Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 21 / 1

Use projectivization by quotients, and setThe fibre over $t$ parametrizes one-dimensional quotients of $E_t$, so it is ${\mathbb{P}}^{n-1}$. The standard line bundles areThus $L$ is the fiber class of a rational normal scroll of the ruling and $M$ is the universal quotient line bundle of ${\pi }^{\ast }E$. In additive divisor notation we also write $L,M$ for their first Chern classes. For $n\ge 2$ these generate the Picard group; for $n=1$, $F={\mathbb{P}}^1$ and $M=a_{1}L$. This fixes the sign convention for the rational normal scroll.

Put $s={\sum }_i{a}_i$. The projective bundle formula for Chow groups givesHere the first relation comes from the one-dimensional base, and the second is the projective-bundle relation with $c_1(E)=sL$ and higher Chern classes zero on the base. Integration over a fibre gives $L{M}^{n-1}=1$. Therefore all top-degree mixed intersection numbers areIn particular, for integers $u,v$,For the canonical divisor of a smooth variety, the relative Euler sequencehas determinant ${\mathcal{O}}_F(nM-sL)$. Combining its dual determinant with $K_{{\mathbb{P}}^1}=-2L$ gives

A common twist does not change the abstract projective bundle. Choose an integer $b$ such that $d_i=a_i+b\ge 1$ for every $i$. The complete linear system of divisors $|M+bL|$ hasIts map to ${\mathbb{P}}^N$, where $N=s+nb+n-1$, has coordinate blocksHere $[s_0:t_0]$ is a point of the base and the ${\lambda }_i$ are quotient coordinates in a local trivialization of $E(b)$. These expressions have the correct transition factors and define the global map. Any nonzero block recovers the base point through its degree-$d_i$ rational normal curve. Once that point is known, the blocks recover the projective fibre coordinates. On the open where the first coordinate of a nonzero block is nonzero, the ratio of the next coordinate to the first recovers $t_0/s_0$; the analogous last-coordinate chart recovers $s_0/t_0$. The remaining coordinate ratios then recover the fibre coordinates regularly. Thus the map identifies $F$ with its closed image and is a projective embedding, not merely a pointwise injection.

The hyperplane class on this image is $M+bL$, soConsequently every such natural embedding hasThe tautological choice $b=0$ is an embedding when all $a_i>0$. If the $a_i$ are only nonnegative, $|M|$ is basepoint-free but can contract the projective subbundle arising from zero summands; one must distinguish that scroll image from the smooth abstract scroll.

Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes \mathcal{O}(c)$. Tensoring all the fibres of a vector bundle by the same line bundle leaves their one-dimensional quotient spaces unchanged. HenceThis proves one implication in the isomorphism classification of abstract rational scrolls.

For the converse first take $n\ge 3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre ${\mathbb{P}}^{n-1}$ giveswhere $h$ is the hyperplane class and $h^2\ne 0$ because $n-1\ge 2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An isomorphism $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, sofor an automorphism $\sigma$ of ${\mathbb{P}}^1$.

On a fibre, the isomorphism preserves the positive generator ${\mathcal{O}}_{{\mathbb{P}}^{n-1}}(1)$ of its Picard group. Consequentlyfor some integer $c$. Push forward along the ruling, using ${\pi }_{\ast }\mathcal{O}(M)=E$ and the projection formula for sheaves:Every automorphism of ${\mathbb{P}}^1$ preserves each $\mathcal{O}(d)$. The uniqueness of the splitting degrees in the Birkhoff–Grothendieck theorem now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,The first difference counts the degrees with $a_i+t\ge 0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.

For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal{O}\oplus \mathcal{O}(a)$, $a\ge 0$. Its surface is the Hirzebruch surface ${\mathbb{F}}_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The nef cone is generated by $L$ and $B+aL$: a nef divisor $xL+yB$ must have $y\ge 0$ from intersection with a fibre and $x-ay\ge 0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge 0$ has squarewhich vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. Isomorphisms may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.

For $n=1$, every scroll is ${\mathbb{P}}^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.