Bounding the modular discriminant of an elliptic curve in the j-invariant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:45:32Z http://mathoverflow.net/feeds/question/48544 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48544/bounding-the-modular-discriminant-of-an-elliptic-curve-in-the-j-invariant Bounding the modular discriminant of an elliptic curve in the j-invariant Ariyan Javanpeykar 2010-12-07T08:07:14Z 2011-07-26T01:34:48Z <p>Consider an elliptic curve <code>$X=\mathbf{C}/ (\mathbf{Z}+\tau \mathbf{Z})$</code>, 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 $X$.</p> <p>Assume $j$ is algebraic, i.e., $X$ can be defined over a number field. </p> <p><strong>Question.</strong> 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?</p> <p>Firstly, one should be able to answer this question <strong>ineffectively</strong>. That is, to give a <strong>yes</strong> or <strong>no</strong> answer to the question. An effective bound (if it exists) might be a bit harder to obtain.</p> <p>I heavily edited this old question. Therefore, the first four comments below might not make sense anymore.</p> http://mathoverflow.net/questions/48544/bounding-the-modular-discriminant-of-an-elliptic-curve-in-the-j-invariant/71239#71239 Answer by Bruno for Bounding the modular discriminant of an elliptic curve in the j-invariant Bruno 2011-07-25T15:48:24Z 2011-07-25T15:48:24Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/48544/bounding-the-modular-discriminant-of-an-elliptic-curve-in-the-j-invariant/71279#71279 Answer by Noam D. Elkies for Bounding the modular discriminant of an elliptic curve in the j-invariant Noam D. Elkies 2011-07-26T00:57:35Z 2011-07-26T01:34:48Z <p>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 $\| \Delta \|$. </p> <p>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 $1 / 10^{100}$).</p>