= Solution
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 gives
$$
\deg E_V=\frac{i}{2\pi}\int_X\operatorname{tr}F_{E_V}=0,
\qquad \mu(E_V)=0.
$$
For 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> is
$$
F_F=B^\dagger\wedge B=-b^\dagger b\,dz\wedge d\bar z,
$$
where $B=b\,dz$ is the <second fundamental form of a holomorphic subbundle>. Consequently
$$
\deg F=-\frac1{2\pi}\int_X i\operatorname{tr}(b^\dagger b)\,dz\wedge d\bar z\le0,
\qquad \mu(F)\le0=\mu(E_V).
$$
The 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>,
$$
\boxed{E_V\text{ is semistable and has degree }0.}
$$
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.
Back to article page
