# Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$

Let $X$ be a smooth projective curve. How do I construct a coherent sheaf $\mathcal{F}$ on $\text{Pic}^n X$ (i.e., the component of the Picard scheme of $X$ parametrizing line bundles of degree $n$) such that$$\mathbb{P}(\mathcal{F}) := \text{Proj}\,\text{Sym}(\mathcal{F})$$equals the $n$th symmetric power $\text{Sym}^nX$?

(Here, "equals" means canonically isomorphic as a scheme over $\text{Pic}^nX$"; notice that $\text{Sym}^nX$ is equipped with a canonical map to $\text{Pic}^nX$ because $X$ is a smooth curve.)

Ideally, the construction of $\mathcal{F}$ should be as canonical as possible.

• Every construction that I know requires choosing a rational point on $X$. Does your curve have a rational point? Does it at least have a zero-cycle of degree $1$? – Jason Starr Aug 27 '15 at 14:46
• I guess what you really need is a Poincare sheaf on $X\times \text{Pic}^nX$. Do you have a Poincare sheaf? If you have a zero-cycle of degree $1$ on $X$, then you have a Poincare sheaf. – Jason Starr Aug 27 '15 at 14:50
• Here is the earliest standard reference I know for this subject: projecteuclid.org/download/pdf_1/euclid.ijm/1255644637 – roy smith Aug 29 '15 at 20:09

There is an obstruction, as alluded to in my comments. If $X$ has genus $0$, for instance (perhaps not the case you are most interested in), then there exists such a sheaf when $n$ is odd if and only if $X$ is isomorphic to $\mathbb{P}^1_k$, where $k$ is your field. So for a conic over $\mathbb{R}$ having no real points, you cannot find such a sheaf when $n$ is odd. For $n$ even, there always exists such a sheaf.
Assume that there exists a Poincare invertible sheaf $\mathcal{P}_n$ on $\text{Pic}^n X\times X$. The sheaf you want on $\text{Pic}^n X$ is $$\mathcal{F} = R\textit{Hom}^0_{ \mathcal{O}_{\text{Pic}^n X}}(R\text{pr}_{1,*}(\mathcal{P}_n),\mathcal{O}_{\text{Pic}^n_X}).$$ But what does that mean? Following is a (somewhat) more explicit construction (used in Eisenbud-Harris, for instance) that depends on some choices.
Assume now that the genus is positive. Let $d$ be a positive integer such that $d(2g-2) + n > 2g-2$, e.g., $d=1$ always works. Let $s$ be a nonzero section of $H^0(X,\omega_{X/k}^{\otimes d})$. Let $D \subset X$ be the zero scheme of $s$. This is an effective Cartier divisor of degree $d(2g-2)$. The short exact sequence on $X$, $$0 \to \mathcal{O}_X \xrightarrow{s} \omega_{X/k}^{\otimes d} \xrightarrow{q} \omega_{X/k}^{\otimes d}|_D \to 0,$$ induces a short exact sequence on $\text{Pic}^n X\times X$, $$0 \to \mathcal{P}_n \xrightarrow{1\otimes \text{pr}_2^*s} \mathcal{P}_n\otimes_{\mathcal{O}} \text{pr}_2^*\omega_{X/k}^{\otimes d} \xrightarrow{1\otimes \text{pr}_2^*q} \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d}|_{\text{Pic}^n X\times D} \to 0.$$ Because of the hypothesis on $d$, the pushforward $$\mathcal{E}^0 := \text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d} \right)$$ is a locally free sheaf of rank $(2d-1)(g-1)+n$, and for every $i>0$, the higher direct image sheaf $R^i\text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d} \right)$ is the zero sheaf. Similarly, because $\text{Pic}^n X \times D$ is finite and flat over $\text{Pic}^n X$ of degree $d(2g-2)$, also the pushforward $$\mathcal{E}^1 := \text{pr}_{1,*}\left( \mathcal{P}_n\otimes_{\mathcal{O}}\text{pr}_2^*\omega_{X/k}^{\otimes d}|_{\text{Pic}^n X \times X} \right)$$ is locally free of rank $d(2g-2)$, and again all the higher direct image sheaves vanish. The pushforward of the map $1\otimes q$ is a morphism of locally free sheaves, $$d^0 : \mathcal{E}^0 \to \mathcal{E}^1.$$ Thus, one model of $R\text{pr}_{1,*}(\mathcal{P}_n)$ is the two-term complex $\mathcal{E}^\bullet$.
Now to form $R\textit{Hom}_{ \mathcal{O}_{\text{Pic}^n X}}(R\text{pr}_{1,*}(\mathcal{P}_n),\mathcal{O}_{\text{Pic}^n_X})$, you just take duals and transposes. Define $\mathcal{E}_0 = \left( \mathcal{E}^0 \right)^\vee$, define $\mathcal{E}_1 = \left( \mathcal{E}^1 \right)^\vee$, and define $d_1$ to be the transpose of $d^0$. Then your sheaf $\mathcal{F}$ is the cokernel of $$d_1: \mathcal{E}_1 \to \mathcal{E}_0.$$
Of course that description of $\mathcal{F}$ is not so canonical, e.g., if your curve has automorphisms, none of $\mathcal{E}_0$, $\mathcal{E}_1$ nor $d_1$ is equivariant for the induced automorphisms of $\text{Pic}^n X$. There are models of these total derived functors that are truly canonical. But if you want an explicit model, typically you have to make choices that remove some of the functoriality.