Let $X, X^{\prime}\colon$ smooth projective curve on $\mathbb{C}$ (genus $\geq 3$),
$M(r,d)\colon$ coarse moduli of stable vector bundles with rank $r\geq2$, and degree $d$ , and
$M(r,\xi)\colon$ coarse moduli of stable vector bundles which have fixed determinant $\xi$ with rank $r$.

In other words, define \begin{eqnarray} \operatorname{det}\colon &M(r,d)&\rightarrow &\operatorname{Jac}^{d}(X)\\ &E&\mapsto&\wedge^{r} E \end{eqnarray} then, \begin{equation} M(r,\xi)=\operatorname{det}^{-1}(\xi) \end{equation}
There is a famous fact, $M(r,\xi)_{X}\cong M(r,\xi^{\prime})_{X^{\prime}}$ then $X\cong X^{\prime}$.

I want to know whether $M(r,d)_{X}\cong M(r,d)_{X^{\prime}}$ implies $X\cong X^{\prime}$.
I think it is possible to prove that by almost the same proof.

But if not, why I must consider fixed determinant ?
Thanks in advance.

  • 3
    $\begingroup$ When $r$ and $d$ are relatively prime, I think this follows from rationalality of the fibers of $M(r,d)\rightarrow\operatorname{Jac}^d(X)$ (which I think is proven in arxiv.org/abs/math/9907068). In that case the map to $\operatorname{Jac}^d(X)$ is the Albanese map of $M(r,d)$ and so if the $M(r,d)$ are the same so are the $M(r,\xi)$. I am not sure what happens when $r$ and $d$ are not relatively prime. $\endgroup$
    – dhy
    Jan 29, 2021 at 18:42

1 Answer 1


As dhy said, this follows if we know that the fibers $M(r,\xi)$ are rational. In fact, it suffices to know that they are unirational - we can deduce that their image in any map to an abelian variety is a point, hence any map from $M(r,d)$ to an abelian variety factors through the determinant map, and thus the Albanese of $M(r,d)$ is $\operatorname{Jac}^d$.

Their unirationality is not so hard to prove.

Take $L$ ample enough that $V \otimes L$ is globally generated for every stable vector bundle of rank $r$ and determinant $\xi$. (It suffices to have $H^1( X, V \otimes L (-P)) =0$ for all points $P$, i.e. it suffices to have $H^0(X, V^\vee \otimes K_X \otimes L^{\vee} (P))$, so it suffices to have $\deg L > 2g-1 + \frac{ \deg xi}{r}$.)

Then among maps $(L^{-1})^{n-1} \to V$, those which have rank $< n-1$ at a point $P$ form a codimension $2$ subset, so those which have rank $<n-1$ at any point form a codimension $1$ subset, and thus there exists a map $(L^{-1})^{n-1} \to V$ with full rank at every point. Hence the quotient is a line bundle, which because $\det V = \xi$, must be $ \xi \otimes L^{\otimes (n-1)}$. So $V$ is an extension of $\xi \otimes L^{\otimes (n-1)}$ by $(L^{-1})^{n-1}$.

Now there is an open subset of $\operatorname{Ext}^1 ( \xi \otimes L^{\otimes (n-1)}, (L^{-1})^{n-1} )$ parameterizing stable vector bundles, which maps to $M(r,\xi)$. By the above argument this map is surjective, so $M(r,\xi)$ is unirational.


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