Let $X$ be smooth projective and connected curve over an algebraically closed field $k$. One knows the description of $k$-valued points of the moduli space $M_X$ of semistable vector bundles of fixed rank and degree on $X$: a point $c \in M_X(k)$, induced by a semistable vector bundle $E$ on $X$, is exactly the $S$-equivalence class of $E$.

Let $f: X \to S$ now be a projective morphism of schemes of finite type over a unversally Japanese ring $R$. There is also a moduli space $M_{X/S}$ parametrizing purely dimensional semistable sheaves on the fibers of $f$.

Maruyama defines an equivalence relation of such families: $E \sim E'$ if (1) $E \cong E' \otimes f^*L$ for some line bundle on $S$ or if (2) there are filrations $E^\cdot, E'^\cdot$ by coherent of the same finite length such that $gr E^\cdot$ and $gr E'^\cdot$ are flat over $S$ and equivalent as in (1), and for each geometric point $s \in S$ the induced filtrations $E_s^\cdot$ and ${E'}_s^\cdot$ are filtrations with stable gradings and same Hilbert polynomial (in other words, $E_s^\cdot$ and ${E'}_s^\cdot$ are Jordan-Hölder filtrations of $E_s$ resp. $E'_s$ with equivalent gradings).

Are two families $E$, $E'$ on $X$ of purely dimensional semistable sheaves parametrized by $S$ equivalent (as defined by Maruyama), if they define the same point in $M_{X/S}(S)$? I assume not, but unfortunately I cannot give a formal argument.

What about points in $M_{X/S}(s)$ for $s \in S$ geometric point? Are these given by S-equivalence classes of purely dimensional semistable sheaves on $X_s$?

[Edit] The situation, where $S = Spec~R$, $R$ a DVR and $X$ a semistable curve over $R$ with smooth generic fiber is already interesting. The base change to $K = Quot~R$ is flat, and hence $M_{X/S}\otimes K$ is the moduli space of vector bundles on $X_K = X \otimes K$, since $M_{X/S}$ uniformly corepresents the corresponding moduli functor. Seshadri states in [1] on p. 211 that $M_{X/S} \otimes k$ ($k$ residue field of $R$) is bijective to the moduli space of semistable sheaves on $X_k = X \otimes k$. Perhaps there is a better reference for this statement, since he does not give a proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.