Let $M$ be a smooth manifold of dimension $\ge 3$, equipped with a conformal structure (or a Riemannian metric). Then, the group of conformal diffeomorphisms is a finite dimensional Lie group.

A proof of this theorem can be found in "Transformation groups in differential geometry" by Kobayashi (Theorem 6.1, pg 143).

Are there other proofs of this claim (perhaps simpler or more readable)?

Kobayashi's theorem also bounds the dimension of the conformal group by $\frac{1}{2}(n+1)(n+2)$, where $\dim M=n$. If I am only interested to know the group is a finite dim Lie group, and don't care about bounding its dimension, is there a simpler proof?

(I am also ready to assume $M$ is compact).

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    $\begingroup$ You might try Gromov and D'Ambra, ihes.fr/~gromov/PDF/1[74].pdf $\endgroup$
    – Ben McKay
    Jan 21, 2018 at 9:42
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    $\begingroup$ It will be not be easy to assemble an actual proof from the various remarks about conformal geometry in Gromov and D'Ambra that are scattered throughout that article, or even to find the crucial information that distinguishes the cases $n=2$ and $n\ge3$. They do remark that the theorem 'appears to be due to É. Cartan', but that's about as close as they come to an actual reference in the text. They even refer to Kobayashi's book TGiDG at a crucial juncture in their discussion, so I'm afraid Ben's advice is going to lead Asaf back to Kobayashi when all is said and done. $\endgroup$ Jan 21, 2018 at 12:30
  • $\begingroup$ I tried to put together a simpler version of Kobayashi's argument, and fixing what seemed to me to be some missing details, in this paper: arxiv.org/abs/2302.14457 $\endgroup$
    – Ben McKay
    Mar 29, 2023 at 18:31

1 Answer 1


Answers to the OP's question depend on where the OP is willing to start. To prove that an abstractly defined group is (i.e., has the structure of) a Lie group, one will have to use something nontrivial, as this is not a trivial task, in general.

For example, É. Cartan's statement that the set of (smooth?, $C^k$?, analytic?) diffeomorphims of a manifold $M$ that preserve a (smooth? $C^k$?, analytic?) parallelization is (i.e., has a natural structure as) a Lie group, was taken to be obvious by him, but was not actually proved to modern standards of rigor until much later. It was the basis of all of his results about the group of symmetries of various geometric structures (including conformal geometry in dimensions at least $3$) being Lie groups. In particular, it implies that the group of self-isometries of a pseudo-Riemannian manifold is a Lie group.

If the OP is willing to assume that the group of self-isometries of a Riemannian manifold is a Lie group, then one can give an argument that the group of conformal symmetries of a manifold of dimension at least $3$ is a Lie group as well using only this, and, moreover, one can immediately see, based on Riemannian geometry, where the assumption that the dimension is at least 3 enters the picture. The idea is as follows:

If $g$ is a metric on an $n$-manifold $M$ and $u$ is a smooth function on $M$, then consider the expression $\mathrm{Ric}(e^{2u}g)$, that computes the Ricci curvature of the conformal metric $g' = e^{2u}g$: $$ \mathrm{Ric}(e^{2u}g) = \mathrm{Ric}(g)-(n{-}2)\left(\mathrm{d}u^2+|\nabla u|^2g\right) - (n{-}2)\,\mathrm{Hess}(u) + (\Delta u)\,g, $$ where $\mathrm{Hess}(u) = \nabla(\mathrm{d}u)$ is a quadratic form that expresses the second covariant derivatives of $u$ and whose trace is $-\Delta u$. Thus, when $n\ge 3$, if one specifies the $1$-jet of $u$ at $p$, there will be a unique extension to a $2$-jet of $u$ at $p$ such that $\mathrm{Ric}(e^{2u}g)(p)=0$. (By contrast, when $n=2$, $\mathrm{Ric}(e^{2u}g) = (2K(g)+\Delta u)\,g$, where $K(g)$ is the Gauss curvature of $g$, so specifying $\mathrm{Ric}(e^{2u}g)(p)=0$ only detemines one coefficient of the $2$-jet of $u$ at $p$ instead of all $3$ coefficients.)

