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On a compact Riemann surface, the degree of a holomorphic vector bundle is the degree of a line bundle of its determinant line bundle. A Hermitian metric on a holomorphic vector bundle and its Chern connection express this degree by the trace-curvature integral. The value is independent of the metric because the normalized trace curvature represents the First Chern class.

The identification with ${\mathbf{P}}^1$ means that $C$ is a smooth plane conic. An arbitrary singular or nonreduced conic over an algebraically closed field cannot be identified with the projective line, so this is a necessary qualification implicit in the requested description.

Choose an equation $\ell =0$ for $L\subset {\mathbf{P}}^3$ and a degree-two equation $q=0$ whose restriction to $L$ defines $C$. Its homogeneous ideal in ${\mathbf{P}}^3$ is $(\ell ,q)$. These equations form a regular sequence: $\ell$ is a non-zero-divisor in the polynomial ring, and modulo $\ell$ the ring is a polynomial ring in three variables, in which the nonzero quadratic $q$ is again a non-zero-divisor. The generator classes therefore form a basis of the conormal sheaf, with their grading shifts:One can verify the absence of relations from the two-generator Koszul complex: a relation between $\ell ,q$ is a multiple of $(q,-\ell )$, whose coefficients become zero modulo ${\mathcal{I}}_C$. This is the normal sheaf of a projective complete intersection calculation. Dualizing the conormal sheaf gives

These twists are restrictions from ${\mathbf{P}}^3$, not intrinsic degree-one twists on ${\mathbf{P}}^1$. A line in the plane cuts the degree-two curve in two points counted with multiplicity, so the degree of a line bundle ${\mathcal{O}}_C(1)$ is two. The Picard group of the projective line then givesEquivalently, the degree-two Veronese embedding $[s:t]\mapsto [s^2:st:t^2]$ pulls each linear coordinate back to a quadratic. Thus the normal bundle of a smooth plane conic isThe same splitting follows from the normal sequence for the two embeddings:Its outer terms are ${\mathcal{O}}_{{\mathbf{P}}^1}(4)$ and ${\mathcal{O}}_{{\mathbf{P}}^1}(2)$. The extension group class lies in ${\mathrm{Ext}}^1(\mathcal{O}(2),\mathcal{O}(4))=H^1({\mathbf{P}}^1,\mathcal{O}(2))=0$, by the preceding Čech cohomology calculation, so the sequence splits. Both methods distinguish the ambient twist from the intrinsic degree on the 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.