PreprintCC BY 4.0arXiv
Lascar Rank in Expansions
Abstract
We prove that the SU-rank of a supersimple theory is no greater than that of any of its supersimple expansions. A proof for finite rank theories already exists in the literature, but non-continuity presents a problem at limit ranks. We prove the non-decreasing of SU-rank in expansions type-by-type by reducing to finding, in the base theory, certain trees of forking extensions in which the formulas and dividing numbers which witness dividing have a sufficiently uniform pattern. In the case of countable ranks, we can give an especially nice description of this uniformity: for a countable ordinal α, any type of SU-rank at least α in a supersimple theory has a chain of forking extensions in the reverse order type of α.
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BibTeX
@misc{lange2026lascar,
title = {{Lascar Rank in Expansions}},
author = {Michael Lange},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.14698},
eprint = {2609.14698},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.14698},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - Lascar Rank in Expansions
AU - Michael Lange
PY - 2026
DA - 2026/09/13
PB - arXiv
DO - 10.48550/arxiv.2609.14698
UR - https://arxiv.org/abs/2609.14698
AB - We prove that the SU-rank of a supersimple theory is no greater than that of any of its supersimple expansions. A proof for finite rank theories already exists in the literature, but non-continuity presents a problem at limit ranks. We prove the non-decreasing of SU-rank in expansions type-by-type by reducing to finding, in the base theory, certain trees of forking extensions in which the formulas and dividing numbers which witness dividing have a sufficiently uniform pattern. In the case of countable ranks, we can give an especially nice description of this uniformity: for a countable ordinal α, any type of SU-rank at least α in a supersimple theory has a chain of forking extensions in the reverse order type of α.
ER -