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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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3 Answers 3

You are looking at the set of points (P,Q,R) in E^3 with P+Q+R=0 and modding out by S_3.if E is embedded in the plane as a cubic, such a triple genericaly determines a line in the plane and conversely a general line determines a triple of points like that. So your space is birational to (the dual) projective plane. You would have to look at the triples with coincidental points to see if the quotient is smooth and whether the birational map is an isomorphism . Tuis works over any algebraically csed field.

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No genericity condition is needed: any line intersects the cubic in an effective degree 3 divisor summing to zero, and conversely. So the result is isomorphic to the projective plane (it is also the linear system |0+0+0|, where 0 means the group identity). –  MartinG Aug 17 '10 at 8:09

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 $\mathcal O(3\cdot0)$.

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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.

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