Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 152 (Toric Geometry)

For $m=(u,v)\in M\cong {\mathbb{Z}}^2$, membership in the dual cone is equivalent toThe Hilbert basis of a rational cone is $(0,1),(1,1),(2,1),(3,1)$. Therefore the coordinate ring of an affine toric variety is

The diagonal action of ${\mu }_3$ on ${\mathbb{A}}_{x,y}^2$ fixes precisely the monomials whose total degree is divisible by three. Its invariant ring is generated byThe assignmentidentifies this invariant ring with the ring in part i. Taking spectra gives the diagonal cyclic quotient singularity of order three

Insert the ray generated by $w=(0,1)$ between $v_1=(1,0)$ and $v_2=(-1,3)$. Sincethe subdivided fan is smooth and gives a toric resolution of singularities. The unique exceptional divisor is $D_w$. Asthe self-intersection formula for a toric surface divisor gives $D_w^2=-3$.

Let $w$ be the primitive generator of the new ray $\tau$, and let $u,v$ be its adjacent primitive rays in the smooth subdivided fan. Smoothness gives a unique integer $a$ withBecause $\tau$ is new, it lies in the interior of an original two-dimensional cone. Its adjacent refined rays lie on opposite sides of it inside that cone, which forces $a>0$. Therefore $D_{\tau }^2<0$.

Let $v_1=(1,0)$ and $v_2=(1,m)$ generate the distinguished cone. If a complete fan had exactly one further ray with primitive generator $w=(a,b)$ and were smooth away from $x_{\sigma }$, the other two cones would be smooth. Completeness puts $w$ on the other side of the two boundary rays, so after choosing cyclic orientation $b=-1$. Smoothness would requirehence $ma=-2$. This has no integer solution for $m\ge 3$. Thus no proper toric variety can satisfy the three conditions from the PDF.

For the rays $v_1=(1,0)$, $v_2=(0,1)$, and $v_3=(-a,-b)$, the determinants of consecutive pairs have absolute valuesBy the smoothness criterion for a toric variety, $\mathbb{P}(1,a,b)$ is smooth exactly when $a=b=1$.

The toric divisor class sequence isThe two characters giveThusgenerated by $[D_3]$.

An invariant Weil divisor is Cartier precisely when its local linear support functions are integral on every maximal cone. Applying this to the three cones shows that a class $n[D_3]$ is Cartier exactly when both $a$ and $b$ divide $n$. Since $\gcd (a,b)=1$, this means $ab\mid n$. Thereforeand its map to the class group identifies it with $ab\mathbb{Z}\subseteq \mathbb{Z}$. This is the divisor class and Picard groups of a weighted projective plane.

The Cox ring has one variable for each ray and is $\mathbb{C}[x,y,z]$, graded byThe irrelevant locus is the common zero of all three variables. The quasitorus associated with the class group is ${\mathbb{C}}^{\ast }$ and acts byThe Cox construction therefore identifies the closed points of $\mathbb{P}(1,a,b)$ with

A toric morphism is induced by a lattice map carrying each source cone into a target cone. Compose a hypothetical fan map from the fan of ${\mathbb{P}}^2$ to the product fan with either coordinate projection to the fan of ${\mathbb{P}}^1$. If the three primitive source rays are $r_1,r_2,r_3$ with $r_1+r_2+r_3=0$, their scalar images must have each adjacent pair in one half-line. This is impossible for three numbers summing to zero unless all are zero. Both coordinate projections of the lattice map vanish, so the map itself is zero and the toric morphism is constant. This proves no nonconstant toric morphism from the projective plane to the product of projective lines.

For $D=k{D}_1$, the lattice polytope of a toric divisor consists of $(m_1,m_2)$ satisfyingThus $m_2=0$ and $-k\le m_1\le 0$. Its lattice points index a basis of global sections, so

For every maximal cone, one endpoint of the interval of lattice points from part a realizes the corresponding Cartier datum. The toric basepoint-free criterion therefore makes $\mathcal{O}(k{D}_1)$ basepoint-free for $k\ge 0$. Equivalently, part d identifies it with the pullback of the globally generated bundle ${\mathcal{O}}_{{\mathbb{P}}^1}(k)$.

The divisor $D_1$ is a fiber of the ruling and has $D_1^2=0$. Hence $(k{D}_1{)}^2=0$ for every $k$. An ample divisor on a complete surface has positive self-intersection, so no $\mathcal{O}(k{D}_1)$ with $k\ge 0$ is ample. These results are summarized by multiples of a fiber on the first Hirzebruch surface.

Let $E_1$ be the invariant point divisor of ${\mathbb{P}}^1$ corresponding to the ray ${\mathbb{R}}_{\ge 0}$. Under $(u_1,u_2)\mapsto u_1$, the ray $(1,0)$ maps primitively to that ray, while no other ray maps into its interior. The toric pullback formula givesequivalently $D_1={\pi }^{-1}(E_1)$ as divisors.

The assertion is false. The Picard group of the Hirzebruch surface is freely generated by the negative section $S$ and a fiber $F=D_1$. Pullbacks from ${\mathbb{P}}^1$ form only the subgroup $\mathbb{Z}F$. For example, $\mathcal{O}(S)$ cannot be a pullback: its restriction to a fiber has degree $S\cdot F=1$, whereas every pullback from the base restricts trivially to every fiber.