Let $X$ be a smooth projective surface, $\mathcal{E}$ a stable torsion free Higgs sheaf of degree 0. Consider the following short exact sequence:
$$0\rightarrow \mathcal{E}\rightarrow \mathcal{E}^{\*\*}\rightarrow \mathcal{S}\rightarrow 0$$ where $\mathcal{E}^{\*\*}$ is the double dual of $\mathcal{E}$. Then $\mathcal{S}$ is a skyscraper sheaf.


Q1: Why is the second Chern class $c\_2(\mathcal{S})\leq 0$?


Q2: We know that since $E$ stable, so $E$ admits a Hermitian-Yang-Mills metric, so is $E^{\*\*}$. Is it easy to explain that $c\_2(E^{\*\*})\geq0$, given the existence of Hermitian-Yang-Mills metric?



Second Chern class of Stable Higgs sheaf
Q1: in fact, for a skyscraper sheaf $\mathcal{S}$ on a surface one has $c\_2(\mathcal{S})=- \mathrm{length}(\mathcal{S})$: this follows for instance from example 15.3.1 in Fulton's *Intersection theory*.


Q2: It depends what you call "easy"... It is a nontrivial computation due originally to Lübke, and extended to the case of Higgs bundles by Simpson (Proposition 3.4 in *Constructing variations of Hodge structure using Yang-Mills theory
and applications to uniformization*. J. Amer. Math. Soc. 1 (1988), 867–918).


