# Square of an elliptic curve and projective plane

Let's assume one takes $E = \mathbb{C}^* / \langle p \rangle$ an elliptic (Tate) curve over the complex field ($p = e^{2 \pi i \tau}$ where $1, \tau$ are the 2 periods in additive notation; $\Im \tau > 0$). On this take points $u_1, u_2, u_3$ such that $u_1 u_2 u_3 = 1$ and then mod out by the action of the symmetric group $S_3$. So we essentially have a hypersurface in $E^3$ - a copy of $E^2$ with coordinates $(u_1, u_2)$ and we mod out by permuting $u_1, u_2$ and $1/u_1 u_2$ (the $u_i$'s are zeros and their reciprocals poles of an elliptic function - essentially the only one up to constant with these zeros and poles).

The question: is this quotient space $\mathbb{P}^2$? I believe the answer is yes, but I can't see a way of using theta functions or other gadgets to explicitly give the isomorphism (whereby a theta function I mean $$\theta_p(x) = \prod_{l \ge 0}(1-p^l x)(1-p^{l+1}/x)$$ which reduces to the Jacobi theta via the triple product identity).

Finally, does this work over other fields (reals, finite fields, other reasonable fields)?

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What we have here is a special case of the following (well-known) construction: Starting with a smooth and proper curve $C$ we may consider its symmetric power $S^nC=C^n/\Sigma_n$. It (because we are dealing with a smooth curve) is also equal to Hilbert scheme of effective divisors of degree $n$ and is always smooth. Mapping an effective divisor to the corresponding line bundle gives a map $S^nC\rightarrow \mathrm{Pic}^n(C)$ to the part of the Picard scheme of degree $n$ line bundles. The fibre over a line bundle $L$ is the projective space associated to the space of sections of $L$. If $n>g(C)$ then this is the projective bundle associated to $\pi_\ast\mathcal L$, where $\mathcal L$ is the universal line bundle on $C\times\mathrm{Pic}^n(C)$ (and $\pi$ the projection). In the special case considered in the question $n=3$ and $g=0$ and we are considering the fibre over $\mathcal O(3\cdot0)$.
Consider some $a,b,c\in E$. Then $a\oplus b\oplus c=0_E$ in the group $E$ iff $a+b+c = 3\cdot 0_E$ in $\mathrm{Pic}^3(E)$ iff $a,b,c$ are colinear in the complete linear system $|\mathcal{O}_E(3\cdot 0_E)|\cong\mathbb{P}^2$. I.e. you "unordered triplets" scheme is ${\mathbb{P}^2}^*$. This is true over any field.