# How do conformal maps affect curvature?

Let $(\overline{M}^{n+1}, \langle \cdot, \cdot \rangle)$ be a riemannian manifold with riemannian connection $\overline{\nabla}$ and consider $M^n \subset \overline{M}$ an orientable hypersurface with unit normal vector field $\nu: M \to T \overline{M}$. Given a conformal diffeomorphism $f: \overline{M} \to \overline{M}$ say, with conformal factor $\mu^2 \in C^{\infty}(\overline{M}, \mathbb{R}_+^*)$, i.e.,

\begin{align*} \langle Df(p) \cdot v_1, Df(p) \cdot v_2\rangle = \mu^2(p) \langle v_1, v_2 \rangle, \quad \forall p \in \overline{M}, \, \forall v_1, v_2 \in T_p \overline{M}, \end{align*} how can we relate the principal curvatures of $M$ at a point $p$ with those of $f(M)$ at the point $f(p)?$

If $\mu = 1$, i.e., if $f$ is an isometry, can we say that the correspondent principal curvatures are equal?

I'm doubtful that there's a simple relationship between the principal curvatures. Let's compare the shape operators. First, observe that the unit normal vector field $\tilde{\nu}_{f(p)}$ to $f(M)$ at $f(p)$ is given by $$\tilde{\nu}_{f(p)} = \frac{1}{\mu(p)} df_p(\nu)$$ Taking $\overline{\nabla}$ of both sides yields $$\overline{\nabla}_{df(\cdot)} \tilde{\nu} = - \frac{ \overline{\nabla}_{\cdot} \mu}{\mu^2} df(\nu) + \frac{1}{\mu} \overline{\nabla}_{\cdot} \big( df(\nu) \big) \\ =- \frac{ \overline{\nabla} \mu}{\mu} \tilde{\nu} + \frac{1}{\mu} \big(\overline{\nabla} df \big) (\nu) + \frac{1}{\mu} df \big( \overline{\nabla} \nu \big)\\ = - \frac{ \overline{\nabla} \mu}{\mu} \tilde{\nu} + \frac{1}{\mu} \big(\overline{\nabla}_\nu df \big) + \frac{1}{\mu} df \big( \overline{\nabla} \nu \big)$$ where the last equality follows from the fact that $\overline{\nabla} df = Hess(f)$ is symmetric. Note that by taking $\overline{\nabla}$ of $\langle df(\nu), df(\nu) \rangle = \mu^2$, one can see that $\frac{ \overline{\nabla} \mu}{\mu} \tilde{\nu}$ is the normal component of $\frac{1}{\mu} \big(\overline{\nabla}_\nu df \big)(\cdot) = \frac{1}{\mu} \big(\overline{\nabla}_{\cdot} df \big)(\nu)$. Therefore, the shape operator $\tilde{S} = \overline{\nabla} \tilde{\nu} : Tf(M) \to Tf(M)$ is the self-adjoint endomorphism given by $$\tilde{S} \circ df = \frac{1}{\mu} \Big( proj_{Tf(M)} \circ \overline{\nabla}_\nu df + df \circ S \Big)$$ $$\text{i.e.}\quad \tilde{S} \big(df(X)\big)= \frac{1}{\mu} \Big( \big( proj_{Tf(M)} \circ \overline{\nabla}_\nu df \big)(X) + df (S (X)) \Big) \qquad \forall X \in TM$$ Without extra assumptions on $Hess(f)$, I don't see how to relate the eigenvalues of $\tilde{S}$ to those of $S$.

Perhaps an informative example might be to consider the Mobius transformation $$z \mapsto \frac{z-i}{z+i}$$ in a neighborhood of the real line in $\mathbb{C}$. This is a conformal transformation that takes the real line which has principal curvature 0 in $\mathbb{C}$ to a circle which has nonzero principal curvature.

• When $\mu$ is a constant function, the hessian of $f$ vanishes and the above formula shows that the principal curvatures differ by a factor of $\mu$. In particular, if $f$ is an isometry then the principal curvatures are equal. – Maxwell Stolarski Nov 22 '17 at 20:01