Doubt in zeroes of Klein J Function

Question

While studying Analytic number theory from Tom M Apostol Modular functions and Dirichlet series in number theory I am unable to think about a conclusion of theorem 2.7 .

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I have doubt in how there is a triple zero at $\rho$ and J($\tau$) -1 has a double zero at $\tau$ = i.

Apostol writes that it's a consequence of Theorem 2.4 which is If f is modular and not identically zero, then in closure of fundamental region $R_\Gamma$ , number of zeroes of f equal to number of poles.

Can someone please explain how to derive this result.

Answer 1

Define $\,D(\tau) = E_4(\tau)^3-E_6(\tau)^2,\,$ then $\, J(\tau) = E_4(\tau)^3/D(\tau)\,$ where $\,E_4\,$ and $\,E_6\,$ are Eisenstein series. At the value $\,\rho=(1+\sqrt{-3})/2\,$ we have $\,E_4(\rho)=0\,$ and $\,E_6(\rho)\ne 0.\,$ Thus $\,J(\tau)\,$ has a triple zero at $\,\rho.\,$ At the value $\,i=\sqrt{-1}\,$ we have $\,E_4(i) \ne 0\,$ and $\,E_6(i) = 0.\,$ Since $\,J(\tau)-1 = E_6(\tau)^2/D(\tau)\,$ the function $\,J(\tau)-1\,$ has a double zero at $\,i.\,$

Note that $\,D(\tau)=2^{12}\,\eta(\tau)^{24}\,$ and $\,\eta\,$ is nonzero in $H$. Thus $D$ is nonzero in $H$ and therefore not both $\,E_4\,$ and $\,E_6\,$ are zero.

Note that $$E_4(\tau) = (\eta(\tau)^{24} + 256\,\eta(2\tau)^{24})/(\eta(\tau)\eta(2\tau))^8$$ whose numerator splits into distinct linear factors one of which is $\,\eta(\tau)^8 - 2^{8/3}(1-\rho)\, \eta(2\tau)^8\,$ which is zero if $\,\tau=\rho\,$ and this is why $\,E_4\,$ has a simple zero at $\,\rho.$

Note that $$E_6(\tau) \!=\!(\eta(\tau)^{16} \!-\! 512\,\eta(\tau)^8\tau(4\tau)^8 \!-\! 8192\,\eta(4\tau)^{16})\\ (\eta(\tau)^8 \!+\! 32\,\eta(4\tau)^8) / \eta(2\tau)^{12} $$ whose numerator splits into distinct linear factors one of which is $\,\eta(\tau)^8 \!-\! 2^6(4\!+\!\sqrt{18})\, \eta(4\tau)^8\,$ which is zero if $\,\tau=i\,$ and this is why $\,E_6\,$ has a simple zero at $\,i.$

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Note that the Wikipedia Dedekind eta function article Special values section lists the values

$$\eta(i) = \frac{\Gamma(\frac14)}{2^{\frac78}\pi^{\frac34}}$$ $$\eta(2i) = \frac{\Gamma(\frac14)}{2^{\frac{11}8}\pi^{\frac34}}$$

Using these values, verify that $\,E_4(i) = \frac3{64}\frac{\Gamma(1/4)^8}{\pi^6} \approx 1.4557 \ne 0$ and similarly for $\,E_6(\rho).$

Answer 2

• $J-a$ has only one simple pole on the modular curve (which is a compact Riemann surface) thus it has only one simple zero.

• $J(i)$ is the $J$-invariant of (the isomorphism class of) the complex torus $\Bbb{C}/(\Bbb{Z}+i\Bbb{Z})$, the only one with complex multiplication by $\Bbb{Z}[i]$. The Weierstrass function gives an isomorphism to an elliptic curve $E:y^2=4x^3+Ax+B$. Since the elliptic curve $E':y^2=x^3-ax-a,(a,b)=(1,0)$ has complex multiplication by $\Bbb{Z}[i]$ (sending $(x,y) \to (-x,iy)$) it means $E\cong E'$ and hence $J(i)=J(E)=J(E')=\frac{4a^3}{4a^3+27b^2} =1$.

• For $J(e^{2i\pi /3})=0$ it is the same with $\Bbb{C}/(\Bbb{Z}+e^{2i\pi/3}\Bbb{Z})$ and $y^2=x^3-1$ having CM by $\Bbb{Z}[e^{2i\pi/3}]$ (sending $(x,y)\to (e^{2i\pi /3}x,y)$)

• The chart from a small disk $|z|<\epsilon$ to $SL_2(\Bbb{Z})i$ is not $z\to SL_2(\Bbb{Z})(i+z)$ but $z\to SL_2(\Bbb{Z})ie^{z^{1/2}}$, this is because $i$ is fixed by $\tau\to -1/\tau$, thus $J-1$ has a simple zero at $SL_2(\Bbb{Z})i$ means that $J(z)-1$ has a double zero at $i$.

• There are 3 elements of $SL_2(\Bbb{Z})$ fixing $e^{2i\pi /3}$ thus $J$ has a simple zero at $SL_2(\Bbb{Z})e^{2i\pi /3}$ means that $J(z)$ has a triple zero at $e^{2i\pi /3}$.