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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?

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  • $\begingroup$ It is not unthinkable that $\mu(p)\ne1$ at a fixed point $p$: just try a homothety of the plane. Questions like this are better suited for MSE. $\endgroup$
    – Alex Degtyarev
    Commented Mar 3, 2016 at 17:00
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    $\begingroup$ If $f$ has fixed points (pre more generally finite orbits), you need the eigenvalues of $Tf$ to have modulus 1. For instance $x\mapsto 2x$ is conformal on $\mathbb{R}^n$ but isn't an isometry for an metric. $\endgroup$
    – Thomas Richard
    Commented Mar 3, 2016 at 17:07
  • $\begingroup$ In this level of generality, no. But with a few more assumptions perhaps yes. $\endgroup$
    – Ryan Budney
    Commented Mar 3, 2016 at 18:51

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The conformal map $z \mapsto z/(1+z)$ on the Riemann sphere fixes the origin and leaves every tangent vector at the origin invariant. A non-trivial isometry can't do that, because it would then commute with the exponential map.

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