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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 $F$.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.