Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 15 / 4

For a nonzero holomorphic vector bundle $F$ on the curve, its degree of a holomorphic vector bundle and slope of a holomorphic vector bundle areWe obtain a bound independent of the particular holomorphic subbundle by fixing a Hermitian metric on a holomorphic vector bundle $E$ once and for all. Let $D_E$ be its Chern connection, $F_E=D_E^2$ its Chern curvature, and $\omega$ any positive area form on the compact Riemann surface. WriteHere $K$ is a smooth Hermitian endomorphism of $E$. By compactness, its largest eigenvalue is bounded above by some finite constant $C$ on the whole curve.

Let $F\subseteq E$ be a holomorphic subbundle of positive rank $r$, with the induced metric. Use the smooth orthogonal splitting of a vector subbundle $E=F\oplus F^{\perp }$. In this splitting the connection has block formwhere $D_F$ is the projected Chern connection and $B$ is the second fundamental form of a holomorphic subbundle. The latter has type $(1,0)$: the $(0,1)$ part of $D_E$ preserves the holomorphic subbundle. The adjoint $B^{†}$ includes conjugation of the form factor and has type $(0,1)$. Squaring this block matrix givesThis derives the curvature formula for a holomorphic subbundle in the column-vector convention used here. Locally write $B=bdz$. ThenThe degree of a holomorphic vector bundle is the trace-curvature integral, since the trace is the curvature of the determinant line bundle. HenceThe last integrand is a nonnegative real two-form and is independent of the local unitary frame and coordinate. Since $K\le C{I}_E$, the compressed trace is at most $rC$ pointwise. ThereforeThis is the subbundle slope bound from ambient Chern curvature. The right side depends only on $E$ and the initially chosen metrics, and is finite. It proves the requested uniform upper bound for every positive-rank subbundle, without assuming any stability property of $E$ or using a bound that depends on $F$.

Take the compact Riemann surface $X={\mathbb{P}}^1$ and the holomorphic vector bundle $E={\mathcal{O}}_X\oplus {\mathcal{O}}_X$. For every positive integer $m$, use homogeneous coordinates $[Z_0:Z_1]$ and the mapThe two sections $Z_0^m,Z_1^m$ never vanish together on ${\mathbb{P}}^1$. Thus this map has rank one at every point; locally one of its two components is invertible in suitable line-bundle frames, so its image is a genuine holomorphic subbundle and its quotient is a locally free sheaf. More explicitly, there is an exact sequenceExactness follows fibrewise from the absence of a common zero and hence also locally as a sequence of vector bundles. The image $F_m$ is isomorphic to ${\mathcal{O}}_X(-m)$ and has rank one, soThus there need not be a lower bound for subbundle slopes, even for a fixed trivial rank-two bundle. Merely multiplying a section of one fixed subline by a function with zeros would not suffice: that could give only a subsheaf with a torsion quotient. The nowhere-simultaneously-zero pair above avoids that problem and realizes unbounded negative degrees of line subbundles.