𝜂(𝛕) is defined throughout the upper halfplane, where it has period 24: DedekindEta[1 + 𝝉] == i^(1/6) DedekindEta[𝝉] By what does it scale if we only take half-steps? The formula for general 𝛕 becomes unclear. E.g., for 𝛕 := i√3, DedekindEta[1/2 + I Sqrt[3]] == (i^(1/12) ((1 + Sqrt[2]) (Sqrt[2] + Sqrt[3]))^(1/4) DedekindEta[I Sqrt[3]]) /(2^(3/16) (1 + Sqrt[3])^(3/8)) Presumably even worse for smaller steps. It's like particle physics—the closer you look, the more things crawl out of the woodwork. Trying for 1/3 of a step, (I DedekindEta[1/3 + I Sqrt[3]]^24)/DedekindEta[I Sqrt[3]]^24 == (1/59049)(10200 3^(1/6) + 1410 2^(1/3) 3^(1/6) - 25440 2^(2/3) 3^(1/6) + 11640 3^(5/6) - 8060 2^(1/3) 3^(5/6) - 5240 2^(2/3) 3^(5/6) - \[Sqrt](2682652239 + 568094400 2^(1/3) - 2140366320 2^(2/3) - 317440080 3^(1/3) - 499582620 2^(2/3) 3^(1/3) + 720388400 3^(2/3) - 704200460 2^(1/3) 3^(2/3) + 881386920 6^(1/3) - 159126160 6^(2/3))) To prove my point, I need to simplify the 24th root of this RHS. Gleep —rwg