Orders of bounded and strongly unbounded lattice type

TL;DR

This paper investigates the classification of orders based on their lattice types, establishing conditions under which they are of bounded or strongly unbounded lattice type, and relates these properties to hypersurface rings and their branched covers.

Contribution

It proves that orders of bounded lattice type are actually of finite lattice type and characterizes strongly unbounded lattice type via the $ ext{ extonehalf}$-length invariant, extending to hypersurface rings.

Findings

Orders of bounded lattice type are of finite lattice type.

Infinite non-isomorphic lattices with same $ ext{ extonehalf}$-length imply strongly unbounded lattice type.

Hypersurface rings have bounded or strongly unbounded lattice type iff their double branched covers do.

Abstract

Brauer and Thrall conjectured that a finite-dimensional algebra over a field of bounded representation type is actually of finite representation type and a finite-dimensional algebra (over an infinite field) of infinite representation type has strongly unbounded representation type. These conjectures, now theorems, are our motivation for studying (generalized) orders of bounded and strongly unbounded lattice type. To each lattice over an order we assign a numerical invariant, $\underline{\h}$-length, measuring Hom modulo projectives. We show that an order of bounded lattice type is actually of finite lattice type, and if there are infinitely many non-isomorphic indecomposable lattices of the same $\underline{\h}$-length, then the order has strongly unbounded lattice type. For a hypersurface $R=k[[x_0,...,x_d]]/(f)$, we show that $R$ is of bounded (respectively, strongly unbounded) lattice type if and only if the double branched cover $R^{\sharp}$ of $R$ is of bounded (respectively, strongly unbounded) lattice type. This is an analog of a result of Kn\"{o}rrer and Buchweitz-Greuel-Schreyer for rings of finite mCM type. Consequently, it is proved that $R$ has strongly unbounded lattice type whenever $k$ is infinite.