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A line subbundle is an injective map from a line bundle into a vector bundle whose quotient is locally free. Equivalently the injection splits locally. On a smooth algebraic curve, saturation of a rank-one coherent subsheaf produces a saturated line subbundle on a smooth curve.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

Cover the projective line by $U_0=\mathrm{Spec}k[t]$ and $U_1=\mathrm{Spec}k[t^{-1}]$. On their overlap the coordinate is invertible. Choose frames $e_0,e_1$ of the twisting sheaf on projective space $\mathcal{O}(n)$ with $e_1=t^n{e}_0$. The vanishing of quasi-coherent cohomology on an affine scheme applies to both charts and their intersection, so the Čech cochain complexcomputes the sheaf cohomology. Its kernel identifies with $k[t]\cap t^{n}k[t^{-1}]$, and its cokernel isFor $n\ge 0$, the intersection has basis $1,t,\dots ,t^n$; for $n<0$ it is zero. In the quotient, powers $t^j$ with $j\ge 0$ or $j\le n$ disappear. The surviving basis for $n\le -2$ is $t^{n+1},\dots ,t^{-1}$; for $n\ge -1$ none survives. HenceThis is the Čech cohomology of twists on the projective line.

By the preceding question, $\omega ={\Omega }_{{\mathbb{P}}^1/k}^1=\mathcal{O}(-2)$. In the $dt$ frame its degree-one Čech cohomology is one-dimensional, represented by $t^{-1}dt$. DefineThis is well defined: differentials regular on $U_0$ have nonnegative powers, and those regular on $U_1$ have powers at most $-2$ when expressed using $dt$. Evaluation and the cup product give natural pairingsFor $E=\mathcal{O}(n)$ with $n\ge 0$, the basis element $t^a$ in $H^0(E)$ pairs with $t^{-a-1}dt$ in $H^1(E^{\vee }\otimes \omega )$ and with no other element of this dual Laurent basis. The pairing matrix is a permutation matrix, hence invertible. For $n\le -2$ the same computation exchanges degrees zero and one. For $n=-1$ both sides vanish. Thus the pairings are perfect for every line bundle on ${\mathbb{P}}^1$.

Here we used the classification of line bundles on the projective line: they are $\mathcal{O}(a)$ for integers $a$. One direct justification is that every line bundle has a nonzero rational section and hence is associated with a divisor on an algebraic curve. The finite closed point defined by an irreducible polynomial $p(t)$ is linearly equivalent to $(\mathrm{deg}p)\infty$, using the principal divisor of $p(t)$. Thus every divisor is equivalent to its degree times $\infty$, proving the classification over arbitrary $k$.

To extend the pairing to every locally free coherent sheaf $E$, induct on rank, without assuming the splitting theorem of the next question. A one-dimensional subspace of the generic fibre, saturated in $E$, gives an exact sequencewith $L$ a line bundle and $Q$ a lower-rank vector bundle. This saturated line subbundle on a smooth curve exists because its stalk and its torsion-free quotient are free over the discrete valuation rings of the curve. The dual sequence, tensored with $\omega$, is exact as well, since the original sequence splits locally.

The two long exact sequences in sheaf cohomology, after dualizing one, are compatible with the evaluation pairings. In Čech cohomology, this compatibility follows by lifting a section locally: the differences of its lifts represent its connecting class, and contracting those differences with a dual cocycle gives the transpose connecting map, with the usual sign. Thus the connecting maps are adjoint; changing one vertical map's sign gives a commuting diagram. Its rows areThe maps for $L$ and $Q$ are isomorphisms by the line-bundle calculation and induction. Exactness then makes the maps for $E$ isomorphisms, by the Five lemma (or by comparing the relevant kernels and cokernels). All groups are finite-dimensional because the curve is projective and the sheaves coherent. Therefore residue duality on the projective line provesThese isomorphisms are natural in $E$. Higher sheaf cohomology vanishes by the same two-chart complex.