For an elliptic curve $E$, how can we describe canonical divisor or addition $\operatorname{Sym}^2(E) \to E$ as projectivization of a vector bundle?

Question

Let $E$ be an elliptic curve. Then I know that $S = \operatorname{Sym}^2(E) = E \times E / S_2$ is a smooth surface, and if we choose an origin $P_0 \in E$, we get a $\mathbb P^1$-bundle $$\pi: S \to E, \{P,Q\} \mapsto P+Q.$$ So $S$ is a ruled surface over $E$. I would like to know determine a canonical divisor on $S$, in particular I wonder if it is a multiple of the diagonal $\Delta \subset S$. One strategy to understand $S$ and $\pi$ better would be to determine a vector bundle $\mathcal E$ on $E$ such that $S = \mathbb P(\mathcal E)$, but I don't know how to do that.

Any help would be appreciated :)

Answer 1

The canonical bundle of $ S $ is given by the Cartier divisor $ \frac{-1}{2} \Delta $ and it doesn't require looking at the $ \mathbb{P}^1 $-bundle. Let $ q : X = E \times E \rightarrow S $ be the quotient map and this is a ramified covering of degree $ 2 $ ramified exactly over the diagonal $ \Delta $ in $ S $. The pushforward $ q_* \mathcal{O}_X $ is locally free of rank $ 2 $ and splits as $ \mathcal{O}_S \oplus \mathcal{N} $ for some line bundle $ \mathcal{N} $ on $ S $ which squares to $ - \Delta $. Now Serre's relative duality using the relative dualizing sheaf $ \omega_q = \omega_X \otimes q^*\omega_S^{\vee} $ gives $$ (q_* \mathcal{O}_X)^{\vee} = q_* \omega_q = q_*q^* \omega_S^{\vee} = \omega_S^{\vee} \otimes q_* \mathcal{O}_X $$ where the first equality is relative duality, the second equality uses the fact that $ X $ has trivial canonical bundle and the last equality is the projection formula. Taking determinants on both sides, $$ \mathcal{N}^{\vee} = (\omega_S^{\vee})^2 \otimes \mathcal{N} $$ and rearrange to get $ \omega_S^2 = \mathcal{N}^2 = \mathcal{O}_S(-\Delta) $ so that the canonical class is given by $ -\Delta / 2 $.

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Addendum: The natural injective map $ s : \mathcal{O}_S \rightarrow q_* \mathcal{O}_X $ has a splitting obtained from the trace map $ t : q_* \mathcal{O}_X \rightarrow \mathcal{O}_S $ given by specifying it on an affine open $ U = \operatorname{Spec} A $ in $ S $ and $ q^{-1}(U) = \operatorname{Spec} B $ as simply the trace map $ B \rightarrow A $. Observe here that $ q $ is a finite morphism (quasi-finite + proper) i.e. $ B $ is finite over $ A $ as a module, so the trace map is well-defined. Then $ ts $ is just multiplication by $ 2 $ on $ \mathcal{O}_S $ so $ \frac{1}{2} t $ is a splitting of $ s $. Indeed we're assuming that the characteristic is not $ 2 $.

If $ \mathcal{N} $ is the cokernel of $ s $, it is coherent and from the above splitting $ q_* \mathcal{O}_X = \mathcal{O}_S \oplus \mathcal{N} $. For each point $ s \in S $, considering the $ k(s) $-dimensions of the fibers on both sides, we get that $ \dim_{k(s)} \mathcal{N}(s) = 1 $ so $ \mathcal{N} $ is indeed a line bundle.

Answer 2

To identify the rank 2 bundle on $E$ corresponding to the map $\pi$ note that the map $$ E \cong \{P\} \times E \hookrightarrow E \times E \to S $$ is a closed embedding and its composition with $\pi$ is the isomorphism $$ E \to E, \qquad Q \mapsto P + Q. $$ Therefore, its image $D_P \subset S$ is a section of $\pi$. Moreover, for $P_1 \ne P_2$ it is easy to see that $$ D_{P_1} \cap D_{P_2} = \{P_1 + P_2\} \in S $$ is a single point and the intersection is transverse. Since all $D_P$ are obviously algebraically equivalent, it follows that the degree of the normal bundle $\mathcal{O}_S(D_P)\vert_{D_P}$ of $D_P$ is equal to $1$, hence we have an exact sequence $$ 0 \to \mathcal{O}_S \to \mathcal{O}_S(D_P) \to \mathcal{O}_{D_P}(P') \to 0 $$ for a point $P' \in E \cong D_P$. Pushing it forward along $\pi$ we obtain $$ 0 \to \mathcal{O}_E \to \pi_*(\mathcal{O}_S(D_P)) \to \mathcal{O}_E(P') \to 0. $$ Therefore, the rank 2 bundle associated to $\pi$ (the middle term above) is an extension of a line bundle of degree $1$ by a line bundle of degree zero. It remains to note that if this extension is trivial, then $\pi$ has a unique section with normal bundle of degree $1$, which is definitely not true in our case (since sections of the the form $D_P$ cover all $S$), and that a non-trivial extension of this form is unique up to twist and translation.

Answer 3

Here is a different perspective on how to show $\omega_{\operatorname{Sym}^2(E)} \cong \mathcal O(-\frac 1 2 \Delta)$. Without motivation this might appear to be a bit over the top, but actually occurs in that way in what I'm studying.

Take K3 surface $S$ that admits an elliptic fibration $\pi: S \to \mathbb P^1$, such that $E$ occurs as a fiber over a point $P \in \mathbb P^1$[1]. Then $\pi$ induces a fibration on the Hilbert scheme, given as the composite map $$f:\operatorname{Hilb}^2(S)\to \operatorname{Sym}^2(S)\to \operatorname{Sym}^2(\mathbb P^1) \cong \mathbb P^2.$$ One can check that the fiber over $2P \in \mathbb P^2$ is actually $X = \operatorname{Sym}^2(E) \cup \mathbb P(T_S |_E)$, with the $\operatorname{Sym}^2$-component having multiplicity $2$. So if we restrict to a line $L \subset \mathbb P^2$, which intersects the locus of double points in $2P$ transversely, then $X$ is a divisor in $N = f^{-1}(L)$ with trivial normal bundle. So by the adjunction formula, $\omega_X = \omega_N|_X$. But as $\operatorname{Hilb}^2$ is holomorphically symplectic, $\omega_N = f^*\mathcal O(1)|_N$, which is also trivial in a neighbourhood of $X$. So $\omega_X = \mathcal O_X$.

On the other hand, $\mathcal O_N(X) = \mathcal O(\mathbb P(T_S|_E)) \otimes \mathcal O(\operatorname{Sym}^2(E))^{\otimes 2}$. So in total, $$\omega_{\operatorname{Sym}^2(E)} = \mathcal O(\operatorname{Sym}^2(E))|_{\operatorname{Sym}^2} = \mathcal O(- \frac 1 2 \mathbb P(T_S|_E)|_{\operatorname{Sym}^2} = \mathcal O(-\frac 1 2 \Delta).$$

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[1] For example take $E \subset \mathbb P^2$ by a Weierstraß equation, and then deform the equation to obtain $S$ as a divisor of type $(3, 2)$ in $\mathbb P^2 \times \mathbb P^1$