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The complex tautological line bundle is a holomorphic subbundle of the trivial bundle with fibre ${\mathbb{C}}^{n+1}$. A global holomorphic section therefore gives $n+1$ global holomorphic functions on compact connected Complex projective space. The maximum modulus principle makes them constant. The constant vector must lie in every projective line through the origin, whose intersection is zero for $n\ge 1$. For $n=0$, the base is one point and the section space is instead $\mathbb{C}$.

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.

The flat unitary connection on $E_V$ is its Chern connection, since it preserves the induced Hermitian metric and its $(0,1)$ part is the Dolbeault operator. Its curvature is zero. The trace-curvature formula for the degree of a holomorphic vector bundle therefore givesFor any positive-rank proper holomorphic subbundle $F\subset E_V$, equip $F$ with the induced metric and use the block-connection calculation of Question 4(a). Since the ambient curvature vanishes, its Chern curvature iswhere $B=bdz$ is the second fundamental form of a holomorphic subbundle. ConsequentlyThe term being integrated is nonnegative, so its sign gives the inequality for every such subbundle, not just for parallel subbundles. By the definition of a semistable holomorphic vector bundle,If semistability is formulated using coherent subsheaves instead, saturate a subsheaf on the smooth curve. The saturation has the same rank, at least as large a degree, and is a subbundle because a torsion-free sheaf on a smooth curve is locally free. The inequality for the saturated subbundle therefore also bounds the original subsheaf. Finally, equality for a subbundle forces $B=0$, so it is parallel; invariant subspaces of $V$ correspond to these degree-zero parallel subbundles. Semistability is the conclusion required here, rather than stability for every unitary representation.

A positive-rank holomorphic vector bundle on a compact Riemann surface is semistable when every nonzero proper holomorphic subbundle has slope at most its own. Equivalently one tests coherent subsheaves: saturation on a smooth curve preserves rank, increases degree, and gives a subbundle with torsion-free quotient. Strict inequality defines stability, which is stronger.