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Let $\mathcal{E}$ be a stable sheaf on a smooth complex projective threefold $X$ and $Ext^k_0(\mathcal{E},\mathcal{E})$ be the traceless Ext groups, defined by the kernel of the trace map $$ Ext^k(\mathcal{E},\mathcal{E})\rightarrow H^k(X,\mathcal{O}). $$ It is known that automorphism, deformation and obstruction space of the sheaf $\mathcal{E}$ with fixed determinant is governed by $Ext^k_0(\mathcal{E},\mathcal{E})$ for $k=0,1,2$ respectively (see for example the famous paper by R. Thomas).

Question Assume now that $X$ be a Calabi-Yau threefold ($K_X\sim0$ and $H^1(X,\mathcal{O})=0$).

  1. The triviality of $H^1(X,\mathcal{O})$ claims that there is no continuous deformation of line bundles. So it seems to me that deformation of the sheaf $\mathcal{E}$ always fixes its determinant. On the other hand we have $0=Ext^k_0(\mathcal{E},\mathcal{E})\ne Ext^k(\mathcal{E},\mathcal{E})=\mathbb{C}$ (for stable sheaves).
  2. Since we have $H^k(X,\mathcal{O})=0$ for $k=1,2$, the trace maps for $k=1,2$ are trivial and we always have $Ext_0^k(\mathcal{E},\mathcal{E})=Ext^k(\mathcal{E},\mathcal{E})$. Doesn't this also suggest that Def-Obs theory of sheaves does not see whether one fixes determinant or not?

I may totally misunderstand something (to be honest, I am beginner and don't really understand Def-Obs theory (with fixed determinant)), but I would really appreciate it if someone kindly answer my questions.

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  • $\begingroup$ What is k in your first question (0 ?)? It does not seem true that automorphism of a sheaf with fixed determinant is controlled by Ext_0^0(E,E) because this can be 0 but the identity is surely an automorphism. For your other remarks, I think you are right : if there is no deformation with non-trivial determinant, then the Def-Obs theory can not see anything else ... Anyway, what is really important as difference between Ext and Ext_0 is the fact that Ext_0^0(E,E)=0 and Ext_0^3(E,E)=0. $\endgroup$
    – user25309
    Commented Nov 11, 2012 at 16:40

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