For $t_j=j/3$, the three central radial directions are proportional to $(1,t_j,t_j^2)$. Their determinant is the nonzero Vandermonde product
$$
(t_2-t_1)(t_3-t_1)(t_3-t_2).
$$
Choose $c>0$ small enough that every triple $\omega_j\in\Sigma_j$ still has determinant bounded away from zero. An invertible linear transformation sends the three central directions to the standard basis; it sends the given tubes to comparable tubes with directions in fixed small neighborhoods of $e_1,e_2,e_3$ and distorts volume and dimensions by fixed factors.

Applying part a after this transformation gives, uniformly in the three families,
$$
\left\|\prod_{j=1}^3
\left(\sum_{T\in\mathcal T_j}\chi_T\right)^{1/3}
\right\|_{3/2}
\lesssim_\epsilon R^{1+\epsilon}
\prod_{j=1}^3|\mathcal T_j|^{1/3}.
$$
This is the required moment-curve version; its uniformity is precisely the transversality supplied by the <moment curve> and the Vandermonde determinant.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 3 / b / Solution

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The degree-$d$ moment curve is $\gamma(t)=(t,t^2,\ldots,t^d)$. Tangent or radial directions from separated parameter intervals are quantitatively transverse by the Vandermonde determinant.


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