Why do you say the moduli space of stable maps to a smooth projective Calabi-Yau is an orbifold? In general it's not even equidimensional and it's highly singular.

actually the same argument, together with the property that $\mathrm{Li}_{-a}(x)=(-1)^{a+1}\mathrm{Li}_{-a}(\frac{1}{x})$ shows that $f_m(a_1,\ldots,a_n)$ is a $\mathbb{Z}_2$-homogeneous polynomial!

Let me elaborate a little. I meant $i^s \mathrm{Li}_{-s}(e^{ix}) = \sum_{k>0} (ik)^s e^{ikx} = \left(\frac{1}{1-e^{ix}} \right)^{(s)} = \delta_+^{(s)}(x)$.

Thank you very much for your answer. Actually that is what I started with. To provide some motivation, I was trying to compute, in a completely formal way, the product $\delta_+^{(a_1)}(x)\ldots\delta_+^{(a_n)}(x)$ as a linear combination of $\delta_+(x)$ and its derivatives, where $\delta_+=\sum_{k\geq 0}e^{ikx}$, or more precisely, the positive-Fourier-modes part of the Dirac delta distribution. Such distribution can of course be expressed as polylogarithms, as you suggest, and my question becomes: how to express your product of poylogs as a linear combination of other polylogs?

Yes i meant precisely that. I'll think about it a little more, but this is already more than satisfactory. Thanks again.

let me elaborate: one cool thing about the equation [\Pi,\Pi]=\Pi^\sharp(\phi) for the integrability of antisymmetric case is that one can put $\Pi=\Pi_0+\epsilon \Pi_1$ and get a compatibility condition between $\Pi_0$ and $\Pi_1$ in order for $\Pi$ to generate an integrable distribution (this is the idea of bihamiltonian systems). In the fully Poisson case it reduces to $[\Pi_0,\Pi_1]=0$ of course, but if we start with a Poisson tensor $\Pi_0$ but perturb it to anything integrable (but still antisymmetric) we get a nice and explicit system of equations involving $\Pi_0$, $\Pi_1, \phi$.

great, thank you Robert! one question though: what about the linearity properties of you differential operator? I see that it is homogeneous of dregree r+2 with respect to multiplication of $\phi$ by a constant (even a function). what about linear combinations of tensors $a\phi+b\varphi$?

I was also wandering why you needed symmetry. I'll wait for your upgrde then! Thank you!

Thank you also for this reply. This is an extremely direct test indeed.