations, and might not be necessary or desirable in the general case; however, it seems reasonable), the restriction of the metric on $\bar M$ to $M$ produces a conformally equivalent metric to the standard metric on $M$, and such that all the conformal self-maps of $M$ which "should exist" can be legitimately interpreted as conformal self-maps of $\bar M$. I don't know that this actually implies $\bar M$ is compact, but certainly if $M$ is compact then the sorts of problems we run into for Minkowski space can't happen. We can then just define the conformal group of an arbitrary manifold $M$ to be the conformal group, in the "mathematics definition", of $\bar M$. A construction of $\bar M$ would thus be a perfectly reasonable answer to this question, though at this point I'm still not sure that any space $\bar M$ need exist in general. I briefly tried to do this myself, but I was not able to come up with anything general enough to be satisfactory, nor was I able to prove that the few cases I know of this construction are unique.

I do know that this approach works for Minkowski space in some sense. Specifically, we can complete Minkowski space in this sense by including it into the projective light cone (i.e. the space of all null lines through the origin) in 6-dimensional Euclidean space with $(4,2)$ signature. The particular embedding is well-known and not so important for my purposes. However, this doesn't generalize particularly well when presented in this way. The reason why it works, at least intuitively for me, is that $\operatorname{SO}(4,2)$ naturally acts on this space, which to me seems like a big coincidence that can't be expected to continue to work if we consider arbitrary spaces. I considered using Penrose diagrams to try to do the equivalent construction for all of the cases I care about, but these only work for asymptotically flat spaces as far as I know and that misses some cases of interest for me (e.g. AdS), plus a general construction would be far more satisfying.

I also considered the possibility that physicists may be using the term inconsistently outside the context of Minkowski space. That is, there may be a number of constructions of "conformal groups", all of which give the same $\operatorname{SO}(4,2)$ on Minkowski space. That would actually be more interesting for my purposes if it's true.

  • 1
    $\begingroup$ Re your comment about special conformal transformations - you're right in that they're not defined on all of Minkowski space. Easiest to see is the inversion which sends the null cone at the origin off to infinity. So you need to compactify Minkowski space ("add a null cone at infinity") to get a manifold on which all these conformal transformations work consistently. $\endgroup$
    – twistor59
    Commented May 7, 2013 at 7:39
  • $\begingroup$ Related: physics.stackexchange.com/q/108472/2451 $\endgroup$
    – Qmechanic
    Commented Mar 31, 2021 at 0:47

1 Answer 1


Fist, let me just say that this book will answer (almost) all of your questions with beautiful precision and wonderful detail; I highly highly recommend that you read the chapter on the conformal group from which I am stealing the following information.

Let me repeat one of the definitions that directly answers your question in the case of the pseudo-Euclidean manifolds $\mathbb R^{p,q}$;

Definition. The conformal group $\mathrm{Conf}(\mathbb R^{p,q})$ is the connected component containing the identity of the group of conformal diffeomorphisms of the conformal compactification of $\mathbb R^{p,q}$.

With this definition, one can show that for $p+q>2$, the group $\mathrm{Conf}(\mathbb R^{p,q})$ is isomorphic to either $\mathrm{SO}(p+1, q+1)$ or $\mathrm{O}(p+1, q+1)/\{\pm 1\}$ if $-1$ is in the connected component of $\mathrm O(p+1, q+1)$ containing $1$.

  • $\begingroup$ Sorry shouldn't the last line read "$O(p+1,q+1)/ \{\pm 1 \}$"? $\endgroup$
    – user7757
    Commented Jun 24, 2014 at 20:57
  • $\begingroup$ The link to the book no longer works and I would like to know what it is. $\endgroup$
    – Okazaki
    Commented Jun 26, 2017 at 16:08
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    $\begingroup$ @Okazaki The link works fine for me, but here is the book information anyhow: A Mathematical Introduction to Conformal Field Theory, Martin Schottenloher. $\endgroup$ Commented Jun 26, 2017 at 16:52

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