# Can a conformal map be turned into an isometry?

Let $f: (M, g) \to (M, g)$ be a conformal diffeomorphism of the riemannian manifold $(M, g)$, with

$$g(f(p))(Df(p) \cdot v_1, Df(p) \cdot v_2) = \mu^2(p) g(p)(v_1, v_2), \quad \forall p \in M, \, \forall v_1, v_2 \in T_p M,$$

for a certain function $\mu \in C^{\infty}(M)$. Is it possible to conformally change the metric of $M$ so as to $f$ become an isometry?

Explicitly, does there exist a metric $\tilde{g} = \alpha g$ in $M$ such that

$$\tilde{g}(f(p))(Df(p) \cdot v_1, Df(p) \cdot v_2) = \tilde{g}(p)(v_1, v_2), \quad \forall p \in M, \, \forall v_1, v_2 \in T_p M \, \text{ ?}$$

Plugging $\tilde{g} = \alpha g$ in the above equation, we obtain that $\alpha$ must satisfy

$$\alpha(p) = \mu^2(p) \alpha(f(p)), \quad \forall p \in M.$$

Can we continue?

Theorem. Let $C(M)$ be the conformal group of a Riemannian manifold $M$ with $dim(M)=n \ge 2$. If $M$ is not conformally equivalent to $S^n$ or $E^n$, then $C(M)$ is inessential, i.e. can be reduced to a group of isometries by a conformal change of metric.

This theorem has a long and complicated history, you can find its proof and the historic discussion in

J. Ferrand, The action of conformal transformations on a Riemannian manifold. Math. Ann. 304 (1996), no. 2, 277–291.

In view of this theorem, whenever $f: (M,g)\to (M,g)$ is a conformal automorphism without a fixed point, then there exists a positive function $\alpha$ on $M$ such that $f: (M,\alpha g)\to (M,\alpha g)$ is an isometry. I will prove it in the case when $(M,g)$ is conformal to the sphere and leave you the case of $E^n$ as it is similar.

Every conformal automorphism $f$ of the standard sphere which does not have a fixed point in $S^n$ has to have a fixed point $p$ in the unit ball $B^{n+1}$. (I am using the Poincare extension of conformal transformations of $S^n$ to the hyperbolic $n+1$-space in its unit ball model.) After conjugating $f$ via an automorphism $q$ of $S^n$ (sending $p$ to the center of the ball), we obtain $h=q f q^{-1}$ fixing the origin in $B^{n+1}$, which implies that $f\in O(n+1)$ and, thus, preserves the standard spherical metric $g_0$ on $S^n$. Now, use the fact that $g_0$ is conformal to $g$ and $q^*(g_0)$ is conformal to $g$ as well. qed

• What does it mean "reduced to a group of isometries by a conformal change of metric"? I don't know if the positive function $\alpha$ is valid for any conformal automorphism or there is a function $\alpha$ depending on the automorphism.
– José
Commented Sep 1 at 19:50
• @José: Reduced means that all conformal automorphisms act isometrically. If you are interested you should read her paper. Commented Sep 1 at 20:21
• Thank you, I will. This is terribly interesting.
– José
Commented Sep 1 at 20:42
• Is there any statement, in this sense, about the conformal group $C(M_1,M_2)$? I mean, is it expectable a statement of the form "if there exists a conformal diffeomorphism $f$ between two Riemannian spaces $(M_1,g_1)$, $(M_2,g_2)$ (under some hypothesis) then there exists some function $\alpha$ such that $f$ is an isometry between $(M_1, \alpha g_1)$ and $(M_2, g_2)$?
– José
Commented Sep 2 at 16:01
• @José: That much is obvious (unlike Ferrand's theorem); and $Conf(M_1, M_2)$ is not a group. Commented Sep 2 at 16:19

This not always possible if $f$ has a fixed point. Consider $R^n$ endowed with the euclidean metric. Let $f$ defined by $f(x)=2x$. The map if is conformal. Suppose that there is a function $g$ which turns the euclidean metric $\langle \cdot,\cdot\rangle$ to a metric invariant by $f$. For every $u,v$ in the tangent space of $0$, you would have $g(0)\langle 2u,2v\rangle =g(0)\langle u,v\rangle$. This is clearly impossible.

• I don't get where you took the $n$ from. But yes, if $f$ has a fixed point, it is not always possible. What if $f$ has no fixed points? Commented Mar 3, 2016 at 17:12