= Solution
A <Riemannian metric> gives a smooth fibrewise isomorphism $g^\flat:TM\to T^*M$, defined by $g^\flat(v)=g(v,\cdot)$. Its inverse is smooth because the inverse of a smoothly varying invertible matrix is smooth. Define the <Riemannian gradient> by
$$
\boxed{V_\sigma=(g^\flat)^{-1}(d\sigma)=\operatorname{grad}_g\sigma.}
$$
It is a smooth <vector field> satisfying $g(V_\sigma,Y)=d\sigma(Y)=Y(\sigma)$. Nondegeneracy also gives uniqueness. In coordinates its components are $V_\sigma^i=\sum_jg^{ij}\partial_j\sigma$.
Multiplication of $g$ by the globally defined smooth positive function $e^{2\sigma}$ gives a smooth symmetric two-tensor $\widetilde g$. For any nonzero tangent vector $v$,
$$
\widetilde g(v,v)=e^{2\sigma}g(v,v)>0.
$$
Thus \b[$\widetilde g=e^{2\sigma}g$ is a Riemannian metric]. This <conformal rescaling of a Riemannian metric> changes lengths but preserves angles.
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