This is probably very elementary, yet I am not sure how to approach this.

Suppose $f$ is a conformal map between Riemannian manifolds, which maps geodesics to geodesics. (i.e if $\alpha$ is a geodesic, then $f \circ \alpha$ is a geodesic).

Is it true that $f$ must be a "scaled isometry"? (Does there exist a constant $\lambda$ such that $df\in \lambda O(n)$ )

Note: We need to assume the domain is connected, of course.

  • 1
    $\begingroup$ This sounds reasonable. Classically, maps sending geodesics to geodesics (respecting their natural parameterization) are called affine. Given two connections $\nabla, \nabla'$ the condition that the identity map between them is affine is that $(\nabla-\nabla')_X X=0$ for every vector field $X$ on the manifold. The condition for conformality of two metrics in terms of the LC connections is more complicated, see e.g. anhngq.wordpress.com/2010/05/16/…. Now, you can put these two conditions together and see what you get. $\endgroup$ Feb 10, 2017 at 6:37
  • $\begingroup$ One more thing about conformal maps in the case of self-maps: With two exceptions (flat $R^n$ and the round sphere) a conformal self-map of a Riemannian manifold $(M,g)\to (M,g)$ becomes an isometric self map after conformal rescaling of the metric (by a nonconstant function of course) $(M, \lambda(x)g)\to(M, \lambda(x)g)$ . $\endgroup$ Feb 10, 2017 at 6:40
  • $\begingroup$ @MoisheCohen Thanks! I was able to complete the argument you sketched (reduction of the general case to a general diffeomorphism, and then reduction of this case to the identity map). BTW, I am more used to the term "affine" as describing a map which preserve the connection, not just its geodesics. This is a stronger condition, as can be observed for instance in the answer to this question: math.stackexchange.com/questions/1411613/… $\endgroup$ Feb 12, 2017 at 10:03

1 Answer 1


It appears the answer is positive, i.e every conformal geodesics-preserving map is a scaled isometry. Here is a proof (based on Moishe Cohen's comment):

First, we prove this for the special case of the identity map $\operatorname{Id}:(M,g) \to (M,\tilde g)$. That is, suppose that every $g$-geodesic is also a $\tilde g$-geodesic, and that $\tilde g =e^{2f}g$ for some $f \in C^{\infty}(M)$.


$\tilde \nabla _X Y = \nabla _X Y + (X f )Y + (Y f )X - g(X,Y) \operatorname{grad}f \tag{1}$.

It's easy to see that the condition "$g$-geodesic $\Rightarrow$ $\tilde g$-geodesic" is equivalent to $\tilde \nabla _X X = \nabla _X X$ for every vector field $X$.

Combining this with $(1)$, we get $$2(X f )X - g(X,X) \operatorname{grad}f=0 \tag{2}$$ (for every $X$).

Now, let $p \in M$. We want to prove $\operatorname{grad}f(p) =0$.

Inserting $X=\operatorname{grad}f$ into $(2)$ we get

$$ \|\operatorname{grad}f\|^2 \operatorname{grad}f=0,$$

so $\operatorname{grad}f=0$. Since $M$ is connected $f$ must be constant.

Reduction of the general case to the case of the identity map:

Let $\phi:(M,g) \to (N,h)$ be a conformal geodesic map. First, suppose that $\phi$ is a diffeomorphism.

Since $\phi^{-1}:(N,h) \to (M,\phi^*h)$ is an isometry, it is confromal and geodesic. Hence, the composition

$$ \operatorname{Id}=\phi^{-1} \circ \phi:(M,g) \to (M,\phi^*h)$$ is confromal and geodesic, hence by the special case, it is a scaled isometry, i.e $\phi^*h=\lambda g$ for some $\lambda \in \mathbb{R}$. This shows $\phi$ is a scaled isometry.

If $\phi$ was not a diffeomorphism, we proceed as follows:

$\phi$ is a local diffeomorphism (inverse function theorem), hence it is a "local scaled isometry", in other words it is conformal with scaling factor which is a locally constant function (by the reasoning above for the diffeomorphism case). Since $M$ is connected, this scaling factor must be globally constant.


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