The depth-one realization conjecture for the weight-five Grassmannian polylogarithm

About 7 years old · traced to

Let Gr⁡5\operatorname{Gr}_5 denote the weight-five Grassmannian polylogarithm, let Alt⁡10\operatorname{Alt}_{10} denote alternation over ten points, and let r⁡3±\operatorname{r}_3^\pm be the invariant and anti-invariant combinations of the triple ratio under swapping its two triples. Consider the combination

1640 Gr5+53Alt⁡10[2I4,1(cr⁡(136∣2459),r⁡3+(12∣345,678))+I4,1(cr⁡(346∣1279),r⁡3−(12∣345,678))].\frac{1}{640}\,\mathrm{Gr}_5+\frac{5}{3}\operatorname{Alt}_{10}\left[2I_{4,1}(\operatorname{cr}(136|2459),\operatorname{r}_3^+(12|345,678))+I_{4,1}(\operatorname{cr}(346|1279),\operatorname{r}_3^-(12|345,678))\right].

Weight-five depth-one conjecture. (i) There exists a formal linear combination Q5(v1,…,v10)\mathcal{Q}_5(v_1,\dots,v_{10}) of rational functions on Conf⁡10(5)\operatorname{Conf}_{10}(5) such that Li⁡5(Q5(v1,…,v10))\operatorname{Li}_5(\mathcal{Q}_5(v_1,\dots,v_{10})) equals the displayed combination modulo products. (ii) Assuming (i), the function φ ⁣:GL⁡5(C)10→R\varphi\colon\operatorname{GL}_5(\mathbb{C})^{10}\to\mathbb{R} defined by

φ(g1,…,g10)=L5(Q5(g1v,…,g10v))\varphi(g_1,\dots,g_{10})=\mathcal{L}_5(\mathcal{Q}_5(g_1v,\dots,g_{10}v))

is a bounded measurable 99-cocycle whose continuous cohomology class is a non-zero rational multiple of the Borel class b5(5)∈Hcts9(GL⁡5(C),R)b_5^{(5)}\in H^9_{\mathrm{cts}}(\operatorname{GL}_5(\mathbb{C}),\mathbb{R}). This is motivated by the conjectured structure of the motivic Lie coalgebra in weight five: the corrected Grassmannian expression is expected to admit a depth-one description and realize the Borel regulator class, but the existence of Q5\mathcal{Q}_5 and the subsequent cohomological identification are not established here.

References

Primary source

Steven Charlton, Herbert Gangl and Danylo Radchenko, “Explicit formulas for Grassmannian polylogarithms”, arXiv:1909.13869 (2022).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.