Ordinary sheaf Bogomolov Gieseker discriminant inequality

We establish a section bound on S which uses no semistability of any restriction. Fix a real number u and let B be ANY slope-semistable vector bundle with ${\mu }_H(B)=u$. Let G be the coherent image of $H0(B)\otimes O_S$ -> B, of rank k. It is globally generated and torsion-free, H0(G)=H0(B), and $Hc1(G)\le ku$. If $k<\mathrm{rank}(B)$ this is semistability; if $k=\mathrm{rank}(B)$, the determinant of its inclusion has effective zero divisor and gives the same inequality. If $u<0$, global generation makes the determinant of G effective of nonnegative H-degree, a contradiction unless $G=0$. Thus h0(B)=0 in that case.

Suppose $u\ge 0$. Choose once an integer $a>0$ such that aH is very ample and $a{H}^2>u$. Semistability and step1.1 imply H0(B(−aH))=0. Choose a smooth C in |aH| avoiding the finite singular set of G. Such choices exist directly from the very ample embedding: the incidence of hyperplanes containing a surface tangent plane has dimension one less than the full dual projective space, and finitely many point-incidence hyperplanes are also proper subsets. The sequence for G and C therefore gives an injection H0(G)->H0(C,G|C). The latter is a globally generated bundle of rank k and degree aH $c1(G)\le aku$. For ANY globally generated rank-k bundle V on a smooth curve, choose k−1 general generating sections with no rank drop: at a fixed point rank drop has matrix codimension two, and union over a one-dimensional curve still has positive codimension in the section parameter space. Its quotient is the line bundle det V. Hence $h0(V)\le k-1+h0(\det V)\le k+degV$. The last bound follows by evaluation at deg(V)+1 distinct points; a nonzero line section cannot vanish at more points than its degree. We conclude $h0(S,B)\le \mathrm{rank}(B)(1+au)$. The constant depends only on u,H, not on B or its rank. The chosen curve may vary with G while its degree aH is fixed; no sheaf restriction theorem was used.

Let E be a slope-semistable bundle of rank r on S. For $m\ge 1$ set $R_m=\mathrm{binomial}(m+r-1,r-1)$, and $B_m=\mathrm{End}({\mathrm{Sym}}^{m}E)$. The attached GIT lemma [F1] makes $B_m$ semistable of slope zero and $\mathrm{rank}{R}_m^2$. Its trace pairing identifies it with its dual. Twisting by $K_S$ preserves semistability and shifts slope by the FIXED number $H{K}_S$. Apply steps1.1–2.1 first at $u=0$ and then at $u=H{K}_S$. Serre duality [F3] gives constants C0,C2 independent of m with $h0(B_m)\le C0R_m^2,h2(B_m)\le C2R_m^2$. One may take C0=1; if $H{K}_S<0$ take C2=0, otherwise choose a as in step2.1 and $C2=1+aH{K}_S$.

We compute the discriminant coefficient exactly. Use formal Chern roots x1,...,xr. On the full flag bundle E has line quotients; the induced symmetric-power filtration has the monomial line quotients indexed below. Additivity of the Chern character and injectivity in the splitting principle therefore justify computing the actual bundle character by these formal weights. The weights of ${\mathrm{Sym}}^m$ are $\sum {\alpha }_i{x}_i$ for all ${\alpha }_i\ge 0$ with $\sum {\alpha }_i=m$. Symmetry and the product generating function ${\prod }_i(1-t_i{)}^{(-1)}$ give ${\sum }_{\alpha }{\alpha }_i=R_{m}m\mathrm{/}r,{\sum }_{\alpha }{\alpha }_i({\alpha }_i-1)=2R_{m}m(m-1)\mathrm{/}(r(r+1)),{\sum }_{\alpha }{\alpha }_i{\alpha }_j=R_{m}m(m-1)\mathrm{/}(r(r+1))$ for $i\mathrm{\ne }j$. For example the last two follow by differentiating in $t_i$ twice, or once in each of $t_i,t_j$, and then taking the coefficient of total degree m. Consequently $c1({\mathrm{Sym}}^{m}E)=R_{m}mc1(E)\mathrm{/}r,ch2({\mathrm{Sym}}^{m}E)=R_{m}m(m+r)ch2(E)\mathrm{/}(r(r+1))+R_{m}m(m-1)c1(E{)}^2\mathrm{/}(2r(r+1))$. Since $ch2(\mathrm{End}V)=2\mathrm{rank}(V)ch2(V)-c1(V{)}^2$ and c1(End V)=0, substitution gives $ch2(B_m)\mathrm{/}{R}_m^2=-m(m+r)\mathrm{\Delta }(E)\mathrm{/}(r^2(r+1))$. These are equalities of rational Chow zero-cycles, so taking degrees is legitimate. They include rank one, where a line bundle has $\mathrm{\Delta }=0$.

