# Bounding the modular discriminant of an elliptic curve in the j-invariant

Consider an elliptic curve $X=\mathbf{C}/ (\mathbf{Z}+\tau \mathbf{Z})$, where $\tau$ is an element in the complex upper half plane. We define $$\Vert \Delta\Vert(X) = (\Im \tau)^6 \vert q\prod_{k=1}^\infty (1-q^k)^{24}\vert,$$ where we write $q=\exp(2\pi i \tau)$ as usual. This is called the modular discriminant of $X$.

Assume $j$ is algebraic, i.e., $X$ can be defined over a number field.

Question. Can the function $\log\Vert \Delta \Vert(-)$ (on the moduli space of elliptic curves over $\mathbf{C}$) be bounded (from above or below) in terms of the (height of the) $j$-invariant?

Firstly, one should be able to answer this question ineffectively. That is, to give a yes or no answer to the question. An effective bound (if it exists) might be a bit harder to obtain.

I heavily edited this old question. Therefore, the first four comments below might not make sense anymore.

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It doesn't seem that the morphism $\pi$ play a role in your question, or am I missing something? – François Brunault Dec 7 '10 at 12:09
Actually it probably does. I first thought I had solved the question simply by taking the Weierstrass function and using that the modular discriminant is s^2(s-1)^2 but now I doubt this being correct. That's why I removed it again. The point is that the degree of \pi should come into play at some point, right? – Ariyan Javanpeykar Dec 8 '10 at 8:20
Ow maybe I removed it too quickly and it was actually correct.... – Ariyan Javanpeykar Dec 8 '10 at 8:24
Why do you take the absolute value of the imaginary part of $\tau$? It looks like it's already a positive real (which is more than I can say for that product in $q$). – S. Carnahan Dec 8 '10 at 9:12
Height only makes sense if $j$ is algebraic. Your formulation implies $X$ is any elliptic curve over $\mathbb{C}$. – Felipe Voloch Jul 25 '11 at 17:38

The new $\| \Delta \|$, defined as $\mathop{\rm Im}(\tau)^6$ times the absolute value of the usual modular form $\Delta$, is invariant under the full modular group $\Gamma = {\rm PSL}_2({\bf Z})$ acting on the upper half-plane $H$. This $\| \Delta \|$ is nonzero and continuous on the quotient $H / \Gamma$, and approaches zero exponentially as $\tau$ approaches the one cusp of $H / \Gamma$. Hence $\|\Delta\|$ is uniformly bounded above, without any hypothesis on $j$; and $\|\Delta\|$ is bounded below if we have an upper bound on $|j|$. The latter bound is completely effective, namely $$\| \Delta \| \gg (\log|j|)^6 / |j| {\rm\ \ \ \ as\ \ \ \ } |j| \rightarrow \infty,$$ and indeed $\| \Delta \| \sim C (\log|j|)^6 / |j|$ for some universal constant $C$, which is $(2\pi)^{-6}$ if I did this right. Now if you bound the height of $j$ from above then you impose an upper bound on the absolute value of any conjugate of $j$, and thus on $\| \Delta \|$.

Whether and how this lower bound depends on the height of $j$ then hinges on which flavor of height you're using, i.e. whether you normalize according to the degree $[{\bf Q}(j) : {\bf Q}]$, and whether you take logarithms. There is no such bound in the other direction: large height of $j$ does not force small $\| \Delta \|$ because it does not force $j$ to have a large conjugate (e.g. $j$ could be $1 / 10^{100}$).

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How does one obtain the effective lower bound for $\Vert \Delta\Vert$ in terms of the absolute value of $j$? – Ariyan Javanpeykar Jul 26 '11 at 9:10
@Ariyan: As $j(\tau) \rightarrow \infty$ with $\tau$ in the usual fundamental domain, $q \rightarrow 0$ with $q \sim 1/j$. Thus $|q| \sim 1/|j|$. Also Im$(\tau) = \log(1/|q|)/(2\pi) \sim \log|j|/(2\pi)$. This accounts for two of the factors of $\|\Delta\|$, and the remaining factor $\prod_{n=1}^\infty (1-q^n)^{24}$ approaches $1$ as $q \rightarrow 0$. This gives the asymptotic formula for $\|\Delta\|$ in terms of $|j|$, and all the error estimates are readily seen to be effective. – Noam D. Elkies Jul 26 '11 at 10:42
Thank you very much. This is a great answer. Do you think there's any hope in making the uniform upper bound on $\Vert \Delta \Vert$ explicit? – Ariyan Javanpeykar Jul 26 '11 at 13:00
You're welcome!Getting an explicit upper bound on $\|\Delta\|$ is a calculus exercise. We may assume $\tau = x+iy$ with $y^2 \geq 3/4$ (fundamental domain). Then the factor $\prod_{n=1}^\infty (1-q^n)^{24}$ of $\|\Delta\|$ is at most $\prod_{n=1}^\infty (1+e^{-\sqrt{3}\pi n})^{24}$, while the rest is $y^6 e^{-2\pi y}$ which is maximized at $y = 3/\pi$, etc.It seems (and is probably known) that in fact the max occurs at the sharper corner $\tau = (\pm 1 + \sqrt{3}i)/2$ of the fundamental domain, where $\|\Delta\| = .002+ = (2\pi^2/9)^6 / \Gamma(2/3)^{36}$. – Noam D. Elkies Jul 27 '11 at 1:55

As it stands, I think this question is still too vague to be answerable in generality. What kind of expression are you permitting for the bound? Certainly one can construct an artificial bound which doesn't even involve $j$ at all, which would be completely silly and certainly not what you have in mind.

As a first step, it's easy to see that no rational function of $j$ bounds $\Delta$. Indeed, if $R$ is a rational function such that $|\Delta| \leq |R(j)|$ everywhere on $X=\mathbb{H}/PSL(2, \mathbb{Z})$, then the bounded meromorphic function $\Delta/R(j)$ must be a constant. But this implies that $\Delta$ has weight $0$, which is false.

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Thank you for this very important remark. It made me realize that I'm actually looking for a bound on the logarithm of the modular discriminant. Sorry for the vagueness. I'm just interested in this particular discriminant $\log \Vert \Delta \Vert$, because from an Arakelov-theoretic point of view it coincides up to a constant with the Faltings delta invariant. – Ariyan Javanpeykar Jul 25 '11 at 16:52