When determinant line bundle is holomorphically trivial

Question

I'm learning the deformation theory of holomorphic structure over given smooth vector bundle by the book Smooth four - manifolds and complex surfaces.

However, when talk about holomorphic vector bundle, they always request that it should have trivial determinant.

I wonder if there is a good geometric meaning of holomorphic bundle with trivial determinant? I know a line bundle is holomorphic trivial iff it is induced by a global meromorphic function. And if the transition functions have determinant $1$, the determinant line bundle is automatically trivial, but both of these explanation seems are not so good.

Any interpretation about why we consider the holomorphic bundle with trivial determinant is also welcome.

Thank you for your answer!

Answer 1

Let $M_{0,d}(C)$ denote the moduli space of ($S$-equivalence classes of) stable vector bundles of degree $0$ and rank $r$ over a fixed curve $C$ of genus $g$. There is a map $\det:M_{0,d}(C) \to \operatorname{Jac}^0(C)$ to the degree $0$ component of the Jacobian of $C$ given by $E \mapsto \det E$. The Jacobian acts on itself via tensor product and hence on the higher rank moduli space via the induced action on the fibers of $\det$. Thus the fibers are isomorphic, and fixing $\det E \cong \mathcal O_C$ just focuses attention on the fiber over the identity element of the Jacobian (which, perhaps also of note, is the unique degree $0$ bundle that has a global section).