# Torelli theorem for stable vector bundle

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 ?

• 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.
– dhy
Jan 29, 2021 at 18:42

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 $$\operatorname{Jac}^d$$.
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 $$ at any point form a codimension $$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.