Questions about the “universal elliptic curve” over the affine $j$-line punctured at 0 and 1728

So my question refers to families of elliptic curves over the $\mathbb{A}^1_\mathbb{C}\setminus\{0,1728\}$ whose fiber above a point $j$ has $j$-invariant equal to $j$ (I understand it's not universal).

Some sources give an equation for such a family, namely $$E_1 := y^2 + xy = x^3 - \frac{36}{j-1728}x - \frac{1}{j-1728}$$

Thanks to TomChurch's comments, I'm revising my questions:

Does $E_1$ admit a nontrivial section? (other than the identity section)

Is there a way to see this family complex-analytically using quotients of the upper half plane?

What I mean is this: Let $\mathbb{H}$ be the upper half plane, and let $\mathbb{H}^\circ$ denote $\mathbb{H}$ punctured at the $SL_2(\mathbb{Z})$-orbits of $i$ and $e^{2\pi i/3}$. Let $\mathbb Z$ act on the product $\mathbb C\times\mathbb H^\circ$ by $$(m,n)\cdot(z,\tau) := (z + m\tau + n,\tau)$$ The quotient $\mathbb Z^2\backslash(\mathbb C\times\mathbb H^\circ)$ is an elliptic curve over $\mathbb H^\circ$. Let $SL_2(\mathbb Z)$ act on $\mathbb C\times\mathbb H^\circ$ by $$\gamma.(z,\tau) := \left(\frac{z}{c\tau + d},\frac{a\tau + b}{c\tau + d}\right)$$ This action descends to an action of $SL_2(\mathbb Z)$ on $\mathbb Z^2\backslash(\mathbb C\times\mathbb H^\circ)$, but as TomChurch noted, $\gamma = -I$ sends $(z,\tau)\mapsto(-z,\tau)$, the fibers of this quotient are actually copies of $\mathbb{P}^1$ and are not elliptic curves.

Is there a similar construction that will actually yield a curve like $E_1$?

• It looks to me like $E_2$ is a $\mathbb{P}^1$-bundle over $\text{SL}_2(\mathbb{Z})\backslash \mathcal{H}^\circ$; for $\gamma=-\text{Id}$ you have $\gamma\cdot(z,\tau)=(-z,\tau)$, so the fiber over $\tau\in \text{SL}_2(\mathbb{Z})\backslash \mathcal{H}^\circ$ is $E_\tau/\langle\pm 1\rangle\simeq \mathbb{P}^1$. (This commment may not make sense in the future if the question is revised.) – Tom Church Nov 14 '14 at 1:52
• @TomChurch Thanks I suppose what I wanted was for SL(2,Z) to act just on the upper half plane. I've edited it to reflect this. – Will Chen Nov 14 '14 at 1:55
• What you had was right (if $\text{SL}_2(\mathbb{Z})$ acts only on $\mathcal{H}^\circ$, it won't descend). The problem is there's no ("fine") universal elliptic curve: Let $T$ be the randomly-chosen $T=\mathbb{C}/\langle m+n(1+i)\rangle$. Let $\mathbb{Z}$ act on $T\times \mathbb{R}$ by $(t,x)\mapsto (-t,x+1)$, and set $X=(T\times\mathbb{R})/\mathbb{Z}$. This is an elliptic curve bundle over $S^1$ with every fiber $\simeq T$, but it cannot be pulled back from your $E_2$ (or any family over the $j$-line). [$X$ is nontrivial because e.g. $H_1(X)=\mathbb{Z}\neq \mathbb{Z}^3=H_1(T\times S^1)$.] – Tom Church Nov 14 '14 at 2:12
• You presumably threw out $i$ and $\rho$ because those curves have extra automorphisms. But every elliptic curve has a non-trivial automorphism, namely $P\to-P$, so there's a problem at every point. Roughly speaking, if you "loop around infinity," your points become negated, which is why the construction doesn't work. (This comment is meant only to help with the intuition. What Tom Church wrote is the formal mathematical formulation.) – Joe Silverman Nov 14 '14 at 3:05
• @TomChurch: I believe your last comment should be made into an answer. – rghthndsd Nov 14 '14 at 14:08

Question 1: Yes, there is a nontrivial section, namely $s: (x,y) = (-1/36, y_0)$ where $y_0$ is either solution of $y^2-y/36=-1/36^3$ (i.e. $y = 1/72 \pm 1/18^{3/2}$). [Note that the numerator $36$ in the equation for $E_1$ is a typo should be $36x$.]
Using the theory of elliptic surfaces we can show that in fact the group of section is infinite cyclic with generators $s$. The formula for $E_1$, considered as an equation in three variables $j,x,y$, defines a rational surface (one can choose any $x,y$ and solve for $j$). We can clear denominators by writing $$(x,y) = \Bigl( \frac{X}{(j-1728)^2},\frac{Y}{(j-1728)^3}\Bigr)$$ to obtain the equivalent equation $$Y^2 + (j-1728)XY = X^3 - 36(j-1728)^3 X - (j-1728)^5$$ with polynomial coefficients, and then use Tate's algorithm to find that the bad fibers at $j=\infty,0,1728$ are of Kodaira types $I_1$, $II$, $III^*$ respectively. The first two fibers make no contribution to the Neron-Severi lattice, and the last contributes $E_7$. Hence the Mordell-Weil group MW of sections is the quotient $E_8/E_7$ with the canonical height given by induced quadratic form; and this is an infinite cyclic group with a generator of height $1/2$. Since $s$ has height $1/2$ our claim follows.
There are three further elliptic surfaces with the same $j$-invariant and good reduction away from $j=\infty,0,1728$; they are obtained from $E_1$ by quadratic twists by $\sqrt{j}$, $\sqrt{j-1728}$, and $\sqrt{j(j-1728)}$. If I did this right, the first of these is an elliptic K3 surface of maximal Neron-Severi rank and trivial MW, and the others are rational elliptic surfaces with MW groups of rank $2$ and $1$ respectively.
For the K3 surface, the fibers at $j=\infty,0,1728$ are of types $I^*_1$, $IV^*$, $III^*$ and contribute $D_5$, $E_6$, $E_7$; this already accounts for a rank-20 subgroup of Neron-Severi, with discriminant $-24$, so the MW rank is zero, and one can show in various ways that there are no torsion sections (for instance the reducible fibers cannot accommodate nontrivial torsion). The remaining twists have singular fibers of type $I^*_1$, $II$, $III$ (giving $D_5 A_1$) and $I_1$, $IV^*$, $III$ (for $E_6 A_1$), which gives trivial torsion and rank $8-(5+1)=2$ and $8-(6+1)=1$ respectively.
• May I ask how you computed the three other elliptic surfaces with the same $j$-invariant? Are these the only other such elliptic surfaces over our twice punctured $j$-line with $j$-invariant $j$? – Will Chen Nov 18 '14 at 4:04
• Yes, they're the only ones. For any elliptic curve $\,E_0$ over a field, the elliptic curves $E$ with $j(E)=j(E_0)$ are precisely the quadratic twists of $E$, unless $E$ has $j$-invariant $0$ or $1728$ which isn't the case here $-$ for us $j(E_0)$ is the function $j \in {\mathbf C}(j)$, which does not equal $0$ or $1728$ in this field. To avoid new punctures we must use quadratic twists by $f(j)$ where $f$ has no zeros or poles outside $\{\infty,0,1728\}$ and there are only four choices including the trivial twist (twists by constants have no effect because ${\bf C}$ is algebraically closed). – Noam D. Elkies Nov 18 '14 at 4:10