Let $(M, g)$ be a compact 2-dimensional Riemannian manifold with genus $\ge 1$. Can $M$ has a conformal Killing vector field $X$ other than Killing vector fields? That is, $L_X g = (\mathrm{div} X) g$ and $\mathrm{div} X\neq 0$.

Note that for genus $\ge 2$, $M$ has no Killing vector fields by this post

Compact surface with genus$\geq 2$ with Killing field

For genus 1, the flat torus has nontrivial Killing vector fields. Does it have other conformal Killing vector fields?

  • $\begingroup$ For the flat torus, the result can be considered in terms of the Riemann surface structure, i.e. looking for holomorphic vector fields on the complex plane invariant under a lattice of translations. Clearly they are multiples $f(z)\partial_z$ of the translation field. But then $f(z)$ is translation invariant, so constant. $\endgroup$
    – Ben McKay
    Jun 1, 2023 at 12:29

2 Answers 2


In 2 dimensions, conformal Killing requires in index notation $$ \nabla_a X_b + \nabla_b X_a = \nabla^c X_c g_{ab} $$ Taking the divergence with respect to the index $b$, you find $$ \Delta_g X_a + \mathrm{Ric}_{ab}X^b = 0 \label{1}\tag{1}$$

Taking the product against $X$ and integrating over the whole manifold you find $$ \int_M - \|\nabla X\|^2 + \mathrm{Ric}(X,X) ~\mathrm{dvol} = 0 $$ As a consequence you have the following generalization of Yano's theorem in reference 1 (also see remark below).

Theorem If $(M,g)$ is a two dimensional compact Riemannian manifold with non-positive Ricci curvature, and $\xi$ a conformal Killing field, then $\xi$ is parallel; furthermore, if $M$ is not Ricci-flat, then $\xi \equiv 0$.

The Theorem above requires that the Ricci curvature be signed; in two dimensions, this is equivalent to having the scalar curvature be signed. Luckily for us, we can make use of the Kazdan-Warner Theorems (see 2 below) which, among other things, state that in non-positive Euler characteristic, every metric is conformally equivalent to one that has constant curvature. As the condition that $\xi$ is a conformal Killing field is conformally invariant, this shows that

Theorem On a closed two dimensional Riemannian manifold with genus $\geq 1$, (a) all conformal Killing vector fields are Killing (in fact parallel) and (b) the conformal Killing field can only be non-trivial when genus $=1$.

Remark That equation \eqref{1} holds for Killing vector fields is one of the main tricks of the Bochner/Yano technique for studying vector fields on manifolds (an analogous trick can also be used to study harmonic vector fields). One of the many coincidences for studying the conformal geometry in 2 dimensions is that equation \eqref{1} also holds for conformal Killing fields in 2D (it does not hold in general in higher dimensions). Yano proved his theorem (ref. 1 below) for Killing fields in general dimensions; in the above we just note that the same argument can be applied also for conformal Killing fields in 2D.

  1. Yano, Kentaro, On harmonic and Killing vector fields, Ann. Math. (2) 55, 38-45 (1952). ZBL0046.15603.
  2. Kazdan, Jerry L.; Warner, Frank W., Curvature functions for compact 2-manifolds, Ann. Math. (2) 99, 14-47 (1974). ZBL0273.53034.

By the uniformization theorem, there is a unique conformally equivalent constant curvature metric up to scaling which is equivalent to a Riemann surface. A conformal Killing vector field would give a 1-parameter family of conformal maps of the corresponding Riemann surface. In the torus/euclidean case, conformal maps lift to conformal maps of the universal cover $\mathbb{C}$ which are similarities which commute with the covering translations (which are translations of $\mathbb{C}$), and hence are also translations. So the conformal maps are isometries. In the negatively curved/hyperbolic case, the conformal map lifts to conformal maps of $\mathbb{H}^2$ (the unit disk in $\mathbb{C}$) and hence are also isometries by the Schwarz-Pick theorem. However the only such isometry commuting with the covering translations is trivial. So there are no non-trivial conformal Killing maps in negative euler characteristic.


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