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Let $j$ be the Klein $j$-invariant (from the theory of modular functions).
Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for any rational function $f$?

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Out of curiosity, why are you interested in such a question? (Lest it not be obvious, I mean absolutely no disrespect in asking. I'm merely interested in whether the (non)existence of such an equation would have interesting consequences.) –  Nick Salter Feb 3 '11 at 23:35
    
(a) other interesting functions satisfy nice differential equations, but there's conspicuously no such equation for j in the usual books. (b) j does satisfy a third-order equation (as one of the other responders pointed out). (c) There is a nice computer graphic of j on Wikipedia, but googling for "C code for the j-invariant" doesn't produce much, and I doubt that using the q-series would be a good way to compute j, so maybe if it satisfied a nice differential equation that would be useful. –  Michael Beeson Feb 4 '11 at 0:59
    
"I doubt that using the q-series would be a good way to compute j" - yes, you can compute the invariant with q-series; just repeatedly apply modular transformations until $\tau$ has large enough imaginary part (and thus a tiny enough $q$), after which you can now use theta functions. –  J. M. May 9 '11 at 2:57
    
In practice, once you've found an equivalent $\tau$ in the fundamental domain you might want to use not $q^{-1} + 744 + 196884 q + \cdots$ but $E_4^3 / q \prod_n (1-q^n)^{24}$ for which you need fewer terms for a good approximation. –  Noam D. Elkies Sep 12 '11 at 17:49

4 Answers 4

up vote 34 down vote accepted

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).

For a rigorous proof:

Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$.

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 $g$.

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.

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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:

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)

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16  
Abstractly one can see this in the following way: let Q be the field of fractions of the ring of quasimodular forms for $\mathrm{SL}(2,\mathbf{Z})$. Then $j \in Q$, Q has transcendence degree three over $\mathbf{C}$ (generated by $E_2, E_4$ and $E_6$), and $Q$ is closed under differentiation since the derivative of a quasimodular form is quasimodular. Hence j and its first three derivatives are algebraically dependent, so j satisfies a nonlinear third degree differential equation. –  Dan Petersen Feb 3 '11 at 21:29

(this is too long for a comment)

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.

       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 )]
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Please tell me how you computed that equation! –  Michael Beeson Feb 4 '11 at 21:32
    
In FriCAS, create a list [1,744,196884,...] containing the first "few" (I used 6000, but much less should suffice) terms of the expansion, and enter guessADE(l, maxDerivative==3, maxPower==6, maxDegree==5, check==skip, debug==true), wait roughly 200 seconds. The settings for maxPower and maxDegree make the guessing much faster. If you have no clue (as I did) and guess without it can take an hour or so. –  Martin Rubey Feb 5 '11 at 8:38

A third-order differential equation for $j(\tau)$ is gotten via the Schwarzian derivative. The result is (1.13) of the paper by Harnad: http://arxiv.org/abs/solv-int/9902013

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