Dice Question Streamline Icon: https://streamlinehq.com

Schmutz’s conjecture: linear growth of the trace set implies arithmeticity

Prove that for any cofinite Fuchsian group Γ, if the trace set Tr(Γ) has linear growth—namely, there exist positive constants C and D such that for all n, the number of trace values with absolute value at most n is bounded by Cn+D—then Γ is arithmetic.

Information Square Streamline Icon: https://streamlinehq.com

Background

Schmutz proposed a strengthening of Sarnak’s conjectural characterization by replacing the bounded clustering property with linear growth of the trace set. A proposed proof contained a gap; Geninska and Leuzinger later fixed part of the argument for nonuniform lattices but did not settle Schmutz’s conjecture.

The present paper proves a different arithmeticity criterion (Theorem A) under a sub-linear growth assumption, but explicitly acknowledges that Schmutz’s conjecture remains unresolved, highlighting the significance of determining whether linear growth alone forces arithmeticity.

References

Conjecture 1.2 (Schmutz [33]). Let Γ be a cofinite Fuchsian group. If Tr(Γ) has linear growth then Γ is arithmetic.

On trace set of hyperbolic surfaces and a conjecture of Sarnak and Schmutz (2410.05223 - Hao, 7 Oct 2024) in Conjecture 1.2, Section 1 (Introduction and Statement of the Main Results)