# Can you conformally map the Lorentzian cylinder to the Lorentzian plane?

When studying CFT in Euclidean signature, for the purpose of radial quantization, we conformally map the Euclidean cylinder to the Euclidean plane (minus the origin, which I ignore).

1. Can one also conformally map the Lorentzian cylinder to the Lorentzian plane (Minkowski space)?

Concretely, in Euclidean signature, as described e.g. here, we can take the Euclidean-signature cylinder $$\mathbb{R} \times S^{d-1}$$ with $$ds^2 = dz^2 + R^2 d\Omega_{d-1}^2$$ and define coordinate transformation $$z = R \log(r/R)$$ to obtain $$ds^2 = R^2 \frac{dr^2}{r^2} + R^2 d\Omega_{d-1}^2 = \frac{R^2}{r^2} \left[ dr^2 + r^2 d\Omega_{d-1}^2 \right]$$, which is conformally equivalent to $$ds^2 = dr^2 + r^2 d\Omega_{d-1}^2$$, i.e. the Euclidean plane.

If we used the same transformation beginning instead with the Lorentzian cylinder $$ds^2 = -dz^2 + R^2 d\Omega_{d-1}^2$$, we would obtain $$ds^2 = -dr^2 + r^2 d\Omega_{d-1}^2$$. However, this is not the usual Lorentzian plane, which has metric $$ds^2 = -dt^2 + d\vec{x}^2$$. So at least with this approach, it's not clear whether the Lorentzian cylinder may be conformally mapped to the Lorentzian plane. (I believe a local patch of the Lorentzian cylinder may be conformally mapped to the Lorentzian plane, I'm interested in global mappings.)

If the Lorentzian cylinder is conformally equivalent to the Lorentzian plane, then they share the same conformal group, but if it's not, then I'm further curious:

1. What is the conformal group of the Lorentzian cylinder?

(I admit this is a math question, but the sort that's likely answered by a physicist with CFT experience.)

• I wouldn't call myself an expert on this, but I do not think they are conformally related. My only intuition on this is that such a map must mix time and space coordinates since the time coordinate on the cylinder needs to map to the radial (space) coordinate on plane. Weyl transformations do not do this. Mar 30 at 22:25

You cannot conformally map the Lorentzian cylinder $$\mathbb{R} \times S^{d-1}$$ to the Lorentzian plane $$\mathbb{R^{1,d-1}}$$even if you allow the domain and range to be missing finitely many points, e.g. by mapping to $$\mathbb{R^{1,d-1}} \backslash \{0\}$$. By conformal mapping, I mean a diffeomorphism for which the metric pulls back to a conformally equivalent metric (i.e. a metric related by Weyl transformation).
I found a simple argument. On $$\mathbb{R} \times S^{d-1}$$, we can choose two null rays that intersect multiple times. (For $$d=2$$, imagine shooting out two null rays in opposite directions from the same initial point, which will meet on the other side of the cylinder.) If there were a conformal mapping $$\mathbb{R} \times S^{d-1} \to \mathbb{R^{1,d-1}}$$, the image of these two null rays would yield two null rays in $$\mathbb{R^{1,d-1}}$$ that intersect multiple times, a contradiction. The same argument holds even if the conformal mapping excludes some points from the domain or range.
Finally, to address question (2): If there were a conformal mapping $$\mathbb{R} \times S^{d-1} \to \mathbb{R^{1,d-1}}$$ for Lorentzian signature, both spaces would have the same conformal group, i.e. $$SO(2,d)/\mathbb{Z}_2$$, known to be the conformal group of $$\mathbb{R^{1,d-1}}$$. However, since such a mapping cannot exist, it's not immediately obvious whether the the conformal group of the Lorentzian cylinder is as large that of the plane. Indeed, the isometry group of the Lorentzian cylinder is somewhat smaller than that of the plane.
It turns out that the conformal group of the Lorentzian cylinder $$\mathbb{R} \times S^{d-1}$$ is actually larger (though locally isomorphic) than that of the plane: it's the universal (infinitely sheeted) covering group of $$SO(2,d)/\mathbb{Z}_2$$. See Some remarks on conformal invariant theories on four-Lorentz manifolds (Tjoe-hian Go, 1974) for discussion of this conformal group, and see Global Conformal Invariance in Quantum Field Theory (Luscher and Mack, 1974) for the field-theoretic aspects.
To understand more concretely how the conformal group acts on Lorentzian $$\mathbb{R} \times S^{d-1}$$, first note that the usual action of $$SO(2,d)/\mathbb{Z}_2$$ on $$\mathbb{R}^{1,d-1}$$ is more carefully understood as an action on the conformal compactification, $$S^1 \times S^{d-1}$$. (See e.g. "A Mathematical Introduction to Conformal Field Theory," Schottenlohur, Chapter 2.) Then the conformal Killing fields on $$S^1 \times S^{d-1}$$ can be lifted to fields on the covering space $$R \times S^{d-1}$$, and flow along these corresponds to the action of the cover of $$SO(2,d) / \mathbb{Z}_2$$. Finally, for concrete expressions of the conformal Killing fields on Lorentzian $$\mathbb{R} \times S^{d-1}$$, see this thesis, Section 2.6.