# Is the determinant map $det:\mathcal{M}(r,d)\rightarrow Pic^d(X)$ on moduli space of semistable vector bundles a fibration?

Let $$X$$ smooth projective curve over $$\mathbb{C}$$, fix a line bundle $$L$$ of degree $$d$$, and let $$\mathcal{M}(r,d)$$ denote the moduli space of semistable vector bundles of rank $$r$$ and degree $$d$$. It is known to be a normal projective variety.

There is a determinant map $$det: \mathcal{M}(r,d)\rightarrow Pic^d(X)$$, where $$Pic^d(X)$$ denotes the isomorphism class of line bundles of degree $$d$$.

Is it true that the $$det$$ map is a fibre bundle? I recently heard this statement, but I can't seem to prove it. Can someone give some idea, or give some reference if possible? Thanks in advance!

• Fix a line bundle $L\in Pic^d(X)$, and consider the étale covering $\pi :Pic^0(X)\rightarrow Pic^d(X)$ given by $M\mapsto M^r\otimes L$. It is easy to see that the pull back of $\ \det\$ under $\pi$ is the second projection $\mathcal{M}_L\otimes Pic^0(X)\rightarrow Pic^0(X)$. Thus $\ \det\$ is an étale fiber bundle, and even more: it becomes trivial after a finite étale covering.
– abx
Dec 9, 2019 at 19:33