If one can construct a coherent sheaf over a smooth projective variety with all chern classes vanishing which is not locally free?
or for any such coherent sheaf $E$ , it must be locally free?
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Sign up to join this communityIf one can construct a coherent sheaf over a smooth projective variety with all chern classes vanishing which is not locally free?
or for any such coherent sheaf $E$ , it must be locally free?
No, that is not true. Let $X$ be $\mathbb{P}^2_k$. Let $p$ be a $k$-point of $\mathbb{P}^2_k$. Denote by $\mathfrak{m}_p \subset \mathcal{O}_X$ be the ideal sheaf of $p$. Let $E$ be $\mathcal{O}(-1)\oplus[\mathfrak{m}_p\cdot \mathcal{O}(+1)]$. Since $\mathfrak{m}_p$ is not locally free, also $E$ is not locally free. But a straightforward computation verifies that the total Chern class of $E$ is just $1$.
One quick way to see this is to observe that, up to choosing homogeneous coordinates $[x,y,z]$ on $\mathbb{P}^2_k$ with respect to which $p$ equals $[1,0,0]$, then the vector space of global sections of $\mathfrak{m}_p\cdot \mathcal{O}(1)$ is $2$-dimensional, generated by $y$ and $z$. These two global sections define an $\mathcal{O}_X$-module homomorphism, $$ \phi:\mathcal{O}_X^{\oplus 2} \to \mathfrak{m}_p\cdot \mathcal{O}(1). $$ In fact, $\phi$ is surjective, and the kernel of $\phi$ is just the $\mathcal{O}_X$-module homomorphism, $$ \psi: \mathcal{O}(-1) \to \mathcal{O}_X^{\oplus 2}, \ \ \psi(s) = (zs,-ys), $$ i.e., just the map coming from the Koszul complex on $\phi$. In fact, you can verify locally that the following is a short exact sequence of $\mathcal{O}_X$-modules, $$ 0 \to \mathcal{O}(-1) \xrightarrow{\psi} \mathcal{O}_X^{\oplus 2} \xrightarrow{\phi} \mathfrak{m}_p\cdot \mathcal{O}(1) \to 0. $$ Thus, in the $K$-group, $[\mathcal{O}_X^{\oplus 2}]$ is in the same equivalence class as $\mathcal{O}(-1)\oplus[\mathfrak{m}_p\cdot \mathcal{O}(1)]$. Therefore the total Chern class of $\mathcal{O}(-1)\oplus [\mathfrak{m}_p\cdot \mathcal{O}(1)]$ equals the total Chern class of $\mathcal{O}_X^{\oplus 2}$, and of course this is just $1$.