This observation can be used to show that, when $n\ge 3$, if we let $C\to M$ be the $\mathbb{R}^+$-bundle of all multiples of $g$, which depends only on the conformal class of $g$, then, on $\pi:J^1C\to M$, the bundle of $1$-jets of sections of $C$, there is a canonical $n$-plane bundle $D\subset T(J^1C)$ that is transverse to the fibers of $\pi:J^1C\to M$ and is such that the $1$-jet graph of a conformal metric $e^{2u}g$ is tangent to the $n$-plane bundle $D$ over a point $p\in M$ if and only if $\mathrm{Ric}(e^{2u}g)(p)=0$. Any conformal transformation of $\bigl(M,[g]\bigr)$ canonically induces a transformation of $J^1C$ that preserves this $n$-plane field $D$. This defines a canonical splitting $T(J^1C) = D\oplus \mathrm{ker}(\pi')$ that can then be used to construct a canonical metric $h$ on $J^1C$ that is preserved by any such transformation. Thus, the group of $[g]$-conformal transformations on $M$ is embedded in the group of self-isometries of $(J^1C,h)$ and hence is embedded as a (closed, it turns out) subgroup of a Lie group. Thus, in this way, it inherits the structure of a Lie group.

By the way, for entirely different reasons, the group of conformal transformations in dimension $2$ is also a Lie group, but it cannot be proved by the method above. Instead, one uses the Uniformization Theorem, which shows that any simply-connected Riemannian $2$-manifold is conformally equivalent to either the disk, the plane, or the sphere (with their standard conformal structures) and then appeals to elementary complex analysis to show that their conformal symmetries are Lie groups.

Finally, I cannot resist pointing out that the group of conformal transformations in dimension $1$ is, nowadays, not regarded as a Lie group, although Lie (and Cartan) regarded it as a Lie group, just infinite dimensional.

  • $\begingroup$ Cartan — who credits the theorem to Liouville for $M=\mathrm E_3$ and Lie for $\mathrm E_n$, projective compactification understood... — emphasizes that for $\mathrm E_2$ it’s also an “infinite” group, right? (Encyclopédie article, p. 30.) $\endgroup$ Jan 22, 2018 at 0:24
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    $\begingroup$ @FrancoisZiegler: That is the language he uses, but the meaning of 'Lie group' has changed since then. For Lie (and Cartan as well) 'Lie group' meant what we would now call a set of 'transformations' (i.e., locally defined diffeomorphisms) with the property of being closed under inverse and composition (where defined) whose elements are characterized as the diffeomorphisms that satisfy a specified system of differential equations (e.g., volume preserving, metric preserving, etc.) 'Finite' meant that an element is locally determined by its Taylor series at one point to some given finite order. $\endgroup$ Jan 22, 2018 at 1:39
  • $\begingroup$ (cont.) Thus, by 'the group of conformal transformations of the plane', Cartan meant the 'group' of diffeomorphisms $f:U\to V$, where $U$ and $V$ are open subsets of the plane, regarded as $\mathbb{C}$, that are either holomorphic or anti-holomorphic, which is an infinite Lie group (in the old sense). Nowadays, when we write 'the group of conformal transformations of the plane', we mean the set of $f:\mathbb{C}\to\mathbb{C}$ that are conformal diffeomorphisms, which is a Lie group (in the new sense) of (real) dimension 4 (with two components, corresponding to orientation preserving or not). $\endgroup$ Jan 22, 2018 at 1:57
  • $\begingroup$ cf en.wikipedia.org/wiki/Pseudogroup $\endgroup$
    – David Roberts
    Jan 22, 2018 at 6:33
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    $\begingroup$ @DavidRoberts: Yes, the notion of 'pseudogroup' was introduced in the mid-20th century as a way to formalize the original ideas of Lie and Cartan, which were worked out before the notion of manifold was formalized as what was then called 'analysis situs' by Poincaré. When Cartan wrote that Encylopédie article in 1915, the formal notion of a pseudogroup was not available as something distinct from what we now call a Lie group. $\endgroup$ Jan 22, 2018 at 9:18

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