Classical and nonhyperelliptic vector-bundle Brill–Noether bounds

Line-bundle Clifford from a valuation calculation. Let C be a smooth connected projective curve of genus $g\ge 1$. Let L have degree $d\in [0,2g-2]$. If $h0(L)=0$ there is nothing to prove. If $h1(L)=0$, Riemann–Roch gives $h0(L)=d+1-g\le d\mathrm{/}2+1$. Otherwise put $p=h0(L)>0$ and $q=h0({\omega }_C{L}^{-1})>0$. At a smooth point choose bases of these two section spaces with strictly increasing vanishing orders $a1<\dots <ap$ and $b1<\dots <bq$: each graded piece of the order filtration is at most one-dimensional. The products s1t1,…,s1tq,s2tq,…,sptq have pairwise strictly increasing orders. They are independent global canonical sections. Thus $p+q-1\le g$. Serre duality and Riemann–Roch give $p-q=d+1-g$, hence $2p\le d+2$. This proves the rank-one bound directly.

The same argument proves $h0({\omega }_C)=g$ and deg ${\omega }_C=2g-2$ from the supplied duality/Riemann–Roch identities. Euler degree for vector bundles agrees with determinant degree: inject a sufficiently negative line bundle of rank-many copies by generic independent sections of a positive twist, avoiding rank drop at the finitely many chosen points. The cokernel has finite length, equal to the determinant zero divisor; Euler additivity proves $\chi (E)=\mathrm{deg}(\det E)+rk(E)(1-g)$.

The low-degree Castelnuovo estimate needed for the equality case. A nondegenerate integral complex curve $B\subset P^r$ of degree 2r, $r\ge 2$, has arithmetic genus at most r+1. A degree-r nondegenerate integral curve has arithmetic genus zero.

Here are the geometric and Hilbert-function details. A general hyperplane meets B at distinct smooth points. Its hyperplane-section monodromy is the full symmetric group: the incidence of ordered distinct smooth pairs lying in a hyperplane is irreducible (over the irreducible pair space its fiber is the projective space of hyperplanes through their line), so monodromy is two-transitive. An ordinary tangent hyperplane supplies a transposition. Such a hyperplane exists by taking a general smooth point p and a hyperplane containing its tangent line but not its osculating plane, and avoiding the finite singular set and other tangencies. In characteristic zero the osculating plane is generically genuinely a plane unless B is a line. For $r\ge 3$, hyperplanes tangent also at a varying second point have one smaller dimension: generically the two tangent lines span a three-space. If all general tangent lines met the fixed tangent line, projection from that line would have zero derivative and would be constant in characteristic zero, putting B in a plane, contrary to nondegeneracy. For a plane curve, the calculation $\mathrm{\ell }(p)\cdot p=\mathrm{\ell }(p)\cdot p^{\mathrm{\prime }}=0$ and hence ${\mathrm{\ell }}^{\mathrm{\prime }}(p)\cdot p=0$ shows that the tangent to the dual curve recovers p; the general tangent line is tangent at just that point, with contact two. A transverse local pencil at an ordinary tangency is $z^2=t$ and exchanges just the two meeting points. Two-transitivity conjugates this transposition to every transposition, proving the claimed symmetric monodromy.

Consequently a general hyperplane section of degree d has its points in linear general position. Indeed it spans the hyperplane: a second linear form vanishing on all those simple points divided by its defining form would be a regular function on the normalization of B and hence constant. Some r of the points are therefore independent. Full symmetric monodromy, and the closed determinant equations for dependence, make every r-subset independent for a general hyperplane.

For $d=2r$, let $h_D(n)$ be the coordinate Hilbert function of this reduced hyperplane section $D\subset P^{r-1}$. Then $h_D(1)=r$, $h_D(2)\ge 2r-1$, and $h_D(n)=2r$ for $n\ge 3$. To prove the latter bounds explicitly, any selected 2r−1 points can be separated by quadratic forms: for each selected point, partition the other 2r−2 into two sets of r−1, choose the hyperplane through each set avoiding that point, and multiply. Cubics separate all 2r points by distributing their other 2r−1 points into sets of r−1,r−1,1. Linear general position provides each avoiding hyperplane. The same construction with $d=r$ gives $h_D(n)=r$ for every $n\ge 1$.

Let $A_B$ be the homogeneous coordinate domain. Multiplication by a general hyperplane form is injective, and $A_B\mathrm{/}(h)$ surjects onto the saturated coordinate ring of D. Thus $h_B(n)-h_B(n-1)\ge h_D(n)$. Summing to a large n and using $h_B(n)=dn+1-p_a(B)$ gives $p_a(B)\le {\sum }_{n\ge 1}(d-h_D(n))$. This is at most r+1 when $d=2r$, and zero when $d=r$. These are the exact estimates asserted, with no general Castelnuovo theorem assumed. The actual genus–delta normalization identity then gives the same upper bound on the normalization genus.

Strict line Clifford on a nonhyperelliptic curve. Suppose C is nonhyperelliptic and $g\ge 3$. For every line bundle of degree $1\le d\le 2g-3$,

$h0(L)\le d\mathrm{/}2+1\mathrm{/}2$.

