Gaussian curvature under a conformal map

Question

This is an exercise from GTM 275, Differential Geometry by Loring W. Tu, page 94.

Let $T:M \rightarrow M'$ be a diffeomorphism of Riemannian manifolds of dimension 2. Suppose at each point $p \in M$, there is a positive number $a(p)$ such that

$${\left\langle {{T_ * }v,{T_ * }w} \right\rangle _{M',T(p)}} = a(p){\left\langle {v,w} \right\rangle _{M,p}}$$

for all $u,v \in T_p(M)$. Find the relation between the Gaussian curvatures of $M$ and $M'$.

In this book, the Gaussian curvature $K$ at a point $p$ of a Riemannian 2-manifold $M$ is defined to be

$${K_p} = \left\langle {{R_p}(u,v)v,u} \right\rangle $$

for any orthonormal basis $u,v$ for the tangent plane $T_pM$.

I'm stuck on how to express the relation between those quantities on $M$ and $M'$ by $T_*$ and $T^*$ without the viewpoint of identifying them as one manifold.

Any help will be appreciated.

Answer 1

For this problem, I find it more helpful to think in terms of curvature forms than directly using the definition $K=\langle R(u,v)v,u\rangle$.

Write $\lambda(p)=1/\sqrt{a(p)}$; to distinguish between the same objects defined on the two manifolds, we mark everything associated with $M'$ with a tilde. At any point in $M$, let $e_1,e_2$ be a local orthonormal frame, then by the conformality of $T$ it is easy to show that $\{\lambda T_*e_i\}=:\{\tilde e_i\}$ is an orthonormal frame on $M'$. Thus, if $\{\theta^i\}$ is the associated dual frame on $M$, $\{\tilde \theta^i\}=\left\{\frac{1}{\lambda}(T^{-1})^*\theta^i\right\}$ is the dual frame corresponding to $\{\tilde e_i\}$.

Now consider the connection forms $\omega$ of the Riemannian connection on $M$, with respect to the orthonormal frame $\{e_i\}$. They satisfy the structural equation $$d\theta+\omega\wedge\theta=0.$$ We can "pull it back" to $M'$ by applying $\frac{1}{\lambda}(T^{-1})^*$ to both sides: \begin{align*} d\left(\frac{1}{\lambda}(T^{-1})^*\theta\right)+(T^{-1})^*\omega\wedge\frac{1}{\lambda}(T^{-1})^*\theta&=0\\ \Longrightarrow d\tilde\theta+(T^{-1})^*\omega\wedge\tilde\theta&=0. \end{align*} Since the connection forms $\tilde\omega$ of the Riemannian connection on $M'$ also satisfy $d\tilde\theta+\tilde\omega\wedge\tilde\theta=0$, we must have $\tilde\omega=(T^{-1})^*\omega$. In particular, $\tilde\omega^1_2=(T^{-1})^*\omega^1_2$, so $$\tilde\Omega^1_2=d\tilde\omega^1_2=(T^{-1})^*d\omega^1_2=(T^{-1})^*\Omega^1_2.$$

All that remains is then a direct calculation: \begin{align*}\tilde K&=\tilde\Omega^1_2(\tilde e_1,\tilde e_2)=(T^{-1})^*\Omega^1_2(\lambda T_*e_1,\lambda T_*e_2)\\ &=\Omega^1_2(\lambda e_1,\lambda e_2)=\lambda^2\Omega^1_2(e_1,e_2)=\frac{1}{a}K. \end{align*}