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? 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 :)
 A: 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.
A: 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 $.

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.
