The musical maps are smooth inverse bundle isomorphisms

Given: A Riemannian metric with coordinate matrix $G$.

Musical isomorphisms: The musical maps for $g$ are $v^{\mathrm{♭}}=g(v,\cdot )$ and its pointwise inverse ${\alpha }^{\mathrm{♯}}$, characterized by $g({\alpha }^{\mathrm{♯}},v)=\alpha (v)$ for all $v$. For the metric in def-riemannian-metric-and-riemannian-manifold, $v^{\mathrm{♭}}=0$ forces $g(v,v)=0$ and hence $v=0$. Thus $\mathrm{♭}$ is injective between equal-dimensional fibres and bijective; this gives the pointwise inverse. Smooth inverse bundle maps are proved in thm-the-musical-maps-are-smooth-inverse-bundle-isomorphisms. On a zero fibre both are the unique map.

Coordinate criterion for a riemannian metric: A tensor $g={\sum }_{i,j}{g}_{ij}d{x}^i\otimes d{x}^j$ is Riemannian exactly when its coordinate matrix $G=(g_{ij})$ has smooth entries and is symmetric positive definite. Under $J=\mathrm{\partial }x\mathrm{/}\mathrm{\partial }y$ it transforms by $G_y=J^T{G}_{x}J$.

Smoothness of a bundle map is equivalent to smooth local matrices: Let $\mathrm{\Phi }:E\to F$ be a fibrewise linear map over a smooth base map $f:M\to N$. Choose local frames $(e_1,\dots ,e_r)$ for $E$ on $U\subseteq M$ and $(u_1,\dots ,u_s)$ for $F$ on $V\subseteq N$ with $f(U)\subseteq V$. Then $\mathrm{\Phi }$ is smooth on $E{\mathrm{\mid }}_U$ if and only if there are smooth scalar functions $a_{ji}:U\to \mathbb{R}$ such that $\mathrm{\Phi }(e_i(p))={\sum }_{j=1}^s{a}_{ji}(p)u_j(f(p))$ for every $p\in U$.