Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 15 / 5

Let $\rho :{\pi }_1(X)\to U(V)$ be the unitary representation. The associated unitary flat holomorphic vector bundle iswhere deck transformations act on the first factor and $\rho$ acts on the second. Local lifts to the universal cover give holomorphic orthonormal frames with locally constant unitary transition matrices. The induced Hermitian metric is therefore well defined, and the induced flat unitary connection is its Chern connection.

Let $s$ be a global holomorphic section. In one of these parallel orthonormal frames write $s=(f_1,\dots ,f_r)$, where every $f_j$ is holomorphic. Its squared norm $u=|s{|}^2={\sum }_j|f_j{|}^2$ is a globally defined smooth function because the transition matrices are unitary. In a local complex coordinate,Choose any smooth conformal metric on the curve, locally $\kappa (z)|dz{|}^2$. Its Laplacian satisfiesOn the compact curve without boundary, Stokes theorem gives ${\int }_X\Delta ud\mathrm{vol}=0$. A continuous nonnegative function with zero integral vanishes everywhere, so each $f_j^{\prime }=0$. Thus the section is locally constant in every parallel frame and is globally parallel. This proves that holomorphic sections of a unitary flat bundle are parallel.

A parallel section is determined by its value $v$ at a chosen base point. Following it around a loop returns the value $\rho (\gamma )v$, so a global parallel section exists exactly when $\rho (\gamma )v=v$ for every $\gamma \in {\pi }_1(X)$. Conversely, such a fixed vector gives the constant equivariant section on $\tilde{X}$, which descends holomorphically. Evaluation and this construction are inverse linear maps, giving the canonical identificationUnitarity is what makes the local squared norms glue and the integrated nonnegative identity available; no assumption that the universal cover is compact is made.

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.