Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 17 / 5 / 1 / Solution

A Riemannian metric gives a smooth fibrewise isomorphism $g^{♭}:TM\to T^{\ast }M$, defined by $g^{♭}(v)=g(v,\cdot )$. Its inverse is smooth because the inverse of a smoothly varying invertible matrix is smooth. Define the Riemannian gradient byIt is a smooth vector field satisfying $g(V_{\sigma },Y)=d\sigma (Y)=Y(\sigma )$. Nondegeneracy also gives uniqueness. In coordinates its components are $V_{\sigma }^i={\sum }_j{g}^{ij}{\partial }_j\sigma$.

Multiplication of $g$ by the globally defined smooth positive function $e^{2\sigma }$ gives a smooth symmetric two-tensor $\tilde{g}$. For any nonzero tangent vector $v$,Thus $\tilde{g}=e^{2\sigma }g$ is a Riemannian metric. This conformal rescaling of a Riemannian metric changes lengths but preserves angles.