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Recall that, for a complex number $\tau$ with positive imaginary part, the $j$-invariant is given by $j(\tau)=1728 \frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}$ where

$g_2(\tau)=60 \sum_{(m,n)\neq(0,0)}(m+n\tau)^{-4}$

and $g_3$ is unimportant. Clearly the zeros of $j$ are the same as the zeros of $g_2$ (and, if we are only considering the zeros, we can forget about the 60 and just consider the sum.)

Is there a way to directly see that $g_2(\rho)=0$, where $\rho=exp(\frac{2}{3} \pi i)$? (e.g. is there an algebraic trick to see that the terms in the sum cancel each other out?)

If not is there a reasonably simple indirect proof?

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I have since worked out the answer to this. So in case anybody comes across the question, I will sketch it below.

Observe $\rho$ has the minimal polynomial $x^2+x+1=0$, so if $\Lambda$ is the lattice $\mathbb{Z}+\rho\mathbb{Z}$, we have $\rho \Lambda = \rho\mathbb{Z}+\rho^2\mathbb{Z}=\rho\mathbb{Z}-\mathbb{Z}-\rho\mathbb{Z}=\Lambda$.

But clearly $g_2(a z)=a^4 g_2(z)$, so $g_2(\rho)=\rho^4 g_2(\rho)$, which gives $g_2(\rho)=0$.

Of course, this doesn't tell us that these are all the zeros (upto $SL_2(\mathbb{Z})$-invariance), but it does at least find the zero.

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  • $\begingroup$ Heh, $-(1+\rho)\Bbb Z$ isn't the same as $-\Bbb Z-\rho\Bbb Z$ but your equality still holds. :-) $\endgroup$
    – whacka
    Apr 14, 2015 at 0:17

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