does the j-invariant satisfy a rational differential equation? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:44:15Z http://mathoverflow.net/feeds/question/54221 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation does the j-invariant satisfy a rational differential equation? Michael Beeson 2011-02-03T18:37:14Z 2011-09-12T17:35:19Z <p>Let $j$ be the Klein $j$-invariant (from the theory of modular functions).<br> Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for any rational function $f$? </p> http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation/54234#54234 Answer by Felipe Voloch for does the j-invariant satisfy a rational differential equation? Felipe Voloch 2011-02-03T20:33:23Z 2011-02-03T20:33:23Z <p>The j-function satisfies a third-order differential equation. I've learned this from an old paper of Daniel Bertrand which I am having trouble locating right now. Maybe is this one:</p> <p>MR0550281 (81i:10042) Bertrand, Daniel Propriétés arithmétiques des dérivées de la fonction modulaire $j(\tau )$. (French) Séminaire de Théorie des Nombres 1977–1978, Exp. No. 22, 4 pp., CNRS, Talence, 1978, 10F37 (10F35)</p> http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation/54235#54235 Answer by Alison Miller for does the j-invariant satisfy a rational differential equation? Alison Miller 2011-02-03T20:33:48Z 2011-09-12T17:35:19Z <p>No. Conceptually, the reason is that $j'(z)$ is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight $2$, so it cannot be expressed in terms of $j$ (weakly holomorphic modular form of weight $0$) and $z$ (not anywhere near being a modular form).</p> <p>For a rigorous proof:</p> <p>Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$.</p> <p>Suppose that the $j$ invariant did satisfy a differential equation of your form. Then we'd have $f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions $z$ and $j(z)$ are algebraically independent (this is just saying that $j(z)$ is a transcendental function). Hence the underlying two-variable rational function $f(x, y)$ satisfies $f(x, y) = f(x, y+1)$. This then easily implies that $f(x, y)$ must be independent of $y$, e.g. $f(x, y) = g(x)$ for some rational function $g$.</p> <p>So our original differential equation must actually take the form $j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight $2$ while the right hand side has weight 0, and a nonzero modular form has a unique weight, so this is impossible.</p> http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation/54312#54312 Answer by Martin Rubey for does the j-invariant satisfy a rational differential equation? Martin Rubey 2011-02-04T13:45:35Z 2011-02-04T13:45:35Z <p>(this is too long for a comment)</p> <p>Here is the explicit equation of order three for the $q$-expansion of $j$ multiplied by $q$. Keep in mind that this does not prove that there is no order one differential equation, so it is not an answer to the question.</p> <pre> n [x ]f(x): 3 4 4 3 5 2 , 2 5 (2x f(x) - 6912x f(x) + 5971968x f(x) )f (x) - 2x f(x) + 3 4 4 3 6912x f(x) - 5971968x f(x) * ,,, f (x) + 3 4 4 3 5 2 ,, 2 (- 3x f(x) + 10368x f(x) - 8957952x f(x) )f (x) + 2 4 3 3 4 2 , 5 (6x f(x) - 20736x f(x) + 17915904x f(x) )f (x) - 6x f(x) + 2 4 3 3 20736x f(x) - 17915904x f(x) * ,, f (x) + 3 2 4 5 , 4 (x f(x) - 1968x f(x) + 2654208x )f (x) + 2 3 3 2 4 , 3 (- 4x f(x) + 7872x f(x) - 10616832x f(x))f (x) + 4 2 3 3 2 , 2 (5x f(x) - 8352x f(x) + 12939264x f(x) )f (x) + 5 4 2 3 , 5 4 (- 2f(x) + 960x f(x) - 4644864x f(x) )f (x) + 1488f(x) - 331776x f(x) = 0 , 2 3 4 f(x)= 1 + 744x + 196884x + 21493760x + O(x )] </pre> http://mathoverflow.net/questions/54221/does-the-j-invariant-satisfy-a-rational-differential-equation/54369#54369 Answer by David Grant for does the j-invariant satisfy a rational differential equation? David Grant 2011-02-04T21:54:37Z 2011-02-04T21:54:37Z <p>A third-order differential equation for $j(\tau)$ is gotten via the Schwarzian derivative. The result is (1.13) of the paper by Harnad: <a href="http://arxiv.org/abs/solv-int/9902013" rel="nofollow">http://arxiv.org/abs/solv-int/9902013</a></p>