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I've been playing around with Riemann surfaces of cubics, and it seems to me that all surfaces obtained as coverings of the Riemann sphere from equations of the form $w^3 = q(z)$, where $q(z)$ is a cubic with three distinct roots, must be isomorphic.

Argument: we have a critical point of multiplicity 3 at each of the roots of $q(z)$. Monodromy around a small counterclockwise circuit about any of these points multiplies $w$ by the same cube root of unity ($\neq 1$). So monodromy around a circuit enclosing all three roots of $q(z)$ leaves $w$ unchanged, so no branch points over $\infty$.

Now we can move the three roots of $q(z)$ to any other positions using a Möbius transformation, so by the previous paragraph, we should be able to establish an analytic isomorphism between the surfaces via continuation.

Is this correct? If so, what is the common J-invariant? And since we can choose $q(z)=z^3-1$, this means that all these curves have CM, right?

References appreciated.

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    $\begingroup$ Your argument seems good, but it's not clear to me how to change $w^3 = z^3 - 1$ into $w^3 = 2(z^3 - 1)$ (i.e., I'm suspicious that your proof really applies only to monic polynomials $q(z)$). Perhaps I haven't thought about it long enough. $\endgroup$ – John Hughes Dec 23 '13 at 23:30
  • $\begingroup$ @John: Good point. However, I think rescaling will handle this. Just replace $w$ with $rw$ and divide the whole equation by $r^3$. Now let $r$ be the leading coefficient of $q(z)$. $\endgroup$ – Michael Weiss Dec 24 '13 at 0:28
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This is a great observation!

Let $E/\mathbb{Q}$ be a curve given by a model $v^3=q(u)$, for some cubic polynomial $q\in\mathbb{Q}[u]$, and assume that the projective closure of this model, i.e., $V^3=W^3q(U/W)$, is smooth, and it has a rational point $P$. Then, $(E,P)$ is an elliptic curve defined over $\mathbb{Q}$. Moreover, $E$ admits an endomorphism $$[\rho] : E\to E$$ that sends $[U,V,W]$ to $[U,\rho V,W]$, where $\rho$ is a primitive third root of unity. Hence, $\operatorname{End}(E)$ is strictly larger than $\mathbb{Z}$, and so $E/\mathbb{Q}$ is a curve with complex multiplication. Further, $$\operatorname{End}(E)\otimes \mathbb{Q} \cong \mathbb{Q}(\rho),$$ and therefore the complex multiplication is by an order in the ring of integers of the quadratic imaginary field $\mathbb{Q}(\rho)=\mathbb{Q}(\sqrt{-3})$. It is a fact that follows from the theory of complex multiplication that all the elliptic curves with complex multiplication by $\mathbb{Q}(\sqrt{-3})$ have $j$-invariant equal to $0$. It follows that all the elliptic curves of the form $V^3=W^3q(U/W)$ are isomorphic over $\mathbb{C}$ (as the $j$-invariant classifies elliptic curves up to isomorphism over $\mathbb{C}$).

For example, consider $E: V^3=U^3-W^3$ (or $v^3=u^3-1$ in affine coordinates), with $P=[1,1,0]$. Then, $E$ is isomorphic to $$E': Y^2Z-9YZ^2=X^3-27Z^3$$ or $y^2 - 9y = x^3 - 27$ in affine coordinates, via the map $\phi:E'\to E$ that sends $$\phi([X,Y,Z])=(Y-9Z,Y,3X),$$ which sends $[0,1,0]$ to $[1,1,0]$. The curve $E'$ is now easily checked to have $j=0$ via the usual formulas, and therefore it has complex multiplication by $\mathbb{Q}(\sqrt{-3})$.

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  • $\begingroup$ Thanks! What reference would you recommend for the result about CM by $\rho$ implying $j=0$? $\endgroup$ – Michael Weiss Feb 26 '14 at 18:42
  • $\begingroup$ See Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", Chapter 2, Example 1.3.2. $\endgroup$ – Álvaro Lozano-Robledo Feb 26 '14 at 19:02

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