= Solution
\b[Vector identity.] Work in <Cartesian coordinates> and use the <Einstein summation convention>. Contraction of the two <Levi-Civita symbols> gives
$$
\begin{aligned}
[\mathbf u\times(\nabla\times\mathbf u)]_i&=\epsilon_{ijk}u_j\epsilon_{klm}\partial_lu_m\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j\partial_lu_m\\
&=u_j\partial_iu_j-u_j\partial_ju_i\\
&=\tfrac12\partial_i(u_ju_j)-[(\mathbf u\cdot\nabla)\mathbf u]_i.
\end{aligned}
$$
Thus the <convective acceleration identity> is
$$
\boxed{(\mathbf u\cdot\nabla)\mathbf u=\frac12\nabla|\mathbf u|^2-\mathbf u\times(\nabla\times\mathbf u).}
$$
Each term is a geometrically defined <vector field>: the <directional derivative> of a vector along itself, the <gradient> of a <scalar field>, and a <cross product> with the <curl>. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate <basis>, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.
\b[Spherical metric.] Let
$$
R(r,\theta,\phi)=(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta).
$$
Differentiating gives the coordinate <tangent vectors>
$$
\begin{aligned}
R_r&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\
R_\theta&=r(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\
R_\phi&=r\sin\theta(-\sin\phi,\cos\phi,0).
\end{aligned}
$$
Their pairwise <dot products> vanish, proving that <spherical coordinates> are <orthogonal coordinates>. Their lengths are the <scale factors of orthogonal coordinates>:
$$
\boxed{h_r=1,\qquad h_\theta=r,\qquad h_\phi=r\sin\theta,\qquad ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta\,d\phi^2.}
$$
These formulas use the usual chart $r>0,0<\theta<\pi$; the origin and polar axis are coordinate singularities.
\b[Azimuthal field.] Write the speed as $U(r)$ to distinguish it from the <velocity field> $\mathbf u=U(r)e_\phi$. Using the <curl in spherical coordinates>,
$$
\nabla\times\mathbf u=\frac{U\cot\theta}{r}e_r-\frac{U+rU'}r e_\theta.
$$
Also $\frac12\nabla|\mathbf u|^2=UU'e_r$. Since $e_\phi\times e_r=e_\theta$ and $e_\phi\times e_\theta=-e_r$,
$$
\mathbf u\times(\nabla\times\mathbf u)=\frac{U(U+rU')}r e_r+\frac{U^2\cot\theta}r e_\theta.
$$
The $UU'$ terms cancel in the <convective acceleration identity>, giving
$$
\boxed{[(\mathbf u\cdot\nabla)\mathbf u]_r=-\frac{U^2}r,\quad[(\mathbf u\cdot\nabla)\mathbf u]_\theta=-\frac{U^2\cot\theta}r,\quad[(\mathbf u\cdot\nabla)\mathbf u]_\phi=0.}
$$
This <centripetal acceleration of an azimuthal flow> points toward the polar axis. A streamline is a <circle> of cylindrical radius $r\sin\theta$, so its acceleration is $-U^2e_\rho/(r\sin\theta)$, with $e_\rho=\sin\theta\,e_r+\cos\theta\,e_\theta$. This independent geometric argument recovers both nonzero components and explains why $U'$ cannot appear: the particle travels at fixed $r$.
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