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 $f$?
[Math] does the j-invariant satisfy a rational differential equation
differential equationsnt.number-theory
Related Solutions
Your questions are a part of what Deninger has been writing about for 20 years. He's proposed a point of view that sort of explains a lot of things about zeta functions. It's important to say that this explanation is more in a theoretical physics way than in a mathematical way, in that, as I understand it, he's predicted lots of new things which he and other people have then gone on to prove using actual mathematics. I guess it's kind of like the yoga surrounding the Weil conjectures before Dwork and Grothendieck made actual cohomology theories that had a chance to do the job (and eventually did). It's pretty clear to me that he's put his finger on something, but we just don't know what yet.
Let me try to say a few things. But I should also say that I never worried too much about the details, because the details he has are about a made up picture, not the real thing. (If he had the real thing, everyone would be out of a job.) So my understanding of the actual mathematics in his papers is pretty limited.
Question 1: He gives some evidence that Euler factors at both finite and infinite places should be seen as zeta-regularized characteristic polynomials. For the usual Gamma function, see (2.1) in [1]. For the Gamma factors of general motives, see (4.1) in [1]. For the Euler factors at the finite places, see (2.3)-(2.7) in [2]. He gives a description that works simultaneously at the finite and infinite places in (0.1) of [2]. Beware that some of this is based on an artificial cohomology theory that is designed to make things uniform over the finite and infinite places. (Indeed, at the risk of speaking for him, probably the whole point was to see what such a uniform cohomology theory would look like, so maybe one day we'll be able to find the real thing.)
Question 2: He expects his cohomology theory to have a Poincare duality which is "compatible with respect to the functional equation". See the remarks and references in [3] between propositions 3.1 and 3.2.
I'd recommend having a look at [3]. It's mainly expository. Also, I remember [4] being a good exposition, but I don't have it in front of me now, so I can't say much. He also reviews things in section 2 of his recent Archive paper [5].
[1] "On the Gamma-factors attached to motives", Invent. Math. 104, pp 245-261
[2] "Local L-factors of motives and regularized determinants", Invent. Math. 107, pp 135-150
[3] "Some analogies between number theory and dynamical systems on foliated spaces", Proceedings of the ICM, Vol. I (Berlin, 1998), pp 163-186
[4] "Evidence for a cohomological approach to analytic number theory", First ECM, Vol. I (Paris, 1992), pp 491-510
[5] "The Hilbert-Polya strategy and height pairings", arxiv.org
Let $K$ be the algebraic closure of the differential field $\mathbb{C}(T)$.
Let $\partial$ denote differentiation w.r.t. $T$. Now $\mathbb{C}(T)[\partial] \subseteq K[\partial]$ are rings of differential operators. Your function $F$ is a solution of $L(F)=0$ where $L \in K[\partial]$ is the differential operator $L = P_{k+1} \partial^{k+1} + \cdots + P_0 \partial^0$. The rings $\mathbb{C}(T)[\partial]$ and $K[\partial]$ (multiplication = composition) satisfy all properties of a Euclidean domain except commutativity. In particular, one can define an LCLM (least common left multiple) which behaves just like an LCM in Euclidean domains.
Let $L_1,\ldots,L_d$ be the conjugates of $L$ over $\mathbb{C}(T)$, obtained by applying ${\rm Gal}(K/\mathbb{C}(T))$ to $P_{k+1},\ldots,P_0$. Now let $M = {\rm LCLM}(L_1,\ldots,L_d)$. Then $M \in \mathbb{C}(T)[\partial]$ and $M$ is right-divisible by $L$. In particular $M(F)=0$.
In summary: Any function $F$ that satisfies a linear differential operator $L$ with algebraic-function coefficients will also satisfy a linear differential operator $M$ with rational-function coefficients. In Maple you can find $M$ with the command DEtools[LCLM](L, `and conjugates`);
Best Answer
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.