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Klein's modular function $J(z)$ is defined and studied in e.g. Apostol's book Modular functions and Dirichlet series in number theory.

Certain specific evaluations are available, for example,

$$J(i) = 1$$

Additionally, it is known that $J(z)$ takes on all complex values.

Question: Can one solve for that $z$ in the fundamental domain

(either explicitly or numerically) which satisfies $J(z) = \mathrm{i}$?

ccorn
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user2052
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2 Answers2

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Yes, it is possible. Here's the way to do it: choose your favorite elliptic curve $E/\mathbf C$ with $j(E) = i$. For example, the curve

$$y^2 +xy = x^3 -36(j-1728)^{-1}x - (j-1728)^{-1}$$

has $j(E)=j$ for $j \neq 0, 1728$ (so, just plug $j=i$). Compute the periods $\omega_1$ and $\omega_2$. Then $\tau = \omega_2/\omega_1$ satisfies $j(\tau)=i$.

Bruno Joyal
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Yes, the inverse function of the $j$-invariant can be expressed in terms of the hypergeometric function. Since,

$$j(\tau) = 1728J(\tau)\tag{1}$$

we can solve for $\tau$ as,

$$\tau = i\frac{{}_2F_1(1/6, 5/6; 1; 1-\alpha)}{{}_2F_1(1/6, 5/6; 1; \alpha)}\tag{2}$$

where $\alpha$ is any root of the quadratic,

$$4\alpha(1-\alpha)=\frac{1}{J(\tau)}\tag{3}$$

Since your post is about $J(\tau) = i$, plugging that into (3), solving for $\alpha$, then plugging either root to (2), we find both,

$$\tau \approx 0.199329+0.7783729 i$$

$$\tau \approx -0.308752+1.2056644 i$$

yield

$$J(\tau) = i$$

Having numerically determined $\tau$, I then tried Mathematica's Recognize function to see if it had a closed-form, but $\tau$ does not seem to be an algebraic number of deg $< 8$.

Tito Piezas III
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  • +1 for the standardized notation. Remark: The second solution for $\tau$ is the sought one, as it is in the fundamental domain. The other one is $-1/\tau$. – ccorn Jan 24 '14 at 23:39
  • Considering the simplicity of $J(\tau) = i$, I first considered that $\tau$ might be algebraic. But it doesn't seem to be. However, perhaps it still has a closed-form in terms of maybe gamma functions and the like. – Tito Piezas III Jan 24 '14 at 23:48
  • I have tried Pari/GP's integer relation algorithms on $\left(X, \ln 2, \ln 3, \ln\pi, \ln\Gamma(1/6)\right)$ with $X\in{\Re\ln\tau,\ln\Re\tau}$ with $200$ bits precision, no hit. But probably $\alpha$ should be mixed in as well. – ccorn Jan 25 '14 at 04:24
  • I also tried the Inverse Symbolic Calculator by using the absolute value of $\tau$, as well as its $\Re$ part. No hit also. By the way, do you know the command for this online Magma calculator to calculate the Galois group? – Tito Piezas III Jan 25 '14 at 04:43
  • I hope this helps – ccorn Jan 25 '14 at 05:15
  • From Zudilin's MO comment: Schneider proved that the values of the modular invariant at non-CM points are transcendental. But then $\tau$ should have degree $2$ over $\mathbb{Q}$. Am I missing something? – ccorn Jan 25 '14 at 12:09
  • @BalarkaSen: That's it: If $\tau$ and $j(\tau)$ are algebraic, then $\tau$ is a CM point. We have to conclude here that $\tau$ is transcendental. – ccorn Jan 25 '14 at 18:36
  • @BalarkaSen: Indeed. (Besides: Typo: $j(\tau)=1728\mathrm{i}$ here). Thanks for the refresher! – ccorn Jan 25 '14 at 19:19