Riemann–Roch [F2] and $c1(B_m)=0$ give $\chi (B_m)\mathrm{/}{R}_m^2=\chi (O_S)-m(m+r)\mathrm{\Delta }(E)\mathrm{/}(r^2(r+1))$. On the other hand h1 is nonnegative, so step3.1 implies $\chi (B_m)\mathrm{/}{R}_m^2\le C0+C2$. If Delta(E)<0 the displayed Euler quotient tends to positive infinity, a contradiction. Thus $\mathrm{\Delta }(E)\ge 0$ for every slope-semistable bundle E on S. This is the ordinary surface Bogomolov inequality, obtained without Hermitian–Einstein or ordinary restriction as a premise.

For the given torsion-free F on S, let $E=F$ double dual. Its quotient $Q=E\mathrm{/}F$ has finite length, and E is locally free by [F5]. It is semistable: intersect any positive-rank subsheaf of E with F; that changes it only at finitely many points and hence preserves rank and degree. For a closed point p, GRR for its closed immersion into S gives ch(k(p))=[p] in degree two and zero in lower degrees. A composition series and Chern additivity therefore give c1(F)=c1(E), ch2(E)=ch2(F)+length(Q). Therefore $\mathrm{\Delta }(F)=\mathrm{\Delta }(E)+2r\mathrm{length}(Q)\ge 0$. This proves the full torsion-free surface clause. The normalized surface discriminant is also nonnegative because $(Hc1{)}^2-2H^{2}r$ ch2 $=[(Hc1{)}^2-H^{2}c{1}^2]+H^2\mathrm{\Delta }(F)$, and the first bracket is nonnegative by the exact surface index theorem [F3].

For the threefold choose a sufficiently high general smooth S in |mH| by the INDEPENDENT ordinary restriction interface [F4]. The restricted sheaf is torsion-free and slope-semistable. Restriction of its coherent Chern character is the usual pullback: multiplication by the Cartier equation is injective on the original torsion-free F, so its derived Cartier restriction has no negative homology; restrict a finite locally free resolution to compute it. Applying step6.1 on S yields $0\le \mathrm{\Delta }(F\mathrm{\mid }S)=mH(c1(F{)}^2-2rch2(F))$. Division by $m>0$ proves the strong threefold inequality. This is the only use of [F4].

Apply the surface Hodge index also to H|S and c1(F)|S. Their intersection numbers are respectively $m{H}^3,m{H}^{2}c1(F)$, and mH $c1(F{)}^2$. Thus $(H^{2}c1(F){)}^2-H^3(Hc1(F{)}^2)\ge 0$. Adding $H^3$ times step7.1's strong discriminant gives $(H^{2}c1(F){)}^2-2H^{3}r(Hch2(F))\ge 0$, exactly the normalized scalar required by the primary tilt construction. Positive normalization of slope by $H^3$, or by $H^2$ on the surface, has no effect on any subobject inequality. No conclusion on singular normal surfaces is drawn.