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Let $C$ be a complex smooth projective curve of genus $g$, $C_d$ its d-fold symmetric product and $C^d$ its d-fold fiber product. Let $L$ a line bundle on $C$, and $L\boxtimes\cdots\boxtimes L$ (d copies) the coresponding line bundle on $C^d$. Since $L\boxtimes\cdots\boxtimes L$ is invatiant under the action of the symmetric group $S_d$, it descends to a line bundle $\widetilde{L}$ on $C_d$.

My question is:

  1. Can $\widetilde{L}$ be expressed by $\theta$ and $x$, here $\theta$ is the class of the pull back of the theta divisor and $x$ is the class of $C_{d-1}$ in $C_d$?

  2. Is $\widetilde{K_C}$ the canonical bundle of $C_d$?

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Here are the answers to your questions.

  1. The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$ To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection.

  2. The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$ EDIT: Here is a justification. If $\pi : C^{d} \rightarrow C_{d}$ is the quotient map, then $\Delta$ is the branch divisor of $\pi.$ By relative duality we have $$({\pi}_{\ast}\mathcal{O}_{C^d})^{\vee} \cong {\pi}_{\ast}(K_{C^d} \otimes {\pi}^{\ast}{K}_{C_d}^{\vee}) \cong {\pi}_{\ast}\pi^{\ast}(\widetilde{K}_{C} \otimes K_{C_d}^{\vee}) \cong \widetilde{K}_{C} \otimes {K}_{C_d}^{\vee} \otimes {\pi}_{\ast}\mathcal{O}_{C^d}$$ We then must have that $\mathcal{O}_{C_d}({\Delta}/2) \cong \det(\pi_{\ast}\mathcal{O}_{C^d})^{\vee} \cong \widetilde{K}_{C} \otimes K_{C_d}^{\vee}.$

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Thank you very much! – Messi Jun 28 '13 at 7:48

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