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Is there a database that has all the known particular values of the $j$-invariant?

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What do you mean by "known"? For any $\tau\in\mathbb C$ with $\text{Im}(\tau)>0$, one can compute $j(\tau)$ to as much precision as one's computer allows, but presumably that's not what you mean. In general, if $\tau$ is algebraic and $[\mathbb Q(\tau):\mathbb Q]\ge3$, then $j(\tau)$ is transcendental over $\mathbb Q$, so you need to explain what would constitute "knowing" the value. When $\tau$ is quadratic over $\mathbb Q$, the associated ellitpic curve has CM, and $j(\tau)$ generates the Hilbert class field of $\mathbb Q(\tau)$. In that case, one can in principal determine the field and then write $j(\tau)$ in terms of a basis for that field. Is that what you mean? If so, I;m sure that many examples have been worked out over the years, but I'm not aware offhand of a place where they've been compiled. Although presumably they've been done for all imaginary quadratic fields of small class number. There's a sample computation for $\tau=\frac{1+\sqrt{-15}}{2}$ in my Advanced Topics in the Arithmetic of Elliptic Curves book (Example II.6.2.2), where it's shown that $$ j\left(\frac{1+\sqrt{-15}}{2}\right) = -52515-85995\frac{1+\sqrt{5}}{2}. $$ (The field $\mathbb Q(\sqrt{-15})$ has class number 2, and its Hilbert class field is $\mathbb Q(\sqrt{-15},\sqrt5)$.)

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  • $\begingroup$ Thank you. I will check you book. $\endgroup$
    – Wenzhe
    Commented Dec 25, 2020 at 22:13
  • $\begingroup$ Any hint on why $j(\tau)$ is automatically transcendental away from CM points ? $\endgroup$
    – reuns
    Commented Dec 26, 2020 at 0:11
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    $\begingroup$ @reuns That's a classical, but by no means easy, theorem due to Theodor Schneider, i.e,. he proved that if $\tau$ is algebraic, but not quadratic imaginary, then $j(\tau)$ is transcendental over $\mathbb Q$. Schneider, of course, is also famous for the Gelfond–Schneider theorem, which is also a result in transcendence theory and which solved Hilbert's 7th problem. $\endgroup$ Commented Dec 26, 2020 at 2:56
  • $\begingroup$ The "Grothendieck period conjecture" implies that 4 periods (by integrating $dx/(2y+x), xdx/(2y+x)$ on 2 classes forming a basis of $H_1(E(\mathbb C),\mathbb Z)$) for a non-CM elliptic curve $E$ with equation $y^2 + xy =x^3-36x/(j-1728)-1/(j-1728)$ with $j=j(\tau) \in \overline{\mathbb Q}$ are algebraically independent over $\overline{\mathbb Q}$, so the ratio of 2 periods (by integrating $dx/(2y+x)$) is transcendental and $\tau \in SL_2(\mathbb Z)$-orbit of the ratio is transcendental. See some text about Schneider's results in Waldschmidt's "Transcendence of periods: the state of the art". $\endgroup$ Commented Dec 29, 2020 at 19:25
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Any (finite) database containing explicit expressions for j-invariants of elliptic curves with CM can be extended by adding j-invariants of isogenous elliptic curves. Given an elliptic curve $E$ in its Weierstrass form and a finite subgroup $F$ of it, a classic paper of Velu provides explicit equations for $E':=E/F$ and the isogeny $E\rightarrow E'$. Now suppose we are working over $\Bbb{C}$ and we know that $E$ is isomorphic to $\frac{\Bbb{C}}{\Bbb{Z}+\Bbb{Z}\tau}$, hence the knowledge of the special value $j(\tau)$. The $j$-invariant of $E'$, which may be computed explicitly using its equation, then yields another special value $j(\tau')$ of the modular $j$-function where $\tau'$ is a period of $E'$. Alternatively, one may start from the target curve and goes up to obtain the $j$-invariant of an elliptic curve above it. To do this, suppose a Legendre form $y^2=x(x-1)(x-\lambda)$ for a CM elliptic curve $\frac{\Bbb{C}}{\Bbb{Z}+\Bbb{Z}\tau}$ is provided ($\lambda$ is an algebraic number). In other words, suppose we have $j(\tau)=256\frac{(\lambda^2-\lambda+1)^3}{(\lambda^2-\lambda)^2}$ in our database. Consider the isogeny $\frac{\Bbb{C}}{\Bbb{Z}+\Bbb{Z}(2\tau)}\rightarrow\frac{\Bbb{C}}{\Bbb{Z}+\Bbb{Z}\tau}$. By analyzing possible Legendre forms for $\frac{\Bbb{C}}{\Bbb{Z}+\Bbb{Z}(2\tau)}$, one can show its $j$-invariant $j(2\tau)$ belongs to $$\left\{16\frac{(u+\frac{1}{u}+14)^3}{(u+\frac{1}{u}-2)^2}\,\Big|\,u\in\left\{\lambda,1-\lambda,1-\frac{1}{\lambda}\right\}\right\}.$$ So there are three candidates for $j(2\tau)$, each in the form of an explicit algebraic number. Approximating $j(2\tau)$ numerically via the $q$-expansion, one can pick the correct expression for $j(2\tau)$ among them and add it to the database. The details of this approach for computing $j(2\tau)$ in terms of $j(\tau)$ can be found in this paper. An analogous method exists for $j(3\tau)$. So starting with for example $j(i)=1728$, for any two positive integers $m$ and $n$, an exact expression for $j\left(2^m3^ni\right)$ can be obtained. For instance $j(2i)=66^3$ and $j(3i)= 64(387+224\sqrt{3})^3(97−56\sqrt{3})$.

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