Let $E$ be a torsion-free sheaf on a smooth projective variety $X$ over $\mathbb{C}$. Let $H$ be an ample line bundle on $X$. Then we say $E$ is stable if $\mu_{H}(F)<\mu_{H}(E),\,\forall 0\neq F \subset E,$ (where slope of torsion free sheaf $F$ is defined by $\mu_{H}(F)=\frac{c_1(F)\cdot H}{rk(F)}$.)

I feel when $X$ is a smooth projective surface, $E$ is stable if and only if $\mu_{H}(F)<\mu_{H}(E),\forall 0 \neq F \subset E$, (where $F$ is a vector bundle). This is because $F^{**}$ is vector bundle on the surface $X$ with $\mu_{H}(F^{**})=\mu_{H}(F)$.

Question: 1) Is the above reason correct?

2) For higher-dimensional smooth projective variety(>2) can we say $E$ is stable if and only if $\mu_{H}(F)<\mu_{H}(E),\forall 0 \neq F \subset E$, (where $F$ is a vector bundle)

I.e. is it enough to consider $F$ a vector bundle to define stability? (The same logic does not go through in this case, since $F^{**}$ is not locally free for any given torsion free sheaf $F$).


1 Answer 1


The above statement on surfaces is correct if you suppose $E$ to be locally free, otherwise you should pay attention to torsion free subsheaves as well. I am thinking about the direct sum $E = I_p \oplus \mathcal{O} (-1)$ of the ideal sheaf $I_p$ of a point and the line bundle $\mathcal{O} (-1)$ on the projective plane. Then $E$ is torsion free, and should have no subsheaf which is torsion free and of higher slope, however it does have the subsheaf $I_p$ which destabilizes it.

In the same way, on a higher dimensional variety, the condition (2) is correct if you suppose that $E$ and $F$ are reflexive sheaves (i.e. they are isomorphic to their bidual sheaves), which for surfaces is the same as being locally free.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.