Let $X$ be a smooth curve, and $E$ a rank $r$ vector bundle over $X$, $E$ is said very stable if every nilpotent map $$u:E\rightarrow E\otimes K_X$$ is zero (nilpotent means that the composition $E\rightarrow EK_X\rightarrow\cdots\rightarrow EK_X^r$ is zero)

Is there any sufficient condition for vector bundle to be very stable?


  • $\begingroup$ Do you have any reference about very stable vector bundle? What is the motivation for this definition? $\endgroup$
    – Giulio
    Apr 6, 2016 at 8:45
  • $\begingroup$ these bundles appears as a particular case in the study of Hitchin systems, you can see the paper of Gerard Laumon, "Un analogue globale du cone nilpotent" for example. $\endgroup$
    – Z.A.Z.Z
    Apr 6, 2016 at 9:02

1 Answer 1


This is not a complete answer but only deals with the rank 2 case with trivial determinant: Let $u=\Phi\neq0$ be a nilpotent Higgs field, then $\Phi^2=0,$ so the kernel line bundle $L\subset E$ (assuming $\Phi$ is not identically zero) contains the image of $\Phi.$ Hence, $\Phi$ is uniquely determined by a holomorphic section $\phi\in H^0(X;L^2K).$ This shows that a sufficient condition of a holomorphic rank 2 bundle with trivial determinant is that it does not contain any holomorphic line bundle $L$ such that $H^0(X;L^2K)\neq\{0\}.$

In particular, in the case of a surface of genus 2, stable holomorphic bundles of rank 2 with trivial determinant are uniquely determined by the divisor of holomorphic line subbundles in $Pic_{-1}(X)$ by Narasimhan-Ramanan. Therefore, in this situation the aforementioned condition is very natural. In particular it also follows from Narasimhan-Ramanan that the set of very-stable rank 2 bundles with trivial determinant over a genus 2 surface is $\mathbb CP^3$ without the Kummer surface (of $X$, corresponding to strictly semi-stable bundles) and without several copies of $\mathbb CP^2$ corresponding to the space of non-trivial extensions $$0\to S^*\to E\to S\to 0$$ of the 16 spin bundles $S$ on $X.$

  • 1
    $\begingroup$ Note that the nilpotent Higgs field $Phi$ gives rise to a holomorphic map $$L^*=E/L\to LK,$$ hence a section of $H^0(X;L^2K).$ $\endgroup$
    – Sebastian
    Apr 6, 2016 at 10:06
  • $\begingroup$ yeah it is triviality of determinant which intervene here! thanks $\endgroup$
    – Z.A.Z.Z
    Apr 7, 2016 at 7:51

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