Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 152 / 4

Let $q:N_{\mathbb{R}}\to N(\tau {)}_{\mathbb{R}}=N_{\mathbb{R}}/\mathrm{span}(\tau )$ be the quotient map. The images $q(\sigma )$ for $\tau ⪯\sigma$ satisfy the face and intersection axioms, so $\mathrm{Star}(\tau )$ is a fan.

To prove completeness, take $\bar{v}\in N(\tau {)}_{\mathbb{R}}$, choose a lift $v$, and choose $w$ in the relative interior of $\tau$. Since $\Sigma$ is complete, each $v+mw$ lies in some cone of $\Sigma$. There are finitely many cones, so one cone $\sigma$ contains $v+m_{j}w$ for an unbounded sequence $m_j$. Closedness gives $w\in \sigma$. Because $w$ lies in the relative interior of the cone $\tau$ and $\sigma \cap \tau$ is a face of $\tau$, this forces $\tau ⪯\sigma$. Finally,Thus every point of the quotient lies in the support of $\mathrm{Star}(\tau )$, proving that it is a complete fan.

Solved by gpt-5.6-sol high.

Let a complete fan in a $d$-dimensional real vector space have ray generators $v_1,\dots ,v_r$. Every cone of the fan lies in the positive hull $\mathrm{cone}(v_1,\dots ,v_r)$. If $r<d$, these vectors do not span the space. If $r=d$, either they still fail to span, or they are linearly independent and their positive hull is a proper strictly convex cone. In either case their positive hull cannot be the whole vector space, contrary to completeness. Hence $r\ge d+1$.

The bound is attained: take rays throughand cones generated by every proper subset of these rays. This is the simplex fan, which is complete and has $d+1$ rays. Therefore the minimal number is $d+1=\dim N_{\mathbb{R}}+1$.

Solved by gpt-5.6-sol high.