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Let $C$ be a projective curve defined over a field $K$, and let $C^{(2)}$ and $J$ be its symmetric square and Jacobian, respectively.

There is a natural map $C^{(2)}\hookrightarrow J$, defined as follows: let $O\in J$ be a point of degree $2$, then map $P\in C^{(2)}\mapsto P - O$.

My question is: when is this map an embedding.

I understand the question is somehow related to the number of global sections of a degree 2 line bundle on $C$. I don't understand in what way. A reference would be appreciated as well.

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  • $\begingroup$ It is not an embedding if and only if $C$ admits a linear series of degree 2 and dimension $>0$, that is, if and only if $C$ is hyperelliptic. $\endgroup$
    – abx
    Commented Sep 11, 2023 at 15:52
  • $\begingroup$ Why is it the case? Do you have any references? $\endgroup$
    – kindasorta
    Commented Sep 11, 2023 at 15:59
  • $\begingroup$ Any book on curves — e.g. Arbarello et al. But at least for the injectivity, work it out by yourself, it follows directly from the definitions. $\endgroup$
    – abx
    Commented Sep 11, 2023 at 16:11
  • $\begingroup$ First off: many thanks. I am away from my computer, if it is not too much of a burden could you find the reference in Arbarello et al? $\endgroup$
    – kindasorta
    Commented Sep 11, 2023 at 16:32
  • $\begingroup$ As far as injectivity goes, the exercise should be something like: assume there is an effective degree 2 divisir $P+Q$ on $C$, and another one $R+S$, which are linearly equivalent, then there is a non-trivial degree 2 automorphism of $C$? $\endgroup$
    – kindasorta
    Commented Sep 11, 2023 at 16:34

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Assume that the genus $g$ of $C$ is $\geq 2$, so that $J$ is at least $2$-dimensional. The map $\phi : C^{(2)} \to J$ is a closed inclusion if and only if $C$ is not hyperelliptic.

More generally, the following are equivalent:

(1) $\phi: C^{(r)} \to \text{Pic}^r(C)$ is a closed embedding.

(2) $\phi: C^{(r)} \to \text{Pic}^r(C)$ is injective on $\mathbb{C}$-points.

(3) $H^0(C, \mathcal{O}(x_1 + x_2 + \cdots + x_r)) = \mathbb{C}$ for all effective divisors $x_1 + x_2 + \cdots + x_r$ of degree $r$.

I'll do the easy implications and cite Milne for the hard one.

$(1) \implies (2)$ is by definition.

$(2) \Longleftrightarrow (3)$: We have $\phi(x_1, x_2, \cdots, x_r) = \phi(y_1, y_2, \ldots, y_r)$ if and only if the divisors $x_1+x_2+\cdots+x_r$ and $y_1+y_2 + \cdots + y_r$ are rationally equivalent. This occurs if and only if there is some $f \in H^0(C, \mathcal{O}(x_1 + x_2 + \cdots + x_r))$ with zeroes at $(y_1, y_2, \ldots, y_r)$ (with multiplicity). So asking that $\phi(x_1, x_2, \cdots, x_r) \neq \phi(y_1, y_2, \ldots, y_r)$ for all distinct points of $C^{(r)}$ is the same as asking that $H^0(C, \mathcal{O}(x_1 + x_2 + \cdots + x_r))$ have no nonconstant global sections for any $(x_1, x_2, \ldots, x_r) \in C^{(r)}$.

Now, suppose that both $(2)$ and $(3)$ hold (since we know that they are equivalent); we want to show $(1)$. Since $C^{(r)}$ is smooth and projective, map from $C^{(r)}$ will be a closed embedding if and only if it is injective and injective on tangent spaces, and $(2)$ shows that the map is injective. It remains to check injectivity on tangent spaces.

So, given $x_1$, $x_2$, ..., $x_r \in C$, we want to show that $$D\phi : T_{(x_1, x_2, \ldots, x_r)} C^{(r)} \to T_{\phi(x_1, x_2, \ldots, x_r)} J \qquad (\ast)$$ is injective. Theorem 5.1 of Milne's notes on the Jacobian shows that the kernel of $(\ast)$ has dimension $\dim H^0(C, \mathcal{O}(x_1+x_2+\cdots+x_r))-1$. So, using $(3)$, we are done.

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  • $\begingroup$ So, what is the relation with hyperellipticity? $\endgroup$
    – kindasorta
    Commented Sep 11, 2023 at 16:48
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    $\begingroup$ By definition (taking $g \geq 2$ and $K$ algebraically closed to avoid awkward cases), a curve $C$ is hyperelliptic if and only if there is a degree $2$ rational map $f : C \to \mathbb{P}^1$. $\endgroup$ Commented Sep 11, 2023 at 16:53
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    $\begingroup$ In the comments above, you refer to order $2$ automorphisms of $C$. The logical implications are (a) If $f:C \to \mathbb{P}^1$ is a degree $2$ rational map, then there is an order $2$ automorphism $\sigma : C \to C$ so that, for all $x \in C$, we have $f^{-1}(f(x)) = x+\sigma(x)$ (equality of divisors). (b) If $\sigma : C \to C$ is an order two automorphism, then there is an algebraic curve $C/\sigma$, and $C \to C/\sigma$ is degree $2$. However, the curve $C/\sigma$ may or may not be $\mathbb{P}^1$. (continued) $\endgroup$ Commented Sep 11, 2023 at 17:15
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    $\begingroup$ So having a degree $2$ map to $\mathbb{P}^1$ is the stronger property and (at least for $g \geq 2$ and $K$ algebraically closed), that is the definition of "hyperelliptic". $\endgroup$ Commented Sep 11, 2023 at 17:16
  • $\begingroup$ Thank you very much for this clarification! Actually, I am interested in the case where the base is an arbitrary number field/ring of integers. How does the definition of "hyperelliptic" changes in this setting? $\endgroup$
    – kindasorta
    Commented Sep 11, 2023 at 17:31

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