= Solution
A <fiber metric> on a real <vector bundle> $E\to M$ is a smoothly varying positive-definite <inner product> on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth <partition of unity> $\{\rho_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a <locally finite family of subsets> of supports, and $\operatorname{supp}\rho_i\subset U_i$; the usual Hausdorff second-countable <smooth manifold> hypotheses ensure this theorem applies. Transfer the Euclidean <inner product> to each local <vector bundle trivialization>, obtaining $h_i$, and set
$$
h_x(v,w)=\sum_i\rho_i(x)(h_i)_x(v,w).
$$
Each weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every <vector bundle trivialization>. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a <fiber metric>, with no orientability or triviality assumption.
A <vector bundle morphism> covering the identity is a smooth map $F:E'\to E''$ that preserves base points and is a <linear map> on each fiber. Its induced map on the <module of smooth sections> is $s\mapsto F\circ s$, and is $C^\infty(M)$-linear. We prove the converse by constructing <bundle morphisms from maps of smooth sections>.
First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a <smooth bump function> $\chi$ supported there with $\chi=1$ near $x$. Then $\chi s=0$, so $\chi\alpha(s)=\alpha(\chi s)=0$, and hence $\alpha(s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.
Choose a local frame $e_1,\ldots,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j e_j$ on $V$. A second bump extends each $a_j$ to a global smooth function $\widetilde a_j$ agreeing near $x$. By locality and $C^\infty(M)$-linearity,
$$
\alpha(s)(x)=\sum_j\widetilde a_j(x)\alpha(t_j)(x)=0.
$$
Every fiber vector $v\in E'_x$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha(s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are <linear maps>, and locally their matrix columns are the smooth sections $\alpha(t_j)$ in a frame of $E''$. Thus $F$ is smooth, is a <vector bundle morphism>, and satisfies $\alpha(s)=F\circ s$. Fiberwise evaluation also proves uniqueness.
Finally apply the <fiber metric> construction to the <tangent bundle>. A <Riemannian metric> $g$ defines the <musical isomorphism>
$$
\boxed{\flat_g:TM\longrightarrow T^*M,\qquad v\longmapsto g(v,\cdot).}
$$
Positive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence \b[$TM$ and $T^*M$ are isomorphic as real smooth vector bundles on every such manifold.] The isomorphism depends on the chosen metric and is not canonical.
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