It suffices to exclude equality $2h0(L)=d+2$ in the classical bound. If h0 or h1 is at most one the endpoint degree equations already exclude such equality. Otherwise put $p=h0(L)\ge 2$, $q=h1(L)\ge 2$; equality gives $d=2p-2$ and $g=p+q-1\ge p+1$. Both L and ${\omega }_C{L}^{-1}$ are basepoint-free: removing any base point keeps its h0 but lowers its degree by one, contradicting §1.

Consider the map of L into $P^{p-1}$. If $p=2$ it is a degree-two map to P1, contrary to nonhyperellipticity. Otherwise put $r=p-1\ge 2$. If the map is birational onto its image of degree 2r, §2 gives $g\le r+1=p$, contradiction. Its degree e is therefore at least two. A nondegenerate curve in $P^r$ has degree at least r (its general hyperplane section spans $P^{r-1}$); hence $2r=e$ deg(image) forces $e=2$ and image degree r. Section 2 makes that image's normalization genus zero. A smooth genus-zero complex curve is P1: Riemann–Roch gives a degree-one function after choosing one point. Thus C has a degree-two map to P1, again a contradiction. This proves the strict bound.

The maximal-proper-slope induction, with all quotient bounds. For a semistable bundle E of rank $n\ge 2$ choose a saturated proper subbundle E1 of MAXIMAL slope, and put $E2=E\mathrm{/}E1$. Such a maximum exists: eventual generation of E* embeds E into $O_C(m{)}^N$. Taking exterior powers bounds the degree of every rank-r subbundle by r m deg $O_C(1)$. There are only finitely many ranks and their degree slopes form discrete bounded-above sets. Saturation raises degree and makes the quotient a bundle.

E1 and E2 are semistable. A destabilizing subbundle of E1 would contradict maximality. For E2, a proper subbundle G of slope greater than $\mu (E2)\ge \mu (E)\ge \mu (E1)$ would have a proper preimage in E of slope greater than μ(E1), another contradiction.

Fix $a=0$ or1 and $U=2g-2-a$. Suppose $\mu (E)\in [a,U]$ and $h0(E)>a$ n, $h1(E)>a$ n (for $a=0$ these are simply nonzero). Evaluation at any a distinct points has target dimension a n. A nonzero section in its kernel gives a saturated line subbundle of degree at least a, so $\mu (E1)\ge a$. Applying the same observation to $E\ast {\omega }_C$, using Serre duality, gives a nonzero quotient line Q of E with degree $q\le U$. Its rank-(n−1) kernel has slope at most μ(E1). Thus $d-q\le (n-1)\mu (E1)$, and, writing $r1=rk(E1)$,

$\mu (E2)=(d-r1\mu (E1))\mathrm{/}(n-r1)\le [q+(n-1-r1)\mu (E1)]\mathrm{/}(n-r1)\le U$.

Also $\mu (E1)\le \mu (E)\le U$ and $\mu (E2)\ge \mu (E)\ge a$. Both smaller-rank factors therefore lie in the SAME closed slope interval [a,U]. This is the missing precise inequality behind the short source phrase 'one sees easily'.

For $a=0$, the non-special cases follow from Riemann–Roch; otherwise induct with the rank-one result §1, sum $h0(E)\le h0(E1)+h0(E2)$, and add their degrees/ranks. It gives $h0(E)\le d\mathrm{/}2+n$ for every semistable bundle of slope[0,2g−2].

For $a=1$ on a nonhyperelliptic C, if $h0(E)\le n$ then $h0(E)\le (d+n)\mathrm{/}2$ because $d\ge n$. If $h1(E)\le n$, Riemann–Roch gives $h0(E)\le d+n(2-g)\le (d+n)\mathrm{/}2$ because $d\le n(2g-3)$. In the remaining case the preceding induction applies with §3 as its rank-one base, yielding $h0(E)\le d\mathrm{/}2+n\mathrm{/}2$ for every semistable bundle of slope[1,2g−3]. No tensor-product semistability or unitary representation theorem is used.

The exact BN consequences. For a smooth C of genus $g\ge 2$, the neighborhood of g−1 is inside[0,2g−2], so the rank-independent first estimate gives $B{N}_C\le (g-1)\mathrm{/}2+1$ by taking the defining limsup. On a nonhyperelliptic curve $g\ge 3$, use[1,2g−3] to obtain $B{N}_C\le g\mathrm{/}2$. The genus-one first estimate follows separately: negative slope has no sections, positive slope has $h1=0$ by duality and semistability, and slope zero has $h0\le \mathrm{rank}$ by the $a=0$ induction. Thus $B{N}_C\le 1$ there as well.

If ${\omega }_C$ is very ample then C is nonhyperelliptic. Indeed a degree-two map $\pi :C\to P1$ would give $L=\pi \ast O(g-1)$, of degree2g−2 and at least g sections. Riemann–Roch makes $h1(L)\ge 1$, so Serre duality supplies a nonzero map $L\to {\omega }_C$ of degree zero, necessarily an isomorphism. Then every canonical section is pulled back (both spaces have dimension g), contradicting separation of the two points of a general fiber. This proves the exact canonical-very-ample implication used in the weighted X6,6 correction, without a gonality theorem imported as a premise.