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One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_*(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity and Riemann-Roch, for $n$ large this is a vector bundle . $E$. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^*u_*L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$.

But it is not clear to me why $E=\text{some line bundle}\otimes P_n$.

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One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_*(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity and Riemann-Roch, for $n$ large this is a vector bundleby the Riemann Roch. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^*u_*L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$.

But it is not clear to me why $E=\text{some line bundle}\otimes P_n$.

1

One way to see that there is a vector bundle $E$ over $J(C)$ with $C^{(n)}\cong \mathbb{P}(E)$ is using semi continuity. Consider the closed immersion $C^{(n-1)}\hookrightarrow C^{(n)}$. This is a divisor on a smooth variety and so corresponds to a line bundle $L$. We take the push forward $u_*(L)$ where $u:C^{(n)}\to J(C)$. Now using semi-continuity, for $n$ large this is a vector bundle by the Riemann Roch. In order to give a morphism $\phi:C^{(n)}\to \mathbb{P}(E)$, it suffices to check that the natural map $u^*u_*L\to L$ is surjective. This is easy to see by checking it over fibers of $u$. Also it is easy to check that $\phi$ is an isomorphism and that the pullback of $\mathscr{O}(1)$ is $C^{(n-1)}$, by looking at the fibers of $u$.

But it is not clear to me why $E=\text{some line bundle}\otimes P_n$.
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